No postulate of light is violated in Galilean transformation.

[geistkiesel]

We don't "omit" the observer's speed. In the observer's frame, he is at rest.

Why does the moving observer always consider herself at rest when measuring the speed of light? Convenience I will surmise.


"

Mass shrinking" is not part of SR. In fact, the mass of a particle is an invariant, as it is the norm of the 4-momentum.

Would length contraction sound better than mass shrinking?

"From where you sit", eh? Then it's time for all of you to get up and move to a new chair, from which you can see the blackboard better.

Some of my more pleasant moments in school was to be able to ask whoever was chalking up the board at the moment, "Where did you get that 2?"
Hey tom, where did you . . .aw fagida abowdit!

 

[wespe]

no i mean THIS

A <- <-B

What the heck is this? Why is the arrow on the right side of A? The arrows are supposed to show a velocity vector. Do you mean like this?

<-A <-B
________

If so, their speeds are in the same direction according to the third observer, and then they are stationary wrt each other.

you can't forget that speed is defined taking a distance over a time. if A covers .9999x300,000km towards B in one second and B travels .9999x300,000km towards A in one second then if A is defined as stationary, B travels .9999x300,000km + .9999x300,000km in one second towards A.
simply so.

In which frame are all these measurements made? If it's the Earth frame, the diagrams below fit your description.

A <-B
_______

or

A-> B
_______

A does approach B at .9999c per second, and B does approach A at .9999c per second, all according to the Earth frame. But they don't approach at nearly 2c.

You didn't like this one, but again:
A-> <-B
________

According to the Earth frame, they approach each other at nearly 2c per second (separation speed). But that doesn't mean they will see each other approaching at nearly 2c. According to SR, their relative speed will still be below c.

as stated above you only believe so because you're completely negating one "observer's" velocity instead of adding it to everything else in the universe when defining that observer as "stationary"

that didn't make any sense. I'm not "defining" observer stationary. Any observer IS stationary wrt itself, what's so difficult to understand?

and as far as particle accelerators go, if we took a linear accelerator capable of accelerating a particle to .9999c and then launched that accelerator into space with a relative velocity of .9999c to the earth, then shot a particle down the tube <pointing away from earth> how fast would it be going <from Earth's frame> ?

A little over 0.9999c but still below c, according to SR. How can you think you can predict the result of such an experiment while you sit on your a##.

and don't bother using the "formula", i want you to tell me LOGICALLY how you feel about the situation. parroting someone elses formulas does not demonstrate you know what you're talking about...

How I FEEL? Nature doesn't have to conform to your or my feelings. If you want to know how nature acts, look at the experiments. You totally frustrated me now with your ignorance and arrogance and incapability to learn. Bye, Ram.

 

[geistkiesel]

I think it's funny that you deman experiments to prove SR, but you have none to back yourself.

I gave at the office and didn't keep enough for expeiments, hence uncle is going to pop for this one. Do you mind? Is this all you have to offer for the post I just completed? Whoooooosh!

You are very helpful, thank you for sharing all that with us.

 

[geistkiesel]

I think it's funny that you deman experiments to prove SR, but you have none to back yourself.

I just demanded the experiment one I just described in the post above. It does not back SR, it dumps SR in the trash bin. What are you talking about?

 

[quantumdude]

Why does the moving observer always consider herself at rest when measuring the speed of light? Convenience I will surmise.

Not only does she consider herself at rest when measuring the speed of light, she considers herself at rest period, because she is inertial.

"
Would length contraction sound better than mass shrinking?

Well, since length contraction is actually a prediction of the theory, while "mass shrinking" is a stupid figment of some retard's imagination, yeah, I guess it would sound better!. :rofl:

 

[Doc Al]

yes,geistkiesel, distance = speed x time: So what?

Tom, somethiong completely different. You are the math guru so you figure it out. I found the following in a website' But before that I had developed an expression from an analysis of the Einstein train gedunken. The train moving right detects the B photon coming from the front then the A photon from the rear. Using t1 as the time the B photon was detected measured from the instant the O' observer was at the midpoint of A and B sources in the stationary frame (t2 the time the photons arrived simultabeously at the midpoint of A and B in the stationary frame) and t3 the time of detection of the A photon by O' in the moviong frame as t3, I derived an expression that

t3 = t1(C + V)/(C - V)

from the following:

A_________________M__________________B
                 t0 t2
____________A______O'_____B_______A_____
                          t1<-------------          
 -------------------------------->t3

t0 the time the O' obsever was at M and the time the photons emitted at A and B. t1 the time B was detected by the moving O'. t2 the time the A and B photons arrived simultaneously at M from A and B. and t3 he time the A photon detected by O'.
From symmetry considerations A and B are equidistant = t1v from M when B is detected. The frame is moving at velocity v. t1 measured from t0 = 0. Therefore the A photon to be detected at t3 it must travel
c(t3-t1) = 2vt1 + v(t3 - t1)
ct3 - ct1 = 2vt1 = Vt3 - V t1
t3(C - v) = t1(C + v)
t3 = t1(C + v)/ (C - v).

I've explained all this before, but you still don't get it. Yes, amazingly, your expression t3 = t1(c + v)/(c - v) is correct. (It's just d = vt.) But only if all the time measurements are made in the O frame, not in the O' frame:

t = 0 is the time of emission according to the O frame
t = t1 is the time that photon B hits O' according to the O frame
t = t3 is the time that photon A hits O' according to the O frame ​

That's the only way you can derive that result. It says nothing whatsoever about measurements made in the O' frame. It says nothing about "simultaneity". It's just a trivial expression relating times t1 and t3, both of which are O-clock times.

By the way, here's a simpler way to derive that result. Note that all measurements are O-frame measurements. Let L be the distance from the midpoint to either A or B. By the time the photon from B hits O', O' has moved towards B, so:
(1) ct1 + vt1 = (c + v)t1 = L
And by the time the photon from A hits O':
(2) ct3 = L + vt3 --> (c - v)t3 = L
Combine (1) and (2) and you're done. So what?

 

[ram1024]

A does approach B at .9999c per second, and B does approach A at .9999c per second, all according to the Earth frame. But they don't approach at nearly 2c.

You didn't like this one, but again:
A-> <-B
________

According to the Earth frame, they approach each other at nearly 2c per second (separation speed). But that doesn't mean they will see each other approaching at nearly 2c. According to SR, their relative speed will still be below c.

why are you so confused? A can travel east at 20 miles per hour. B can travel west at 20 miles per hour. both cars have a speed limit of 20 miles per hour.

you're telling me the maximum speed they can approach each other is 20 miles an hour, which is patently untrue given if they're 40 miles apart and they both drive towards each other they can collide in an hour.

by any definition if you take A as a stationary object, B now travels 40 miles an hour towards A.

"but light is different..."

BS.

 

[Doc Al]

A can travel east at 20 miles per hour. B can travel west at 20 miles per hour. both cars have a speed limit of 20 miles per hour.

Viewed from an observer at rest with the road, what you say is perfectly true. All speeds are given with respect to the road.

you're telling me the maximum speed they can approach each other is 20 miles an hour, which is patently untrue given if they're 40 miles apart and they both drive towards each other they can collide in an hour.

According to observers at rest with the road, the two cars approach each other at 40 miles per hour.

by any definition if you take A as a stationary object, B now travels 40 miles an hour towards A.

Ah, now you want to switch frames and view things from A's frame. Better do it right: use the relativistic addition of velocities formula. For low speeds only it does turn out that $V_{B/A} = V_{B/road} + V_{road/A}$.

"but light is different..."

Everything is different, not just light. Welcome to the 20th century!

 

[geistkiesel]

I've explained all this before, but you still don't get it. Yes, amazingly, your expression t3 = t1(c + v)/(c - v) is correct. (It's just d = vt.) But only if all the time measurements are made in the O frame, not in the O' frame:

t = 0 is the time of emission according to the O frame
t = t1 is the time that photon B hits O' according to the O frame
t = t3 is the time that photon A hits O' according to the O frame ​

No and nly if you invoke some unwanted and unwarranted SR theory. This gedunken precedes Sr and should, like AE ofered it, be able to stand on its own two 1/3meters.

No doc, the times are measured using the passenger wrist watches. I specifically instructed all the passenger to first get a WWV time hack and use only the watches they wear. Of course the stationary observers do the same, but we aren't comparing stationary and moving frame clocks here. C is the speed of light c = 3x 10^8m/s. v is measured with train watches and distance measuring devices. t3 and t1 are moving frame times. This is how I set up th experiment. If you want to conduct the same experiment using stationary frame times, go for it, but only your opinion insists on using stationary times, your opinion, only.

On what possible ground do you infer, that I infer, the times weren't moving frame times?

Also, at the velocities used here, you could never prove your SR time dilation SR stuff anyway, could you?. Gotcha Doc.

Anyway, some time in this experiment I want to do the impossible: Stop the train and accelerate the embankment.Uh, oh, I have to set v = 0, right? Then my expression becomes, t3 = t1(c + 0)/(c -0) or t3 = t1 , which is contradicted by measurement that t3 > t1, Hence the passengers conclude that the train ain't stopping for no SR theory.

That's the only way you can derive that result. It says nothing whatsoever about measurements made in the O' frame. It says nothing about "simultaneity". It's just a trivial expression relating times t1 and t3, both of which are O-clock times.

Doc the times are from watches on the passengers on he trains, OK? Why are you persisting to swap my times on me? This isn't a Hidden "frame swap" of which you are getting such a rep for inserting into people's posts, is it?

By the way, here's a simpler way to derive that result. Note that all measurements are O-frame measurements. Let L be the distance from the midpoint to either A or B. By the time the photon from B hits O', O' has moved towards B, so:
(1) ct1 + vt1 = (c + v)t1 = L
And by the time the photon from A hits O':
(2) ct3 = L + vt3 --> (c - v)t3 = L
Combine (1) and (2) and you're done. So what?

So what? Doc, check this out.
http://www.mathpages.com/rr/s4-08/4-08.htm
I thought I left you a message that I had found a twin expression in the link regarding simultaneity. I was actually asking youd to take a look at what I had found so you could "mentor" me on the comparison. Did you not get my message? Doc the expression is a crucial aspect of simultaneity, you had better look, before you leap.

 

[geistkiesel]

Viewed from an observer at rest with the road, what you say is perfectly true. All speeds are given with respect to the road.

I used radar instrumentation in both cars, where the return signal is translated into relative speed, as when the fuzz is setting behind the big sign waiting for you to speed through her trap, her gun measures your velocity respecto hers, when she is stationary. But moving radar can be used, even though it is translated into velocity wrt the ground. I know, a Texas Ranger clocked me at 92mph between Dallas and FT. Wrth and he only gave me a warning. He showed me the 92.

B]According to observers at rest with the road[/B], the two cars approach each other at 40 miles per hour.

[Ah, now you want to switch frames and view things from A's frame. Better do it right: use the relativistic addition of velocities formula. For low speeds only it does turn out that $V_{B/A} = V_{B/road} + V_{road/A}$.


Everything is different, not just light. Welcome to the 20th century!

But I don't want to use relativistic addition of velocities formulas.

 

[geistkiesel]

evenj if she accelerated?

Not only does she consider herself at rest when measuring the speed of light, she considers herself at rest period, because she is inertial.

If she measures her acceleration to .1c and then a partner on the stationary Earth frame sends her a light beam, that has a standard 10^-8m wavelength. She is heading into the oncoming beam. and measures the time it takes one lambda to pass her eye at 10^-8/3.3x10^8 = .3030 x 10^-16 sec., for a determined frequency of 3.3003 10^16/sec. If I hadn't used her velocity I would have calulated .3333x10^-16sec. for one wavelength to pass, for a frequency of 1/.333= 3.3003x10^16/sec. which is shorter than it physicall is measured. She calculates then .9090x10^8m for the wave length.

If she determines her relative velocity she calculates he wavelength as 3.3]3.3 = 10^-8m, he correct wavelength. She still gets her red shift , but she isn't imposing physical errors in measurement into the equation.
Calcukating the wavelength change by .3/3.3003 = .091x10^-8m. Add this to her calulated wavelength or ,9090 + .091 = 10^-8, the correct wavelength. I haven't changed the speed of light.



Well, since length contraction is actually a prediction of the theory, while "mass shrinking" is a stupid figment of some retard's imagination, yeah, I guess it would sound better!. :rofl:[/QUOTE]

Ah yoo saeng i um ray tahded? :yuck: Yes, but check with ram1024, you've only got another day of SR theorey left, unless he generously extends the deadline.

 

[Doc Al]

No and nly if you invoke some unwanted and unwarranted SR theory. This gedunken precedes Sr and should, like AE ofered it, be able to stand on its own two 1/3meters.

Show me what "unwanted and unwarranted" SR theory I invoked. It's just D = VxT, as seen by the O frame. That's all you're doing.

No doc, the times are measured using the passenger wrist watches. I specifically instructed all the passenger to first get a WWV time hack and use only the watches they wear. Of course the stationary observers do the same, but we aren't comparing stationary and moving frame clocks here. C is the speed of light c = 3x 10^8m/s. v is measured with train watches and distance measuring devices. t3 and t1 are moving frame times. This is how I set up th experiment. If you want to conduct the same experiment using stationary frame times, go for it, but only your opinion insists on using stationary times, your opinion, only.

Laughable. For one, I'll bet you haven't the foggiest notion of what the O' wristwatch time will read when that B photon hits him. It's certainly not what you call "t1". Go ahead, calculate it (in terms of L and v). I'll wait. (I'll give you a hint down below.)

Second, if you really meant that t1 and t3 are O' times, then your equation makes no sense. Remember you can't just ASSUME that O and O' clocks agree.

On what possible ground do you infer, that I infer, the times weren't moving frame times?

You are welcome to use the O' times, but then you'll have to use the O' frame. Can you handle it? But then your equation, which uses O-frame times, won't work any more. I'm probably giving you too much credit. You really don't have any idea what you're doing.

Also, at the velocities used here, you could never prove your SR time dilation SR stuff anyway, could you?. Gotcha Doc.

You still haven't graduated from Einstein's Train Gedanken academy yet. You are a long ways away from calculating time dilation.

Doc the times are from watches on the passengers on he trains, OK? Why are you persisting to swap my times on me? This isn't a Hidden "frame swap" of which you are getting such a rep for inserting into people's posts, is it?

If those times are O' watch times, then your equation makes no sense. You are the one "swapping frames": you try to use O' times in an equation derived using O times.

So what? Doc, check this out.
http://www.mathpages.com/rr/s4-08/4-08.htm
I thought I left you a message that I had found a twin expression in the link regarding simultaneity.

That's an article discussing relativity--too advanced for you. There is nothing wrong with your equation, it's just not what you think it is.

I'll give you a big hint. What time does the O frame say that photon B hits O'? For this, you can use your reasoning, since only O-frame measurements are involved. vt1 + ct1 = L, so t1 = L/(v + c). That's the time that the B photon hits O' according to the O clocks. Your assignment: figure out what the O' wristwatch reads when B photon hits him.

 

[Doc Al]

I used radar instrumentation in both cars, where the return signal is translated into relative speed, as when the fuzz is setting behind the big sign waiting for you to speed through her trap, her gun measures your velocity respecto hers, when she is stationary. But moving radar can be used, even though it is translated into velocity wrt the ground.

Right. Note that the speed is always measured with respect to the radar, then translated to be with respect to the ground. If the speeds were fast enough, that translation would require knowledge of SR.

But I don't want to use relativistic addition of velocities formulas.

I know, I know. It hurts so bad!
Pity that nature doesn't care what you want or don't want.

 

[geistkiesel]

She cares for me.

Right. Note that the speed is always measured with respect to the radar, then translated to be with respect to the ground. If the speeds were fast enough, that translation would require knowledge of SR.

I know, I know. It hurts so bad!
Pity that nature doesn't care what you want or don't want.

Nature cares what I want and don't want, that's why Mother Nature exposes herself so I can see how she dances, how she twirls and shows a flash or two of a well turned ankle, with just the slightest tinge of a blush on her cheeks. She is always there with crooked smile, but straight teeth, and a bent finger beckon to, slowly now, "come hither dear geistkiesel". She let's me spend the night with her, every beautiful night. And what a sweet dream it is.

 

[quantumdude]

Tom: We don't "omit" the observer's speed. In the observer's frame, he is at rest

Ram: which means the speed of ALL objects in the universe acquire the observer's velocity, LIGHT INCLUDED.

Wrong. I can't even imagine how you could end up at this ridiculous conclusion.

why would light behave differently?

Differently than what?

and don't say the data says so, because the data does NOT include the addition of observer motion.

The data says so. Deal with it.

 

[quantumdude]

Tom: Janus proved that you have no idea of what the postulate of the speed of light says, or what "invariant" means.

Geistkiesel: Janus proved nothing . He merely quoted gospel and verse.

Incorrect. He quoted the relativistic velocity addition formula, which quite clearly differs from the Galilean velocity addition formula. The former preserves the speed of light in every frame, and the latter does not. This is a simple, direct contradiction to your false claim in the title of this thread. Silly wabbit.

Tom: This is so badly mistaken, I can hardly believe that anyone who claims to have studied any physics could say it.

Giestkiesel: "I know I am but what are you?"

Let me know when you graduate from the 4th grade.

When light is reflected between two parallel mirrors that are stationary each reflection follows the exct same path, i.e. one line. Of course, there is always side radiation, but you know of what I speak.

OK, so, you’ve got a light beam bouncing up and down between two mirrors.

When the mirrors move sideways in a direction parallel with the mirrors, the reflected light does not change [position. Remember, the source of the light, like each reflection from the mirror adds no light velocity in the direction of motion of the mirrors. The light, therefore, remains fixed like a perfect z-axis, stationary, absolute zero velocity.

In which frame are you talking about? Your statement here is only true in the rest frame of the mirrors. For someone who is in the original rest frame of the mirrors, the light doesn’t move straight up and down, it moves in a “V” pattern, like this:

\         /\          /\
 \       /   \       /   \ 
  \    /      \    /       \
   \ /          \/           \  …..etc.

A perfect and absolute frame of refer3enhce.

What is so perfect and absolute about it?

The reflections continues up and down. Eventually the mirrors will come to an end and unless some adjustments are made the left side of the apparatus will simply crash into the z-axis light beam.

The reflected light beam will not leave the mirror system. It will travel along with it.

Measuring side reflections is an arbitrary piece of physics sloth.

Huh?

The mirror system can measure where each returning relfection is located along the mirror surface, therefore the pulses can be cycled to start over on the right hand side of the apparatus, depending on the measured velocity.

It is really an engineering problem tom.

No, it’s an experimental physics problem.

I have degrees in physics and engineering. I don’t need you to tell me the difference between the two.

The diection of motion of the apparatus wrt to the z-axis can be detemined in a number of ways, one I just mentioned. Another, for the x-axis is with a modified ram1024 buoy

Goodie.

.Don't ever try to tell me tom_mattson that I cannot use the z-axis I just created and use it as an absolute reference frame.

I wouldn’t dream of it, but only because you have yet to define what an “absolute reference frame” is.

 

[quantumdude]

Notice, that inspite of claims
SR,
Galilean transformations,
Lorentzian transformations,
simultaneity,
time dilation
c measured as c in all inertial frames (with time dilation of course)
and length contraction
It all works as designed.

Poor, poor giestkiesel still doesn’t understand the difference betweeen real experiments and thought experiments.

You SR theorists have clouded your brilliant minds with theoretical molasses. Just think where we would all be if the simplicity of physics had been working for the past hundred years instead of that SR crap.

We? When precisely did you become part of we, anyway? You aren’t a scientist, and you probably never will be one. If we reject SR, then modern particle physics and astronomy is impossible to do.

It doesn't work you say? Prove it.

I mean prove it.

Get off your lazy arse and prove that it does work.

Don't send the robot Janus over here with his mantras, or you tom. Show it won't work in an experiment. This is all that will convince me. You are wasting your breath and good finger muscle by attempting to convince me otherwise.

Oh, dearie me! You think that I have any interest whatsoever in convincing you of anything? No, boy, I am not here to convince you.

I’m here to correct you.

I read your galilean post tom and in fact I was impressed which I believe I mentioned.

Tell the truth: You didn’t understand one bit of that post, did you?

The only thing is it doesn't get us there from here.Those watching may be impressed by your singular and collective brilliances, but I need an experiment. The one shown here is simple enough isn't it? Do this one. Our mutual uncle can afford it. Dig into his pockets.

I understand very well that a good physicist is necessaily a good proposal writer, right?

Yeah, right, I could get funding for this one.

“Dear Uncle Sam. I need a few thousand dollars so that I can convince some idiot on the internet who I don’t even know that SR is really, really right. I know you’ve already spent millions upon millions on experiments that show decisively that SR is correct, but trust me, this experiment will really make the difference. After this, I won’t bug you about it again. Until the next idiot comes along, that is…”

Do it yourself.

The physics is extremely simple. The speed of light is independent of the motion of the source. Therefore the the moving apparatus can and does keep a running track of the velocity of the light with respect to the inertial frame. Heresey? Get used to it.

Not heresy, just a simple non-sequitir.

News flash: The word "therefore" only belongs in the last line of a deductively valid argument.

I did learn something I didn't know earlier. That simultaneity is not operational at distances. What distances? What needed adjustment that you discarded or diosconrived your precious simultaneity? It certainly wasn't at the generosity of SR theorists was it? Does that mean we get absolute time back? Wow, just think of it! Or is that only at distance also?

Pure gibberish. If you want a response, then explain yourself.

 

[geistkiesel]

random expression find.

tom_mattson - this is for you, specifically.
I

http://www.mathpages.com/rr/s4-08/4-08.htm

Tom I saw the expression

t3 = t1(C + V)/(C - V)

in the link I was browsing: The Breakdown of Simultaneity. I had derived the expression when posting the Einstein train and station gedunken. t2 refers to the arrival of the first photon from B and t3 the arrival of the second photon from A. A and B having been simultaneously emitted into the stationary frame just as O' was at the midpoint of the A and B photon sources at M in the stationary frame. I used the times and v from the moving frame, but at this juncture I don't think it critical which frame times were used in the gedunken. Doc Al seems to think, or at least he says, screams actually, that I used stationary times when I didn't, even though I wasn't concerned with SR formalities in the first place as you are well aware. See my thread, you've been there before, 'No SR postulates violated in galilean transfomation' for a more complete showing of how and why I got what I did.

I was impressed with myself on this rather cheap random find. The link, an educational link, and not biased in a geistkiesel perspective, is a bit involved and to shorten the time to when I will become proficient enough to use it in my arguments against SR, could you take a look and cut some corners for me? You might be familair with the concepts now, I dunno. I went looking for simultaneity and distance having just learned that distance can effectively kill Simultaneity. How come you never mentioned this to me? Was it supposed to be a secret? Did you think I would never stumble across the concept? Were you purposefully stunting my educational development? Just kidding!

 

[Doc Al]

Doc Al apologizes to geistkiesel!

Yes, it's true. I've been so intent on ripping up everything you say, that perhaps I've been too harsh.

Regarding:

t3 = t1(C + V)/(C - V)

in the link I was browsing: The Breakdown of Simultaneity. I had derived the expression when posting the Einstein train and station gedunken. t2 refers to the arrival of the first photon from B and t3 the arrival of the second photon from A. A and B having been simultaneously emitted into the stationary frame just as O' was at the midpoint of the A and B photon sources at M in the stationary frame. I used the times and v from the moving frame, but at this juncture I don't think it critical which frame times were used in the gedunken. Doc Al seems to think, or at least he says, screams actually, that I used stationary times when I didn't, even though I wasn't concerned with SR formalities in the first place as you are well aware. See my thread, you've been there before, 'No SR postulates violated in galilean transfomation' for a more complete showing of how and why I got what I did.

Since it is trivially true that t3 = t1(C + V)/(C - V) when t3 and t1 are times measured using O-frame clocks, perhaps I just ASSumed that geistkiesel had mixed up his clocks. The question I should have asked is: Is his answer correct?

Well, yes it is! t'3 = t'1(C + V)/(C - V), where t'1 and t'3 are measured using O'-frame clocks. So, perhaps I misjudged your (obscure) reasoning.

If so, I humbly apologize! :uhh: (Yes, I will need years of therapy.)

But my question remains: So what?

 

[geistkiesel]

Yes, it's true. I've been so intent on ripping up everything you say, that perhaps I've been too harsh.

Regarding:

Since it is trivially true that t3 = t1(C + V)/(C - V) when t3 and t1 are times measured using O-frame clocks, perhaps I just ASSumed that geistkiesel had mixed up his clocks. The question I should have asked is: Is his answer correct?

Well, yes it is! t'3 = t'1(C + V)/(C - V), where t'1 and t'3 are measured using O'-frame clocks. So, perhaps I misjudged your (obscure) reasoning.

If so, I humbly apologize! :uhh: (Yes, I will need years of therapy.)

But my question remains: So what?

So I owe you a drink, tha's so what!

 

[geistkiesel]

Incorrect. He quoted the relativistic velocity addition formula, which quite clearly differs from the Galilean velocity addition formula. The former preserves the speed of light in every frame, and the latter does not. This is a simple, direct contradiction to your false claim in the title of this thread. Silly wabbit.

Let me know when you graduate from the 4th grade.

Tom when I add an observers speed to the speed of an oncoming light beam, I do not change the velocty of the light beam. I make my chicken scratches on a piece of paper. I do so in order to count the measured frequency that the unperturbed wave length passing the plane of my eyes. I cannot see how anything gets perturbed, that shouldn't be perturbed. I am making caluclations I am not compressing the oncoming wave train.

Actually I was in the 4th grade three times. Once because the 1st through 6th grade was in one room and I was a third grader and twice more when I moved in the middle of the 4th grade term. Did they graduate kids from the 4th grade when you went to school? Wow you must be really old.

OK, so, you’ve got a light beam bouncing up and down between two mirrors.

In which frame are you talking about? Your statement here is only true in the rest frame of the mirrors. For someone who is in the original rest frame of the mirrors, the light doesn’t move straight up and down, it moves in a “V” pattern, like this:

\         /\          /\
 \       /   \       /   \ 
  \    /      \    /       \
   \ /          \/           \  …..etc.

What is so perfect and absolute about it?

Misguided road runner chasing Wiley Coyote! Here is some 5th grade physics everybody and Watch Wiley. He's got a rope tied around his waist, the other end tied around a log suspended over a canyon. He hears a "be-beep" and dives over the side of the cliff, lining up his spear with his shish-kabob lunch about to meet him as it comes running aorund the cuirve. Whoops, the rope is about 6 feet too long.

A photon once emitted is not dependent on the source of the photon for any subsequent motion, correct? Say yes. Does not the photon remain Newtonianly moving invariantly in astraight line?

Also, we do not need any "moving observers frame." or "stationary observer frame" rhetoric and here is why. The mirrors are parallel and hence the photon is emitted perpendicular to the plane of the mirrors and it will move in a straight line as if the mirrors were non existent, right? [We discount side lobe radiated photons. You holding a flash light and pointing it away from me does not hide those photons from my eyes. I can clearly see the line of the beam.]

So the mirrors move while the photon is in flight to the opposite mirror. When it reaches the opposite mirror for sure it will not strike the same spot is struck in the previous visit, or when the the mirrors were stationary with wrt to the invariant trajectory of the emitted beam, our z-axis..

Again, once emitted the photon knows noting of the apparatus, correct?

There might be some perturbation by the mirror surface especially at high velocities, but these can be compensated for. Let us refrain, if possible from using any off perpendicular radiation that is seen coming from all straight line moving beams like flashlights, ok? We know they are there, but to use those off radiating beams corrupts our invariant strainght line, absolutely invariant z-axis defining beam, OK? Let's shield the effing beam, as bestwe can if we have to.

The mirrors are still parallel and eventually the left side of the apparatus will catch up to our z-axis, unless we adjust for it. If we place along the x axis, the direction of motion, photon counters on each mirror surface and count the number of counters the beam strikes/second in a rapidly reflecting frequency the motion to the right can be calculated, can it not, knowing the x spacing of the counters? Don't we ahve a battery of computer whizzes/hackers? [We had a summer intern come into a computer group using a Vax size off brand system. He was 18 years old. Wihthin a half hour of coming through the door to the facilty he had everybody's password.].

If we detect the left side catching up and about to swamp the z-axis we simply start the process as Yogi Berra would say, " its deja vu all over again" from the far right hand length of the mirror system. Software, high impedence photon detectors, physicists with open creative minds and thinking abilities, and the most important: The willingness to trust the invariant laws of physics even when contradicting a cast in concrete theory as "this little puppy will never be modified by scurrilous dissidents of the likes of geistkiesel et al".

Remember, the only "law of physics" we invoke here is the one that everyone in these threads use in virtually every post. The motion of the emitted photon remain directed in a straight line, exactly straight, forever, until acted upon by a force not considered in the time span of "forever".

{A slight digression: The michelson-morley experiment makes the same mistake. The beam moving transverse to the frame motion should be allowed to reflect perpendicularly to the straight line motion of the beam moving parallel with the motion of the frame. At most the half-silvered miorrors will have to be lengthened, but there aren't any technical prohibitions to this. Using a side reflection of the downward directed beam gets you nothing but confusion, and time dilation before it is called for. Make an MM calculation assuming 90 degree reflection of the MM transverse reflected beam and tell us all what you get.}

The reflected light beam will not leave the mirror system. It will travel along with it.

Only if you use a side lobe of our z-axis flashlight beam. We restart the beam, at the right as the left end of the apparatus is about to strike the up/down beam.

If you do not agree, you will have to show me why the z-axis beam, will ever decide to taking a saw toothed motion. Do not confuse side lobe photon reflection, which I admit will be reflected forward as well as aft, right?, why not use the aft reflected beams also, together with the photon counters on both surfaces, as well as forward saw toothed beams so you can have a triple whammy error correcting system comparing up/down, reflected forward, and reflected aft as checks of resolution etc.Coded pulses can take care of the accounting problems.
geistkiesel

If we let our parallel miror system get up to .99 c. in the plus x direction, we might need a healthy few hundred rapidly relfecting beams for ultra high frame velocities, but so far we do not need any time dilation because we aren't making the same mistake that those that preceded us made in forgetting about our faithful invariant z-axis and substituting off radiated photons for that axis, when they mistook the saw tooth reflection as the same beam as our frame independent z-axis, right? One law of physics and we get all this and it is painless isn't it? Especially if you aren't buying into any of this advanced 5th grade foolishness.

I wouldn’t dream of it, but only because you have yet to define what an “absolute reference frame” is.

Ah yes, the plot thickens, just as I suspected: engineering and physics, it is all becoming so clear now.

Absolute reference frame is a slam dunk engineering project actually. I used to shoot craps and then one day I got n idea that changed me from a worhtless gambling man tio an obsessed "technician" I won't use the "P" word or even the "E" word, you might get upset.

I saw a die as a six sided reflecting inertial frame. We had a thread recently where photons were emitted from the midpoint of moving and stationary frames, recall? We'll just consider the moving frame to the right. Extending forward and aft are two reflecting mirrors that reflect the emitted photons back to

1. the source that is moving to the right.
2. The mirror from the left will strike the left moving photon first while the right moving photon has yet to catch up with the receding right reflecting mirror.(First time segment completed).
3. The left moving photon will be reflected back to what was the midpoint (stationry frame) which has dirfted with the frame velocity to the right, while the right moving photon now catches up to the right moving mirror. (Second time segment completed).
4. The once left moving photon now heading to the right is now two time segment to the right from where it was reflected by the left mirror. The right moving photon is now relfected one time segment back to the left arriving simultaneously with the original left photon and original observer.(Third time segment completed).
5. Summary: the left photon makes one left segment, two right segment legs. The right photon makes two right segment, one left segment legs.
6. Conclusion: initializing the three axes and using feedback control we can zero any xyz motion wrt invariant xyz photon axes, AKA an absolute zero velocity reference frame, and to hell with all the stellar objects moving hither and thence in the universe, our reference frame is absolute zeror velocity wrt absolute invariant photon axes.

 

[ram1024]

the moving parallel mirrors is pure genius, Geist

on many levels

 

[ram1024]

here's a fun spin on your mirrors, geist.

Two circular mirrored platters, parallel and rotating either in the same direction, or if you want to get cheeky, in opposite directions.

bounce the beam between those and then let her rip with high RPMs. the circumference of the circle enscribed by the beam multiplied by the RPM, would be your added velocity to light's "constant" speed taking the relative inertial frame of the platter. rotating the opposing platter in the opposite direction would increase the relative velocity applied even more.

 

[quantumdude]

Tom when I add an observers speed to the speed of an oncoming light beam, I do not change the velocty of the light beam.

You do change the velocity of the light beam, relative to that observer.

I make my chicken scratches on a piece of paper. I do so in order to count the measured frequency that the unperturbed wave length passing the plane of my eyes.

...except you haven't made any real measurements.

I cannot see how anything gets perturbed, that shouldn't be perturbed. I am making caluclations I am not compressing the oncoming wave train.

...except you haven't done any real calculations.

Misguided road runner chasing Wiley Coyote! Here is some 5th grade physics everybody and Watch Wiley. He's got a rope tied around his waist, the other end tied around a log suspended over a canyon. He hears a "be-beep" and dives over the side of the cliff, lining up his spear with his shish-kabob lunch about to meet him as it comes running aorund the cuirve. Whoops, the rope is about 6 feet too long.

Been sniffin' glue again?

A photon once emitted is not dependent on the source of the photon for any subsequent motion, correct? Say yes. Does not the photon remain Newtonianly moving invariantly in astraight line?

3 questions:


1. What does it mean for a photon to be dependent on the source? Do you mean the photon's speed? It's velocity?
2. What does it mean for a photon to remain Newtonionly moving invariantly in astraight line? Does it mean that the speed is the same? The velocity? The trajeectory?
3. Are you just using a random word generator to construct your posts?

Also, we do not need any "moving observers frame." or "stationary observer frame" rhetoric and here is why.

Yes, you do need it. Velocities, speeds, and trajectories are determined from a particular reference frame. Measured and calculated values of observables for a frame pertain to that frame only.

So the mirrors move while the photon is in flight to the opposite mirror. When it reaches the opposite mirror for sure it will not strike the same spot is struck in the previous visit, or when the the mirrors were stationary with wrt to the invariant trajectory of the emitted beam, our z-axis..

Again, once emitted the photon knows noting of the apparatus, correct?

No, not correct. The photons collide with the mirrors, and so there will be a component of momentum parallel to the mirrors.

 

[geistkiesel]

Absolute zero velocity frame design, summary, recap, quesion/answers..

You do change the velocity of the light beam, relative to that observer. i

I disagree. I have notbooks of calulations and no evidence of ever having alteed any wavelengths..

...except you haven't made any real measurements.

that is correct, I have no lab, so sorry. Anyway, when I get around laboratories, I always hea glass tinkling in the background as I walk up to the door. Do you get my drift? Tools are things tems to drop on the floor and toes. Labs are places where knuckles get scrapped. Labs are places where people look at you as if you shouldn't be there. Just kidding.

...except you haven't done any real calculations.

Are you referring to some unreal calculations? What kind of calculations are you referring? Something you are unable to handle yourself, or that you do not understand without me providing some numbers? or is this a test question?

Been sniffin' glue again?

Not my thing, but if you have some reccomendations, I am always willing to listen. What I've heard about glue doesn't place it in any interest zonesof mine. Actually, I used to laugh at women mostly, who talked about being chocolate junkies. I understand those ladies completely. Anyway this is a petty sticky subject matter.

3 questions:

1. What does it mean for a photon to be dependent on the source? Do you mean the photon's speed? It's velocity?
2. What does it mean for a photon to remain Newtonionly moving invariantly in a straight line? Does it mean that the speed is the same? The velocity? The trajectory?
3. Are you just using a random word generator to construct your posts?

Answers to TM interrogatory.

1. Once emitted the motion of a photon is independent of the motion of the siource of the of the photon. Specificallt e source and photon velocities are completely independent of each other. The velocity of the photon s independent of the velocity of the source.
2. A particle motionoess or moving in a straight line will continue in that motion until acted upon by an outside source. I forget the number, but it is the first three of Newtons laws of motion. Or in a modern version computed motions will contiue until acted upon by programs not included the algorithmic structure ofthe computed motion.#2 a.b.c. yes,yes,yes.
3. Interesting question. In the sense that the probability that all the air molecules in your present room will all point to the wall in front of you right now is the same probability that they will point as they are right now. Some might call the described system random, I don't know, but I do use the dictionary, which is a random word generator of sorts. Is this what you meant? probably not. What the hell, the answer to your question number 3 is yes.

Yes, you do need it. Velocities, speeds, and trajectories are determined from a particular reference frame. Measured and calculated values of observables for a frame pertain to that frame only.

I was thinking in terms of SR imperatives. A form of gun-shy reaction. Do you want to see all my SR posting scars? It isn't a pretty sight.

What I was writing at the moment didn't need a background of frame identification, moving vs stationry in order to give physical value to what was being described. I supposed I assumed that the document would take a latin turn and res ipsa loquitor. Having described the absolute zero velocity frame, all your requiements are ullfilled are they not, those requirements stated and implied in your statement? I agree with what you say.

No, not correct. The photons collide with the mirrors, and so there will be a component of momentum parallel to the mirrors.

I thought I mentioned somewhere that at higher velocities some photon/mirror interfacing effects would have to be dealt with. Thinking about it now there will always be some scatter at the mirror surface/photon interface, which can corrected for. This is in the category of "side lobe'" radiation, not a fatal interference. The more optically perfect the surface the less the "random'' scattering effects. One method that can be explored for perfecting mirror surfaces is through vapor deposition of Au onto single crystals of cleaved surfaces of NaCl, for instance. The atomic spacing of salt and gold are very close. Under the right conditions huge areas of defect free zones can be deposited, perfect zero-defect areas defined as flat planes of single crystals of gold are manufactured. (heat gold wire on a tungsten filament, heat the base single crystal NaCl to ~330 degree C, expose the heated wire and evaporate.) The flat zones are single planes of Au, no hills and valleys and gorges that the "best" polished surface are always left with. However, there are potential hills on the suface of the NaCl until sufficient numbers of Au atoms have arrived and the potential hills flatten and disappear, theoretically speaking.
In any event there will always be sufficient numbers of perpendicular photons to do the job. The photons would have to be stimulated for emission periodocally in any event, LED emjitters/surfaces is always a possiibility. Also, by pulsing the up/down photons with identification codes, the accounting burdens can be materially minimized and scatteing short falls effectively eliminated.

 

[quantumdude]

Geistkiesel: Tom when I add an observers speed to the speed of an oncoming light beam, I do not change the velocty of the light beam.

Tom: You do change the velocity of the light beam, relative to that observer.

Geistkiesel: I disagree. I have notbooks of calulations and no evidence of ever having alteed any wavelengths..

First of all, altering we weren't talking about altering the wavelengths. We were talking about adding velocities. And second, you quite clearly do change the velocity of light relative to the moving observer, using the Galilean velocity addition formula. It's in your first post.

Geistkiesel: I make my chicken scratches on a piece of paper. I do so in order to count the measured frequency that the unperturbed wave length passing the plane of my eyes.

Tom: ...except you haven't made any real measurements.

Geistkiesel: that is correct, I have no lab, so sorry.

So you admit that the part in red is a falsehood then?

Geistkiesel: I cannot see how anything gets perturbed, that shouldn't be perturbed. I am making caluclations I am not compressing the oncoming wave train.

Tom: ...except you haven't done any real calculations.

Geistkiesel: Are you referring to some unreal calculations?

I am referring to the bogus calculations you do regarding the propagation of light, which completely exclude Maxwell's equations.

Tom: What does it mean for a photon to be dependent on the source? Do you mean the photon's speed? It's velocity?

Geistkiesel: Once emitted the motion of a photon is independent of the motion of the siource of the of the photon. Specificallt e source and photon velocities are completely independent of each other. The velocity of the photon s independent of the velocity of the source.

Do you realize that that is the speed of light postulate of SR, and that it is not consistent with the Galilean velocity addition formula you used in your first post?

Tom: What does it mean for a photon to remain Newtonionly moving invariantly in a straight line? Does it mean that the speed is the same? The velocity? The trajectory?

Geistkiesel: A particle motionoess or moving in a straight line will continue in that motion until acted upon by an outside source. I forget the number, but it is the first three of Newtons laws of motion. Or in a modern version computed motions will contiue until acted upon by programs not included the algorithmic structure ofthe computed motion.#2 a.b.c. yes,yes,yes.

Do you realize that the light does not continue moving in a straight line? The impetus that alters its motion is provided by the mirrors.

And as for your belief that you do not need to specify a reference frame in order for your statements to be physically meaningful, that is not true. The photon trajectory, for instance, is going to look different in every frame.

 

[quantumdude]

why are you so confused?

He's not. You are.

A can travel east at 20 miles per hour. B can travel west at 20 miles per hour. both cars have a speed limit of 20 miles per hour.

Yes, both cars can have a speed of 20 mph relative to the ground. No one disputes this.

you're telling me the maximum speed they can approach each other is 20 miles an hour,

No he isn't. Do you even read what anyone else writes? It seems that you do not. Janus posted the relativistic velocity addition formula on the first page of the thread. SR does not predict that the velocity of one car relative to another is 20 mph. It predicts that the relative speed is just a shade under 40 mph, and that for small speeds the difference between the actual relative speed and 40 mph is too small to detect.

which is patently untrue given if they're 40 miles apart and they both drive towards each other they can collide in an hour.

by any definition if you take A as a stationary object, B now travels 40 miles an hour towards A.

Not exaclty. As I said, B travels at just under 40 mph towards A, and vice versa.

"but light is different..."

BS.

You haven't learned a thing. No one here says that the kinematics of light is any different from the kinematics of cars. In both cases, the SR velocity addition formula holds.

 

[ram1024]

You haven't learned a thing. No one here says that the kinematics of light is any different from the kinematics of cars. In both cases, the SR velocity addition formula holds

then let's hear your logical reasons for why 5 + 5 = 4.999239582394...

 

[quantumdude]

then let's hear your logical reasons for why 5 + 5 = 4.999239582394...

I never said it did.

Next stupid question?

 

[ram1024]

that "stupid question" was derived from your "stupid answer" thankyouverymuch

Not exaclty. As I said, B travels at just under 40 mph towards A, and vice versa.

 

[quantumdude]

that "stupid question" was derived from your "stupid answer" thankyouverymuch

No, it wasn't. It came straight from your brain to your keyboard, and was unfortunately entered to PF to become yet another waste of bandwidth in this circus of a thread.

Persistent, hard-headed use of the Galilean velocity addition formula is your hangup, not mine. When I say that the velocity of one car as measured by the other is less than 40 mph, I am not saying that 20+20 does not equal 40. When you put words in my mouth like that, you make yourself look like a moron. And if you act like a moron, then don't be surprised when people treat you like one.

The SR velocity addition formula is in Janus' post. That is the correct way to add velocities in a Minkowski spacetime, which is an excellent approximation to physical spacetime for sufficiently low energy and momentum densities (IOW, when we can ignore GR effects).

 

[ram1024]

complete horsecrap. the only reason you ascribe to this "SR velocity addition" is because it conforms to your candy-land fantasy universe. there is NO logical reason a 20 mile per hour car approaching another car also traveling in the opposite direction at 20 miles per hour should NOT be able to consider himself stationary with the other car approaching at 40 miles per hour.

Let us look at the definitions of the terms we're using and we'll lay the groundwork for who is being completely unreasonable.

Speed: Covering a certain distance over period of a certain time.

20 miles per hour: given an HOUR worth of time, the object would travel 20 miles.

so here you have 1 car traveling 20 miles in one hour and the other car traveling 20 miles per hour. suppose they start 40 miles apart at the same time pointed right at each other. at 1 hour what happens? oh lo and behold, they freaking crash. not "from either of the driver's point of view they didn't crash because the other car only traveled 99.9999% of the distance towards him". they DO crash.

what was the total distance? 40 miles
what was the time for the experiment? 1 hour

what is 40 miles in 1 hour? 40mph = the speed at which they approached each other.

Sorry Tom, that filth won't fly.

 

[russ_watters]

complete horsecrap. the only reason you ascribe to this "SR velocity addition" is because it conforms to your candy-land fantasy universe. there is NO logical reason a 20 mile per hour car approaching another car also traveling in the opposite direction at 20 miles per hour should NOT be able to consider himself stationary with the other car approaching at 40 miles per hour.

Enter me to snap this thread back to reality...

Ram, you're absolutely right. It doesn't make logical sense in a Galilean universe. But what the others have been trying to tell you and you refuse to accept is that we do not live in a Galilean universe. Ram, I know its hard to accept, but that is reality.

All this bickering really has a simple reason and a simple solution. Someone's worldview is wrong. How do we know its your worldview that's wrong and not ours? Experimentation and data say so. Simple as that.

So, have you looked at any of the actual experiments we've posted? Now, the last time someone brought up actual experiments, you made a left turn and accused the experimenters of falsifying their own results. I shouldn't have to tell you that that isn't a good way to learn, nor is it a good way to win an argument. You have another shot at that choice. You can choose to ignore reality because you don't like it or choose to look at the experiments rationally and objectively, and accept what they say about reality.

 

[Janus]

complete horsecrap. the only reason you ascribe to this "SR velocity addition" is because it conforms to your candy-land fantasy universe. there is NO logical reason a 20 mile per hour car approaching another car also traveling in the opposite direction at 20 miles per hour should NOT be able to consider himself stationary with the other car approaching at 40 miles per hour.

Let us look at the definitions of the terms we're using and we'll lay the groundwork for who is being completely unreasonable.

Speed: Covering a certain distance over period of a certain time.

20 miles per hour: given an HOUR worth of time, the object would travel 20 miles.

so here you have 1 car traveling 20 miles in one hour and the other car traveling 20 miles per hour.

Relative to what? The ground, I assume, as measured from the ground or from the car whose speed you're measuring.

suppose they start 40 miles apart at the same time pointed right at each other.

As measured from where?, the ground or the cars? If this 40 miles is measured from the ground, then the cars will not be 40 miles apart as measured from either car, nor will the cars start at the same time according to either car..

at 1 hour what happens? oh lo and behold, they freaking crash.

As long as all measurements are made from the ground frame, yes

not "from either of the driver's point of view they didn't crash because the other car only traveled 99.9999% of the distance towards him".

Nobody said they wouldn't.

they DO crash.

Yes, they do crash according to all frames, and they crash at the same point on the ground according to all frames. The end results are not disputed by anybody. The cars just won't agree that their relative speed to each other was 40 mph or that both cars started at the same moment.

what was the total distance? 40 miles

As measured from the ground

what was the time for the experiment? 1 hour

As measured from the ground

what is 40 miles in 1 hour? 40mph = the speed at which they approached each other.

as measured form the ground

 

[quantumdude]

complete horsecrap. the only reason you ascribe to this "SR velocity addition" is because it conforms to your candy-land fantasy universe.

Wrong. I subscribe to it because it is in agreement with experiments done in the real world, which I cordially invite you to come back to.

there is NO logical reason a 20 mile per hour car approaching another car also traveling in the opposite direction at 20 miles per hour should NOT be able to consider himself stationary with the other car approaching at 40 miles per hour.

Actually, there is a logical reason. The reason is that the data say so!

Let us look at the definitions of the terms we're using and we'll lay the groundwork for who is being completely unreasonable.

That's easy: You are.

Speed: Covering a certain distance over period of a certain time.

20 miles per hour: given an HOUR worth of time, the object would travel 20 miles.

so here you have 1 car traveling 20 miles in one hour and the other car traveling 20 miles per hour. suppose they start 40 miles apart at the same time pointed right at each other. at 1 hour what happens? oh lo and behold, they freaking crash. not "from either of the driver's point of view they didn't crash because the other car only traveled 99.9999% of the distance towards him". they DO crash.

what was the total distance? 40 miles
what was the time for the experiment? 1 hour

what is 40 miles in 1 hour? 40mph = the speed at which they approached each other.

Sorry Tom, that filth won't fly.

What is your major malfunction? You just keep doing the same thing over and over.

1. Assume that Galilean relativity is correct.
2. Devise a thought experiment, predicated on #1.
3. Conclude that Galilean relativity is correct.

Can't you see how idiotic that is? You make no reference whatsoever to real experimental results.

And furthermore, you talk about logic? You haven't the foggiest notion of what logic is. The logic of SR comes from the fact that Maxwell's equations and Galilean relativity cannot both be true. Now we know that Maxwell's equations are right (ignoring quantum effects), and we know that Galilean relativity is wrong from independent experiments. But all of this is lost on you, because rather than look at any of the experimental results or logic, you would just rather sit there with your fingers in your ears, shouting at the top of your lungs that SR is wrong. It is really pathetic.