SR Simultaneous Lines Drawn in the Sand [Archive]

geistkiesel

Jun8-04, 04:47 AM

The Einstein Train Gedunken

The observer O’ in the moving frame at M’ arrives at M, the midpoint of A and B in the stationary frame, at t0’ when A and B emit photons.

Later a photon from B is detected when M’ is at t1’; the photon from A is detected when M’ is at t2’ in the moving frame. For convenience the velocity of the moving frame is v = 1 and dt = t2' – t1'.
t0'
--------------------------------------t0'|--------t1'-|----------|-t2'
||||||||||||---------------------------M’------------------------------------||||||||||||||| -> motion

-----A-----------------------------------M------------------------------------------B

There are added sections of photo-sensitive strips ||||| such that A and B afre guaranteed to be within a section length when the photons are emitted at A and B. The midpoint of the photo-sensitive strips was determined using the same techniques used to determine M the midpoint of A and B. The strips are numbered starting from the inside positions and then consecutively to the ends of the sections. Each equally numbered pair of photo-sensitive strips have a common midpoint at M’. The resolution of the strips is in the sub-micron range.

As photons are emitted at A and B the photo-sensitive strips located within one photon wave length of A and B, or less, are exposed.

The postulates of special relativity theory state that the laws of physics and the measure of the constancy of the speed of light are invariant in all inertial frames. From special relativity theory observers in the moving frame conclude the events of the emitted photons were not simultaneous in the moving frame.

Are the emitted photon events that are simultaneous in the stationary frame simultaneous in the moving frame?

1.Comments on experimental arrangements or conditions are gratefully accepted.

2.Comments on explicit or implicit stipulations and/or assumptions are gratefully accepted.

3. Other comments..

Doc Al

Jun8-04, 11:48 AM

The Einstein Train Gedunken

The observer O’ in the moving frame at M’ arrives at M, the midpoint of A and B in the stationary frame, at t0’ when A and B emit photons.
I'll rephrase this a bit. In the stationary frame (O), M is the midpoint of two light sources A and B. Lights A and B are switched on at the exact moment that a moving observer O' passes M as observed in the stationary frame. All observations in the stationary frame confirm that the lights were turned on at the same time. For simplicity, let's call that time t = 0.

Since the entire issue is whether or not observers in the moving frame will agree with the stationary frame's assessment that the lights were turned on simultaneously, we must take care to make no assumptions about simultaneity in the moving frame.

Einstein's argument is simple. Rather than add things (like length measurements) which may only serve to confuse the issue, let's deal with Einstein's argument itself (as I recall it).

Einstein reasons thusly:

(1) If an observer is exactly between two lights when they emit pulses simultaneously (according to the observer), then
(2) the pulses from both sources will be detected (received) by the observer simultaneously.

Do you not agree then, that if statement (1) is true, that statement (2) must follow?

Let us also stipulate, per Einstein, that since the speed of light is an invariant in any frame, this argument applies regardless of the speed of the observer. It equally applies to the stationary observer or to the moving observer.

There is no question that for the stationary observer both statements are satisfied. But what can we conclude about the moving observer?

Einstein argues that, from the viewpoint of the stationary observers, O' is moving towards B and away from A. Thus light from B must reach O' before light from A. Do you agree with this? Do you agree that every observer (in any frame) must agree that the light from A and B reached O' at different times?

If you agree with this conclusion, which follows from the invariance of the speed of light, then you agree that statement (2) of Einstein's argument is denied.

Since Einstein's argument is of the form:
If A, then B.
not B,
thus not A.
We are forced to conclude that statement (1) is not true! Thus O' must conclude that the lights were switched on at different times (according to his observations).

That's Einstein's gedanken experiment. Can you point out an error in that argument?
Later a photon from B is detected when M’ is at t1’; the photon from A is detected when M’ is at t2’ in the moving frame. For convenience the velocity of the moving frame is v = 1 and dt = t2' – t1'.
t0'
--------------------------------------t0'|--------t1'-|----------|-t2'
||||||||||||---------------------------M’------------------------------------||||||||||||||| -> motion

-----A-----------------------------------M------------------------------------------B

There are added sections of photo-sensitive strips ||||| such that A and B afre guaranteed to be within a section length when the photons are emitted at A and B. The midpoint of the photo-sensitive strips was determined using the same techniques used to determine M the midpoint of A and B. The strips are numbered starting from the inside positions and then consecutively to the ends of the sections. Each equally numbered pair of photo-sensitive strips have a common midpoint at M’. The resolution of the strips is in the sub-micron range.

As photons are emitted at A and B the photo-sensitive strips located within one photon wave length of A and B, or less, are exposed.
Not sure what you are measuring with these photo-sensitive strips.
The postulates of special relativity theory state that the laws of physics and the measure of the constancy of the speed of light are invariant in all inertial frames. From special relativity theory observers in the moving frame conclude the events of the emitted photons were not simultaneous in the moving frame.
Right!
Are the emitted photon events that are simultaneous in the stationary frame simultaneous in the moving frame?
No.

ram2048

Jun8-04, 11:58 AM

ram2048

Jun8-04, 12:04 PM

Einstein reasons thusly:

(1) If an observer is exactly between two lights when they emit pulses simultaneously (according to the observer), then
(2) the pulses from both sources will be detected (received) by the observer simultaneously.

are you sure that's what he's reasoning? because he would have to assume light travels instantaneously in order for that to be true.

as another observation, if a person arrives AT the midpoint M at the exact time to be hit by photons from A and B, NO MATTER what direction speed acceleration, we can safely assume the photons were emitted from the sources simultaneously, even if they were not calibrated to do so.

but this is simply an observation of speeds and distance midpoints known by any elementary school student

Doc Al

Jun8-04, 12:18 PM

are you sure that's what he's reasoning? because he would have to assume light travels instantaneously in order for that to be true.
Obviously the pulses are emitted and detected at different times. :smile:
as another observation, if a person arrives AT the midpoint M at the exact time to be hit by photons from A and B, NO MATTER what direction speed acceleration, we can safely assume the photons were emitted from the sources simultaneously, even if they were not calibrated to do so.
Not true. If you get hit by two bullets simultaneously, can you conclude that they were fired simultaneously? Of course not: it depends on where they were when they were fired.
but this is simply an observation of speeds and distance midpoints known by any elementary school student
Time to move to a better school district.

ram2048

Jun8-04, 01:05 PM

Not true. If you get hit by two bullets simultaneously, can you conclude that they were fired simultaneously? Of course not: it depends on where they were when they were fired.

if the bullets ALWAYS travelled the same speed AND i was at the MIDPOINT distance between the shooters WHEN HIT.

yes, yes i could

Doc Al

Jun8-04, 03:53 PM

if the bullets ALWAYS travelled the same speed AND i was at the MIDPOINT distance between the shooters WHEN HIT.
Nope. It doesn't matter where the shooters are when you are hit, it only matters where they were when they pulled the trigger.

Think about it. By the time the bullets reach you, the shooters have moved.

ram2048

Jun8-04, 05:03 PM

the shooters don't move. they're stationary, the observer is the only thing moving in my example

sorry if you were confused by that

Pergatory

Jun8-04, 05:06 PM

This is the same as the other two "SR questions of the century" that have been posted earlier on this forum. I'll admit I didn't read those all the way through cuz they were so long! But I'll take a shot at this one.

Using M as the point of origin (so that AM = MB = AB/2, and A = -B), at the point when observer O reaches t1, the photon from B will also be at t1 (by definition) but the photon from A will be at -t1. Given the same duration of local time, both photons will have travelled an equal distance regardless of the frame of reference. At the point when observer O and photon A reach t2, photon B will be at -t2 (the photons have both passed M).

(1) If an observer is exactly between two lights when they emit pulses simultaneously (according to the observer), then
(2) the pulses from both sources will be detected (received) by the observer simultaneously.

#2 is true only if the observer is exactly between the lights when the photons are DETECTED. It does not matter where the observer is when they are emitted. In fact, the observer does not even need to exist when they are emitted. Look at the stars tonight. Consider how long it has been since the light you are seeing has been emitted. Have you really been alive that long? The events you are seeing are not occuring, they have already occured in the past.

Time is local, so simultaneity is a matter of your frame of reference.

ram2048

Jun8-04, 05:19 PM

#2 is true only if the observer is exactly between the lights when the photons are DETECTED. It does not matter where the observer is when they are emitted

exactly, which was the point i was trying to make, but i left out that the emitters do not move <woops>

baffledMatt

Jun8-04, 05:46 PM

Have you ever looked at the so-called 'pole-vaulter paradox'? That's an interesting one as it shows you explicitely how according to SR you must lose simultaneity or face paradox.

Matt

Doc Al

Jun8-04, 06:25 PM

#2 is true only if the observer is exactly between the lights when the photons are DETECTED.
Incorrect. In fact, exactly the opposite is true. I don't care where the lights are when the photons are detected--they need not even exist by that time. :smile:
It does not matter where the observer is when they are emitted.
It matters if you wish to make a deduction about whether the pulses are detected simultaneously. Which is the entire point.
In fact, the observer does not even need to exist when they are emitted.
The thought experiment assumes that both observers have existed forever traveling at the same speed. The point is that statement (2) follows from statement (1). That's all.

Look at the stars tonight. Consider how long it has been since the light you are seeing has been emitted. Have you really been alive that long? The events you are seeing are not occuring, they have already occured in the past.
True, but I don't see the relevance to Einstein's argument.
Time is local, so simultaneity is a matter of your frame of reference.
Now that I agree with! :smile:

Pergatory

Jun8-04, 06:34 PM

Incorrect. In fact, exactly the opposite is true. I don't care where the lights are when the photons are detected--they need not even exist by that time. :smile:

It matters if you wish to make a deduction about whether the pulses are detected simultaneously. Which is the entire point.

My apologies for not being clear. What I meant by the light, is A and B. Not the physical light, you're right, it could be removed once the photons have been emitted, it does not matter.

My point is that in order for the photons to arrive at the observer at the same time, the observer must be equidistant from both originating locations at the point in time when both photons are observed. It does not matter where the observer is or is not when the photons are emitted. THAT was my point.

Please don't waste your time by further tearing apart minor details of my responses, and focus solely on the point I'm trying to prove in response to the point you've tried to make. If we lose focus, it will turn into another 4-page discussion and there will be another identical thread created in a few days discussing the same principal.

Doc Al

Jun8-04, 06:37 PM

the shooters don't move. they're stationary, the observer is the only thing moving in my example

From the observer's point of view, the shooters are moving. Regardless, all that matters for the current argument (in analogy with Einstein's) is where the shooters were (with respect to the observer) at the moment the guns were fired. If the observer (victim?) was exactly between the shooters at the time they fired, then we can deduce that the bullets will arrive simultaneosly. (These are special photon bullets, of course, that always travel the same speed with respect to all observers. :smile: )

Doc Al

Jun8-04, 06:49 PM

My point is that in order for the photons to arrive at the observer at the same time, the observer must be equidistant from both originating locations at the point in time when both photons are observed. It does not matter where the observer is or is not when the photons are emitted. THAT was my point.
I believe I understood your point the first time. You're still wrong. The position of the lights at the moment the photons are detected is irrelevant.

Please focus on the argument at hand: If statement (1) is true, then statement (2) is true.

Please don't waste your time by further tearing apart minor details of my responses, and focus solely on the point I'm trying to prove in response to the point you've tried to make. If we lose focus, it will turn into another 4-page discussion and there will be another identical thread created in a few days discussing the same principal.
I am completely focused. The point of yours that I am "tearing apart" is not a minor detail.

jdavel

Jun8-04, 08:13 PM

ram2048

Jun8-04, 08:56 PM

that makes NO sense.

A and B flash. observer M is in the middle WHEN they flash. He immediately accelerates to light speed in the direction of B.

Photon from A never reaches him.

Photon B is intercepted halfway to B

jdavel

Jun8-04, 09:56 PM

ram2048,

Sounds like you have a real solid understanding of this theory! :wink:

ram2048

Jun8-04, 10:19 PM

what part of photons travelling the speed of light is hard to understand?

if you travel AWAY from a clock AT LIGHT SPEED and look to see what time it says what do you see?

NOTHING. no new photons are hitting your eyes from that direction :grumpy:

Tom Mattson

Jun8-04, 10:27 PM

ram2048

Jun9-04, 12:15 AM

fine. i AM a photon

i'm looking behind me at a clock :P

geistkiesel

Jun9-04, 12:44 AM

I'll rephrase this a bit. In the stationary frame (O), M is the midpoint of two light sources A and B. Lights A and B are switched on at the exact moment that a moving observer O' passes M as observed in the stationary frame. All observations in the stationary frame confirm that the lights were turned on at the same time. For simplicity, let's call that time t = 0.
Sure, but also as to the moving observers their clocks are t' = 0. The two '0' are the same.

Since the entire issue is whether or not observers in the moving frame will agree with the stationary frame's assessment that the lights were turned on simultaneously, we must take care to make no assumptions about simultaneity in the moving frame.

Right. We only define simultaneity as "the photons emitted at the same time in the frame." This is the question, not the answer. OK?

Einstein's argument is simple. Rather than add things (like length measurements) which may only serve to confuse the issue, let's deal with Einstein's argument itself (as I recall it).

Einstein reasons thusly:

(1) If an observer is exactly between two lights when they emit pulses simultaneously (according to the observer), then
(2) the pulses from both sources will be detected (received) by the observer simultaneously.

Do you not agree then, that if statement (1) is true, that statement (2) must follow?

Your observer here is the stationary observer, right? No ambiguity here, right?

I agree. The sequence goes like this: 1.The stationary observer is at the midpoint of A and B 2. Photons are emitted from A and B simultaneously. 3. Later, thw photons from A and B arrive at M (the observer) at the same time, simultaneously. You logic is correct. At this point characteristics of the speed of light have no special significance as long as what we are calling photons move with the same speed to the mindpoint. The photons could be crawling ants.

Let us also stipulate, per Einstein, that since the speed of light is an invariant in any frame, this argument applies regardless of the speed of the observer. It equally applies to the stationary observer or to the moving observer.

No stipulation. I already stipulated that a moving observer using SR theory will conclude the photons were not emitted from A and B simultaneously in the moving frame. This is my stipulation. Can you agree to this?

There is no question that for the stationary observer both statements are satisfied. But what can we conclude about the moving observer?

Einstein argues that, from the viewpoint of the stationary observers, O' is moving towards B and away from A. Thus light from B must reach O' before light from A. Do you agree with this? Do you agree that every observer (in any frame) must agree that the light from A and B reached O' at different times?

This is effectively the same as stated above. The answer is yes, whether we are using crawling ants emitted from A and B, or light photons. Yes.

If you agree with this conclusion, which follows from the invariance of the speed of light, then you agree that statement (2) of Einstein's argument is denied.

Since Einstein's argument is of the form:
If A, then B.
not B,
thus not A.
We are forced to conclude that statement (1) is not true! Thus O' must conclude that the lights were switched on at different times (according to his observations).

That's Einstein's gedanken experiment. Can you point out an error in that argument?

I didn't stipulate to Einstein's postulate regarding the speed of light. I stipulated that SR theory would predct the photons were not emitted at A and B in the movinng frame. Do you have a problem with this? I am not going to debate the truth or falsity of the 'speed of light' postulate. If you say that the SOL postulate is fundamental to the SR derivation of simultaneity, so be it.

I am scrutinizing the operation of the postulates. I am not testing the factual reality of that postulated. Do you see the difference?

The difference goes like this: simultaneity(SR) is placed in an enclosed logical box that cannot be entered and the contents modified, by cleverness, wit, dishonesty or mistake, or stipulation.

Not sure what you are measuring with these photo-sensitive strips.

Right!

No.

The photo-sensitive strips in the moving frame are arranged such that the photons emitted at A and B immediately expose the closest ps-strips in the sections, which are located within one wave length of the photon source when the moving frame passes by. Using the same laws of physics that determined M the midpoint of A and B, the moving frame scientists have determined the midpoint M' of all co-numbered ps-strips at each end of the measuring rod.

russ_watters

Jun9-04, 12:45 AM

fine. i AM a photon Fine. I am Elvis.

Saying it doesn't make it true and a thought experiment is only useful if it conforms to the theory it is used to describe.

geistkiesel

Jun9-04, 12:51 AM

uhh.

before a photon even hits the moving observer he's already moved to t'1 (b photon). then he's already moved to t'2 before the second one hits him (a photon)

all motion aside, put an observer at t'0 t'1 and t'2 and which observer does the photon from B hit first?

the one from t'2 of course.

it's quite rudimentary, i am not understanding the confusion behind all this
The sequence of events: T0 the moving frame M' = M in the stationary frame at t0' = t0.
You cannot do as you say. The photon from B strikes the moving observer at t1', later the photon from A, behind, strikes the moving observer at t2'. You are mixing spatial location with timed points.

geistkiesel

Jun9-04, 01:00 AM

are you sure that's what he's reasoning? because he would have to assume light travels instantaneously in order for that to be true.

as another observation, if a person arrives AT the midpoint M at the exact time to be hit by photons from A and B, NO MATTER what direction speed acceleration, we can safely assume the photons were emitted from the sources simultaneously, even if they were not calibrated to do so.

but this is simply an observation of speeds and distance midpoints known by any elementary school student

Doc Al is ambiguous. he is constantly observed to holding little factual tidbits so he can confuse the issue and distrct the flow of the thread. He isn't being honsest is what I am saying. He really is saying that an observer at M in the stationary frame will see the photons arrive at M simultaneously. I agree with your second paragraph. But some using SR theory calculate the photons that arrived at M simultaneously were NOT emitted simultaneously in the moving frame. In fact, before my exile, Doc Al and I discussed this issue in another thread.

geistkiesel

Jun9-04, 01:10 AM

Obviously the pulses are emitted and detected at different times. :smile:

Not true. If you get hit by two bullets simultaneously, can you conclude that they were fired simultaneously? Of course not: it depends on where they were when they were fired.

Time to move to a better school district.

Not true. Bullets fired from 1 km to the midpoint at the instant bullets fired from fired from 2 kim to the midpoint reach a point 1 km to the midpoint, will arrive at the same time. The same holds for photons emitted at the same locations as those launching the bullets. The farthest emitted photon will arrive simultaneously with the nearest emitted photon.

This isn't trivial for everybody?

geistkiesel

Jun9-04, 01:24 AM

Nope. It doesn't matter where the shooters are when you are hit, it only matters where they were when they pulled the trigger.

Think about it. By the time the bullets reach you, the shooters have moved.

This is what I mean abiout dishonesty. After the bullets are launched, or photons are launched the location of the point of emitted entities is still the same point.

ram2048 - you must see the Doc Al 'confuse and distract methodology'

read the first pot again. The issue under disussion is whether the photons that were emitted simultaNeously when M' of the moving frame was at M in the stationary frame, which is the instant the photons were emitted at A and B were also emitted simultaneously in the moving frame, THIS IS THE THE SAME TIME THEY WERE EMITTED IN THE STATIONARY FRAME. SR THEORY CALCULATES THE PHOTONS WERE NOT EMITTED SIMULTANEOUSLY IN THE MOVING FRAME.

Doc Al knows this, he is consciously throwing distractions and bull **** into the conversation. Doc Al is not being honest, do you get it everybody?

geistkiesel

Jun9-04, 01:32 AM

This is the same as the other two "SR questions of the century" that have been posted earlier on this forum. I'll admit I didn't read those all the way through cuz they were so long! But I'll take a shot at this one.

Using M as the point of origin (so that AM = MB = AB/2, and A = -B), at the point when observer O reaches t1, the photon from B will also be at t1 (by definition) but the photon from A will be at -t1. Given the same duration of local time, both photons will have travelled an equal distance regardless of the frame of reference. At the point when observer O and photon A reach t2, photon B will be at -t2 (the photons have both passed M).

#2 is true only if the observer is exactly between the lights when the photons are DETECTED. It does not matter where the observer is when they are emitted. In fact, the observer does not even need to exist when they are emitted. Look at the stars tonight. Consider how long it has been since the light you are seeing has been emitted. Have you really been alive that long? The events you are seeing are not occuring, they have already occured in the past.

Time is local, so simultaneity is a matter of your frame of reference.

Pergatory, you have just finished you probationary period. Step up to the big time. You have said it all perfectly. I would only add the observation that SR theory does not agree with your assesment of SR theory. SR theory says that the moving observer will detect the photons as not being emitted simultaneously in the moving frame.

Of course the photons do not arrive simultaneously, but this is not sufficient to prove they weren't emitted simultaneously.
See the link for proof. (http://frontiernet.net/~geistkiesel/index_files/)

There is a simple method described in the link to determine if the phoons were emitted simultaneously in the moving frame.

geistkiesel

Jun9-04, 01:48 AM

My apologies for not being clear. What I meant by the light, is A and B. Not the physical light, you're right, it could be removed once the photons have been emitted, it does not matter.

My point is that in order for the photons to arrive at the observer at the same time, the observer must be equidistant from both originating locations at the point in time when both photons are observed. It does not matter where the observer is or is not when the photons are emitted. THAT was my point.

Please don't waste your time by further tearing apart minor details of my responses, and focus solely on the point I'm trying to prove in response to the point you've tried to make. If we lose focus, it will turn into another 4-page discussion and there will be another identical thread created in a few days discussing the same principal.

What you say is true. The moving observer who was at the midpoint of the photons when they were emitted later observes the photons arriving at different times as he is moving towards one of the photons and away from the other.

SR theory says, that the different arrival times at the moving observer means the photons were emitted at differnt time in the moving frame. This is different than saying his mere preception is the photons were emitted at diifferent times. SR theory has the PHYSICAL EMISSION of the photons into the moving frame not being simultaneous. This is fundamental SR.

Doc Al is being dishonest.

ram2048

Jun9-04, 01:49 AM

geistkiesel

Jun9-04, 01:51 AM

geistkiesel

Jun9-04, 01:57 AM

Fine. I am Elvis.

Saying it doesn't make it true and a thought experiment is only useful if it conforms to the theory it is used to describe.
it is also useful if the thiought proes the theory wrong.

geistkiesel

Jun9-04, 02:07 AM

so. here's what i'm getting from everyone trying to confuse the issue...

if i'm in the middle of A and B, and they emit photons towards me, no matter WHAT i do, move left right up down side to side stand on my head, WHATEVER...

i will always get hit by photons A and B at the same time.

is THAT what they're saying?
No. If you are the middle of he photons when they are emitted and you reamin there th ephiotons will arrive at the saem time. If you move like the experiment here, the photons will arrive at difeent times. Bullets, photons or craw;ing ants, the same is true. Keep everything in a straight line. It is the simplest of simple.

Doc Al is throwing useless bull **** around to distract.
Read the first post in this thread. As the moving frame passes the stationary frame when M' in the moving frame was at M in the stationary frame, the photons were emitted in the stationary frame simultaneously, and the photons were detected at this very instant simultaneously in the moving frame.

geistkiesel

Jun9-04, 02:14 AM

Pergatory,

Doc Al is right.

If two lights flash simultaneously (in my frame) and the distance from me to each one is the same at the time of the flashes, then light from the two flashes reaches me at the same time. That's what it means for light speed to be a constant with respect to all observers.

Your bullet analogy lets you down because the bullets are traveling at constant speed with respect to the shooters, but if you are moving wrt the shooters, the bullets are moving at different speeds wrt you.

If you want to get to the point where constant light speed and all its consequences seem more intuitive, stop thinking about bullets!

Read the first post in this thread. As the moving frame passes the stationary frame when M' in the moving frame was at M in the stationary frame, the photons were emitted in the stationary frame simultaneously, and the photons were detected at this very instant simultaneously in the moving frame. Do you understand?

ram2048

Jun9-04, 02:15 AM

No. If you are the middle of he photons when they are emitted and you reamin there th ephiotons will arrive at the saem time. If you move like the experiment here, the photons will arrive at difeent times. Bullets, photons or craw;ing ants, the same is true. Keep everything in a straight line. It is the simplest of simple

so they DON'T hit you at the same time if i move left or right.

which is what i thought, so why the arguments if everyone agrees?

geistkiesel

Jun9-04, 02:30 AM

ram2048,

Sounds like you have a real solid understanding of this theory! :wink:

In the first post of this thread the moving observer at M' was at M the midpoint of sources of photons when the photons were emitted simultaneously. These photons were instantaneously detected in the moving frame as the were emitted in the stationary frame. The moving frame detected the simultaeous emission of the photons in the moving frame. .

Using the detected arrival times of the photons in the moving frame, SR theory calculates that the photons were not emitted simultaneouslly in the moving frame. THE PHOTONS WERE DETECTED SIMULTANEOUSLY IN THE MOVING FRAME AT THE INSTANT THE PHOTONS WERE EMITTED IN THE STATIONARY FRAME.

All you theorists can argue all you want about SR theory, speed of light, Einstein, laws of physics, constancy of the speed of light foreever. The fact of the observation the photons were emitted simultaneously in the moving frame remains invariant under any theoretical perturbation.

Your problem is one of embarrassment when you ponder: "How could I ever have accepted special relativity in the first place?" Like I said, it is your problem, you solve it.

geistkiesel

Jun9-04, 02:40 AM

Not sure what you are measuring with these photo-sensitive strips.

quote geistkiesel: "Were the photons emitted simultaneously in the stationary frame also emitted simultaneously in the moving frame?"

No.

In the first post of this thread the moving observer at M' was at M the midpoint of sources of photons in the stationary frame when the photons were emitted simultaneously. These photons were instantaneously detected in the moving frame as they were emitted in the stationary frame. The moving frame detected the simultaeous emission of the photons in the moving frame.

This is an experimental observation.

And you say no they were not emmitted in the moving frame simultaneously.

geistkiesel

Jun9-04, 02:43 AM

Pergatory,

Doc Al is right.

If two lights flash simultaneously (in my frame) and the distance from me to each one is the same at the time of the flashes, then light from the two flashes reaches me at the same time. That's what it means for light speed to be a constant with respect to all observers.

Your bullet analogy lets you down because the bullets are traveling at constant speed with respect to the shooters, but if you are moving wrt the shooters, the bullets are moving at different speeds wrt you.

If you want to get to the point where constant light speed and all its consequences seem more intuitive, stop thinking about bullets!

In the first post of this thread the moving observer at M' was at M the midpoint of sources of photons in the stationary frame when the photons were emitted simultaneously. These photons were instantaneously detected in the moving frame as they were emitted in the stationary frame. The moving frame detected the simultaeous emission of the photons in the moving frame.

What are you talking about " ...seeming more intuitve . . ?.Aren't observations intuitive enough for you?

ram2048

Jun9-04, 03:04 AM

geistkiesel

Jun9-04, 05:14 AM

why must you use a moving frame and a stationary frame?

i think that's what is throwing me off

do one experiment where he's stationary and THEN do the exact same thing when it's moving

combining the two creates unnecessary confusion :P

1) A -> ________________M________________ <- B
1) A ____ ->____________M____________<- ____ B
1) A __________ ->______M______<- __________ B
1) A _________________ xMx _________________ B

2) A -> ________________M________________ <- B
2) A ____ ->______________M__________<- ____ B
2) A __________ ->__________M__<- __________ B
2) A ________________ ->______Mx ___________ B
2) A ______________________ ->__Mx _________ B
2) A ____________________________xMx _______ B

so i'm still not seeing the contradiction. everyone AGREES that the photons don't hit him at the same time in the moving frame. How hard would it be to do the calculations in reverse for the moving frame setup and acquire that the beams WERE emitted simultaneously BECAUSE of the frame shift, extrapolate it using M's velocity towards B.

Why do we need two frames?
Because SR theory says that events that are simultaneous in the stationary frame are not simultaneous in the moving frame. We are talking about the same physical event. We must do the experiment with two frames, one stationary and one moving to test the theory. Hence we have the photo-sensitive strips to measuee the emission of the photons.
Here is the contradiciton that everyone is having such a difficult time with.

[b]"Were the photons that were emitted simultaneously in the stationary frame also emitted simultaneously in the moving frame?"[/b ]

There was the record of the phosensitive strips attached to the moving frame that were exposed just when the photons were emitted in the stationary frame. These photo-sensitve strips were within a photon wave length of the emitted photons at A and B in the stationary frame when they wee exposed.

SR theory says, the photons were not emitted simultaneously in the moving frame. SR Theory gets to work backwards in time and redo the physical events that occured there.

Hurkyl

Jun9-04, 06:27 AM

why must you use a moving frame and a stationary frame?

Because we're disagreeing about what SR says when we analyze the exact same events from different frames.

Specifically, Geistkiesel is asserting that both of these diagrams are representing the exact same sequence of events:

A M B
A\ M /B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \M / B
A M / B
A M / B
A M / B
A M / B
A M/ B

A M B
A\ M /B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \M/ B

We have two relatively stationary light sources (A and B), and an observer who starts in the middle and moves towards A.

The first diagram depicts how things look in the rest frame of the lights, if the lights are activated simultaneously.

The second diagram depicts how things look in the rest frame of the observer, if the lights are activated simultaneously.

However, there is a very important difference between the two diagrams; in the first diagram the photons do not meet M at the same event, however in the second diagram the photons do meet M at the same event.

The conclusion is that these diagrams cannot possibly represent the same events. Among the possible assumptions we can abandon, abanding that of absolute simultaneity is by far the most reasonable; the emission of photons is simply simultaneous in one frame but not the other.

Doc Al

Jun9-04, 07:59 AM

Sure, but also as to the moving observers their clocks are t' = 0. The two '0' are the same.
You cannot arbitrarily set all clocks in O' to read t'=0. Doing so requires assuming that simultaneity is independent of the reference frames. Since that's what we're trying to discover, we can't just assume it.
No stipulation. I already stipulated that a moving observer using SR theory will conclude the photons were not emitted from A and B simultaneously in the moving frame. This is my stipulation. Can you agree to this?
Absolutely not! I don't want anyone making assumptions or stipulations about simultaneity. The entire point of Einstein's simple argument is to deduce the nature of simultaneity, not make assumptions about it.
I didn't stipulate to Einstein's postulate regarding the speed of light. I stipulated that SR theory would predct the photons were not emitted at A and B in the movinng frame. Do you have a problem with this? I am not going to debate the truth or falsity of the 'speed of light' postulate. If you say that the SOL postulate is fundamental to the SR derivation of simultaneity, so be it.
Once again, the point of Einstein's argument is to deduce the relativity of simultaneity, not assume it. The invariant speed of light is fundamental to any gedanken experiment concerning light. Without that, there is nothing to discuss.
The difference goes like this: simultaneity(SR) is placed in an enclosed logical box that cannot be entered and the contents modified, by cleverness, wit, dishonesty or mistake, or stipulation.
Don't be cute. Re-read my post and tell me exactly where you think Einstein's argument fails. If you can't accept the starting point--the invariant speed of light--then there is no point in continuing.

Doc Al

Jun9-04, 08:07 AM

Doc Al is ambiguous. he is constantly observed to holding little factual tidbits so he can confuse the issue and distrct the flow of the thread. He isn't being honsest is what I am saying.
Please tell me exactly where I am being ambiguous. Refer to my post #2 in this thread. And dishonest? Come now, geistkiesel, have you run out of intellectual ammunition this early in the game?
He really is saying that an observer at M in the stationary frame will see the photons arrive at M simultaneously.
Anyone curious as to what I actually said, can read my own words in post #2.
I agree with your second paragraph. But some using SR theory calculate the photons that arrived at M simultaneously were NOT emitted simultaneously in the moving frame. In fact, before my exile, Doc Al and I discussed this issue in another thread.
We have "discussed" this many times. I am giving geistkiesel a golden opportunity to clearly and unambiguously point out the flaws in Einstein's argument. Is he up to the challenge?

Doc Al

Jun9-04, 08:17 AM

Doc Al is throwing useless bull **** around to distract.
Read the first post in this thread.
Re-read my response in post #2.
As the moving frame passes the stationary frame when M' in the moving frame was at M in the stationary frame, the photons were emitted in the stationary frame simultaneously,
Right!
and the photons were detected at this very instant simultaneously in the moving frame.
Now what possibly can you mean by this garbled statement? Are you talking about photons being detected by O'? Or O' deducing that they were emitted simultaneously? Speak clearly and stop wasting people's time.

Doc Al

Jun9-04, 08:18 AM

Read the first post in this thread. As the moving frame passes the stationary frame when M' in the moving frame was at M in the stationary frame, the photons were emitted in the stationary frame simultaneously, and the photons were detected at this very instant simultaneously in the moving frame. Do you understand?
I defy anyone to understand your point. See my last post. Talk sense, man!

Doc Al

Jun9-04, 08:36 AM

In the first post of this thread the moving observer at M' was at M the midpoint of sources of photons when the photons were emitted simultaneously. These photons were instantaneously detected in the moving frame as the were emitted in the stationary frame. The moving frame detected the simultaeous emission of the photons in the moving frame.
Assuming observers in the moving frame were posted at the right positions, then they WILL detect the photon emissions. But what makes you think that they will detect them SIMULTANEOUSLY? Since that is the point we are arguing, please give us your argument. (Note how Einstein's simple argument does not involve multiple moving observers. Why not deal with that argument directly?)
Using the detected arrival times of the photons in the moving frame, SR theory calculates that the photons were not emitted simultaneouslly in the moving frame.
We all agree that the moving observer detects the photons as arriving at different times. Using that fact, plus simple assumptions of the invariance of light speed, Einstein deduces that the moving observer must conclude that the photons were emitted at different times. Every observation made in the moving frame--including the direct observation of the photon emissions by the moving observers--must agree with this conclusion!
THE PHOTONS WERE DETECTED SIMULTANEOUSLY IN THE MOVING FRAME AT THE INSTANT THE PHOTONS WERE EMITTED IN THE STATIONARY FRAME.
Again, you merely ASSume, where Einstein argues.
All you theorists can argue all you want about SR theory, speed of light, Einstein, laws of physics, constancy of the speed of light foreever. The fact of the observation the photons were emitted simultaneously in the moving frame remains invariant under any theoretical perturbation.
I truly believe that you believe this. But you are still wrong. Rather than add questionable assumptions to Einstein's gedanken experiment, please deal directly with Einstein's simple argument.
Your problem is one of embarrassment when you ponder: "How could I ever have accepted special relativity in the first place?" Like I said, it is your problem, you solve it.
Geistkeisel, I am making a special effort in this thread to be nice. Why don't you do the same?

geistkiesel

Jun9-04, 08:57 AM

I defy anyone to understand your point. See my last post. Talk sense, man!

The instant photons were simultaneously emitted from A and B in the stationary frame the photo-sensitve strips in the moving frame were exposed (|||) at both ends of the moving frame, at locations equally spaced from M'. This is the point, the only point. For your convenience we give another picture.

|||----M'----||| -->moving frame-->
-A-----M-----B-| XX stationary frame XX

This is the picture the instant photons were emitted from A and B in the stationary frame.

Said another way, the instant the photo-sensitive strips were exposed in the moving frame by photons emitted in the stationary frame..

Is this enough "sense talk, man"?

I title this: A Simultaneous Line in the Sand for Doc Al.

If you were from Texas, or had spent any time in Texas, maybe, just maybe you would be able to understand. And this even though it is common knowledge that Texans have half their brains tied behind their back, leaving one loose piece of gray matter to rattle around inside their skull cavities.

You defied anyone to understand the point. OK, someone from Texas, splain it to Doc Al.

geistkiesel

Jun9-04, 09:07 AM

Again, you merely ASSume, where Einstein argues.

I truly believe that you believe this. But you are still wrong. Rather than add questionable assumptions to Einstein's gedanken experiment, please deal directly with Einstein's simple argument.

Geistkeisel, I am making a special effort in this thread to be nice. Why don't you do the same?

You don't know how to be nice. There were no added assumptions that corupted Einstein's gedunken. You just haven't realized that you've lost this game.

As from me to you, this is as nice as it gets.

russ_watters

Jun9-04, 10:40 AM

it is also useful if the thiought proes the theory wrong. If the thought process of the theory is wrong, then it can easily be shown to be wrong through experimentation. You you can't prove that a theory is wrong if you don't address what the theory says. You are stating (assuming, as Doc says) the theory is wrong and building a thought experiment around how you think the universe should work, then offering it up as a proof that the theory is wrong. Sorry, science doesn't work that way.

Depending on the theory you apply to the thought experiment, the outcome is different. Which is right and which is wrong? Well, that's a question answered by experimentation.

What's funny about this is you think you're making an argument against Relativity, but what you are actually doing is demonstrating you don't even understand the scientific method, much less Relativity. The other guys here aren't so much defending Relativity as trying to explain to you what it says.

Doc Al

Jun9-04, 11:18 AM

Doc Al

Jun9-04, 11:28 AM

There were no added assumptions that corupted Einstein's gedunken. You just haven't realized that you've lost this game.
You ASSumed that:

(1) The marks on the photo-sensitive strips (caused by the photon emissions) in the moving frame are equally spaced from the point M' (which passed M at the exact moment that the clock at M read t=0): True!
(2) That the moving observers detect the photon emissions as happening simultaneously: Not true!

As long as you insist on adding these assumptions, there is no point in continuing the discussion.

However, if you would like to discuss Einstein's actual argument--which you refer to constantly but obviously fail to grasp--have at it. If you truly understand Einstein's point, this should be no problem--since he makes fewer assumptions than you do.

baffledMatt

Jun9-04, 11:38 AM

The instant photons were simultaneously emitted from A and B in the stationary frame the photo-sensitve strips in the moving frame were exposed (|||) at both ends of the moving frame, at locations equally spaced from M'. This is the point, the only point. For your convenience we give another picture.

|||----M'----||| -->moving frame-->
-A-----M-----B-| XX stationary frame XX

This is the picture the instant photons were emitted from A and B in the stationary frame.

Said another way, the instant the photo-sensitive strips were exposed in the moving frame by photons emitted in the stationary frame..

But how did you determine the distance between the photosensitive strips? If you made it the same length as A->B whilst in the moving frame then you have a problem because in the stationary frame this distance will be observed to be shorter than A->B due to Lorentz contraction. Thus in the stationary frame it is impossible for them to fall over A and B at the same time.

Ok, so you determined the length when the moving frame was actually stationary. But this doesn't help you either because when they start moving the moving observer will observe the distance between A and B to contract. Hence, again the strips cannot both be over the lights at the same instant in this frame.

The only way out of this mess is to accept the loss of simultaneity. This is why I suggested looking up the 'pole vaulter paradox' earlier because it points this out quite beautifully.

Matt

ram2048

Jun9-04, 11:46 AM

makes no sense for SR to resolve that the photons were NOT emitted simultaneously.

it's a "Given" part of the experiment that they WERE emitted simultaneously as outlined in Einstein's original set up

there's a billion ways we can PROVE that they were emitted simultaneously hooking them up to synchronized clocks, photosensitive strips, whatever. that's not part of the argument, though. It's not even a possibility.

i think you SR supporters are somehow taking what einstein said the wrong way.

wespe

Jun9-04, 12:06 PM

makes no sense for SR to resolve that the photons were NOT emitted simultaneously.

it's a "Given" part of the experiment that they WERE emitted simultaneously as outlined in Einstein's original set up

[QUOTE=ram2048]it's a "Given" part of the experiment that they WERE emitted simultaneously as outlined in Einstein's original set up
No it's not given or automatically assumed. It is deduced, because the stationary midpoint observer received the lights at the same time. The midpoint observer just happens to receive the lights at the same time, that's the setup. Only because of this you can deduce that they were emitted simultaneously *in the stationary frame*, you can' assume anything else. You can't analyse this like you have a bird's eye view and like you can see everything at the same time. But if you want a graphical view according to SR, check this link:

http://casa.colorado.edu/~ajsh/sr/simultaneous.html

ram2048

Jun9-04, 12:12 PM

yeah i saw that one from the other thread.

i still think the moving observer KNOWING he is moving towards B can extrapolate using his velocity/acceleration backwards to determine that they were emitted simultaneously.

i believe this is the proper way to come to the conclusion in the experiment, not calculate forward and find out that your perceptions are skewed, calculate backwards and find out where "reality" is.

ram2048

Jun9-04, 12:15 PM

wespe

Jun9-04, 12:20 PM

yeah i saw that one from the other thread.

i still think the moving observer KNOWING he is moving towards B can extrapolate using his velocity/acceleration backwards to determine that they were emitted simultaneously.

Yes, but the moving observer in fact considers himself stationary, and considers the embankement moving. If he extrapolates the values of his own measurements (his own speed = zero), he finds the SR result. Why would he think his measurements are faulty and correct them according to stationary frame's measurements? It's mutual. You can't determine who is really really really moving. Even if you assume there's a preffered frame, you can't detect your speed relative to it by any experiments.

wespe

Jun9-04, 12:31 PM

ram2048

Jun9-04, 12:44 PM

wespe

Jun9-04, 12:50 PM

you've got to be kidding!

that can't be the way it works. that's completely backwards thinking :D

are you telling me when they plan a mission to mars they calculate that mars is travelling closer, not the probe/lander/vehicle is travelling towards it? :|

That looks equally valid to me, results of the calculations would be the same. But I guess you would normally plan a mission from a third frame's perspective (the earth).

geistkiesel

Jun9-04, 12:52 PM

[QUOTE=wespe]No it's not given or automatically assumed. It is deduced, because the stationary midpoint observer received the lights at the same time. The midpoint observer just happens to receive the lights at the same time, that's the setup. Only because of this you can deduce that they were emitted simultaneously *in the stationary frame*, you can' assume anything else. You can't analyse this like you have a bird's eye view and like you can see everything at the same time. But if you want a graphical view according to SR, check this link:

http://casa.colorado.edu/~ajsh/sr/simultaneous.html[/QUOTEYour link is bogus, unproved crap. Did you gothrough the maths that justifies what the link says?
No
you are talking silly. The photons were emitted simultaneoulsy and exposed the photo-sensitive deectors in the moving iframe as they were emitted. The fact that the photons were emitted at the same time has nothing to do with the obsever at M seeing teh photons later. This is given in the problem ()"relativity" pages 25 -27). . In this therad SR is assumed to predict that the photons were not emitted similtaneously.

Wespe, why do you come barging in here not even half prepared? You're screwing up a decent thread. Where do you get all that "you can't assume anything" crap?" prove it. You come here with your stupid useless intervention totally ignorant of what is happening, but so full of answers. Go away.

geistkiesel

Jun9-04, 01:02 PM

Yes, but the moving observer in fact considers himself stationary, and considers the embankement moving. If he extrapolates the values of his own measurements (his own speed = zero), he finds the SR result. Why would he think his measurements are faulty and correct them according to stationary frame's measurements? It's mutual. You can't determine who is really really really moving. Even if you assume there's a preffered frame, you can't detect your speed relative to it by any experiments.

BS. It is given he knows his velocity and can measure it as he crosses M when the hotons are emitted. You're just as ignorantly dishonest as Doc Al, your mentor.. Otherwise the propblem is nothing. Yes you can. Blue/red analyses. Also each frame can emit 1 second inteval pulses. The stationary (slower) frame will read the moving frame pulses as slower than his two seconds. Likewise, the moving frame will read the stationary frame pulse rate as faster than his own. You say no absolute time? prove it.

Why ae you making things up? Showing off your childish television learned physics? Just to sabatoge and get even for your embarrassing loss awhile back?

Answer the question of the photo-sensitive strips being exposed when the photons are emitted simutaneously. These strip are located equidistant from the midpoint M' in the moving frame. Get up tp speed Wespe or get off the bus. Asswer this one question posed here. can you do this or do you want to measure the SOL first?.

baffledMatt

Jun9-04, 01:09 PM

Your link is bogus, unproved crap. Did you gothrough the maths that justifies what the link says?
No
you are talking silly.

and I was just about to say exactly the same thing about your link.

Answer the question of the photo-sensitive strips being exposed when the photons are emitted simutaneously. These strip are located equidistant from the midpoint M' in the moving frame.

Errrm, hello?

*pointing to the last post I made*

Matt

geistkiesel

Jun9-04, 01:17 PM

in any case this setup is different from cerulean and vermilion by virtue of being contained within 2 other bodies (the photon emitters) which have a known property of being "fixed" in the universe which provides excellent reference points to capture movement data to work backwards from.

you wouldn't have that luxury in the example in that link
Ram2048 Beware wespe. This is theone that started the other SR ain't working thread until surrenduring to frustration. He taks alot of you can't do this and that. The photons were emitted simultaneously when M' was at M, THis is a given , You don't ahve to prove it. The velocity of the moving frame is determined whjen passing through M. Cerulean, aslo, has no proof of anyhing. However, it shows the inanity of SR. A physical event occured and by them ere fact that an observer was thrown in at some time later that observer can make a physical alteration of the sequence of the emitted photons just because he is an observer, never having any contact whith any of the mechanics of thephoton, their sources, the photons themselves. This is how inane SR really is.

Wespe pisses me off. he has a limited extent of thinking range which he is unwilling to extend. All thos things he said about the experiment of this thread is bogus. he hasn't read the thread and knows not what is happening. Your figure is correct. I would remind you that th ethread stipulates that SR will predict the photons weren'y emitted simultaneously,

Have your quiestioners answer, among other questions, how can the photons be detected simultaneously in the moving frame, simultaneousl and marked and identified as equidistant from the midpoint of those exposed strips? No clocks needed, AN instantaneous event. The midpoint of he moving frame was determined the same way as the midpoint in the stationary frame, by the laws of physics that are invariant in all inertial frames., by the photo-sensitive strip, yet later be calculated as having occured nonsimultaneously?

Im gonna be off line for a few hours. Myself, I am not equipped to the taking of prisoners. Shoot anybody who surrenders.

wespe

Jun9-04, 01:21 PM

Your link is bogus, unproved crap. Did you gothrough the maths that justifies what the link says?

It just shows graphically that it is possible for two frames to have relative simultaneity, when you abandon the idea that time is absolute.

Noyou are talking silly. The photons were emitted simultaneoulsy and exposed the photo-sensitive deectors in the moving iframe as they were emitted.

You say "emitted simultaneously" but you don't say in which frames. You just assume it is in both frames. That is the argued point but you start by assuming it.

The fact that the photons were emitted at the same time has nothing to do with the obsever at M seeing teh photons later. This is given in the problem ()"relativity" pages 25 -27). . In this therad SR is assumed to predict that the photons were not emitted similtaneously.

The only facts in an experiment are the measurements. Don't assume "something is so and so in reality". You can not dictate what reality should be.

Wespe, why do you come barging in here not even half prepared? You're screwing up a decent thread. Where do you get all that "you can't assume anything" crap?" prove it. You come here with your stupid useless intervention totally ignorant of what is happening, but so full of answers. Go away.

I'm just saying "don't assume anything except the measurements". The measurement is not "they were emitted simultaneously in stationary frame", that is the conclusion. The measurements are "M received the photons simultaneously", "M is equadistant from A and B", "speed of light is constant".

You have no math and no answers. When you are disproven by calculations you say "maybe so" "calculations must be flawed" "reality is like in fact this". And you didn't stop posting in my thread even I asked you nicely, why should I go away?

geistkiesel

Jun9-04, 01:24 PM

But how did you determine the distance between the photosensitive strips? If you made it the same length as A->B whilst in the moving frame then you have a problem because in the stationary frame this distance will be observed to be shorter than A->B due to Lorentz contraction. Thus in the stationary frame it is impossible for them to fall over A and B at the same time.

Ok, so you determined the length when the moving frame was actually stationary. But this doesn't help you either because when they start moving the moving observer will observe the distance between A and B to contract. Hence, again the strips cannot both be over the lights at the same instant in this frame.

The only way out of this mess is to accept the loss of simultaneity. This is why I suggested looking up the 'pole vaulter paradox' earlier because it points this out quite beautifully.

Matt

I don't like you Matt, you've been paying attention. They determined the midpoint using the same laws of pysics as the stationary frame did. Also, the strips of photo-sensitive material is fractions of a micron wide, thousand of strips in a section. Each strip<< photon wave length.. The sections holding the strips are long enough so that shrinking in the moving frame is accounted for. Each pair of A', B' strips is equidistant fronm the midpoint M'. Assume the moving platform will shrink 100 units from velocity considerations. We make the sections at each end of A' and B' 200 units long er, totalling 400 units overkill. with the strips covering the whole 200 units. Overlap, overkill.

geistkiesel

Jun9-04, 01:29 PM

and I was just about to say exactly the same thing about your link.

Errrm, hello?

*pointing to the last post I made*

Matt
matt, I answered another post of yours with the answer, bsically the midpoint of the strips was determio=ned by the same laws of physics thatadetermined M in the stationary frame. Each strip is ifractio s of a micron wide. Thousand in a section . The two section much longer 9calulated fro SR)to assure that the stips will overlap A and B whe the moving frame passes? OK uMMRRR?

jdavel

Jun9-04, 01:35 PM

baffledMatt

Jun9-04, 01:37 PM

They determined the midpoint using the same laws of pysics as the stationary frame did.

What do you mean by this? Did they determine the length whilst the moving frame was actually moving or not?

Also, the strips of photo-sensitive material is fractions of a micron wide, thousand of strips in a section. Each strip<< photon wave length.. The sections holding the strips are long enough so that shrinking in the moving frame is accounted for. Each pair of A', B' strips is equidistant fronm the midpoint M'. Assume the moving platform will shrink 100 units from velocity considerations. We make the sections at each end of A' and B' 200 units long er, totalling 400 units overkill. with the strips covering the whole 200 units. Overlap, overkill.

Hmm, if you like. But by this argument you are only able to determine the level of simultaneity to an accuracy set by the length of your strips. Do you see what I'm getting at? Say you do the experiment with strips of length 200 units and find that they do in fact both detect the emitted light. You say that this proves that the emission was simultaneous. However, if the emissions at A and B were in fact separated by a time interval t < 200 * v then you would still have detection on both strips even though the emission was not simultaneous. So to prove real simultaneity you MUST make your strips of vanishing length.

Matt

EDIT: changed some numbers to avoid possible confusion

geistkiesel

Jun9-04, 01:45 PM

It just shows graphically that it is possible for two frames to have relative simultaneity, when you abandon the idea that time is absolute.
nothinmg was abanded. Th estrips were placed to coincide with A and B within a wave length of theemitted photons. The very instant the photons were emitted, what 10-6 of a second as an instant. You are the one that is asserting time is not absolue, prove it.

You say "emitted simultaneously" but you don't say in which frames. You just assume it is in both frames. That is the argued point but you start by assuming it.
emitted simultaneously in the moving frame. Detected as the photons were emitted the photo-sensitive strips attached to the moving frame.one wave length from the sources at A and B.

The only facts in an experiment are the measurements. Don't assume "something is so and so in reality". You can not dictate what reality should be. The photons were exposed as the photons were enmitted. This means the photons were emitted simultaneously in the moving frame because that is when they were detected. I can dictate the truth when I see it. YOu know nothing. You have a childs view of SR and you are talking to me like they talked to you at one time. It sounds like you were beaten intio shape.

I'm just saying "don't assume anything except the measurements". The measurement is not "they were emitted simultaneously in stationary frame", that is the conclusion. The measurements are "M received the photons simultaneously", "M is equadistant from A and B", "speed of light is constant". See there you go. You are either consciously lying or you haven't read the thread. The given is that the photons were emitted just as M' passed through M. Tere was no wait to determine this. The ones putting on the experiment did this. Do you have that yet after having it drilled in what you call a head a few dozen tuime? Why are you so carelss?" We all err. But with you its cronic. You have to make up crap to win don't you, just like Doc Al taught you. He is so proud.

You have no math and no answers. When you are disproven by calculations you say "maybe so" "calculations must be flawed" "reality is like in fact this". And you didn't stop posting in my thread even I asked you nicely, why should I go away?

Evry word out of your mouth is a lie. I stipulated at the beginning that SR would predict the phoytons were no emitted simultanoulsy. Go back to ANdorra or from wheever you originated. So whaT if i didn'tstop posting in "your" thread. So sue me. Actually i am cleaning up your unfinished business that you ran from like a frightend child. Why should you go away? because you bore me.

geistkiesel

Jun9-04, 01:48 PM

If the thought process of the theory is wrong, then it can easily be shown to be wrong through experimentation. You you can't prove that a theory is wrong if you don't address what the theory says. You are stating (assuming, as Doc says) the theory is wrong and building a thought experiment around how you think the universe should work, then offering it up as a proof that the theory is wrong. Sorry, science doesn't work that way.

Depending on the theory you apply to the thought experiment, the outcome is different. Which is right and which is wrong? Well, that's a question answered by experimentation.

What's funny about this is you think you're making an argument against Relativity, but what you are actually doing is demonstrating you don't even understand the scientific method, much less Relativity. The other guys here aren't so much defending Relativity as trying to explain to you what it says.

read the opening thread.Istipulated SR would predict nonsimultaneity.

ram2048

Jun9-04, 02:00 PM

non simultaneity is a stupid way to conclude that. there. i said it :D

a better way to conclude it is they WERE simultaneous (which IS reality) and extrapolate how much your perceptions were skewed (Time dialation) based on the distance covered

you wouldn't be able to do so in Vermillion and Ceruleon because there's no universal referance frame (my new postulate). but on a moving train "gedunken" you can. if you have the tools available to make conclusions that are correct, might as well use them.

ludicrous as it is they have every right to claim non-simultaneity because that's the way they want to conclude it. i highly doubt that line of thought will get very far though. It's not so much a paradox as the ignoring of pertinent data (fixed absolute locations) and drawing the wrong conclusions from there.

baffledMatt

Jun9-04, 02:08 PM

a better way to conclude it is they WERE simultaneous (which IS reality) and extrapolate how much your perceptions were skewed (Time dialation) based on the distance covered

So are you saying that in this example the 'stationary' observer was the one who observed reality, yet the 'moving' observer saw a skewed reality?

How do we know who's reality is the 'correct' one?

Matt

jdavel

Jun9-04, 02:11 PM

ram2048 said, "It's not so much a paradox as the ignoring of pertinent data (fixed absolute locations)...."

Fixed absolute locations? Even Newton didn't believe in that! No wonder your'e having trouble with 20th century phyiscs; you haven't learned 17th century physics yet!

ram2048

Jun9-04, 02:29 PM

Fixed absolute locations? Even Newton didn't believe in that! No wonder your'e having trouble with 20th century phyiscs; you haven't learned 17th century physics yet!

so maybe you guys have been led astray since the 17th century? man that's wacky :D

So are you saying that in this example the 'stationary' observer was the one who observed reality, yet the 'moving' observer saw a skewed reality?

How do we know who's reality is the 'correct' one?

i haven't given my universal relativity postulate in this thread yet?

let's just say by majority rules ;D every other "thing" in the universe has the moving observer moving relative to itself, but every other "thing" has NOTHING else but the observer moving relative to it (for simplicity sake). The observer has the EVERYTHING else moving in relation to him.

on a scale of which is likely and which is no, the moving observer being stationary while the universe moves is "infinitely unlikely" whereas JUST the observer moving is "infinitely likely"

so now that we've determined which is moving it's simple to say which has the skewed perception ;D

wespe

Jun9-04, 02:31 PM

baffledMatt

Jun9-04, 03:00 PM

let's just say by majority rules ;D every other "thing" in the universe has the moving observer moving relative to itself, but every other "thing" has NOTHING else but the observer moving relative to it (for simplicity sake). The observer has the EVERYTHING else moving in relation to him.

What makes you think reality would make this choice? Also, is there a way that I can (at least in principle) determine whether my reality is the correct one? i.e. how do i determine whether I am the 'moving' or 'stationary' observer?

on a scale of which is likely and which is no, the moving observer being stationary while the universe moves is "infinitely unlikely" whereas JUST the observer moving is "infinitely likely"

But everything in the rest of the universe is also moving, it's not just a case of here is the moving person and here is the static 'rest of the universe'.

Consider putting your thought experiment on its head. In other words, what if I called the observer who is recieving the light the 'stationary' one and the one who is emitting the pulses the 'moving' observer. By your reasoning the stationary observer has the correct reality so in fact the pulses really were not simultaneous and the fact that the moving observer saw them emitted simultaneously was just due to his 'skewed reality'. This allows me to change what reality is just by rewording the problem. Does this not bother you?

Matt

jdavel

Jun9-04, 03:00 PM

ram2048

Evidently you don't want to understand this theory. The nice thing about physics is that if you don't want to understand it, you won't!

ram2048

Jun9-04, 03:45 PM

Consider putting your thought experiment on its head. In other words, what if I called the observer who is recieving the light the 'stationary' one and the one who is emitting the pulses the 'moving' observer. By your reasoning the stationary observer has the correct reality so in fact the pulses really were not simultaneous and the fact that the moving observer saw them emitted simultaneously was just due to his 'skewed reality'. This allows me to change what reality is just by rewording the problem. Does this not bother you?

you can't call the moving observer "stationary" in my postulate. compared to the universal referance frame he IS moving it is "infinitely likely" that it is correct. you can't do the math on EVERYTHING in the universe, but if need be you CAN do the math on alot of the things that would matter (anything local and earth based) keep in mind that for every object you add to the "universal frame" you're also adding a slew of necessary calculations, as you would need to calculate not only that object's reference to the observer, but also that object's reference to every other thing you're allowing to be in that frame. it's this kind of "triangulation" that allows me to pinpoint who is perceiving reality as it conforms to the "majority" and determine who is the moving body with the perception shift and who is not.

the more objects you add to your universal reference frame in your calculations, the more "true" your results will become. in this case we only need relative reality calculated for maybe global scale. that way the results would be something everyone on earth would agree upon. if you needed galaxy-wide, or universal-wide reality, calculations would be notably more complex and tedious :D

geistkiesel

Jun9-04, 04:17 PM

Geistkiesel,

You are thinking simultaneity as something absolute, something you can measure. Then shouldn't everyone get the same mesurements of this absolute thing? Describe the method that measures simultaneity, and how everyone can get the same value. Einstein describes his method, and the observers get different measurements in the experiment according to that method, so they conclude simultaneity is relative. If your method of measurement is the formula in that infamous link, it has been invalidated mathematically in the other thread (which you responded "maybe so").

All the above also applies to time. Einstein says time is what a clock measures, and there is experimental support for time dilation, therefore the conclusion is that time is relative. If you claim that time is absolute, describe a method which measures time the same for all observers.

For this experiment, you won't get anywhere with blue/red shifts. Suppose, the light sources were co-moving with M'. There is no blue/red shift for M'. Still, the first pair of emitted photons get detected at the same time by M, which is the experiment setup. That's because speed of light does not depend on source speed, which is experimentally supported.

Geistkiesel, what is your alternative to relativity? Do you really have a theory, any math, any experimental support? I don't think so.

what specific post invalidated the famous link?

supose instead of the light co-moving with the moving frame we finish the current experiment. Let the stationary and moving frame send sinals at one second intervals. The faster frame's signals will be slower, the slower frames signal faster, and each will know who is moving.

Your assumption that the stationry frame gets the signals the same as when the photons were emitted in the stationry frame may not be as you say. Inface the link you referred to in doctordicks thread about cerulean etc.? the moving frame here, according to SR and the link you referenced says your wrong. what thread and which specific post invalidated the famous link?

Are you submitting to the panel what Einstein said about "time is what a clock measures"? Check the literature there are tons of different ideas abot time and clocks, most disagreeing with AE. Again whicy post invalidated the famous link, which tyoy haven't read or understood, have you, and do you? Why are you so incompetent as to attempt to negate a mathematical formalism of which you are completely ignorant?

Now I know whay they were all laughing at you. I thought you had some balls there for a while, guess I was wrong, wasn't I?

ram2048

Jun9-04, 04:24 PM

wow geist... harsh!

maybe not the best way to hold a discussion, but it's interesting in a JerrySpringer-esque kinda way :D

baffledMatt

Jun9-04, 04:25 PM

you can't call the moving observer "stationary" in my postulate. compared to the universal referance frame he IS moving it is "infinitely likely" that it is correct.

So are you saying that in order for you to make any predictions for this thought experiment you need to know what the rest of the universe is doing so that you can determine which observer is 'stationary'? wow.

If there is any meaning to your universal reference frame then you should be able to determine which observer is in it simply from the results of the experiment. The results are: One observer sees simultaneous emission, the other doesn't. How do you determine from this which one is really 'moving' and therefore is not seeing true reality?

Imagine I described the thought experiment to you but just specified that one observer is moving relative to another. If there is anything to what you are saying there should be some observation they can make which will tell them who is in the universal frame. SR says that there is in fact no such observation so we have no choice but to treat each frame equally. What do you say?

Through all this you are also assuming that there is a whole load of universe 'out there' which is in the same inertial frame. Hubble told us many many years ago that this is far from the truth. Everything is moving with respect with each other so really you should conclude that every object in the universe is equally likely to be in the universal frame.

Matt

baffledMatt

Jun9-04, 04:35 PM

Why are you so incompetent as to attempt to negate a mathematical formalism of which you are completely ignorant?

Now I know whay they were all laughing at you. I thought you had some balls there for a while, guess I was wrong, wasn't I?

Look now, let's try to keep this civil eh? If you want to have a discussion please have the courtesy to do it in a polite and mature manner.

Wespe is trying to do you a favour by sorting out your misunderstanding - he could quite easily let you rot in your own ignorance. This is something you should be grateful for!

Matt

ram2048

Jun9-04, 04:35 PM

So are you saying that in order for you to make any predictions for this thought experiment you need to know what the rest of the universe is doing so that you can determine which observer is 'stationary'? wow.

not so, i said in order to do so with universally accepted accuracy you would have to poll the entire universe on its opinion and make a conclusion from there as the what the majority of objects would say. think of it in terms of triangulation it's a good way of looking at it.

If there is any meaning to your universal reference frame then you should be able to determine which observer is in it simply from the results of the experiment. The results are: One observer sees simultaneous emission, the other doesn't. How do you determine from this which one is really 'moving' and therefore is not seeing true reality?

if you allow me to include the rest of the universe i could determine to ridiculous and astounding accuracy which one is moving, but with just the two participants you can't. even adding a third you can't. or a fourth. but the more you add the closer you get to being able to describe exactly what is taking place. you get to the point where everything in the world is included in your reference frame (which is how we perceive the world) and it's as simple as instinct to determine which one is moving on a globally accepted scale.

Imagine I described the thought experiment to you but just specified that one observer is moving relative to another. If there is anything to what you are saying there should be some observation they can make which will tell them who is in the universal frame. SR says that there is in fact no such observation so we have no choice but to treat each frame equally. What do you say?

absolutely true. luckily we live within the universe and we will never EVER be forced into a situation where this is the case.

Through all this you are also assuming that there is a whole load of universe 'out there' which is in the same inertial frame. Hubble told us many many years ago that this is far from the truth. Everything is moving with respect with each other so really you should conclude that every object in the universe is equally likely to be in the universal frame.

and if you knew where everything was and had a huge computer you could accurately calculate how every universal object was moving within the whole and determine its perception and reality.

baffledMatt

Jun9-04, 04:52 PM

not so, i said in order to do so with universally accepted accuracy you would have to poll the entire universe on its opinion and make a conclusion from there as the what the majority of objects would say. think of it in terms of triangulation it's a good way of looking at it.

When you ask each observer, how many of the answers do you think will be in exact agreement?

if you allow me to include the rest of the universe i could determine to ridiculous and astounding accuracy which one is moving, but with just the two participants you can't. even adding a third you can't. or a fourth. but the more you add the closer you get to being able to describe exactly what is taking place. you get to the point where everything in the world is included in your reference frame (which is how we perceive the world) and it's as simple as instinct to determine which one is moving on a globally accepted scale.

Then I suppose your answer to my question 'how do you determine this universal frame' will be 'I define it that way'. Ok, have it your way. If you want you can define the 'correct' frame which is this funny democratic average of all the frames in the universe. However, imagine I decide on a different definition, say that where I am sitting is the only stationary frame and everything else revolves around me. Will there be any way you can tell me which of our definitions is correct? will there be any observations which will differ between the universe with my definition and the one with yours?

absolutely true. luckily we live within the universe and we will never EVER be forced into a situation where this is the case.

Not so. I put you into space in a spaceship with no windows. Is there any experiment you can carry out (without looking outside of course) which will allow you to determine whether you are moving or not?

Matt

geistkiesel

Jun9-04, 04:54 PM

What makes you think reality would make this choice? Also, is there a way that I can (at least in principle) determine whether my reality is the correct one? i.e. how do i determine whether I am the 'moving' or 'stationary' observer?

But everything in the rest of the universe is also moving, it's not just a case of here is the moving person and here is the static 'rest of the universe'.

Consider putting your thought experiment on its head. In other words, what if I called the observer who is recieving the light the 'stationary' one and the one who is emitting the pulses the 'moving' observer. By your reasoning the stationary observer has the correct reality so in fact the pulses really were not simultaneous and the fact that the moving observer saw them emitted simultaneously was just due to his 'skewed reality'. This allows me to change what reality is just by rewording the problem. Does this not bother you?

Matt

I don't know if I can satisfy the elevated level the conversation just went, but let me try. For the current experiment (we'll move to grander scales in a bit) you want to know if you can tell who is moving and who is stationary (or moving slower) right?

Both observers send out 1 second pulses from their respective clock timing circuits. The slower frame's pulse rate is less than the fastest frame's pulse rate and vice versa, right? This was off the top of my head a few days ago. I have used it and have had no reply,but is seems like it should work.

Assuming like I did in the opening post that when the photons were emitted in the stationary frame the moving frame's photo-sensitive strips (ps_strips) were exposed by a few extra photons at A and B in the stationary frame, while some of the ps-strips were going by within one photon wavelength from the sources. The photons exposed the ps-strips right then at both ends of the moving frame, in fractions of 10^-6 seconds. The strips are numbered and a #10 in the forward end is the same distance from M' as a #10 in th erear position. This was at the instant that M' was at the M when the photons were emitted. Some have suggetsed that no, the moving frame cannot guarantee I can find the midpoint of the ps_strip. My counter, which I havwen;'t seen challenged, is I used the same law of physics that allowed the stationry framwe to determine themisddle of he photon sources at A and B. at M. M is gien in the experimebnt. M is not measured by the photons in the experiment reaching M after the same travel time form the sources. This may have been the methos used, but this was stricly pre-current -experiment.

Can you see my position when I want to say that the photons were mesured in the moving frame equal distances from the midpoint of gthe moving frame when the pulses were emitted by A and B. Just can you see my reasonaning. I will not hold you to an adnmiission that I am right and you are adnmitting error. Only can yo see why I might say that?

If you do then hear this. I have stipulated that SR will predict that which I justs described differently. SR will predict the photons were not emitted simultaneously inthe moving frame. This is stipulated.

I am saying I do not have to argue time dilation, shrinking absolte space and time , noe of that. Why? I have agreed in the results of SR. Whi amI or you for that matter to be chosen as the obne who arbitrates what is physical law? Me saying SR will predict such and such doesn't make the poustulates true, the posstulates stand ot dall on their pwn merist.

Anyway, we have a comflict, I am sure you can envision. but if my observation by reading the ps-strips three weeks after the experiment does bring a nontrivial element into the discssion. You an get all the SR theorists on the planet and vote on the issue and I the anti-SR theorists al 5 of us, and I stipulate your vote would win, slam dunk win. But physical law isn't detemined by political means. Therefore, because the exposure of the ps-strips at the instant the photons were emitted appears contradictory, can you point to a physical experimental reality, fact or conditon that would unambiguously negate what I am calling a reasonable observation? Can you negate the observation with other than SR theory.

It is like the Ptolemy model of circles within cirlces of the solar system with the earth at the center and all stellar matter revolving around the earth. If you were Galilleo and offered a contradicting observation to the Dr, Ptolemy who countered your observation that "the circles within circles model has worked for 2000 years, ergo your observation is void", you wouldn't accept that would you?

I can't force anyone to anything in this thread, except to look at the problem as I do and try to beat it on its own terms. If SR is so overwhelming "real" as some seem to think, it should be a "slam dunk" trivial exercise to defeat my observations model on its own merits. AT least you should try shouldn't you?

baffledMatt

Jun9-04, 04:59 PM

Errm, slow down cowboy! - like really, I guess you're typing too fast which is causing you to make a lot of typos and they can make it difficult to understand you.

Matt

ram2048

Jun9-04, 05:11 PM

When you ask each observer, how many of the answers do you think will be in exact agreement?

the majority :| the majority will always define reality for the minority.

next time you rear-end a parked car with no driver, i want you to argue with the cop that from a relativistic standpoint it's just as real that the parked car hit YOU so it should be equally to blame.

do that for me :D

Then I suppose your answer to my question 'how do you determine this universal frame' will be 'I define it that way'. Ok, have it your way. If you want you can define the 'correct' frame which is this funny democratic average of all the frames in the universe. However, imagine I decide on a different definition, say that where I am sitting is the only stationary frame and everything else revolves around me. Will there be any way you can tell me which of our definitions is correct? will there be any observations which will differ between the universe with my definition and the one with yours?

when 6 billion people come tell you that the world doesn't revolve around you, you have to at least be open to the possibility that they may be right...

Not so. I put you into space in a spaceship with no windows. Is there any experiment you can carry out (without looking outside of course) which will allow you to determine whether you are moving or not?

can i detect anything at all? gravity fluctuations, "radon waves", acceleration deceleration, centripetal inertia or whatever?

if not then this is a completely encapsulated universe, and whatever takes place here has no bearing on the grander universe, and vice-versa.

but within that spaceship, which is now "the universe" certain things will stand out as "fixed locations". bulkheads computer terminals, hatches whatever. using those locations in your "relativistic calculations" it's very easy to determine motion within the ship relative to the whole ship.

wespe

Jun9-04, 05:14 PM

what specific post invalidated the famous link?

the famous link is the one you keep posting around
http://frontiernet.net/~geistkiesel/index_files/sim_fix_einstein/index.html

the invalidation I was referring to is in the "why relativity is wrong" thread.
http://www.physicsforums.com/showpost.php?p=227893&postcount=209

You give Hurkyl your formula to determine simultaneity, Hurkyl calculates and says "Since the observer is moving, this equation clearly cannot hold", then you say "maybe so""

supose instead of the light co-moving with the moving frame we finish the current experiment. Let the stationary and moving frame send sinals at one second intervals. The faster frame's signals will be slower, the slower frames signal faster, and each will know who is moving.

Huh? I don't see how this works, it's all mutual. You can't determine who is "really" moving. If you can prove you can do that, well, do so.

Your assumption that the stationry frame gets the signals the same as when the photons were emitted in the stationry frame may not be as you say. Inface the link you referred to in doctordicks thread about cerulean etc.? the moving frame here, according to SR and the link you referenced says your wrong. what thread and which specific post invalidated the famous link?

Are you submitting to the panel what Einstein said about "time is what a clock measures"? Check the literature there are tons of different ideas abot time and clocks, most disagreeing with AE. Again whicy post invalidated the famous link, which tyoy haven't read or understood, have you, and do you? Why are you so incompetent as to attempt to negate a mathematical formalism of which you are completely ignorant?

What is wrong with Einstein's definitions? Experiments agree with him. If you don't, please define your version of time, or define a method to measure time. Also please define simultaneity and how to measure simultaneity. Present math that predicts results of these measurements of all observers.

baffledMatt

Jun9-04, 05:18 PM

I'll have to think a little about your last post (and if you could sort out the typos that would be a great help) but here is a little digression:

It is like the Ptolemy model of circles within cirlces of the solar system with the earth at the center and all stellar matter revolving around the earth. If you were Galilleo and offered a contradicting observation to the Dr, Ptolemy who countered your observation that "the circles within circles model has worked for 2000 years, ergo your observation is void", you wouldn't accept that would you?

don't forget that 100 years ago it was Newtonian mechanics that was playing the role of Ptolemy (it had been working for almost 300 years!) and special relativity was the crazy new theory which all the stuffy old professors were trying to dismiss.

All that has happened is that the original sceptics of relativity have died and so many of their 'paradoxes' and 'simple demonstrations of the silliness of SR' died with them. Nobody records these ideas because in the day they were crushed by people like Einstein, so of course when people like you come up with similar arguments it feels all new and original. The only thing which will kill relativity now is something really wild and new, like trying to apply it with QM in the middle of a black hole.

I'm not saying that healthy scepticism of SR is a bad thing, indeed, accepting it blindly would mean the end of your days as a scientist! However, do you really think that there is any simple argument against it which you can come up with which probably hasn't allready been thought of? considering just how many great minds there were working on SR at the start of the 20th century I would think this very unlikely.

Anyway, back to our discussion...

Matt

geistkiesel

Jun9-04, 05:29 PM

When you ask each observer, how many of the answers do you think will be in exact agreement?

Then I suppose your answer to my question 'how do you determine this universal frame' will be 'I define it that way'. Ok, have it your way. If you want you can define the 'correct' frame which is this funny democratic average of all the frames in the universe. However, imagine I decide on a different definition, say that where I am sitting is the only stationary frame and everything else revolves around me. Will there be any way you can tell me which of our definitions is correct? will there be any observations which will differ between the universe with my definition and the one with yours?

Just a slight interjection here. Aren't you both assuming you each know the direction to the "absolute frame"? Isn't there a way to solve this problem without being so universally esoteric? i don't want to deny any of you your say, but it seems slightly over the edge of relevence.
Before we get to your discussion points let us first dispose of the trivial.

1. Stipulation:SR theory predicts the photons emitted in the stationary frame in our experiment will predict the same photons were not emitted simultaneously in the moving frame.

2. PS-strips (photo-sensitive strips) located within 1 photon wave length of the sources of the photons emitted in the stationary frame are exposed (in the moving frame) as the photons are emitted. Super fast film, ps-strips << micorn wide, Each ps-strip locatable to the midpnt M' with mirror image ps-strip at other end within any mutuallly agreeable resolution.

Something's gotta give.
Can SR defeat the observation ( that appears contradictory)?

Can the observation defeat SR.

If no observer in a moving frame then no problem, no discussion. Is the presence of the moving frame itself analyzable as a source of physical force that guarantees the photons will not be emitted simultaneously in the moving frame. As far as I can determine the alleged effect is purely one running directly from the postulates of SR exclusively.

Can you see my problem of understanding how a physical event, supported by my famous ps-strips can be modified, not by any real or imagined force, not even any quantum mechanical "nonlocal force channels" are implied or even speculated. The modification, as I call it, is brought into the reality of the universe by the mere presence of an observer. A million observers all moving at different angles all arrive at the same conclusion of nonsimultaneity, and all would have different calculated times for the emission of the photons.

If no moving platform we have a simple simultaneous event of photon emission. With observres, the only expeiemntal difference in the two situations, we have physical modification of an obsevable manifestly existent only by the presence of a moving platform.

If we look first only at the first postulate of SR that the laws of physics are invariant in all inetial frames, will someone please explain how in one frame an event is simultaneous but in another frame, inertial to be sure, that same physical event subject to the laws of physics being invariant in all inertial franes is an event that iss not simultaneous. ie the event in the moving frame is variant to the laws of physics in the moving frame? Maybe the rule is, there is just one exception to the invariance of physical law?

This is much lengthyer than your post and I wouldn't exoect an off the cuff answer. Pick and choose what you think is most important.

Not so. I put you into space in a spaceship with no windows. Is there any experiment you can carry out (without looking outside of course) which will allow you to determine whether you are moving or not?

Matt
what is wrong with looking outside?

ram2048

Jun9-04, 05:42 PM

geistkiesel

Jun9-04, 05:49 PM

ram2048

Evidently you don't want to understand this theory. The nice thing about physics is that if you don't want to understand it, you won't!
jdavel please do us all one large favor, please.

Take two unambiguous subjects for analysis.

1.The opening thread stipulated that SR theory would predict the simultaneously emitted photons in the stationary frame would not be simultaneous in the moving frame. This given.

2. Photo-sensitve strips in the moviong frame were exposed by the subject photons when the photons were emitted. The locations of the exposed strips were efefctively exactly at A and B in the srationary frame an each srip, faction of a micron thick are sufficient in numbers on both ends of themoving platform such that the photons are guiarant3edd to be exposed whatever the shrinking of he moving platform. Finally, the ps-stips are identified in pairs where each pair is equal in distance from the midpoint M' of the moving frame, which was determined by the same physicallaws tha assured us that M was themidpoint of A and B in thre stationary frame. This is given,. We do not use the simultaneous arrival iof the photons as the source of the information taht M is the mispoint of the sources, nor in the calculations that SR used in the consklusion that were made, though theinforamtion couls be used. No reason to exclude it. But this was determined long before the experiemnt ran..

Refreshing your memory of the experiment, when M' was at M, the midpoint of the phioton sources at A and B, photons were emitted simulatneously in the stationary frame.

You can see the apparent copntradiction: Emitted photons simulatneously in the stationary platferom detected effectively instantaneously in the moving frame at the same instant the photons were emitted in the stationary frame.

Give us your veyr very best shot at analyzing the situation. Does SR defeat observation, pe se? IS there any physical law connecting the difference in simultaneity conclusion in the different frames? If physical laws arei invariant in inertial frames why is this exception allowed to continue in the language undiscussed?

baffledMatt

Jun9-04, 05:59 PM

Both observers send out 1 second pulses from their respective clock timing circuits. The slower frame's pulse rate is less than the fastest frame's pulse rate and vice versa, right? This was off the top of my head a few days ago. I have used it and have had no reply,but is seems like it should work.

No, if they are sending the same 1 second pulses to each other they will each observe exactly the same thing - either a slowing down of the pulses (red shift) or a speeding up (blue shift) depending whether they are moving toward each other or away from each other.

Some have suggetsed that no, the moving frame cannot guarantee I can find the midpoint of the ps_strip. My counter, which I havwen;'t seen challenged, is I used the same law of physics that allowed the stationry framwe to determine themisddle of he photon sources at A and B.

I still don't see exactly what you mean by this. Could you perhaps give me the dumbed down version of how exactly this is done? It's no good you saying 'using the same physics' since we evidently don't agree on exactly what the physics is!

Can you see my position when I want to say that the photons were mesured in the moving frame equal distances from the midpoint of gthe moving frame when the pulses were emitted by A and B. Just can you see my reasonaning. I will not hold you to an adnmiission that I am right and you are adnmitting error. Only can yo see why I might say that?

I can see why you might think that - I remember having exactly these sorts of problems with SR. The way I would always go about solving these problems though was to sit down and just calculate the thing. To illustrate:
Let's make some definitions:
x : space coordinate in 'stationary' frame. x = 0 is the midpoint between A and B, which are located at x = -L and x = L respectively.

v : velocity of the moving observer relative to the stationary observer.

Ok, so now I will calculate what the 'stationary' observer sees. Note I am not going to use any SR as such. We first calculate the location of the point where the 'moving' observer and the light from A coincide. This happens at a time t_A after the light is emitted from A and the point where they meet we denote d_A. By equating the distance the light moves in this time to the distance the moving observer moves we have:

vt_A = d_A
ct_A = d_A + L
so
\frac{v}{c}(L + d_A) = d_A
and
d_A = \frac{\frac{v}{c}L}{(1-\frac{v}{c})}

Similarly, we can derive d_B which is the point where the moving observer meets the light from B
vt_B = d_B
ct_B = (L-d_B)
so
\frac{v}{c}(L-d_B) = d_B
and again
d_B = \frac{\frac{v}{c}L}{(1+\frac{v}{c})}

Evidently, d_A \neq d_B so the photons do not hit the moving observer at the same time/place.

I am saying I do not have to argue time dilation, shrinking absolte space and time , noe of that. Why? I have agreed in the results of SR. Whi amI or you for that matter to be chosen as the obne who arbitrates what is physical law? Me saying SR will predict such and such doesn't make the poustulates true, the posstulates stand ot dall on their pwn merist.

Ok, now repeat the calculation above, but for the moving observer. The only SR result I want you to use is that each observer must observe the same value for the speed of light. If you do this you should see that you need things like time dilation and lack of simultaneity to resolve the observations.

Matt

Alkatran

Jun9-04, 06:05 PM

geistkiesel

Jun9-04, 06:09 PM

i'm going to sum up my arguments for the time being and take an interlude.

it is entirely valid for the moving observer to claim the photons were not simultaneous because that is his perception of reality.

it's entirely valid for any stationary observer to say the photons WERE simultaneous because that is their perception of reality.

when you get 5.99 billion people to say, "yeh dude, those were simultaneous" you must be open to the possibility that perhaps your perception of reality is wrong BECAUSE you were the one moving, NOT everyone else

possibly the reason you guys are complaining so much is because it is TOO trivial :D

-peace for now

Amen, a moment of silence for ram2048 conclusions, if you please.

In my country, a free country, we have the freedom of thought to perceive as we do, nay as we will. One is allowed the freedom, yes just one person in a crowd of 70,000 say, is allowed the unconditional freedom to perceive pink elephants doing the moon walk, where all others, the 49,999 all perceive the running of the Kentucky Derby. Why do the people scream as they do when watching horse races, you might ask? Because tt makes the horses run faster.

ram2048, take a well deserved few days off. Perchance the bubbling thoughts in the myriad of vortexes in the mind will produce an moe enlightened way of describing what is so clear to others, namely this writer, and what is becoming clear to the host body of scientists.

Tis trivial is it? as a physics problem, I agree. However, the economics of the trivial physics problem poses a nontrivial barrier: Who will cover all the professional funeral expenses of the SR disenfrancised? I am not familiar with the insuance coverage in the theoretical physics industry, but surely these matters must have been discussed in the past, you know, a sudden demise of a preciously held physical theory, like when an observational guillotine suddenly falls foom on a high and . . . .?? Surely everybody is covered, aren't you?

baffledMatt

Jun9-04, 06:15 PM

it is entirely valid for the moving observer to claim the photons were not simultaneous because that is his perception of reality.

it's entirely valid for any stationary observer to say the photons WERE simultaneous because that is their perception of reality.

Ok, I can safely say we agree here :smile:

when you get 5.99 billion people to say, "yeh dude, those were simultaneous" you must be open to the possibility that perhaps your perception of reality is wrong BECAUSE you were the one moving, NOT everyone else

The laws of nature are not a democracy! If you want to have one frame more 'correct' than another there must be a real physical difference so that you can determine which frame this is.

Why can't both perceptions be correct? You have the idea that there is some 'true' reality of whether or not these events were simultaneous. SR is trying to tell us to forget this notion, there is no way we can tell which frame is 'correct', so we shouldn't try to consider it.

Matt

geistkiesel

Jun9-04, 06:16 PM

But to you they really weren't simultaneous and those 5.99 billion people (who, understanding relativity,) admit that they're simultanity may not be your simultanity.

Alkatran, pray tell man, is simultaneity a characteristic of humanity? Or, on the other hand, is simultaneity a righteous physical phenmomenon? Must one understand SR befoe simultaneity is physically real? In the same sense does my undersatnding of the laws of gravity the assurance I need in order not to drift over to Bakersfield California for instance? I have to know and understand the laws of gravity, you're suggesting in order to avoid Bakersfield, California? My god man, think of the consequences.

baffledMatt

Jun9-04, 06:45 PM

2. PS-strips (photo-sensitive strips) located within 1 photon wave length of the sources of the photons emitted in the stationary frame are exposed (in the moving frame) as the photons are emitted. Super fast film, ps-strips << micorn wide, Each ps-strip locatable to the midpnt M' with mirror image ps-strip at other end within any mutuallly agreeable resolution.
Something's gotta give.

Ok, so we have these film strips and after the experiment they will have two little dots on them corresponding to the points A and B when the photons were emitted. In the moving frame we measure the distance between these points using a metre ruler and come up with a length. We have (using my previous notation) that in the stationary frame the distance between A and B is 2L. However, due to Lorentz contraction the stationary observer will observe the moving observer to measure this distance to be 2L\gamma

NB
\gamma = \frac{1}{\sqrt(1-\frac{v^2}{c^2})} > 1

This is because he sees the moving observer trying to measure a distance of 2L with a shorter (contracted) measuring stick. So he is observing the moving observer making a measurement larger than 2L.

The moving observer on the other hand sees the distance between A and B contracted. Hence, naively you might think he will measure a length of 2L/\gamma. This cannot be because what each observer sees must be the same (remember that we just had the stationary observer seeing the moving observer measure a longer length), so somehow the moving observer must also measure 2L\gamma. This happens because he does not observe the photon emissions to be simultaneous. Hence there is a little bit of time between the first and the second photon, which makes his measured length larger. This will ensure that his observation coincides exactly with that of the stationary observer.

Matt

ram2048

Jun9-04, 08:18 PM

ok here we go.

suppose in the future we all live in space and drive spacecars around (whee fun)

suppose i get into an accident with another driver in a spacecar who happened to be stopped at a spacestopsign.

now on his UPS (universal position system) his data is telling him that by triangulating the positions of key points in the solar system he was "globally accepted" to be at rest in relation to the rest of the solar system.

MY UPS has the entire universe moving in relation to me.

when the cop comes by in his space cop car, do you think my defense of "relatively, he rammed me as well" is going to stand up in court?

what if spacecars were only built in such a way that they can only go FORWARDS, leaving absolutely NO DOUBT that i DID rear-end the other guy?

Hurkyl

Jun9-04, 08:59 PM

Alkatran

Jun9-04, 09:29 PM

Alkatran, pray tell man, is simultaneity a characteristic of humanity? Or, on the other hand, is simultaneity a righteous physical phenmomenon? Must one understand SR befoe simultaneity is physically real? In the same sense does my undersatnding of the laws of gravity the assurance I need in order not to drift over to Bakersfield California for instance? I have to know and understand the laws of gravity, you're suggesting in order to avoid Bakersfield, California? My god man, think of the consequences.

An understanding and acceptance of relativity would probably help in understanding (or convincing, depending on your point of view.. how relative) that my simultanity isn't your simultanity.

I only said that even if those billions of people perceived the event, they wouldn't say it was simultaneous for others if they believed in relativity.

geistkiesel

Jun9-04, 11:30 PM

Ok, so we have these film strips and after the experiment they will have two little dots on them corresponding to the points A and B when the photons were emitted. In the moving frame we measure the distance between these points using a metre ruler and come up with a length.

No we use reflected light pulses and set the ps-strips the same way that M was established as the midpoint od A and B. As we have an excess of ps-strips whose width << lamba(photon), i.e. sub micron ranges, SR shrinking is not a concern. We can rig it so that the number of ps-strips are two or three orders of magnitude less than the photon wave length. Overlap and overmesure o an overkill degree.

We have (using my previous notation) that in the stationary frame the distance between A and B is 2L. However, due to Lorentz contraction the stationary observer will observe the moving observer to measure this distance to be 2L\gamma
\gamma = \frac{1}{\sqrt(1-\frac{v^2}{c^2})} > 1

This is because he sees the moving observer trying to measure a distance of 2L with a shorter (contracted) measuring stick. So he is observing the moving observer making a measurement larger than 2L.

Not quite. As we have an excess of ps-strips, we assure ourselves that whatever shrinking occurs, one strip will be colocated at A and B within a minimum acceptable errror. The numbered ps-strips guarantees the location of mesurements being equal distant from M' established by relected laser measurements. The frame knows nothing of stationary observers, perceptions or even that an experiment is being conducted. What the stationary observer sees does not imply the use of gamma for any useful purpose. WE want to deermine if the simultabeous emission of photons mitted in the stationary frame are emitted simultaneously in the moving frame. And this is all is it not?

The moving and stationary observers each have the planet's best SR and Dissident SR physicists to assure the most rabid sceptic, either way, that A = A' , M = M' and B = B' when the photons are emitted and detected by the ps-strips to a resolution << perturbations due to shrinking effects. Remember, we are only verifying simultaneous emission of photons in the moving frame or not. We need not concern ourselves what observes are noting about the frames cordinates.

The moving observer on the other hand sees the distance between A and B contracted. Hence, naively you might think he will measure a length of 2L/\gamma. This cannot be because what each observer sees must be the same (remember that we just had the stationary observer seeing the moving observer measure a longer length), so somehow the moving observer must also measure 2L\gamma.

No quite. We are not concerned with what observers see. The human element is totally irrelevant. No one is making corrections based on seeing another accross the way doing anything like making a measurement.

This happens because he does not observe the photon emissions to be simultaneous. Hence there is a little bit of time between the first and the second photon, which makes his measured length larger. This will ensure that his observation coincides exactly with that of the stationary observer.
I don't see this at all. If the stationaray observer, knowing the photons are emiotted simultaneously in his frame, as a given, (OK for this we'll put a stationary observer at A and B, temporarily,), how pray tell can the stationary obsever also see the photons emitted a little time later, just to stay synchronuzed to SR theory?in between? It can't be done baffledMatt, there is no such stationary observer in the universe. this cannot be done.

If you insist on this as strenuously as you are able look what you are arguing. The mere fact that the moving frame is present as we described it places a physical delay in photon emission where if no moving frame passes, the photons are emitted simultaneously. They are still emitted simultabneously, but again, two times. Do you suspect a conservation of energy problem here baffledMatt,? A physical event forced on the physical laws by postulated imperative? Is there any phsyical law, such as that SR descriibed that is invariant under all this? You are left with a shrug of the shoulders and " that's the way it is", arent't you? Also, the stationary observer has a certificate of the experimental conditions that the photons are emitted simultaneously in the stationary frame. How can the mere presence of the moving frame alter the sequence? Can it be both ways? Human observers within eyelashes of each other see the same physical event occuring at different times? When the observers see the photon and raise their hand in recognition, you say each observer raises their hand at diffeent moments, and the stationary observer sees this? Incredible, isn't it baffledMatt? While they are staring each other in the face? Bam, the photons are emitted, then bam, they are emitted again? This is SR theory applied to force experimental observation being consistent with theory, it is called a mathematical contrivance with no physical analog attached. These results are looked at weeks after the experiment. No observer has "seen" any other observer measure anything. read some post of those supporting SR theory. Sk oyuself which ae useful, and which are useless, which ae supportinmg for pure professional or personal reasons only. Which haven't a clue to what is occuring.

Only the ps_strips and ergo their mutual distance to M' when the photons were emitted at t = 0 are of concern.

baffl4ed Matt, you keep referring o what " . . . the stationary observer sees the moving observer.." etc
If you agree that when M' is at M and this is the instant the photons are emitted from A and B is it not allowable to either insert the value '0' in a clock located at MM', a colocated coordinate set, say 0,0 in both frames? For confifdence we would want to 0 all clocks at all observation points. But for the ps-strip exposures I see no need for any clocks.

Certainly we can minimize the error sufficiently to negate any shrinking or time dilation problems.

I will accept your professional sense of honesty and let you iron out the engineering technicalities, it is your budget.

You want to go over this again, so be it. there is a ton of stuf in here baffled matt, a on. too much for mortals to digest in so short a time, unless one has an ephiphonous variance in their outlook. I have to tell you I am not equipped to take any prisoners.

geistkiesel

Jun9-04, 11:47 PM

wow geist... harsh!

maybe not the best way to hold a discussion, but it's interesting in a JerrySpringer-esque kinda way :D

In an otherwise uneventful military hitch, I was talked to in what you would really call "harsh". No one manifested any overt symptoms of pathological eactions to this even though the nature of the conversation was intensely maintained for three months. I was taught to speak by an official of the US Givernment, a cruel and sadistic drill instructor. I speak "official" English.

I would apologize to Wespe if it weren't for the admonition by America's greatest thesbian of all time, John Wayne, who said, and very seriously too,: "Don't apologize, its a sign of weakness." Actually this probably wasn't original JW, but who cares at this juncture. Hell, I'll claim it as a Geistkiesel original and dare anyone to challenge me, well not anyone, but a lot of 'ones'. It is like my number 1 baseball hero, the worlds greatest, slickest and most effective baseball pitcher of all times, Leroy "Satchel" Paige, remarked" , "'When your enemy is stronger than you, walk him."

geistkiesel

Jun9-04, 11:53 PM

An understanding and acceptance of relativity would probably help in understanding (or convincing, depending on your point of view.. how relative) that my simultanity isn't your simultanity.

I only said that even if those billions of people perceived the event, they wouldn't say it was simultaneous for others if they believed in relativity.
Then you're telling me I need not fear and uncontrolled spontaeous visit to bakerfield, Thats a relief.

I can comprehend believing in god and the tooth fairy and especially Santa Claus, I 've seen Santa Claus, but what physical significance do we place, ina scientific sense, what any number of people believe? I thought that was the reason scientists split off from the dictates of religious mythologicval dogma.

geistkiesel

Jun10-04, 12:04 AM

Keep in mind that the principle of relativity is part of classical mechanics; it was not invented by Einstein for the purpose of developing SR.

And to drive home the cosmological point about there being no good point of reference, consider these:

(1) Suppose in your future that we also considered a similar accident occuring on the streets of future earth. In this case, neither car was stationary with respect to the solar system!

(2) The guy who stopped was not at rest with respect to the rest of the universe. And, for instance, you might have been at rest with respect to the Milky Way. The important point is that a particular frame of reference was chosen (in this case, "at rest WRT the solar system"), so the traffic law is defined relative to this choice.

(3) What is "at rest in relation to the rest of the solar system" anyways? Things are moving in all sorts of directions, accelerating all over the place.

Not from the laws of physics do we intuit no absolute reference point but from the sheer technological chore of finding such a point. Even if the best instumentation conceivable were at our disposal and we were able to measure out to 20 billion light years and found a most perfect spot, it is the stellar entiities beyond 20 billion light years that would screw up the perfection. In any event we might perceive a need for a contacted and practical perfect rest point for some perceived need, for some finite duration of time. Ergo to within useful limits I see nio rule or law of physics preventing anyone from determining a relative point that does the job for the situation under consideration. i wouldn't tell any SR theorist about it though, no way. Read the posts, see what you would be subjecting yurself to?

jdavel

Jun10-04, 12:18 AM

geistkiesel

Jun10-04, 12:22 AM

geistkiesel

Jun10-04, 12:39 AM

ram2048 said, "it's a "Given" part of the experiment that they WERE emitted simultaneously as outlined in Einstein's original set up"

Not quite! They were simultaneous in the stationary frame, but, as it turns out, they weren't simultaneous in the moving frame. That's not an assumption; it's a conclusion based on the assumption that c is the same in both frames.

Here's "Einstein's original set up":

IN THE STATIONARY FRAME THE FLASHES ARE SIMULTANEOUS and are equidistant from the moving observer at the time of the flashes. So, the stationary observer sees light from the two flashes reach the moving observer at different times.

The moving observer agrees, light from the two flashes reached him at different times. Then he looks at the marks that the flashes made on his train, sees they are at equidistant from his seat on the train. Since he knows that c was the same for the light of both flashes, he concludes that THE FLASHES WERE NOT SIMULTANEOUS IN THE MOVING FRAME.

You are 100% absolutely correct. I did not intend to imply differently. In the opening post of this thread I stipulated that SR would predict the events were not simultaneous in bot he frames.

Simulatneity is not measured by your, AE's gedun
jken, it is measured by whether the photons emitted at A and B simultaneously were emitted into the moving frame simultaneoulsy, and the events of the staggered arrival is used as "proof" that the event were not emitted simultaneously in the moving frame. I recognize there is a considerable difference in AE'a trivial example and modern use, but his model is used tioday to emophacize and explain points right?

But look at what AE was using to justify loss of simultabneoty: the reception of photons at different times after the photons were emitted. I see no "relativity" implications here. Take the situation one step farther than AE took. The different arrival times could have been considered by an alert and acute scientist.'" HMMM, two photons different times and places. If the photons were emitted simultaneously and I were moving then I would see one before the other. Or if the photons were emitted at different times I might see them arriving simultaneously at the moving fame or at different times and not know which was emitted first. AE didn't ahve to stop at the naive point in the analysis that SR theorists are quoting this very day, yours not the first.

geistkiesel

Jun10-04, 12:46 AM

geistkiesel said, "Give us your veyr very best shot at analyzing the situation."

Sorry, the situation you're describing is too complicated to explain in one "shot". And your insistence that SR is wrong would doom to certain failure any attempt of mine to convince you otherwise!

But how about this? I'll explain it one step at a time. As long as you understand, and say that you believe, each step, I'll give you the next one. At the end you'll understand why SR is right and you were wrong.

You can agree to this by starting a new thread on the Relavitity Forum with the title "Why isn't the simultaneity of two events absolute?"

I'll be waiting!
I didn't say Sr was wrong for th e purposes of this thread. I stipulated SR would pedict no simultaneity. What I would ask, is look at the gedunken used as a template for this thread. Discuss in a professsional way the apparent pardox of he observed emission simultaneously being observed nonssimultaneously in the moving frame. Use the measurement condition o of the thread. I've been living SR for a while now, I do not need another proof. I am convicned that you will come up with what eveybody else came up with, Talk about the thread hypo, i f you please,,and if you are able.

geistkiesel

Jun10-04, 12:55 AM

and I was just about to say exactly the same thing about your link.

Errrm, hello?

*pointing to the last post I made*

Matt
baffledMatt can you point out to me by argument or otherwise whe e mylinkfails in your estimation? (http://frontiernet.net/~geistkiesel/index_files/) Elaborate or brief, just specifics please. Myself, I thought it original which it will probable remain , who would want to confiscate silly physics? Any hint to a specific flaw(S) would be greatfully appreciated. Thanx G

geistkiesel

Jun10-04, 01:04 AM

No, if they are sending the same 1 second pulses to each other they will each observe exactly the same thing - either a slowing down of the pulses (red shift) or a speeding up (blue shift) depending whether they are moving toward each other or away from each other.

I still don't see exactly what you mean by this. Could you perhaps give me the dumbed down version of how exactly this is done? It's no good you saying 'using the same physics' since we evidently don't agree on exactly what the physics is!

I can see why you might think that - I remember having exactly these sorts of problems with SR. The way I would always go about solving these problems though was to sit down and just calculate the thing. To illustrate:
Let's make some definitions:
x : space coordinate in 'stationary' frame. x = 0 is the midpoint between A and B, which are located at x = -L and x = L respectively.

v : velocity of the moving observer relative to the stationary observer.

Ok, so now I will calculate what the 'stationary' observer sees. Note I am not going to use any SR as such. We first calculate the location of the point where the 'moving' observer and the light from A coincide. This happens at a time t_A after the light is emitted from A and the point where they meet we denote d_A. By equating the distance the light moves in this time to the distance the moving observer moves we have:

vt_A = d_A
ct_A = d_A + L
so
\frac{v}{c}(L + d_A) = d_A
and
d_A = \frac{\frac{v}{c}L}{(1-\frac{v}{c})}

Similarly, we can derive d_B which is the point where the moving observer meets the light from B
vt_B = d_B
ct_B = (L-d_B)
so
\frac{v}{c}(L-d_B) = d_B
and again
d_B = \frac{\frac{v}{c}L}{(1+\frac{v}{c})}

Evidently, d_A \neq d_B so the photons do not hit the moving observer at the same time/place.

Ok, now repeat the calculation above, but for the moving observer. The only SR result I want you to use is that each observer must observe the same value for the speed of light. If you do this you should see that you need things like time dilation and lack of simultaneity to resolve the observations.

Matt

I ain't gona do your calculations for you. Hurkyl conned me into one of those. But then he complimented me on my calualtions, which showed simple way to determine if thephotons e were whatever. I think he screwed up, by complimenting me on something that effectiely undermined his specific point there. My argument assumes SR. so the easy out is the rationalization for te apparent paradox of the simulatneous emission of photons in a stationary frame turn magically into nonsimultaneity by the mere presen e of an inertial frame other than stationary in the vicinity.

geistkiesel

Jun10-04, 01:17 AM

baffledMatt, modify my two clocks with the same frequency. The new model was loaded on one ship, Its frequency is 10^10hz. Now they can tell which is which and who is moving slower or faster? assuming they each had acceleration history to share or acceleration history available in some form. I suppose they wouldn't need the different clock speeds then would they?

ram2048

Jun10-04, 01:28 AM

still seems silly that SR prefers the conclusion that simultaneity is at fault and not personal perspective. it's much more logical to conclude that personal reality is skewed because of movement and calculate backwards to conform to what the rest of the (stationary) universe sees.

but that's one of the things that's a matter of opinion i think. <shrug>

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

Doc Al

Jun10-04, 04:29 AM

In an earlier post in this thread, I made a sloppy statement. I accused poor geistkiesel of assuming that:

(1) The marks on the photo-sensitive strips (caused by the photon emissions) in the moving frame are equally spaced from the point M' (which passed M at the exact moment that the clock at M read t=0): Not true!
(2) That the moving observers detect the photon emissions as happening simultaneously: Not true!

Yes, he made those assumptions--but I had meant to mark the first one as true. That first statement should have read:
(1) The marks on the photo-sensitive strips (caused by the photon emissions) in the moving frame are equally spaced from the point M' (which passed M at the exact moment that the clock at M read t=0): True!
I had meant to contrast assumption 1 with assumption 2, but I messed it up.

Just for the record, moving frame would record the time and position of the photon emissions (on geistkeisel's photo-sensitive strips) as follows.

In the "stationary" frame (O), photons are emitted at A (x = -L; t = 0) and at B (x = L; t = 0). These flashes are simultaneous in the stationary frame. (I assume that A and B are a distance L from the midpoint M.)

In the moving frame (O'), the photon emissions are recorded on the strips at the following postions and times:

The photon emission from B is recorded at:
x' = \gamma L
t' = -\gamma \frac {vL}{c^2}

The photon emission from A is recorded at:
x' = -\gamma L
t' = \gamma \frac {vL}{c^2}

Where, as usual, \gamma = 1/\sqrt{(1 - \frac{v^2}{c^2})}.

I apologize for adding any additional confusion to this discussion. (I will add a note to the earlier posts.)

Of course, as I said before, much of this discussion--with geistkiesel's extraneous assumptions--could be avoided by addressing Einstein's actual argument in his train gedanken experiment. Of couse, geistkiesel is unable (or unwilling) to do that.

Note added: Just because that first assumption happens to be true, does not mean that it is justified and can be merely assumed. On the contrary, both assumptions are arbitrary, unjustified, and most importantly extraneous to the simple argument of Einstein. Einstein makes no such assumptions in his demonstration that simultaneity must be relative.

baffledMatt

Jun10-04, 05:52 AM

baffledMatt can you point out to me by argument or otherwise whe e mylinkfails in your estimation? (http://frontiernet.net/~geistkiesel/index_files/) Elaborate or brief, just specifics please. Myself, I thought it original which it will probable remain , who would want to confiscate silly physics? Any hint to a specific flaw(S) would be greatfully appreciated. Thanx G

The is no physics in your link! You don't make any calculations so how can I judge it? Then I click on the link to the so-called analysis and there is just a tiny little calculation which has already assumed your point is correct.

I have tried to show you calculations in a way to help you understand how the whole thing works but you reply with:

I ain't gona do your calculations for you. Hurkyl conned me into one of those.

I mean, do you want to be helped? Sorry, but physicists are not philosophers. You have to do calculations to really understand things.

so the easy out is the rationalization for te apparent paradox of the simulatneous emission of photons in a stationary frame turn magically into nonsimultaneity by the mere presen e of an inertial frame other than stationary in the vicinity.

Why is this a paradox? where is the magic? You are the one trying to bring in fairies by demanding something 'is' or 'is not' simultaneous.

No baffledMatt, why do you refer to stationary observers seeing what moving observer do.

This is the crucial point. Imagine I am observing someone moving relative to me. I can watch everything he does, see him measure things with his own measuring devises etc. Now, there must be no discrepancy between the measurements he makes and the ones I observe him to make. It's just that the explanation to why he got the result he did will differ between our frames (I'll say that his measuring stick shrunk wheras he will say that the events were not simultaneous.)

As another example, try this one. Imagine we both have clocks, we know that when they are in the same inertial frame that they tick at exactly the same rate - and they stay like that (they are very good clocks!). Now, you are moving at a velocity v relative to me and at the moment you pass me (ie our x coordinates coincide) we synchronize the clocks - make sure they are telling exactly the same time at that point in time.

Now, you are moving relative to me and so I observe your clock as ticking slightly slower. However, since I am also (relativistically speaking) moving relative to you, you shall also observe my clock as going slower than yours. We each observe each other's clock as going slower. But this is fine, there is no disagreement as such because all I can say is "I observe your clock as slow", I don't know what you might be observing.

Ok, now imagine that you break your clock at time T by your clock. This is an event we must both agree on, when your clock had the small hand pointing at T you broke it. Now, I observe your clock as ticking slower than mine so my clock reads \gamma T at the point when you break it. In fact, as soon as I see your clock stop i stop mine also so we must again both agree on this measurement, we both see \gamma T on my clock.

However, now for the 'paradox'. In your frame you were observing my clock as running slower than yours! So you might reason that my clock ought to read T / \gamma. But it doesn't, it reads \gamma T and we can both see this.

Resolve the paradox.

Not quite. As we have an excess of ps-strips, we assure ourselves that whatever shrinking occurs, one strip will be colocated at A and B within a minimum acceptable errror.

But then your establishment of simultaneity will only be accurate to this same error. You say that you will use an excess of strips so this error will in fact be pretty huge. Thus the results will prove nothing.

The numbered ps-strips guarantees the location of mesurements being equal distant from M' established by relected laser measurements.

But how are you performing these measurements?! You are assuming that these are things you can simply measure and will give you your expected result QED. What you are forgetting is that SR is all about measurement! It's a fact that different observers measure different things. But they all observe the same events.

Do you see the subtlety? "The distance between the sources is 2L" is a measurement which observers can disagree on, as is "the emission was simultaneous". But "I measured the distance to be 2L" is an event. All observers will see me make this measurement and observe me getting the result I did. It is only the events which we all agree on.

The frame knows nothing of stationary observers, perceptions or even that an experiment is being conducted.

Yes it does. The frame is trying to measure events which occur in a different frame. I would say that therefore this frame must know a lot about the stationary observer.

Matt

baffledMatt

Jun10-04, 05:59 AM

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

This is then a purely legal issue, but has nothing to do with physics.

Matt

Hurkyl

Jun10-04, 06:14 AM

to conform to what the rest of the (stationary) universe sees.

The universe of which I'm aware is full of objects travelling every which way, constantly accelerating to different directions, and without any nice, global structure into which everything fits.

So, there are two very problematic things about this idea; why should there be such a thing as a "stationary universe", and how do we tell what it sees?

in any case in terms of space and relative fixed points, i think it's imperative that 3 dimensional space be charted in order to be utilized.

when living on earth it's easy to use relative fixed points for reference, we do it intuitively and almost instinctively daily, driving between the lines, walking on the sidewalk, sitting down on a chair instead of missing it and landing on the floor...

in space you're dealing with far less points of reference AND far greater possible speeds, so everything becomes that much bigger of a problem.

seems only a matter of time before we MUST define a system to all conform to in order to share this real estate.

And the point is that a reference frame is chosen, and then everything else is defined relative to that.

Relativity does not say that you're not allowed to choose a reference frame as a "standard"; it merely says that any reference frame would suffice.

geistkiesel

Jun10-04, 06:29 AM

This is then a purely legal issue, but has nothing to do with physics.

Matt

It sounds to me ram2048 is talking about the real estate of space. A need getting our story straight if we are ever going to make some serious ventures into the cosmos.. This is what I read and we all know about perceptions and observations do we not?

geistkiesel

Jun10-04, 06:43 AM

Hurkyl

Jun10-04, 06:51 AM

geistkiesel

Jun10-04, 07:40 AM

I still don't see exactly what you mean by this. Could you perhaps give me the dumbed down version of how exactly this is done? It's no good you saying 'using the same physics' since we evidently don't agree on exactly what the physics is!

baffledMatt, whatever the physics used to measure the midpoint in the stationary frame is used in the moving frame.
You can pick you own method.

measure the midpoint of M using some eflected laser scheme relying on time of flight of the beam photons. Ditto for M'.The M' only allows us to guarantee that the number of an an exposed ps-strip near A will have the same number of the exposed ps-strp at B. The scientists calculate shrinking from a known velocity acihieved through repeated acceleration schemes. Approaching A and B from two directions and taking phtographs of the AA' positions, the moving frame can assure itself that wherever the photon expsoes the ps-strip its eflected number at the other end of the frame will be identically exposed. Observers eye-to-eye 1 wave length apart when the photons are emitted must see the same event. If you want to slow down the brain functions in the moving frame ok, by the photons aren't going to play any special games with the moving frame just because it is there. So if brain slowing is the physical ef fect of nonsimultaneity I can understand that.

I can see why you might think that - I remember having exactly these sorts of problems with SR. The way I would always go about solving these problems though was to sit down and just calculate the thing. To illustrate:
Let's make some definitions:
x : space coordinate in 'stationary' frame. x = 0 is the midpoint between A and B, which are located at x = -L and x = L respectively.

v : velocity of the moving observer relative to the stationary observer.

Ok, so now I will calculate what the 'stationary' observer sees. Note I am not going to use any SR as such. We first calculate the location of the point where the 'moving' observer and the light from A coincide. This happens at a time t_A after the light is emitted from A and the point where they meet we denote d_A. By equating the distance the light moves in this time to the distance the moving observer moves we have:

vt_A = d_A
ct_A = d_A + L
so
\frac{v}{c}(L + d_A) = d_A
and
d_A = \frac{\frac{v}{c}L}{(1-\frac{v}{c})}

Similarly, we can derive d_B which is the point where the moving observer meets the light from B
vt_B = d_B
ct_B = (L-d_B)
so
\frac{v}{c}(L-d_B) = d_B
and again
d_B = \frac{\frac{v}{c}L}{(1+\frac{v}{c})}

Evidently, d_A \neq d_B so the photons do not hit the moving observer at the same time/place.

Ok, now repeat the calculation above, but for the moving observer. The only SR result I want you to use is that each observer must observe the same value for the speed of light. If you do this you should see that you need things like time dilation and lack of simultaneity to resolve the observations.

Matt

The problems you used to have were eleiminated using SR. Without SR there is no paradoxes, none. It osunds like a religious convert, really. "My life turned around when I found Godot", or something similar.
Do you want to argue all the experimetnal results proving the postuilates, while I show all the contradictions, exceptions, flaws. You show me my flaws I show you yours. This really isn't a physics discussion, all that much is it?

I know its my fault.

You needn't show this to me I believe it will work as you say. My doing the calculations will not change anything. There is still the matter of the ps-strips being exposed when the photons are emitted. There is no paradox until SR is applied. then we have the big sticker. Mere presence of observers grossly affecting a physical event, by operation of postulate. Is there any language you can use other than that demanded by Sr that can explain the physical manifestation arising from a simultaneous event by operation of the presence eof observers. Can you do this?

If there are no observers on a moving frame observing is there a nonsimultaneous event occuring? I am not asking a question with any implied determinism demanded in the answer, but the inversion of reason and the operation of physical law don't work in SR, at lexst not heer. I discard it.

geistkiesel

Jun10-04, 07:55 AM

Geistkiesel: here is your experiment drawn as a diagram:

0 0 0
*\ * /*
* \ * / *
AZ \ MN / BY
AZ \ MN / BY
AZ \ MN / BY
A Z \ M N/ B Y
A Z \ M * B Y
A Z \M/N B Y
A Z * N B Y
A Z M\ N B Y
A Z M \N B Y
A Z M \N B Y
A Z M * B Y

Z A NM 0
ZA NM /*
ZA 0 / *
ZA * / BY
0 * / BY
*\ MN / BY
* \ MN/ B Y
AZ \ M* B Y
AZ \M/N B Y
AZ * N B Y
A Z M\N B Y
A Z M * B Y

Legend:
A, M, B: your A, M, B
Z, N, Y: Your A', M', B'
0: A point where a clock read 0
\, /: photons
*: Multiple things at this point (such as two clocks, or a clock and a photon)

The first diagram is the stationary reference frame. The second diagram is the moving reference frame. Space runs from left to right, and time increases as you go downward.

The first diagram was taken directly from your experimental setup. I placed M and N midway between A/Z and B/Y, and simply drew out the time evolution of the system.

To draw the second diagram, I started with the unambiguous fact that both photons meet precisely when they each meet M. I drew the time evolution backwards, and used the fact that N is midway between Z and Y when M meets N to place where Z and Y should be. I then applied the fact that Z meets A and are both set to zero when the left photon is emitted to finish off the left side of the diagram, and similarly for Y and B. I did, however, have to estimate how N lies relative to M. Any other such choice yields a similar diagram.

Your mistake, as everyone is trying to tell you, is made clear from the diagram. In the moving frame, the clocks are not synchronized; you can see that they are all zero at different times. We see that SR can handle this scenario perfectly well, as long as you don't start with the assumption that synchronization in one frame = synchronization in all frames.

(NOTE: In the second drawing, to keep the diagram small, the lexical distance between A and M is 6 and one-third characters)

You are mistating the thread hypothetical. I have never said the frames were sycnchronized between them. I have stated the opposite. Your statement is a total falsehood, conscious or otherwiise. Your diagram , is a perversion on the original. You are trying to sabotage a thread for your own aggrandizment. Do you get a bonus, or is theprize recognition? Your just another phony, and a liar. You made it up Hurkyl

Take your lexical distances and shove them some place that will do you some good. But all I can predict now, until the last star fades from the sky is that Hurkyl is just another common usless liar and is not to be trusted, ever. I am not answering your crap anymore. Our conversation is over.

russ_watters

Jun10-04, 07:56 AM

Why don't you cut with the personal attacks and draw out a diagram of your own? Or is intentional obfuscation your goal?

baffledMatt

Jun10-04, 08:08 AM

The problems you used to have were eleiminated using SR. Without SR there is no paradoxes, none. It osunds like a religious convert, really. "My life turned around when I found Godot", or something similar.
Do you want to argue all the experimetnal results proving the postuilates, while I show all the contradictions, exceptions, flaws. You show me my flaws I show you yours. This really isn't a physics discussion, all that much is it?

The difference between this and religion is that SR is completely logical. You start from the principle of relativity and everything else is deduced from this. So the only thing I got converted to was the principle of relativity - if you want to call that religion then be my guest.

You have not shown us any flaws in SR, only flaws in your interpretation of SR.

There is still the matter of the ps-strips being exposed when the photons are emitted. There is no paradox until SR is applied.

There is no paradox, period! Only gross misunderstandings.

then we have the big sticker. Mere presence of observers grossly affecting a physical event, by operation of postulate. Is there any language you can use other than that demanded by Sr that can explain the physical manifestation arising from a simultaneous event by operation of the presence eof observers. Can you do this?

This depends on your point of view. I do not consider there to be any physical relevance to two things being simultaneous, so the results of SR do not bother me.

Why do you think there is any significance to things being simultaneous? If two events happen at different locations and the same time then there is no physical way the two events are connected - they are outside of each others light cones. So each event has no idea that it is occuring 'at the same time' as another event. Why do you attach so much significance to things which cannot be causally related?

Matt

baffledMatt

Jun10-04, 08:12 AM

You made it up Hurkyl

Take your lexical distances and shove them some place that will do you some good. But all I can predict now, until the last star fades from the sky is that Hurkyl is just another common usless liar and is not to be trusted, ever. I am not answering your crap anymore. Our conversation is over.

Riiight. All this coming from a guy trying to convince us that his own theory - which is devoid of calculations or even physics for that matter - proves that SR is wrong.

Don't you think it would be better to try and show us exactly why you think Hurkyl is lying, rather than all this immature name calling?

Matt

geistkiesel

Jun10-04, 08:24 AM

The is no physics in your link! You don't make any calculations so how can I judge it? Then I click on the link to the so-called analysis and there is just a tiny little calculation which has already assumed your point is correct.

I have tried to show you calculations in a way to help you understand how the whole thing works but you reply with:

I mean, do you want to be helped? Sorry, but physicists are not philosophers. You have to do calculations to really understand things.

Why is this a paradox? where is the magic? You are the one trying to bring in fairies by demanding something 'is' or 'is not' simultaneous.

This is the crucial point. Imagine I am observing someone moving relative to me. I can watch everything he does, see him measure things with his own measuring devises etc. Now, there must be no discrepancy between the measurements he makes and the ones I observe him to make. It's just that the explanation to why he got the result he did will differ between our frames (I'll say that his measuring stick shrunk wheras he will say that the events were not simultaneous.)

As another example, try this one. Imagine we both have clocks, we know that when they are in the same inertial frame that they tick at exactly the same rate - and they stay like that (they are very good clocks!). Now, you are moving at a velocity v relative to me and at the moment you pass me (ie our x coordinates coincide) we synchronize the clocks - make sure they are telling exactly the same time at that point in time.

Now, you are moving relative to me and so I observe your clock as ticking slightly slower. However, since I am also (relativistically speaking) moving relative to you, you shall also observe my clock as going slower than yours. We each observe each other's clock as going slower. But this is fine, there is no disagreement as such because all I can say is "I observe your clock as slow", I don't know what you might be observing.

Ok, now imagine that you break your clock at time T by your clock. This is an event we must both agree on, when your clock had the small hand pointing at T you broke it. Now, I observe your clock as ticking slower than mine so my clock reads \gamma T at the point when you break it. In fact, as soon as I see your clock stop i stop mine also so we must again both agree on this measurement, we both see \gamma T on my clock.

However, now for the 'paradox'. In your frame you were observing my clock as running slower than yours! So you might reason that my clock ought to read T / \gamma. But it doesn't, it reads \gamma T and we can both see this.

Resolve the paradox.

But then your establishment of simultaneity will only be accurate to this same error. You say that you will use an excess of strips so this error will in fact be pretty huge. Thus the results will prove nothing.

But how are you performing these measurements?! You are assuming that these are things you can simply measure and will give you your expected result QED. What you are forgetting is that SR is all about measurement! It's a fact that different observers measure different things. But they all observe the same events.

Do you see the subtlety? "The distance between the sources is 2L" is a measurement which observers can disagree on, as is "the emission was simultaneous". But "I measured the distance to be 2L" is an event. All observers will see me make this measurement and observe me getting the result I did. It is only the events which we all agree on.

Yes it does. The frame is trying to measure events which occur in a different frame. I would say that therefore this frame must know a lot about the stationary observer.

Matt

why all the talk about clocks we aren' using them? . If the moving frame has a physical entity located a wave length from a stationary entity and these entities are identical ps_strips and a photon is enmitted exactly between them exposing the ls-strips and the same event is occuring on the other end of the frames, the distance betwene the exposed ls-strips is identical at the instant the photons were emitted, lorentz terms notwithstanding.. The exposure, a fraction of a mico second is identically located, like mirror images of each other other. Your SR theory that describe other than this is bogus. Mathematical chicken scratches on a piecxe of paper does not substyitue for physical law and reason. I trust you will always go in peace with your theory.
I realize I have my own perceptions, biases and a personal fault or two, but I don't swlallow SR: it is poison for the mind, the scientific equivalent of fascism.

Explain to me Doc Al and Hurkyl flat out lying and double talking, sonfusin. attemopting to get me of on some wild goose chase. Thee men are your colleagues. I don't fault you for your fdifferences, in fact i don't fault you. You are, to be sure instinctively SR, but I haven't detected the corruption seen in the other two mentioned.

These are "mentors" . I have only seen such peversion and dishnesty in the politics in thios country. I wonder whio is pulling the chain around their necks?

Don't try this at home children, please don't try this at home.

Some say we humans are free thinkers, free to choose our own destiny. Those two mentioned chose the way they think act and work, and corrupt, always to corrupt.
Chroot, pulled a post of mine cruitical to some yoyo named carp or crap. Hedidn't think the post was scientific enough. Jesus christ, Now I cannot edit my own profile or dio any thing except woprk in thi porum. No problem, I just don't appreciate being handkled by a jerk with such insensitivity. The man is arbitary and snot nosed brat.

geistkiesel

Jun10-04, 08:26 AM

Why don't you cut with the personal attacks and draw out a diagram of your own? Or is intentional obfuscation your goal?
See the first post in the thread. I am not making personal attacks I am reposing obsevations.

Whats the matter,you dont like my threads?

baffledMatt

Jun10-04, 08:44 AM

If the moving frame has a physical entity located a wave length from a stationary entity and these entities[...]

All the time you are assuming that there are all these physical entities which 'have length X' or 'are at time Y'. How do you know this? How do you know that the distance between A and B is 2L? You measure it of course. But hang on, now you haven't actually determined what this entitie's 'true length' is, you have made a measurement - there is no way you can talk about the 'true length' because there is no measurement free way of determining it. Then what SR tells us is how these measurements will differ between intertial frames.

are identical ps_strips and a photon is enmitted exactly between them exposing the ls-strips and the same event is occuring on the other end of the frames, the distance betwene the exposed ls-strips is identical at the instant the photons were emitted, lorentz terms notwithstanding.. The exposure, a fraction of a mico second is identically located, like mirror images of each other other. Your SR theory that describe other than this is bogus.

You still think that you are determining some true 'real' property using this measurement. What SR tries to tell us is that the only quantities which do in fact have a 'true' property are the invariant ones (such as rest mass).

Until you think very very carefully about the way you are making your measurements you will never understand this.

Mathematical chicken scratches on a piecxe of paper does not substyitue for physical law and reason.

Excuse me, are you a physicist or a philosopher? If you want to talk this way then please move your discussion to the philosophy section of the forum. I'm afraid that mathematics is the only way we know of to build a coherent and consistent model of the world around us.

Your suggestion of using 'physical law and reason' instead of hard mathematics is exactly the kind of thinking that gave us Ptolemy. They reasoned that perfect circles were 'physical law' despite what the mathematics was telling them. Do you really want us to go back to that way of reasoning?

Matt

Doc Al

Jun10-04, 09:00 AM

Explain to me Doc Al and Hurkyl flat out lying and double talking, sonfusin. attemopting to get me of on some wild goose chase. Thee men are your colleagues. I don't fault you for your fdifferences, in fact i don't fault you. You are, to be sure instinctively SR, but I haven't detected the corruption seen in the other two mentioned.
Do you even know what the word "lying" means? It means intentionally saying something that isn't true. Can you point to even one post in which Hurkyl or I said something that wasn't true? Never mind intentionally.
These are "mentors" . I have only seen such peversion and dishnesty in the politics in thios country. I wonder whio is pulling the chain around their necks?
You make wild assertions, and when they are questioned all you can do is sputter and fume. In lieu of reason and argument, you make personal attacks. You bring nothing substantive to the table, geistkiestel.
Chroot, pulled a post of mine cruitical to some yoyo named carp or crap. Hedidn't think the post was scientific enough. Jesus christ, Now I cannot edit my own profile or dio any thing except woprk in thi porum. No problem, I just don't appreciate being handkled by a jerk with such insensitivity. The man is arbitary and snot nosed brat.
You spew the same drivel over and over and over again. You've had dozens of chances to make your point. You have always been free to babble away--in TD, where folks can choose to ignore you and you don't disrupt serious discussions.

This is your last chance, geistkiesel. If you care to keep posting here, among adults, then:
No more name calling
No more personal attacks

geistkiesel

Jun10-04, 10:29 AM

Riight. All this coming from a guy trying to convince us that his own theory - which is devoid of calculations or even physics for that matter - proves that SR is wrong.

Don't you think it would be better to try and show us exactly why you think Hurkyl is lying, rather than all this immature name calling?

Hurkyl lying: res ipsa loquitor.

Matt

---à motion ----> M' = M
A__________|______|______|______|______B
-------------t1-----t0------t1-------t2
These are the critical positions as follows:

When M’ = M, t0 = 0 and photons emitted from A and B.

The photon from B arrives on moving platform at t1
Photon from A arrives at t2.
Dt = t2 – t1 = 1 unit time.
V = 1 unit tme.
Assume temporarily M’ the midpoint of A and B at t0 = 0.

[Note: baffledMatt: is this assumption what you were telling the world was a rejection of SR from the get go? It was a convenient assumption that is properly tested below? Not quite as you reported it publically, is it, but what the heck, mentor, SR theorist, propagandist, rote robotic believer?].

At t1, when photon from B arrives, the A photon is assumed to be at –t1.
During dt = 1 the photon from A arrives at t2.

During dt = 1 the A photon travel c = t1 + t1 + 1
or rearranging t1 = (c – 1)/2 = k.

If t1 = k the photons emitted simultabneously in both frames. If t1 < k A the A photon emitted first, otherwise B emitted first.

Simple enough? Reasonable enough? fundamental physics.
See the thread for relativistic an observed expansion. similar logic and reason and physics.

geistkiesel

Jun10-04, 10:35 AM

Do you even know what the word "lying" means? It means intentionally saying something that isn't true. Can you point to even one post in which Hurkyl or I said something that wasn't true? Never mind intentionally.

You make wild assertions, and when they are questioned all you can do is sputter and fume. In lieu of reason and argument, you make personal attacks. You bring nothing substantive to the table, geistkiestel.

You spew the same drivel over and over and over again. You've had dozens of chances to make your point. You have always been free to babble away--in TD, where folks can choose to ignore you and you don't disrupt serious discussions.

This is your last chance, geistkiesel. If you care to keep posting here, among adults, then:
No more name calling
No more personal attacks

Where is there evidence of maturity and adult behaviour in you people? These men are my colleagues? You are all fired.
You certainly aren't teachers, professors, working to instill curiosity and independence in thinking. Just do it! that is your cfrtyive message. message,
the only message.

I call em like I sees 'em. You want that I hold back, be restrained, give yo a few inches so you can hang some body?

No,.

baffledMatt

Jun10-04, 11:21 AM

---à motion ----> M' = M
A__________|______|______|______|______B
-------------t1-----t0------t1-------t2
These are the critical positions as follows:

When M’ = M, t0 = 0 and photons emitted from A and B.

The photon from B arrives on moving platform at t1
Photon from A arrives at t2.
Dt = t2 – t1 = 1 unit time.
V = 1 unit tme.
Assume temporarily M’ the midpoint of A and B at t0 = 0.

Do you see that by defining dt = 1 and v = 1 you have already specified all the distances in the problem?

For instance, (using my notation again, note that your t1 and t2 are my t_B and t_A).

v (t_B - t_A) = 1 = d_A - d_B = \frac{v}{c} L \left(\frac{1}{1-\frac{v}{c}} - \frac{1}{1+\frac{v}{c}}\right)

using the fact that v = 1 and t_B - t_A = 1.

so:

L = \frac{c^2 - 1}{2}.

Now we can calculate the location of the photon from A when the photon from B hits M' (i.e. at t1)

d = ct1 - L = ct1 - \frac{c^2-1}{2} \neq - ct1

hence:

At t1, when photon from B arrives, the A photon is assumed to be at –t1.
During dt = 1 the photon from A arrives at t2.

you are no longer free to make this assumption - it is wrong.

[Note: baffledMatt: is this assumption what you were telling the world was a rejection of SR from the get go? It was a convenient assumption that is properly tested below? Not quite as you reported it publically, is it, but what the heck, mentor, SR theorist, propagandist, rote robotic believer?].

I have no problem with the midpoint as such. The problem is your assumption that the measured distance of A and B from this midpoint will be the same in each frame. So, in each frame they will indeed see M and M' exactly in between A and B, but if you ask each observer to measure the distance between the two they will disagree.

Matt

geistkiesel

Jun10-04, 01:41 PM

Do you see that by defining dt = 1 and v = 1 you have already specified all the distances in the problem?

I don't think so. 1 is just another symbol. The velocity has to be something I just mae it 1 for convenience. When I make calulation testing t1 I would have to use a eal functional value for he velocity

For instance, (using my notation again, note that your t1 and t2 are my t_B and t_A).

v (t_B - t_A) = 1 = d_A - d_B = \frac{v}{c} L \left(\frac{1}{1-\frac{v}{c}} - \frac{1}{1+\frac{v}{c}}\right)

using the fact that v = 1 and t_B - t_A = 1.

so:

L = \frac{c^2 - 1}{2}.

Now we can calculate the location of the photon from A when the photon from B hits M' (i.e. at t1)

d = ct1 - L = ct1 - \frac{c^2-1}{2} \neq - ct1

hence:

you are no longer free to make this assumption - it is wrong.

How do you calculate A if you don't know if it was emitted simultaneosusly or when it emitted?

Yes I am free to do this as I haven't made any tests yet. I merely continue with the assumption that when M'=M that the photons were emitted simultaneously, which happens to be what happened in this hypo. I make the same claim against you for arbitrarily inserting SR constraints into the problem.

Continuing wih the assumption that the photons were emitted simultaeously when B is detected thenm M = M' at t = 0 is the midpoint hence -t1 the location of the oncoming A photon. The photon reaches t2 during dt = t2 - t1 = t'" if this is more satisfying and let v = v. The distance c moved to arrive at t2 is c(t") = t1(v) + t1(v) + t"(v) or 2(t1v) = ct" - t"(v) or t1v = (ct" - t"v)/2

or t1 = (ct" - t"v)/2v =(t"/2v)(c -v) = k. Now I ask the same question:
Is t1 = k? If so the photons were emitted simultaeously. If t1 < k, the photon left A first, otherwise the photon left B first.
Why did you use the expressions 1 + v/c amd 1 -v/c? was this necessary?

I got, t1 = (ct" - t"v)/2v =(t"/2v)(c -v)

I have no problem with the midpoint as such. The problem is your assumption that the measured distance of A and B from this midpoint will be the same in each frame. So, in each frame they will indeed see M and M' exactly in between A and B, but if you ask each observer to measure the distance between the two they will disagree.

Look again isn't there only the implication that that M' was the midpoint, not necessaily of A and B but of the wave fronts of the photons emitted at somewhere?
This is the divegence point. I say your injection of SR to properly describe the midpoint assumes SR here, does it not?, and therefore, assume they
would not still maintain the charateristic of having the midpoints M = M'? Once you've done this the problem has been hijacked by SR, perhaps hijacked is a tad strong , but you know what I mean.

I think if you looked again I really only assuned the M =M' was the midpoint of the photon wave front, and there is no inference of the A and B location,
Maintaining steadfastly in the moving frame I get an expression that looks the same as yours up to a point.

Well then I just wont ask each observer to measure anything as I see no benefit as this time for anything to measure. Why and what do you measure here?
Now when my expession is rolled out it is aleady to be asked to undergo the test: t1 ?= (t"/2v).(c - v) .

I suspect you don't like the c- v?

baffledMatt

Jun10-04, 03:08 PM

I don't think so. 1 is just another symbol. The velocity has to be something I just mae it 1 for convenience. When I make calulation testing t1 I would have to use a eal functional value for he velocity

No, by specifying v=1 and dt=1 you are setting the time and velocity. distance is velocity multiplied by time, ergo you have also specified distance. Ipso facto your subsequent analysis is erroneous.

How do you calculate A if you don't know if it was emitted simultaneosusly or when it emitted?

I know that they were simultaneous in the stationary frame because that's how the situation was set up. However, this is not to say that they will be simultaneous in the moving frame.

I make the same claim against you for arbitrarily inserting SR constraints into the problem.

I am not inserting SR arbitrarily. I am using results which follow directly from the principle of relativity. I could just as well go through each and every calculation from first principles, but I will still get the same result.

The photon reaches t2 during dt = t2 - t1 = t'" if this is more satisfying and let v = v. The distance c moved to arrive at t2 is c(t") = t1(v) + t1(v) + t"(v) or 2(t1v) = ct" - t"(v) or t1v = (ct" - t"v)/2

Ok, what have you done here. You have equated ct'', which is the distance a photon travels in time t'' to (t1 + t1 + t'')*v, which is the distance the observer moving at velocity v moves in time t1 + t2. What makes you think that these two quantities are equal?! I'm afraid that what you have done here is complete and utter nonsense.

I suspect you don't like the c- v?

This is the least of my concerns!

Matt

baffledMatt

Jun10-04, 03:15 PM

I personally think I am something rare and special [...]

Well, we can't help what you personally think about yourself, but that still doesn't warrant all this hostility.

Besides, if they have been rude to you surely it would be a better show of character not to stoop to their level?

Matt

geistkiesel

Jun10-04, 03:34 PM

Well, we can't help what you personally think about yourself, but that still doesn't warrant all this hostility.

Besides, if they have been rude to you surely it would be a better show of character not to stoop to their level?

Matt
whta else do you want

Hurkyl

Jun10-04, 04:09 PM

Some say we humans are free thinkers

I always find it terribly ironic that those who talk most about "free thought" tend to be those who are least likely to consider that others might actually have a good reason to disagree.

If t1 = k the photons emitted simultabneously in both frames.

I find it curious that you conclude this, because everything up until this statement has involved only a single frame...

baffledMatt

Jun10-04, 05:05 PM

whta else do you want

Politeness would be a good start.

Being civil is not a sign of weakness. And if you stop with all the vulgarity people might be willing to listen more carefully to what you are saying.

Matt

geistkiesel

Jun10-04, 05:55 PM

I always find it terribly ironic that those who talk most about "free thought" tend to be those who are least likely to consider that others might actually have a good reason to disagree.

I find it curious that you conclude this, because everything up until this statement has involved only a single frame...

t1 = (ct" - t"v)/2v =(t"/2v)(c -v) = k. Yes it involved calculation in the moving frame. To get to this point an assumpotion was made that the photons were emitted simultaneously in the moving frame. Having gotten this far the next step is to test the assumption. If the mesured t1 = k the assumpion was true, if t`< k then the A photon was emitted first, otherwise the B photon was emitted first. All the calculations were performed in the moving frame. So what does "I find it curious ..." mean? Do you find the math, the assumptions, the conclusion erroenous, or am I supposed to guess what you mean? If you have a specific objection I suggest you state it as clearly as you are able so we don't have to read each others minds.
Thee were nio SR assumptions expresssed in the derivaion of the expression. as no SR implications were presented in the problem, without the presentaion being in a contrived mode.

Hurkyl

Jun10-04, 06:06 PM

geistkiesel

Jun10-04, 06:12 PM

Geistkiesel: here is your experiment drawn as a diagram:

0 0 0
*\ * /*
* \ * / *
AZ \ MN / BY
AZ \ MN / BY
AZ \ MN / BY
A Z \ M N/ B Y
A Z \ M * B Y
A Z \M/N B Y
A Z * N B Y
A Z M\ N B Y
A Z M \N B Y
A Z M \N B Y
A Z M * B Y

Z A NM 0
ZA NM /*
ZA 0 / *
ZA * / BY
0 * / BY
*\ MN / BY
* \ MN/ B Y
AZ \ M* B Y
AZ \M/N B Y
AZ * N B Y
A Z M\N B Y
A Z M * B Y

Legend:
A, M, B: your A, M, B
Z, N, Y: Your A', M', B'
0: A point where a clock read 0
\, /: photons
*: Multiple things at this point (such as two clocks, or a clock and a photon)

The first diagram is the stationary reference frame. The second diagram is the moving reference frame. Space runs from left to right, and time increases as you go downward.

The first diagram was taken directly from your experimental setup. I placed M and N midway between A/Z and B/Y, and simply drew out the time evolution of the system.

To draw the second diagram, I started with the unambiguous fact that both photons meet precisely when they each meet M. I drew the time evolution backwards, and used the fact that N is midway between Z and Y when M meets N to place where Z and Y should be. I then applied the fact that Z meets A and are both set to zero when the left photon is emitted to finish off the left side of the diagram, and similarly for Y and B. I did, however, have to estimate how N lies relative to M. Any other such choice yields a similar diagram.

Your mistake, as everyone is trying to tell you, is made clear from the diagram. In the moving frame, the clocks are not synchronized; you can see that they are all zero at different times. We see that SR can handle this scenario perfectly well, as long as you don't start with the assumption that synchronization in one frame = synchronization in all frames.

(NOTE: In the second drawing, to keep the diagram small, the lexical distance between A and M is 6 and one-third characters)

I find your diagram confusing and obfuscating. There was a diagram, presented in the opening thread I suggest you deal with the given instead of reinventing the thread language.. No ervyone didn't tell me that because i never made any statement to that effect. If I did use the word "synchroization " it was clear to the meaning that I did not mean it o be cross frame synchronization. Show the signifiicance of your statement in terms of the figure in the opening thread , please. I find it curious you have notcomplained about the clarity of the original figure. So SR can handle the scenario perfectly well. WEll then handle it. I have handled it my way, also perfectly well. Take a few minutes and find a fatal flaw in my analyisi befoe you try to usurp the essence of the thread with what you state, but do not priove, : SR's perfect handling of the scenario.

If you are unable tio explain how photons emitted simulataneioysly to th eeimmediate presence, within a photon wave length of light sensitive strips and not expose thos strip immediately yu are not communicating to me on issues i find critical. If you recall I stipulated ha SR would find he pgotons were not emitted simultabneously. Why are trying to prove what was stipulated?

I suggest you form your argument inline with the given paramjeters of the problem instead of creating diversions, intentionally or negligently.. This is a bad habit you have, chnges he direcion of threads to suit your own purposes, what evre they happenm to be.

geistkiesel

Jun10-04, 06:22 PM

The curious part is that you could derive a conclusion that has nothing to do with the rest of the post. I went back to reread #131 and I can find nothing that says anything about two different reference frames. I can't even find an explicit reference to one reference frame.
So what?
t
What do you think 131 was discussing?. At lest you are beginnng to see that there is only one sginficant frame here. Ths stationary frame provided the simultaneous emission of photons. From that event on the stationary frame iis insconsequential to the physics of the problem, as the relevant events were completed at the instant the photons were emitted by A and B and then immediately exposed the light sensitive strip located within a wavelength of the event, the enmitted phoons.
try looking at the problenm as it you hadn't already assumed the answer, as you have been doing all along. be a physiicist fior this analysis.

Hurkyl

Jun10-04, 06:34 PM

I will I've mentally blended some of the arguments you've made over the past few days; I seem to be responding to the one in the link you kept claiming went undiscussed as opposed to the one in this thread.

geistkiesel

Jun10-04, 06:54 PM

All the time you are assuming that there are all these physical entities which 'have length X' or 'are at time Y'. How do you know this? How do you know that the distance between A and B is 2L? You measure it of course. But hang on, now you haven't actually determined what this entitie's 'true length' is, you have made a measurement - there is no way you can talk about the 'true length' because there is no measurement free way of determining it. Then what SR tells us is how these measurements will differ between intertial frames.

the problem is a hypothetical. I inserted hypothetical measuring paameters consitent withthe laws fo physics. If o hve to lead sem bythe hand on thsi elemnarty level, I am wonder just what you think to=your function is. What you just said sounds as if you are trying to ob scure themeaning of theinvariance of physical lasw in all inertial frames. I described the ls-strips. I describes how pairs of strips were numberd such that the same numbed pairs wee equal distane from he midpoint. Are you telling me SR does not allow finding midpoints of the ls-strips? If you are telling me this I will ignroe you. equal
Themeasurement: using the most accurate stainless steel measuring tape possible tobe constructed, the midpoints ot the numbered ls-striops were located equidistant fro M' n the moving frame.
When the photons were emitted the nearest pair of co=numberd ls-strips were exposed. to the emitted dhotons, situated one wave length from the photons. Are you sayuing this is physially impossible?

[quotwe-baffledMatt]You still think that you are determining some true 'real' property using this measurement. What SR tries to tell us is that the only quantities which do in fact have a 'true' property are the invariant ones (such as rest mass).

Until you think very very carefully about the way you are making your measurements you will never understand this.[/quote]

I am making the exposure of the ls-strips that are placed one wave length from the photons emitted in the stationary frame. If you think this is not a physical possiblity, then say so. I don't rmember saying I was determining a 'real' property. I said I was obtaining a iny mark on an ls_ striop, period. What is so diffiicult in grasping something that is not of a complex nature requiring all the scurrying around.

Are you telling me that SR is so all inclusive and pervasive that it negates the detection of photons at A and B when emitted? you must prove this to me.

Excuse me, are you a physicist or a philosopher? If you want to talk this way then please move your discussion to the philosophy section of the forum. I'm afraid that mathematics is the only way we know of to build a coherent and consistent model of the world around us.

Your suggestion of using 'physical law and reason' instead of hard mathematics is exactly the kind of thinking that gave us Ptolemy. They reasoned that perfect circles were 'physical law' despite what the mathematics was telling them. Do you really want us to go back to that way of reasoning?

Matt
This is just my opinion, but I beleive that mathematical models are to a very large extent a corruption on the progress of phsyics. Ptolemy's system worked you know, satisfactorily even thoufgh groundless a a refledtion of natural dynamics. Thus, Ptolemy's model was a precursor to the coruption that relativity theory has had on the progress of science. Sillines, but this is just m yopinion.

At what point in your life did you come to realize that you had eveything covered? I just arrived at that point myself only lastweek.

geistkiesel

Jun10-04, 07:05 PM

The difference between this and religion is that SR is completely logical. You start from the principle of relativity and everything else is deduced from this. So the only thing I got converted to was the principle of relativity - if you want to call that religion then be my guest.

You have not shown us any flaws in SR, only flaws in your interpretation of SR.

There is no paradox, period! Only gross misunderstandings.

This depends on your point of view. I do not consider there to be any physical relevance to two things being simultaneous, so the results of SR do not bother me.

Why do you think there is any significance to things being simultaneous? If two events happen at different locations and the same time then there is no physical way the two events are connected - they are outside of each others light cones. So each event has no idea that it is occuring 'at the same time' as another event. Why do you attach so much significance to things which cannot be causally related?

Matt

How do events have ideas, about other events?

I attach importance to theoreical constructs that enter the physuical domain as pure mental constructs and purport to gao back in time an alter the eality of events already having taken place. You deny this. I assert it to be true,
It isn't that simulateity is disrupted is is the perturbation after an efecnt ahs occued that is claimed as a real physical pheniomenon. tha disturbs me.

A simultaneous eveny of two lights being emitted in a stationary frame occur. This is all. A million of hese evens thisw is all. Only simulatbeous emission fo phoptons is the simultabeiys event. Now comes an obsever, just an observer that no one is pretending applied any physical or other meansurable force on any physical property of the parametrs constuituing the sinmulatneous event and now we have twop ebvenjt. Only the presence of the observer opeates to construct three events from two events., all ths by the mere pesence of an observer.

geistkiesel

Jun10-04, 07:07 PM

ram2048

Jun10-04, 07:26 PM

The universe of which I'm aware is full of objects travelling every which way, constantly accelerating to different directions, and without any nice, global structure into which everything fits.

So, there are two very problematic things about this idea; why should there be such a thing as a "stationary universe", and how do we tell what it sees?

i've defined a nice way for determining universal perspective but it includes calculating relative motions for EVERYTHING in the universe. suffice to say it's incredibly accurate and infinitely tedious to run ;D

barring that we can use it to triangulate events from the perspective of relevant local bodies and base our reality /time / space off of those calculations to satisfy on a global level. it may not agree with aliens from beta centauri, but who cares what they think?

And the point is that a reference frame is chosen, and then everything else is defined relative to that.

Relativity does not say that you're not allowed to choose a reference frame as a "standard"; it merely says that any reference frame would suffice.

and my relativity states that you can choose whatever reference frame you want, but when your calculations are vastly disagreeing with the majority, there's a high probability that you're the one that's wrong about "reality"

-------------------

to leave this on a different note, does it really make sense for one person travelling fast and having his perceptions skewed to have HIS OWN, valid reality, complete with slowed stuff behind him and fast stuff in front of him, non-simultaneity, and shrinking rulers? i can only bring myself to think that some of the SR people are down the rabbit hole, smokin with the worm.

Hurkyl

Jun10-04, 09:21 PM

to leave this on a different note, does it really make sense for one person travelling fast and having his perceptions skewed to have HIS OWN, valid reality, complete with slowed stuff behind him and fast stuff in front of him, non-simultaneity, and shrinking rulers? i can only bring myself to think that some of the SR people are down the rabbit hole, smokin with the worm.

Yes.

The most important reason, I think is the following: I expect the laws of physics inside my spaceship to be the same whether I'm in a nice leisurely geosynchronous orbit or hurtling between galaxies at a constant speed of 99% of c.

E.G. If I set something at rest (WRT the ship), I expect it to stay at rest until something nudges it. If I toss a ball at the wall, it should bounce back in the "right" direction. And because of Maxwell's laws, if I send a light signal from aft to stern, I expect it to travel at c.

This is the principle of relativity, and it is what is used to derive everything about SR.

Some other reasons you haven't encountered because you are misleading yourself by only considering situations where theres an obvious "right" choice.

Thought experiments can be set up where the universe consists of only a few objects, none having any sort of special significance that would entail using it to define the "right" frame.

The "right" frame can change; if I'm taking a rocket to Mars, I "should" be in Earth's frame at the start, I "should" be in the solar system's frame in the middle, and I "should" be in Mars's frame at the end. It seems silly to pick a point on my journey and say that I suddenly switch frames!

The "right" frame isn't always the easiest frame to analyze. For instance, calculations might be simpler and natural to do in the frame of the spaceship instead of, say, the frame of the solar system.

geistkiesel

Jun11-04, 02:01 AM

Because we're disagreeing about what SR says when we analyze the exact same events from different frames.

Specifically, Geistkiesel is asserting that both of these diagrams are representing the exact same sequence of events:

A M B
A\ M /B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \M / B
A M / B
A M / B
A M / B
A M / B
A M/ B

A M B
A\ M /B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \ M / B
A \M/ B

We have two relatively stationary light sources (A and B), and an observer who starts in the middle and moves towards A.

The first diagram depicts how things look in the rest frame of the lights, if the lights are activated simultaneously.

The second diagram depicts how things look in the rest frame of the observer, if the lights are activated simultaneously.

However, there is a very important difference between the two diagrams; in the first diagram the photons do not meet M at the same event, however in the second diagram the photons do meet M at the same event.

The conclusion is that these diagrams cannot possibly represent the same events. Among the possible assumptions we can abandon, abanding that of absolute simultaneity is by far the most reasonable; the emission of photons is simply simultaneous in one frame but not the other.

The intrinsic fallacies off Special Relativity Theory.

Postulates of Special Relativity Theory
The laws of physics and the measure of the speed of light are invariant in all inertial frames.

Experimental Conditions:
One battery with power for two photons only is connected with a single switch to two lights.
Jill is walking toward Jack who is at the midpoint of the lights.
When Jill reaches Jack the lights are switched on simultaneously.

Contradictory Observations
Jack observes the lights switched on simultaneously.
Jill observes the lights switched on sequentially.

Irrational Basis of Special Relativity Theory.
Violation of conservation of energy principal.
Two energy units available.
Four units of energy claimed.

No rational explanation for the sequential order of photon emissions.
Violation of invariance of physical laws in inertial frames.

baffledMatt

Jun11-04, 02:11 AM

the problem is a hypothetical. I inserted hypothetical measuring paameters consitent withthe laws fo physics.

No, you have no idea what the laws of physics are so you are not qualified to make this statement.

I described the ls-strips. I describes how pairs of strips were numberd such that the same numbed pairs wee equal distane from he midpoint. Are you telling me SR does not allow finding midpoints of the ls-strips?

You will find the midpoint, but each observer will measure a different distance to that midpoint. Why is this so difficult for you to understand?

If you are telling me this I will ignroe you. equal

Go ahead, you'll only be shooting yourself in the foot. You can even believe you won the argument if you like and continue through life oblivious to your ignorance. I don't care.

Themeasurement: using the most accurate stainless steel measuring tape possible tobe constructed, the midpoints ot the numbered ls-striops were located equidistant fro M' n the moving frame.
When the photons were emitted the nearest pair of co=numberd ls-strips were exposed. to the emitted dhotons, situated one wave length from the photons. Are you sayuing this is physially impossible?

It is impossible to get the results you describe. You take it as an assumption that this can be measured, but this assumption is against the laws of physics. Hence, any conclusions you draw from it is equally unphysical.

I am making the exposure of the ls-strips that are placed one wave length from the photons emitted in the stationary frame. If you think this is not a physical possiblity, then say so. I don't rmember saying I was determining a 'real' property. I said I was obtaining a iny mark on an ls_ striop, period. What is so diffiicult in grasping something that is not of a complex nature requiring all the scurrying around.

You have such a high opinion of yourself yet you completely fail to grasp the subtlety of what I am trying to tell you. The measurement is physically possible, but getting the result you are assuming in physically impossible.

This is just my opinion, but I beleive that mathematical models are to a very large extent a corruption on the progress of phsyics. Ptolemy's system worked you know, satisfactorily even thoufgh groundless a a refledtion of natural dynamics. Thus, Ptolemy's model was a precursor to the coruption that relativity theory has had on the progress of science. Sillines, but this is just m yopinion.

Go on, admit it. You just can't do the math can you? It's ok, not everyone is good at maths and there are lots of other things you can be instead of a physicist.

Matt

ram2048

Jun11-04, 06:02 AM

with an obvious velocity limit of C what makes you think that your physics will always be the same?

if you're inside a spaceship travelling 99.99999999999% C and decide to pitch a baseball at 100 mph inside the craft, would it go faster than light speed?

Tom Mattson

Jun11-04, 06:11 AM

geistkiesel,

I deleted your last post. Threats of physical violence are unacceptable.

with an obvious velocity limit of C what makes you think that your physics will always be the same?

That, by itself, does not guarantee that the physics will always be the same. SR rests on two postulates: The constancy of the speed of light, and the relativity postulate, which states that the laws of physics will be the same in every frame. By requiring these two, we arrive at the Lorentz transformation.

if you're inside a spaceship travelling 99.99999999999% C and decide to pitch a baseball at 100 mph inside the craft, would it go faster than light speed?

No, it won't.

Tom Mattson

Jun11-04, 06:42 AM

The intrinsic fallacies off Special Relativity Theory.

Postulates of Special Relativity Theory
The laws of physics and the measure of the speed of light are invariant in all inertial frames.

Experimental Conditions:
One battery with power for two photons only is connected with a single switch to two lights.
Jill is walking toward Jack who is at the midpoint of the lights.
When Jill reaches Jack the lights are switched on simultaneously.

Contradictory Observations
Jack observes the lights switched on simultaneously.
Jill observes the lights switched on sequentially.

The intrinsic fallacies of geistkiesel's crackpot physics

It doesn't recognize the importance of mathematics to physics. (See, it doesn't matter if you can provide a qualitative discussion of a phenomenon if you can't discuss it quantitatively. As Warren Siegel put it, physics is not just about "what comes up must come down". It's also about where and when it comes down.)
It doesn't recognize the importance of experimental work to physics. (Made clear by the fact that not a single real experiment is ever cited).
It assumes that a thought experiment is a valid substitute for #2. (See "experimental conditions" in the quoted post for a good laugh).

Irrational Basis of Special Relativity Theory.
Violation of conservation of energy principal.

No, it doesn't. Relativistic Lagrangians are still invariant under time translations, and so energy is still conserved.

Two energy units available.
Four units of energy claimed.

"Energy units" have nothing to do with any physical theory, and in fact there are many energy units available to all theories. Furthermore, there is nothing wrong with that.

No rational explanation for the sequential order of photon emissions.

That's not true at all. The explanation stems from the postulates. And the postulates cannot be said to be "irrational" just because you personally don't like them.

Violation of invariance of physical laws in inertial frames.

This is the dumbest of all your points. SR is designed to preserve the invariance of physical laws in all frames.

Geistkiesel, it's time for you to put up or shut up. You think that energy conservation is violated in SR? Fine: prove it. You think that SR doesn't preserve the invariance of physical laws in inertial frames? Fine: prove it. And don't just blather on for pages on end, show the mathematical details.

Hurkyl

Jun11-04, 07:01 AM

with an obvious velocity limit of C what makes you think that your physics will always be the same?

One philosophical reason to do so is that given that we, as earthbound folk whirling around the sun, which is circumnavigating the Milky way, which is hurtling towards Andromeda, with both being sucked towards the great attractor, which has who knows what relationship to the next big thing, it is an extraordinarily bold claim that we are lucky enough to (nearly) be in the one frame where the laws of physics look "right".

As a practical matter, experiments confirm that the physics still looks the same, many of astonishing predictions stemming in part from this assumption have been confirmed, and it is an integral part of the most accurate theory known to man.

geistkiesel

Jun11-04, 10:08 AM

One philosophical reason to do so is that given that we, as earthbound folk whirling around the sun, which is circumnavigating the Milky way, which is hurtling towards Andromeda, with both being sucked towards the great attractor, which has who knows what relationship to the next big thing, it is an extraordinarily bold claim that we are lucky enough to (nearly) be in the one frame where the laws of physics look "right".

As a practical matter, experiments confirm that the physics still looks the same, many of astonishing predictions stemming in part from this assumption have been confirmed, and it is an integral part of the most accurate theory known to man.
Hurkyl-
Look at Dayton Miller's papers re Michelson Morely experiments. Beside thefact that DM essentially eproduced MM results, nopt NULL, but a wave shift ~1.20 of what was then predicted.
I am direing this at your statement that the milky Way is "hurtling towards Andrmeda'. In DM's analysis he took great pains to determine the direcion of m,
otion we are heading and came to different conclusion than that suspected in the1930's .It has been awhile since I read the paper, but i recall he ws looking for some motion (earth-solar system-milkyway) that would explain the 1/20wave length from the "expected" shift. Apparently we ain't heading to Andromeda.( or what was geneally believd then)

russ_watters

Jun11-04, 10:28 AM

Contradictory Observations
Jack observes the lights switched on simultaneously.
Jill observes the lights switched on sequentially.
I guess I should have seen this before, but its preplexing in a way - this isn't a contradiction in any version of relativity. It isn't unique to Einstein's. Its should be obvious. What you are showing here is a pretty fundamental misunderstanding of physics that is far more basic than Einstein's version of Relativity.

Haven't you ever played catch with a ball? If you are throwing the ball at a moving person, you need to anticipate where the person is going to be when the ball gets there and aim for that spot, not where the person is now. Thats all forms of relativity, its simultenaety, and its built in to the human brain at a subconscious level (the part of the brain that co-ordinates movement).

Thinking about it more, maybe the issue is speed. In your Jack and Jill example, if you take it literally, the speed of light is so fast compared to the speed they walk, they wouldn't notice the difference - but the difference would exist, nonetheless. The only way they could possibly see the light switch on simultaneously (assuming they were carrying sensitive enough instruments to measure the difference) is if C is infinite. For a thought experiment like that, C is so much higher than the speed they walk at that it may as well be infinite to Jack and Jill. But to an atomic clock and a GPS satellite (a device capable of noticing that C is finite), C is not infinite.

Is that all this issue is about?

baffledMatt

Jun11-04, 10:32 AM

geistkiesel,

I deleted your last post. Threats of physical violence are unacceptable.

Blimey, well I hope it wasn't directed at me.

Well geistkiesel, I'm afraid that I've had enough of your attitude so I'll leave you to wallow in a pit of your own, well, whatever it is you are wallowing in at this moment in time.

Believe what you like, ignore everyone and everything around you. Live in the belief that you are and always will be right about everything. Yell at, hurl abuse at (and threaten?!) people who try to tell you otherwise. I don't care, it's your life.

All I can say is good luck.

Matt

geistkiesel

Jun11-04, 11:28 AM

The intrinsic fallacies of geistkiesel's crackpot physics

It doesn't recognize the importance of mathematics to physics. (See, it doesn't matter if you can provide a qualitative discussion of a phenomenon if you can't discuss it quantitatively. As Warren Siegel put it, physics is not just about "what comes up must come down". It's also about where and when it comes down.)

Wow.! I get it Tom like all the mathematics in your knee jerk propaganda piece! My mistake!!! There, see,there, some mathematics in Mateson's post that I missed: "list=1" . Where can I research this? Von Neumann, Feynman, Ptolemy, The White Rabbit? What does "1" mean?

It doesn't re cognize the importance of experimental work to physics. (Made clear by the fact that not a single real experiment is ever cited).

Well , I guess you're right there. I was posting in thread reagarding an analysis of Einsteins famous impotant experimental work, that "single real experiment" that for the past hundred years tied a few generations of mental powers into sub-human level of performance.

An example, Tom Mateson writes a 'scathing' response to am observvation, citing the need for mathematics, which isn't provided, except for the famous "list = 1" equation, "Stockholm here I come", right Tom Mateson?

The incompetents in this forum who have me picked out for easy pickings have failed to recognize just who is is the "nit" and who is the "picker" on who, who is running experiments and who is living in the fantasy land of sophisticated mathematical theory in physical model development. You haven't arrived yet Tom, you're still in the memorization stage of your scientific development. Your silly post here is a public relation adventure, designed to hold yourself up as a justifiably smug, silly old man, hey I vote for Tom Mateson.

It assumes that a thought experiment is a valid substitute for #2. (See "experimental conditions" in the quoted post for a good laugh).

I saw some more mathematics from the great scientist, Tom Mateson: Heh, Tom Can you prove that "2" there?

Let em see e^(i(pi)) +1 = ? , what was that again? Has anybody seen my calculator, I don'tthink it's 27.

No, it doesn't. Relativistic Lagrangians are still invariant under time translations, and so energy is still conserved.

"Energy units" have nothing to do with any physical theory, and in fact there are many energy units available to all theories. Furthermore, there is nothing wrong with that.

Hey, Tom I read in a supeman comic book about the conservation of energy and it dawned on me that SR theory as it applies to "simultabeity" constructs" is a patently obvious violation of the conservation priinciple".
The thought expeiment limited the energy out put to two photons, which are ultimately absorbed in Pacoima . California. Speacial relativty theory by reconstructing reality with the addition of another set of events, creates two more emitted photons for the moving observer, just because the observer is there. Neat triick. I guess I confused you with the "unit enregy" shorthand. OK the battery had enough power to provide two 27 megawatt bursts of enrgy for the production of photons. After the theorists got their pocket calculators clicking and a clacking, relativity theory produced 2 more 27 megawatt buirst of nonsimutaneous photons for those n the moving frame who just happened to be in the 'hood'. I wish I could do that.

I'm going to rest now for a bit. Two number "27" in the same paragraph, has me bushed and a little light in the head.

That's not true at all. The explanation stems from the postulates. And the postulates cannot be said to be "irrational" just because you personally don't like them.

i like Postulates, I took three of them to lunch just last week. Postulates, those are mental aberrations of theoretical physicicts that substitute for experiemntal results, like the kind i was accused (I plead Guilty yer honor) of not performing.

SO Tom , if one is a fancy pants phsiquest, like yourself where postulations are official substitutes for real experiments I can do that too: I had a note from an engineer who said I could peform as many mental gymnastifications as I could possibly,or even Impossibly, postulate. So here.

This is the dumbest of all your points. SR is designed to preserve the invariance of physical laws in all frames.

Geistkiesel, it's time for you to put up or shut up. You think that energy conservation is violated in SR? Fine: prove it. You think that SR doesn't preserve the invariance of physical laws in inertial frames? Fine: prove it. And don't just blather on for pages on end, show the mathematical details.

I blathered in less than a page. You are reading some one elses blathes.

You ought to get your money back, because your designed invariance flange just cracked and there are all kinds od invariances spewing from special relativity theory all over the immobile moving frame..

Put up or shut up?

Does anybody want to run an Einstein "train experiment" where we measure the simultaneious photon emissions tested with photosensitive devices, as a test of simultaneity , which according to the famous "put up or shut up" physicsist Tom Mateson, should be able to produce two extra photons for any stray observer casually passing through the hood?

Phrased another way: Who is willing to provide assets to directly test SR theory. If the test fails we could be in some serious rank doodoo if we are skirting around reality in some ptolemaic fog. Does everybody understand? to allow existence of SR that is infected with bloated mathematical contrivances some investigative work will be in order right?

The cheapest way to solve the problem, of course, is have Tom Mateson splain it all to everybody. I hear he is a cetificated "experiment bypass" authority. myself, I have to bust knuckles and drop tools on the floor and break glass [just by entering labs!!yeah], and turn dials, and swear a lot. Well actually they don'teven let me do that. They have professionals available that know how all about that. Myself, I like to watch.

Tom Mattson

Jun11-04, 12:26 PM

Wow.! I get it Tom like all the mathematics in your knee jerk propaganda piece! My mistake!!! There, see,there, some mathematics in Mateson's post that I missed: "list=1" . Where can I research this? Von Neumann, Feynman, Ptolemy, The White Rabbit? What does "1" mean?

:rofl: :rofl: :rofl: :rofl: :rofl:

No, the "list=1" only shows up when you quote the post. It's part of the formatting that generates the numbered lists. See how it doesn't actually appear in my post?

Well , I guess you're right there. I was posting in thread reagarding an analysis of Einsteins famous impotant experimental work, that "single real experiment" that for the past hundred years tied a few generations of mental powers into sub-human level of performance.

Einstein never did any experimental work. You'd know that if you had half a clue.

An example, Tom Mateson writes a 'scathing' response to am observvation, citing the need for mathematics, which isn't provided, except for the famous "list = 1" equation, "Stockholm here I come", right Tom Mateson?

What math do you want to see here? Do you want me to write the equation that describes just what an ignorant fool you are? I'm afraid that all the computing power in the world couldn't crunch that one any time soon.

The incompetents in this forum who have me picked out for easy pickings have failed to recognize just who is is the "nit" and who is the "picker" on who, who is running experiments and who is living in the fantasy land of sophisticated mathematical theory in physical model development. You haven't arrived yet Tom, you're still in the memorization stage of your scientific development. Your silly post here is a public relation adventure, designed to hold yourself up as a justifiably smug, silly old man, hey I vote for Tom Mateson.

:uhh:

I saw some more mathematics from the great scientist, Tom Mateson: Heh, Tom Can you prove that "2" there?

Let em see e^(i(pi)) +1 = ? , what was that again? Has anybody seen my calculator, I don'tthink it's 27.

What the hell are you talking about?

Hey, Tom I read in a supeman comic book about the conservation of energy and it dawned on me followed by some extremely stupid remarks

No comment on that part.

Speacial relativty theory by reconstructing reality with the addition of another set of events, creates two more emitted photons for the moving observer, just because the observer is there. Neat triick.

It doesn't "reconstruct reality", it simply describes it. Try to understand the difference.

I guess I confused you with the "unit enregy" shorthand.

You didn't "confuse" me, you simply said it wrong. "Units of energy" are things like Joules, ft-lb, kW-h, and the like.

OK the battery had enough power to provide two 27 megawatt bursts of enrgy for the production of photons. After the theorists got their pocket calculators clicking and a clacking, relativity theory produced 2 more 27 megawatt buirst of nonsimutaneous photons for those n the moving frame who just happened to be in the 'hood'. I wish I could do that.

You don't understand a thing. It's not that there are different photons in each frame, it's that there are different spaitotemporal intervals in each frame.

I'm going to rest now for a bit. Two number "27" in the same paragraph, has me bushed and a little light in the head.

Don't hurt yourself, cupcake.

i like Postulates, I took three of them to lunch just last week. Postulates, those are mental aberrations of theoretical physicicts that substitute for experiemntal results, like the kind i was accused (I plead Guilty yer honor) of not performing.

Again, you don't understand a thing. Postulates aren't used in place of experiments. They are submitted to be tested by experiments. And you know what? The postulates of SR have survived every test put to them.

SO Tom , if one is a fancy pants phsiquest, like yourself where postulations are official substitutes for real experiments I can do that too: I had a note from an engineer who said I could peform as many mental gymnastifications as I could possibly,or even Impossibly, postulate. So here.

You are a retard.

I blathered in less than a page. You are reading some one elses blathes.

You ought to get your money back, because your designed invariance flange just cracked and there are all kinds od invariances spewing from special relativity theory all over the immobile moving frame..

In English please?

Put up or shut up?

Does anybody want to run an Einstein "train experiment" where we measure the simultaneious photon emissions tested with photosensitive devices, as a test of simultaneity , which according to the famous "put up or shut up" physicsist Tom Mateson, should be able to produce two extra photons for any stray observer casually passing through the hood?

Phrased another way: Who is willing to provide assets to directly test SR theory. If the test fails we could be in some serious rank doodoo if we are skirting around reality in some ptolemaic fog. Does everybody understand? to allow existence of SR that is infected with bloated mathematical contrivances some investigative work will be in order right?

Actually, SR is tested every day, in particle accelerators and, with GPS systems, and with nuclear reactors and weapons, and even with humble radios in moving cars. There is a wealth of experimental information available for anyone who wants to see it, and all of it confirms SR and shows Galilean relativity to be wrong. But you don't want any part of it, because you prefer to remain in willful ignorance, screaming anti-SR nonsense like a jackass.

The cheapest way to solve the problem, of course, is have Tom Mateson splain it all to everybody. I hear he is a cetificated "experiment bypass" authority. myself, I have to bust knuckles and drop tools on the floor and break glass [just by entering labs!!yeah], and turn dials, and swear a lot. Well actually they don'teven let me do that. They have professionals available that know how all about that. Myself, I like to watch.

No, the cheapest way for you to solve your problem is to invest the time and energy studying real physics. That's the only way you will break free of these errors you are making.

On that note, I don't see any point in this thread continuing. You obviously have nothing worthwhile to contribute here.