Today Special Relativity dies

[Eyesaw]

Originally Posted by Eyesaw
In particular, I'd like to address your statement here : "The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame)." : Only if you are silly enough to use the front and rear of the train as the distance the two photons from events A and B traveled to the moving observer in lieu of the actual places where the events occurred as marked by the burned wood on the tracks to deduce the time of events.

So your suggestion is :the train observer must prefer the burned wood on the tracks over the burn marks on the train, as the location of events. Why? What is special about the tracks?

Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.




Quote:
Originally Posted by Eyesaw
And only if you are naive enough to think the sequence of actual events happening is equivalent to the sequence of detection of light signals from those events.

Speed of light is constant relative to all observers (do you disagree at this point?). Then, if light from both events travel the same distance (that is, the observer is at the midpoint), the order of detection of light signals can be used to conclude about the sequence of events. Why do you think not?

Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.

 

[wespe]

Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.

I don't understand your point. What kind of difference? Elaborate please..

Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.

Yes, both frog and duck are fried at the same time in the stationary frame. And, both frog and duck will be fried in the train frame, despite at different times. But their frying at different times doesn't mean you can prevent the frog or duck from frying after the other one is fried. That would require sending a message faster than light, which means going back in time (does bring up paradoxes, but FTL is not possible according to SR).

 

[Doc Al]

case #8 -- not so great

You still seem to think that all you need to do to switch a diagram from one frame to another is just slide things over so that the pictures line up with the other observer. If things were that easy then everyone would understand relativity!

You keep ignoring simultaneity, length contraction, and other relativistic effects.

nah we've moved on to Case #8 now, jcsd

lemme diagram it for you

Case #8

     [u](o)                    |                    (o)[/u]
     [u](o)                    |                  (o)[/u]
     [u](o)                    |                (o)[/u]
     [u](o)                    |              (o)[/u]

The observers in this case are both Observers AND Emitters. at the onset of the experiment, ObserverA and ObserverB shine a light towards each other. Simultaneity is verified by an observer at the midpoint of the starting locations of both observers.

So the midpoint is in the frame of Observer A. And the lights are switched on simultaneously according to Observer A, but not Observer B. The diagram shows the view from the Observer A frame.

ObserverB then moves towards ObserverA. Predict who receives a photon first.

According to the A frame clocks, Observer B detects a photon first. But so what? The two events--(1) Observer B detects a photon and (2) Observer A detects a photon--are not causally connected and happen at different locations. Their time order is frame dependent.

     [u](o)                    |                    (o)[/u]
       [u](o)                    |                  (o)[/u]
         [u](o)                    |                (o)[/u]
           [u](o)                    |              (o)[/u]

now all we've done is taken that same experiment from ObserverB's point of view AS IF he was the stationary one.

You think you are showing the view from B's view, but you're not drawing it accurately. So the diagram is useless for drawing any meaningul conclusions. An accurate diagram would show length contraction, time dilation, and most importantly the lack of simultaneity of the two light flashes.

in this case the reverse result is shown, ObserverA receives his photon first. It is also interesting to note that the midpoint observer no longer receives photons simultaneously in this view, further proving that frame shifting is wrong for this type of experiment.

All this demonstrates is that you don't know how to draw the diagram from B's viewpoint. It's not easy.

How can something absolute such as simultaneity AT A POINT be violated by merely changing how you view a situation?

I'm glad you realize that that would make no sense! But the problem lies with your analysis, not with relativity. Both frames agree that the photons arrive at M simultaneously.

 

[Alkatran]

Well, RAM. You've had a few different people say exactly the same thing for an entire page. That's how special relativity sees it. It seems like you have your "own" special relativity that you're trying to disprove to prove your own theory.

You can't change an existing theory to disprove it.

 

[ram1024]

Been wracking my brain for two days trying to find SOME way to get you guys to understand me.

i may have stumbled upon the answer, however, so try and hear me out once more.

How does one measure distance? it is a measurement based upon a constant "movement" over an allotted "time"

speed is the direct proponent of this calculation including both distance AND time.

now when you say the speed of light is constant relative to the observer, that's all well and good, but only if the observer is stationary. stationary in respect to WHAT you may ask. stationary in respect to THE SPEED OF LIGHT says SR. The measurement "method" of SR is to assume that the observer is ALWAYS stationary, and since any "motion" of a light source has no effect, light will always be stationary as well.

think of how anything else moves through the universe however. If something IS moving, then it covers a distance TOWARDS something else.

Hence if i have 3 positions on a line A, B and C. and a person at point C throws a ball to point B while A moves to point B to reach that point the same time the ball reaches there.

the relative velocity of the ball, to ParticipantA is the velocity of the distance C-B over the time it took the ball to get from C-B PLUS the velocity of the distance A-B divided by the time it took ParticipantA to get from A-B. basically A-B/t + B-C/t = A-C/2t

Even taking ParticipantA as stationary, now ParticipantB has a velocity of B-A and the ball has a velocity C-B making the ball's velocity towards A, A-C/2t

now the problem with relativity is you have to abandon this "stationary observer" method of measurement, or account for the problem caused by light being unbound to its source. The problem you run into is as follows.

A is making progress to a point B, reaching there at the same time as a photon from C. The relative velocity of light for A in THIS case is of course easily measured as above. A covers a distance A-B/t and the photon covers a distance C-B/t. the relative velocity of the photon is of course A-C (total distance covered)/2t (total time elapsed)

case1
http://www.imagedump.com/index.cgi?pick=get&tp=93222

now move to make A a stationary observer.

I'll present to you two pictures and let you guess which is right. which is wrong. which is SR prediction.

case2
http://www.imagedump.com/index.cgi?pick=get&tp=93223

case3
http://www.imagedump.com/index.cgi?pick=get&tp=93224

Now looking at these two pictures you can note a few things.

Case2 asserts that A is stationary, it does NOT move, which means B must move relatively to its position and light from C intercepts them both at that location.

Case3 asserts THE SAME THING, except it asserts A is already IN B's position. well then where is B? obviously B has to move relatively to get to A so we move it over to maintain consistancy. C, which is not an object, but light from a source does NOT move, so lo and behold the measurement of light from C to A is the same as it was from C to B in the previous case.

Case3 ignores a very real distance, that is a distance covered from A-B in the time it took light to cover the distance C-B. Case 3 is an incorrect measurement of the speed of light.

 

[Alkatran]

Judging entirely by sight, the first one can't be true because light is exceding c according to the observer. (Assuming the yellow line is a photon)

So, by elimination, the second one is true. But I fail to see your point.

Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.

 

[ram1024]

by "second one" do you mean case3 ?

Are you going to give another argument about the observers being in two places at once? Because I've already told you it's because of their "when" line being skewed.

no, not at all. in case3 the setup applied was that there has to exist relative gains between Participants A and B so if A is already in position B <to keep lightspeed constant> then logic dictates B is closer to C than the original picture. but we know this is NOT the case, so the picture is wrong.

 

[Alkatran]

What in the world are you talking about? In ALL the pictures B is ALWAYS closer to C than A.

 

[ram1024]

BC in case1 is a different distance than BC in case3

simply

 

[Alkatran]

That would be because of *gasp* length contraction. Due to the "when" line being skewed WHICH I ALREADY SAID!

 

[ram1024]

no DUH it's because of length contraction

but the length contraction is wholly based on the wrongness of the picture. Length contraction, Time Dialation, constant light speed are COMPLETELY based off of using the wrong picture to measure distances and times.

look at case2 which is the RIGHT way to measure things. that is the way we measure EVERYTHING else in the universe. why measure light different?

because you're all a bunch of lunatics :D

 

[Alkatran]

Because all measurements of light have shown that that is the only way to measure it. Duh.

You see, it's all math until you apply it to REALITY, that's when it becomes Physics.

 

[ram1024]

no, you could measure it THE CORRECT WAY

craziness.

i'm going to have to make machines to do it properly aren't i. dammitall

 

[Alkatran]

Oh, you're so willing to learn, aren't you? The mere idea that time can be so lined with space drives you mad.

You go and make your machines to measure c, but there's no point, people much more intelligent than you have tried it before. Guess what? They all found out the same thing (except maybe the ones who made mistakes): Einstein was right.

 

[ram1024]

no einstein assumed that the motion of the observer makes no difference which is a ludicrous proposition AT BEST.

all calculation done for speed of light were taken WITH a stationary observer completely discarding that motion/distance

time/space is fine, but not with light as absolute speed defining it.

 

[Eyesaw]

Quote:
Originally Posted by Eyesaw
Actually my suggestion is that you try running from the burn marks on the track until you catch up to the observer inside the train and then tell us if there is a difference between running just from the burn marks on the train to the middle of the train.

I don't understand your point. What kind of difference? Elaborate please..
Quote:
Originally Posted by Eyesaw

Because if a frog and a duck were on the track at the time and place of simultaneous lightning flashes A&B (according to the embankment observer *snort*) , they would both be fried, despite what the train observer using SR's wrong assumptions calculate.

Yes, both frog and duck are fried at the same time in the stationary frame. And, both frog and duck will be fried in the train frame, despite at different times. But their frying at different times doesn't mean you can prevent the frog or duck from frying after the other one is fried. That would require sending a message faster than light, which means going back in time (does bring up paradoxes, but FTL is not possible according to SR).

Reread Einstein’s gendanken in his Special Relativity. The experimental set up is such that the rear and front of the train are located at A&B when the lightning flashes occur, i.e. those events happen simultaneously in the absolute sense for both the platform frame and the train frame. What is not simultaneous from one frame to the other are the reception of the signals from the events to the observers in the different frames. This however is hardly surprising since the train is moving towards one signal and away from the other signal when compared with the observer of the signals on the platform.

What is special about the platform frame is that since the observer is supposed to be at the midpoint of the flashes, she can only receive the signals simultaneously if she was really absolutely at rest- otherwise she would have to conclude that the speed of light measured from two different directions are not the same, contradicting the postulates of SR. So since an absolute rest frame was found it definitely makes life easier to measure all speeds relative to it. In reality the train experiment is of course flawed since the Earth is actually moving through space so that even the platform observer cannot receive the signals simultaneously.

But for amusement purposes, let’s assume that Einstein wasn’t talking about an absolute rest frame in his gedanken. It’s still absurd to use the reception time of signals as a representation of the sequence of how actual events occurred. If we see a star explode today from our telescope, who would be gullible enough to think that it happened today? So why do you make the case for relativity of simultaneity based on the reception time of the signals if in the absence of an absolute rest frame, it would be nigh impossible to determine the actual place in space when the light was produced nor our velocity through space, which surely must affect our determination of the speed of light since if the speed of light was truly constant in vacuo, if we have any motion through space, we will misjudge its true velocity. I don’t think this is a good reason to rid of absolute time do you? Really, which of you SR die hards can claim to have experienced any event that wasn’t absolutely simultaneous for everyone? If events were not absolutely simultaneous, your hand should have fell off from your wrist and your head from your body long before you finished typing your response.

 

[wespe]

... (post#485)

Ram,

I hate to tell you this but (A-B)/t + (B-C)/t = (A-C)/2t is wrong, it equals to (A-C)/t. I hope you won't discuss this or we have serious problems here.

Apart from that, I looked at your first picture. I'm not very familiar with spacetime diagrams. But, the picture looks upside down; conventionally, time axis should increase in the up direction.The angle between red and yellow lines should be 45 degrees (time and space units are chosen so that speed of light is 1). You should label things to avoid any confusion. See http://www.cord.edu/dept/physics/credo/spacetime.html
Also specify the values of speed, time, distance (chose convenient numbers).

Do that and ask what the diagram would look like from A's perspective according to SR. Obviously you don't know the answer. Just ask, don't draw conclusions. I'm not saying I can draw the correct skewed diagram, but I know the angle between the observer and light should always be 45 degrees. Perhaps someone here will help you if you show willingness to learn.

 

[wespe]

Reread Einstein’s gendanken in his Special Relativity. The experimental set up is such that the rear and front of the train are located at A&B when the lightning flashes occur, i.e. those events happen simultaneously in the absolute sense for both

sorry, we can't communicate if what you comprehend is that from reading http://www.bartleby.com/173/9.html

 

[ram1024]

http://www.imagedump.com/index.cgi?pick=get&tp=93385

happy now?

 

[ram1024]

sorry, we can't communicate if what you comprehend is that from reading http://www.bartleby.com/173/9.html

well that IS what it says.

When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment. But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the traveling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possesses this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.

The conclusion is based off of not KNOWING he has moved any distance. Basically because of the distance unknown he has to assume his distances <measured on the train> are correct and that simultaneity is at fault, not any of his measurements.

 

[wespe]

well that IS what it says.

I'm sorry Ram, I can't communicate with you either.

It does NOT say "experimental set up is such that .. those events happen simultaneously in the absolute sense for both frames" as Eyesaw claims. It says the opposite.

Nice that you now corrected your diagram. Hope someone helps you with that. If not, you could try another forum.

 

[ram1024]

i'll break it down into parts with my translations, tell me where i go wrong.

When we say that the lightning strokes A and B are simultaneous with respect to the embankment, we mean: the rays of light emitted at the places A and B, where the lightning occurs, meet each other at the mid-point M of the length A —> B of the embankment

.

Observers taking the Embankment as a reference frame <stationary> will see that light from A and B hit M mid-point simultaneously

But the events A and B also correspond to positions A and B on the train. Let M' be the mid-point of the distance A —> B on the traveling train. Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train

Assigning a point M' for the train reference frame such that at the precise instant any first flash of lightning hits "anywhere" M' coincides with M perfectly.

If an observer sitting in the position M’ in the train did not possesses this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated

If he didn't make any relative motion in relation to the embankment he would receive simultaneous light just as the embankment viewers.

Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.

Reality is defined as REAL motion being made TOWARDS target light B and AWAY from target light A. Naturally this follows that he will see light B before light A.

Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.

BUT if he takes his own reference frame as stationary, to him light would have to cover the SAME DISTANCES to reach him, but yet he receives light in staggered intervals. since light speed is constant, and the distances are the same <because he measures the distances in HIS frame (not including the distances he traveled in the embankment frame) he MUST conclude that God didn't turn the lights on at the same time.

<end interpretation>

 

[geistkiesel]

i'll break it down into parts with my translations, tell me where i go wrong.

.

Observers taking the Embankment as a reference frame <stationary> will see that light from A and B hit M mid-point simultaneously



Assigning a point M' for the train reference frame such that at the precise instant any first flash of lightning hits "anywhere" M' coincides with M perfectly.



If he didn't make any relative motion in relation to the embankment he would receive simultaneous light just as the embankment viewers.



Reality is defined as REAL motion being made TOWARDS target light B and AWAY from target light A. Naturally this follows that he will see light B before light A.



BUT if he takes his own reference frame as stationary, to him light would have to cover the SAME DISTANCES to reach him, but yet he receives light in staggered intervals. since light speed is constant, and the distances are the same <because he measures the distances in HIS frame (not including the distances he traveled in the embankment frame) he MUST conclude that God didn't turn the lights on at the same time.

<end interpretation>

Yes, in a sense. Here Einstein concludes that the mere fact that the moving observer measures the photon arrival at staggered intervals that the passengers on the train "must, therefore come to the conclusion" the photons were not emitted simultaneously in the moving frame. If this is true, your God solution fits the glove like a hand.

And are you familiar with the study that showed the differences in praying practices performed by persons praying in chrches versus those praying in gambling casinos?
The gamblers really mean it!

 

[geistkiesel]

Einstein's gedanken corresponds to your case#3. The moving observer does not see the lightenings at the same time (since they occurred simultaneously in the stationary frame, they can't have occurred simultaneously in the train frame). I am 100% sure this is SR's prediction. Plus, if he saw them at the same time, there would be a paradox, because the non-moving midpoint observer sees them at the same time at another point in space.

Your 100% assurance is not correct.

AE did not invoke SR to come to his conclusion. You are bootstrapping, it is called circuitous reasoning. AE came to the conclusion that because the B photon was detected before the A photon was detected that this alone is why the passengers "must therefore come to the conclusion" that the photons were not emitted simultaneously in the moving frame. Any SR imperatives, even if applicable and they aren't, would be swamped by the time difference between the time the B and A photon were detected. Ene Doc Al agrees with this.

But if you make the moving observer stationary, of course he sees the lightenings at the same time, because he would be in the stationary frame now. That would correspond to case#1.

I don't understand why you say "midpoint DOES move". The midpoint of events never move within any frame, because the location of events never move within a frame. Location of an event in a frame may look like moving from another frame (like burning marks on the train as seen from the embankement), but they would move at the same speed as the train.



For the hundreth time, there is no motion of the observer in the observer's frame, observer is at rest wrt itself. Of course separation speed changes when looked from another frame, but that's irrelevant. What changes in a frame is the simultaneity of events, not how light moves. That's the whole point you are missing.

Looking at these probkem by stopping one frame and moving another is grossly and physically impossible. Where do you get justification for this? Do you kinow? It is from hand me down mathematics isn't it?

This is not a photon problem. As stated above AE came tot he conclusion he diod based purely on the fact the B photon was detected earlier than the A photon. But AE did not discuss passengers at a and b located at A and B just as the photons were emitted and ercorded the train time of the events. The a and b clock are synchribnized wrt the moving frame. This inoformation senet immediately tio O' will arrive just as the A photon arrives. Also, observers loated at M just as the photons A and B arrived at M also record the time of arrival and send thir tiems tio O'. O' now has the arival time of A, the recorded times of the A and B emissions from a and b, and the recorded time the photons arrived simultaneously at M in the stationary frame. The clock information is convincing that the photons arrived in the moving frame simultaneously as measured by twp sets of passengers. Likewise, O' can calculate the simultaneity himself. After a little algebra,

t3 = t1(C + v)/(C - v).

Draw your own time and distance map, this is what you will get, if you keep SR out of the gthe picture, or even if you insist SR will be swamped. Try it out.

t1 is the time of arrival of the B photons measured from when the O' observer was at the midpoint M when the photons were emitted and v the known velocity of the train. t3 is also a measured quantity the time the A photon was detected by O'. calculation can determine the accuracy of the measured vs. calculated time t3.

Again SR affects, if any, will be swamped by the staggered times of the B and A photon arrival.

 

[Doc Al]

Your 100% assurance is not correct.

AE did not invoke SR to come to his conclusion. You are bootstrapping, it is called circuitous reasoning. AE came to the conclusion that because the B photon was detected before the A photon was detected that this alone is why the passengers "must therefore come to the conclusion" that the photons were not emitted simultaneously in the moving frame. Any SR imperatives, even if applicable and they aren't, would be swamped by the time difference between the time the B and A photon were detected. Ene Doc Al agrees with this.

Nonsense. The reasoning that Einstein uses in the train gedanken is SR in action, my friend. What Einstein invokes is the invariant speed of light: a key premise of SR. And what he concludes--The relativity of simultaneity--is a key result of SR.

And what are you babbling about SR "imperatives" being swamped by time differences? (Do you seriously think I agree with this gibberish?)

You, geistkiesel, are the true master of circular reasoning.

 

[geistkiesel]

Let me clarify.

The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.

Correct so far.

[quote =wespe]Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant.[/quote]
Wrong, even from SR theory.

What is being measured is "events simultaneous in the staionary frame are not simultaneous in the moving frame". When M' was at M, the midpoint when the photons were emitted simultaneously does not allow the moving observer to say the moving iobserver was not at M when the photons were emitted simultaneously.This is a given, remember, you can't change the givesn. You can add observers, measuring equipment and the like, but changiong the giovens is a no no. This is one problem that SR invokes when "considering a moving frame stationary, a physical impossibility.

The moving observer concludes the photons were not emitted simultaneously, not from SR theory, but from the simple staggered measurement of the photons, first B then A, nothing else. SR follows this, it does not precede this experiment. You are close to being there.

But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.

About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.[/QUOTE]

 

[geistkiesel]

Let me clarify.

The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.

Correct so far.

Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant.

Wrong, even from SR theory. M' concludes the lighning doesn't strike simultaneously only from his staggered measurement of the B and A photon, nothing else. Stick to the experiment. read it again befopre yopu start showing off your Sr theory knowledge, which has a nother day or two to survive and that is just becasue ram1024 is extending a proferssional courtesy.

What is being measured is "events simultaneous in the staionary frame are not simultaneous in the moving frame". When M' was at M, the midpoint when the photons were emitted simultaneously does not allow the moving observer to say the moving was not at M when the photons were emitted simultaneously.This is a given, remember, you can't change the given. You can add observers, measuring equipment and the like, but changing the givens is a no no. This is one problem that SR invokes when "considering" a moving frame stationary, a physical impossibility.

The moving observer concludes the photons were not emitted simultaneously, not from SR theory, but from the simple staggered measurement of the photons, first B then A, nothing else. SR follows this, it does not precede this experiment. You are close to being there.

But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.

Maybe so, but the moving observer will still detect the B photon before the A photon. The experiment doesn't change.The light apeed deos not depend on the velocity of the source.

About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.

I suppose so.

 

[geistkiesel]

Let me clarify.

The "moving" observer in Einstein Gedanken is M'. When the lightenings strike, M' is at the midpoint according to the embankement. M' then moves to the right, according to the embankement.

Now, according to M', the lightenings don't strike simultaneously. So, when looking from M' frame, we can't actually say "M' was at the midpoint when the lightenings stroke", because that's not a single instant. But we can say "M' was at the midpoint of the locations where the lightenings stroke". Those locations are where the burning marks are made on the train (not the embankement, we are in the train frame now). Then, M' does not move anywhere (in his own frame, according to himself), thus remains at the midpoint of the events the whole time.

About your case #7 (post#249, took some time to find), all observers remain at the same distance from the location of any emitted photons (in accordance to what I wrote above). In other words, the location any event never changes *within a frame*. The event has happened at an instant and its location cannot be carried with the source of the event or whatever.

I brought the point up in response to another post of yours. The observer only determines the photons were not emitted simultaneously from the staggered detection of the B and A photons, nothing more or less. You cannot stop moving frame and consider the other moving. This is SR stuff. This problem is not SR. The staggered detection of the B and A photon will swamp any SFR aspects of the experiment, which there aren't any anyway.

 

[wespe]

OK, one more try for Geistkiesel:

Suppose we have two space-trains passing by each other. They are moving inertially, but we don't know which one had accelerated in the past. No experiment can determine which train is "really" moving. We only know that they have a relative speed wrt each other. Now, suppose two explosions is space occur near the trains (but not tied to any of the trains, just instant explosions, at any two times, we don't know yet). The explosions create burn marks on both trains.

SR's claim is: if there were two observers at the mid point of those burn marks (measured in each train separately), they could determine the simultaneity of the explosions according to their train, by looking at the order they see the explosions, because speed of light is constant and distances are the same.

Now, if one of the observers see the explosions at the same time, the other will not. Because, the burn marks and the midpoints on two trains will have a relative speed wrt each other, and therefore the light from the explosions cannot hit both observers at the same time.

So tell me, if one observer sees them at the same time, and both events were simultaneous in both frames, and speed of light is independent of its source, why won't the other see them at the same time? You can't use the excuse "the trains were moving" or "one of them was stationary", because then you assume aether, and then you can't determine simultaneity in any frame with this method without knowing speed wrt aether. Are you saying that simultaneity is affected by speed wrt aether? So what method can you tell me to determine if two events occurred simultaneously, if we can't determine speed of a frame wrt aether? (of course in fact there is no sign of aether whatsoever)

 

[geistkiesel]

Nonsense. The reasoning that Einstein uses in the train gedanken is SR in action, my friend. What Einstein invokes is the invariant speed of light: a key premise of SR. And what he concludes--The relativity of simultaneity--is a key result of SR.

No SR is the key result of simultaneity, read the history sir.

BS. NASCARS are running invariant in speed in my example and you ignored that example. Eintein does not use any SR postulate in arriving at his conclusion. He uses only the the staggrered arrival of theh B photon and A photon as detected by the O' observer. You have said on many occasions that this is what all agree.

quote Doc Al "The fact (agreed by all) the light from B hits O' before the light from A leads the O' observer to conclude the lights could not flash simultameously."

Which leg standeth thee on oh master mentor?

And what are you babbling about SR "imperatives" being swamped by time differences? (Do you seriously think I agree with this gibberish?)

You, geistkiesel, are the true master of circular reasoning.Quote:
Originally Posted by geistkiesel
I just came up with another proof that the moving observer must conclude the photons were emitted in the moving frame simultaneously with the emitted photons in the stationary frame.
Yeah, "another" one.

Let's cut through the nonsense, once again. Let's say a and b are the observers on the train located right next to the flashing lights at A and B when they flash. When the see the lights flash, they check the time. No need for any "relaying" of clock times anywhere. Assuming, like you did, that all clocks on the train are synchronized then--like it or not--observers a and b will record different times for the two photon emissions.

How do you know this? They must check the times they are on opposite ends of the train, but each immediately relay their findings to O' who receives the data along with the simultaneous observation of the A and B photons arriving at M simultaneously.

You are invoking SR time dilation and mass shrinking right? What else could you be invoking to make your statement that the observers at A and B will not record the same time of the emitted photons. The photons know nothing of SR theory or even that there are two observers dutifully recording the times the photons are emitted. the a and b observers record the same time.

Remember, what you said back in the Lost simultaneoity thread. let me remind you:
quote by Doc Al
"If you understood the Eisntein simple argument you would know that nowhere does "time dilation" or " mass shrinking" enter into it"