Relativity of Simultaneity

[timo]

[math] \frac{1}{\gamma(1+\beta)}[/math]. This was your timing concept for the light ray. Where is your distance?

The space distance is the other entry in the vector I calculated. But it should be rather obvious that it is zero considering it is the distance that O' traveled in the frame in which O' is at rest.

[swansont]

I was saying here, and it should have been clear,

 

you can use algrebra to find a contradiction if a theory has one.

 

So, of course I assumed we were talking about this concept.

 

I see you thought I was questioning algebra when in fact I was questioning the theory.

 

For example.

 

Theory: There is a greatest integer.

 

Let g be the greatest said integer.

 

But, if g is an integer than g + 1 is an integer by the induction hypothesis.

 

Also, g + 1 > g.

 

Thus, there is no greatest integer.

 

Note I use algebra to run a theory into a contradiction.

 

 

Yes, you can find if an assumption is incorrect by finding a contradiction. But the Lorentz transforms are the maths you get by applying the constant c condition. That's already "baked into" the math. It cannot be disproven by a contradiction. What you will disprove is some other assumption, which is external to relativity.

[vuquta]

Why not?

 

You claimed the transformation is faulty. Your explanation as to why is lacking. Either show us why it's faulty or go over the derivation to see that it isn't.

 

Fair, nuff.

 

The defect in the transformation occurs prior to the derivation.

 

Specifically, here is the problem.

 

Step 1

We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

Step 2

In accordance with definition the two clocks synchronize if

tB - tA = t'A - tB

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

 

Step 3

In agreement with experience we further assume the quantity

2AB/(t'A - tA) = c.

to be a universal constant--the velocity of light in empty space.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

 

Therefore, we may conclude

 

t = AB/c.

 

Then, I put this up against the Relativity train embankment experiment and the conclusions are not correct.

 

Now, I said in the post, R of S and LT are in conflict. What I meant specifically was that Einstein used the clock synchronization described above which fundamentally is based on t = AB/c to derive LT.

 

But, R of S disagrees.

 

Here is how.

 

Take M' as stationary, then A and B are moving light sources to M'. According to the light postulate, the motion of the light sources has no bearing on the behavior of light.

 

Now, when the A and B light sources are equidistant from M' and M, they emit.

 

The motion of the light sources makes no difference in the problem for either observer.

Since d is the distance from the light source A and also from B to M', then we should apply the LT logic AM'/c = t and also BM'/c = t.

Oh, but look at this.

That also applies to M.

AM/c = t and also BM/c = t.

Therefore, both M and M', according to t = AB/c will conclude both light strikes were simultaneous.

 

So, here is where we are.

Assume the truth of the train embankment experiment.

Then, t = AB/c is false.

Therefore, LT is false.

 

Assume the truth of t = AB/c. Then M and M' see the strikes as simultaneous and therefore, the conclusion of Einstein that M' sees B before A is false.

Worse, if we assume t = AB/c, then M and M' will both see the strikes as simultaneous at different positions and therefore one or the other or both will measure c for two different light path lengths with equal times for both paths, which is a contradiction.

 

 

.

 

 

 

 

Not when your claims are lacking in the basic principles. In that case, we must go over the basic principles first before we move on to the more complicated case.

This is a simple accusation.

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The space distance is the other entry in the vector I calculated. But it should be rather obvious that it is zero considering it is the distance that O' traveled in the frame in which O' is at rest.

 

your units are wrong for this equation.

If is supposed to be for time, but has the units of speed.

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Yes, you can find if an assumption is incorrect by finding a contradiction. But the Lorentz transforms are the maths you get by applying the constant c condition. That's already "baked into" the math. It cannot be disproven by a contradiction. What you will disprove is some other assumption, which is external to relativity.

 

Agree with all you say, except I believe if have found a contradiction between LT and R of S.

 

Hey, if not, what the heck. Hopefully, I posted above with enough detail to decide one way of the other.

Edited January 23, 2010 by vuquta
Consecutive posts merged.

[swansont]

Agree with all you say, except I believe if have found a contradiction between LT and R of S.

 

Then you have assumed something about simultaneity that is not true.

[vuquta]

Then you have assumed something about simultaneity that is not true.

 

Well, what can that be? I would welcome comments as to the flaw if there is one.

 

I did something to get around all that anyway.

 

I used:

Based on the motion of M', when both the light sources are equidistant to M', both emit light.

 

So, I worked with distances.

 

The frame M is able to use the motion of M' to predict when M and M' will be coincident. So, this part is viable.

 

Next, A and B are equidistant to M as the setup.

 

Therefore, A and B are equidistant to M' when M and M' are coincident.

 

Since the two light sources are equidistant to M' at the time of emission, and the time of their emission is simultaneous in the frame of M, regardless of the motion of the light sources relative to M', M' must conclude d'/c = t for both directions.

[timo]

Your units are wrong for this equation. If is supposed to be for time, but has the units of speed.

What do you think my units for time, distance and speed are?

[vuquta]

What do you think my units for time, distance and speed are?

I have no idea what you are trying to do.

 

You said that equation was for the following,

I.e. the time the moving observer measures between passing through the origin (time coordinate equals zero) and hitting the light of the laser is

[timo]

I have no idea what you are trying to do.

It seems to me that as soon as you don't understand a basic thing you are complicating the picture so that the lack of basics does not show too much. What I am trying to do is give you the coordinates of the important events in one of your examples (in both coordinate systems). From those, a lot of questions you might have on the physics can be answered - provided you are humble enough to sit down with a pencil rather than approaching questions with the attitude that your intuition as a summa cum laude graduate is sufficient to solve problems. But if you do not understand the coordinates I used then that is somewhat pointless.

[vuquta]

It seems to me that as soon as you don't understand a basic thing you are complicating the picture so that the lack of basics does not show too much. What I am trying to do is give you the coordinates of the important events in one of your examples (in both coordinate systems). From those, a lot of questions you might have on the physics can be answered - provided you are humble enough to sit down with a pencil rather than approaching questions with the attitude that your intuition as a summa cum laude graduate is sufficient to solve problems. But if you do not understand the coordinates I used then that is somewhat pointless.

 

 

Your opinions do not concern me since you offered an equation you cannot even explain. Further, the coordinate of A or B is not relevent as long as the two are equidistant to M' at light emission.

 

You see, I defined the problem in terms of distances from the emitter to the receiver.

 

So, at both light emissions, it is time t in M and time τ in M'.

 

Since the light sources are equidistant to M' at emission, and both emit at time t in M and this corresponds to time τ in M', then we apply the logic of LT that (t'a - τ) = d'/c and (t'b - τ) = d'/c. Thus, t'a = t'b.

 

But, this disagrees with the logic of R of S.

[Klaynos]

vuquta just a quick question, do you think that there is some absolute frame from which velocity, time, etc... can be measured?

[mooeypoo]

Your opinions do not concern me since you offered an equation you cannot even explain. Further, the coordinate of A or B is not relevent as long as the two are equidistant to M' at light emission.

You are avoiding the question you were asked, and it seems that verifies timo's estimation that you do that to hide the fact you don't know the answer yourself.

 

You're the one making the claim, it's up to you to explain it. You were asked this question about units (and your approach as to an absolute frame) quite a number of times. Your inability to answer it is not a reflection on timo's lack of knowledge, it's a reflection on yours.

 

Will you ever answer these questions? This diversion tactics, purposeful or not, is getting quite tedious.

 

~moo

[vuquta]

vuquta just a quick question, do you think that there is some absolute frame from which velocity, time, etc... can be measured?

 

The only way this could be true is if when light moves through space at the same speed c regardless of any circumstances, the frame moves through space also somehow, then I imagine a specific v could be true.

 

As to absolute time, no I do not believe there is one specific time.

 

I would think the only possibility of this concept, but this is not absolute time, is that clocks can be synchronized across frames similar to GPS.

 

It would seem to me, absolute time would require you ro prove you know the absolute beginning of the universe and you can prove time flows at one beat only at every point of the universe. I am not sure how this could ever be proven.

 

So, I guess I see time as sort of like the North star. You pick a standard frame and use equations for other frames to adjust clocks to beat at the same rate as that standard.

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You are avoiding the question you were asked, and it seems that verifies timo's estimation that you do that to hide the fact you don't know the answer yourself.

 

You're the one making the claim, it's up to you to explain it. You were asked this question about units (and your approach as to an absolute frame) quite a number of times. Your inability to answer it is not a reflection on timo's lack of knowledge, it's a reflection on yours.

 

Will you ever answer these questions? This diversion tactics, purposeful or not, is getting quite tedious.

 

~moo

 

 

This is a normal debate tactic of yours. You did this to me on the speciific problem I found pitting LT up against R of S.

 

I answered that specifically.

 

Now, I answered this time question already and supplied a link as support. If the poster disagrees, then let the poster supply a disproof since I have already presented a time argument that represents the R of S side of this debate.

 

Here is the link again. Page 17

 

http://www.tc.umn.edu/~janss011/pdf%20files/appendix-SR.pdf

 

If you disagree or the other poster disagrees with the results of this link, since I agree with the link, you are welcome to present a specific counter argument.

 

 

But, that would be simple minded to stop at the point and call the problem resolved.

 

There is also the LT side which specifies t=d'/c.

 

I have dealt specifically with this issue as well.

[mooeypoo]

This is a normal debate tactic of yours. You did this to me on the speciific problem I found pitting LT up against R of S.

That's because you took some convincing to answer that one properly too.

 

timo asked you a question. Are you going to answer it? It's a fundamental question, and it seems would allow all sides to finally get to see where the problem lies.

 

You challenged timo's equation because of units. He asked you, then, to tell him what units he should be using. Your reply was that his question shows he doesn't know -- and you have, by that, evaded the answer.

 

I am simply reminding you that you've evaded it. Regardless of your link, timo's question stands. He asked of the units, you should answer.

[vuquta]

That's because you took some convincing to answer that one properly too.

 

timo asked you a question. Are you going to answer it? It's a fundamental question, and it seems would allow all sides to finally get to see where the problem lies.

 

You challenged timo's equation because of units. He asked you, then, to tell him what units he should be using. Your reply was that his question shows he doesn't know -- and you have, by that, evaded the answer.

 

I am simply reminding you that you've evaded it. Regardless of your link, timo's question stands. He asked of the units, you should answer.

 

The units of time are seconds. Does the poster use other units? And I had a specific question of the poster of where the distance was located in the posters equation. Since the poster could not answer this simple question, what is the point of having a discussion. I am quite sure this will now make your curious of the question I had of the poster,

 

Naturally, I am sure you can understand my reasoning. The thought experiment in this thread is based on distances. The poster offered an equation without a distance.

 

That's because you took some convincing to answer that one properly too.

 

I answered it properly with my thought experiment as I saw it.

However, once I saw the nature of your questions, I then realized what I needed to add to make it clearer. It was already quite clear to me.

 

So, your assessment that I divert is completely false since I have presented iron tight math here and clearly presented thought experiments.

 

Instead of making your false and baseless accusations, perhaps you can make statements like "your argument is not clear".

 

So, the topic of this debate:

Is it logical to pit LT up against R of S and arrive at a contradiction.

 

Thus far, this is holding up as true.

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It seems to me that as soon as you don't understand a basic thing you are complicating the picture so that the lack of basics does not show too much. What I am trying to do is give you the coordinates of the important events in one of your examples (in both coordinate systems). From those, a lot of questions you might have on the physics can be answered - provided you are humble enough to sit down with a pencil rather than approaching questions with the attitude that your intuition as a summa cum laude graduate is sufficient to solve problems. But if you do not understand the coordinates I used then that is somewhat pointless.

 

OK, I will use your stuff and show you where you are wrong.

 

Here are the correct equations.

d'/(λ(c+v)) and d'/(λ(c-v))

 

These translate to

t1 = d'/c √((c-v)/(c+v))

t2 = d'/c √((c+v)/(c-v))

 

In your "lingo", √((c-v)/(c+v)) is 1/(λ(1+ β)) after you do the simplification that λ = 1/(√(1- β² ) and assume v is in terms of light units.

 

Error 1, you forgot d'

Error 2: You only did one of the times for M'. There are two different times in the argument for R of S.

 

But, this in no way addresses the LT requirement that

( tA - τ ) = d'/c

and

( tB - τ ) = d'/c

Thus, tA = tB.

By taking M' as the stationary frame, the light sources were both a distance d' at emission and this LT logic also applies.

Edited January 24, 2010 by vuquta
Consecutive posts merged.

[swansont]

Take M' as stationary, then A and B are moving light sources to M'. According to the light postulate, the motion of the light sources has no bearing on the behavior of light.

 

Now, when the A and B light sources are equidistant from M' and M, they emit.

 

The motion of the light sources makes no difference in the problem for either observer.

Since d is the distance from the light source A and also from B to M', then we should apply the LT logic AM'/c = t and also BM'/c = t.

Oh, but look at this.

That also applies to M.

AM/c = t and also BM/c = t.

Therefore, both M and M', according to t = AB/c will conclude both light strikes were simultaneous.

 

So, here is where we are.

Assume the truth of the train embankment experiment.

Then, t = AB/c is false.

Therefore, LT is false.

 

Assume the truth of t = AB/c. Then M and M' see the strikes as simultaneous and therefore, the conclusion of Einstein that M' sees B before A is false.

Worse, if we assume t = AB/c, then M and M' will both see the strikes as simultaneous at different positions and therefore one or the other or both will measure c for two different light path lengths with equal times for both paths, which is a contradiction.

 

You are comparing two different scenarios. In the Einstein train scenario, M and M' are co-located when M sees the flashes, i.e. the flashes were emitted when M and M' are not co-located.

 

In your scenario, M and M' are co-located when the flashes are emitted. They are not co-located when the flashes are seen. Since you are describing different scenarios, they cannot be compared to "refute" one another.

[vuquta]

You are comparing two different scenarios. In the Einstein train scenario, M and M' are co-located when M sees the flashes, i.e. the flashes were emitted when M and M' are not co-located.

 

In your scenario, M and M' are co-located when the flashes are emitted. They are not co-located when the flashes are seen. Since you are describing different scenarios, they cannot be compared to "refute" one another.

 

In the Einstein train scenario, M and M' are co-located when M sees the flashes, i.e. the flashes were emitted when M and M' are not co-located.

 

 

No, the were co-located when the flashes occured in his stuff.

 

Just when the flashes of lightning occur, this point M' naturally coincides with the point M

http://www.bartleby.com/173/9.html

 

They are co-located in mine as well when the flashes occur.

 

Everyone agrees the flashes will hit M at the same time.

 

Einstein said they will not hit M' at the same time.

 

But, A and B were equidistant to M' at light emission for both A and B.

 

These positions become the light emission points in the frame of M' and the light emission points are stationary to M'.

 

Since they are equidistant to M', then M' must be hit by both at the same time. Relativity does not care about the positions of the moving light sources, from the frame of M', after light emission.

 

But, if they hit M' at the same time and M at the same time, and M' and M are not co-located when hit, then we have a contradiction because we have two light beams meeting at two different points in space.

[swansont]

In the Einstein train scenario, M and M' are co-located when M sees the flashes, i.e. the flashes were emitted when M and M' are not co-located.

 

 

No, the were co-located when the flashes occured in his stuff.

 

Not in the example you linked to here:

http://www.tc.umn.edu/~janss011/pdf%20files/appendix-SR.pdf

 

(pages 6-7)

At exactly the moment that Al and Bob come face to face

with one another (see the solid line connecting them), the two

flashes reach Bob. Al and Bob agree that these two flashes hit Bob at the same time.

 

Al and Bob are co-located when the flashes hit. Hence my confusion.

 

But, OK, we'll use the other scenario

 

 

Just when the flashes of lightning occur, this point M' naturally coincides with the point M

http://www.bartleby.com/173/9.html

 

They are co-located in mine as well when the flashes occur.

 

Everyone agrees the flashes will hit M at the same time.

 

Einstein said they will not hit M' at the same time.

 

But, A and B were equidistant to M' at light emission for both A and B.

 

These positions become the light emission points in the frame of M' and the light emission points are stationary to M'.

 

Since they are equidistant to M', then M' must be hit by both at the same time. Relativity does not care about the positions of the moving light sources, from the frame of M', after light emission.

 

But, if they hit M' at the same time and M at the same time, and M' and M are not co-located when hit, then we have a contradiction because we have two light beams meeting at two different points in space.

 

The flashes are not simultaneous in (M')'s frame, and you can't assume that they are. Even though they are equidistant, they will not hit if the flashes did not occur at the same time, according to M'.

 

According to M, they are simultaneous in his frame, but (M)'s explanation for them not being simultaneous in (M')'s frame is that M' is moving toward one flash, giving a closure speed of c+v, while M' is moving away from the other flash, at a closure speed of c-v. So they both agree that to M', the front flash hits first.

[vuquta]

Not in the example you linked to here:

http://www.tc.umn.edu/~janss011/pdf%20files/appendix-SR.pdf

 

(pages 6-7)

At exactly the moment that Al and Bob come face to face

with one another (see the solid line connecting them), the two

flashes reach Bob. Al and Bob agree that these two flashes hit Bob at the same time.

 

Al and Bob are co-located when the flashes hit. Hence my confusion.

 

But, OK, we'll use the other scenario

 

Got it. Yea, perhaps that was unclear which experiment ws being referenced.

 

 

 

 

 

The flashes are not simultaneous in (M')'s frame, and you can't assume that they are. Even though they are equidistant, they will not hit if the flashes did not occur at the same time, according to M'.

 

According to M, they are simultaneous in his frame, but (M)'s explanation for them not being simultaneous in (M')'s frame is that M' is moving toward one flash, giving a closure speed of c+v, while M' is moving away from the other flash, at a closure speed of c-v. So they both agree that to M', the front flash hits first.

 

OK, we are down to whether the flashes were indeed simultaneous in M' if they are simultaneous in M.

 

Your analysis is correct above, but we are looking at this problem with M' stationary.

 

What is not questioned is that A and B were equidistant from M' when the flashes occured.

 

Now if an M' observer is placed at A at emission and one is placed at B at emission and the clocks are synched in M', the issue is will A' and B' have the same time on their clocks.

 

We know A and B do in M.

 

This is similar to GPS.

 

If GPS1 emits light at 12:00 and GPS2 emits light at 12:00, what is the time of the earth clocks? These emission events will translate to one time in the earth frame.

 

Now, once light starts to travel, then we end up using full LT.

 

I will post a simpler thought experiment.

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Experiment 2

First, take M as stationary.

This time, based on the motion of M', the frame of M calculates a simultaneous emission from A and B such that lights will strike M simultaneously when M' and M are coincident.

 

      Light emits
            M' ->v
A--------------M--------------B

 

        Light received
              M' ->v
A--------------M--------------B

 

 

 

 

 

Now, take the position that M' is stationary.

 

According to LT and SR, the light path for a frame is measured from the emission point in the frame to the receiver, not from the position a moving light source will reside when the receiver is struck by light.

 

Clearly then, when the light is emitted from A and B, the light emission point in the M' frame at the position where A emits is closer to M' than is the light emission point corresponding to the light emission of B. But, M' will receive the light flashes simultaneously since M' will be coincident with M when the light flashes hit by design. Yet, M' calculates two different light path lengths with equal times and will not measure the same speed of light for both paths.

[Sisyphus]

Yet, M' calculates two different light path lengths with equal times and will not measure the same speed of light for both paths.

 

This is incorrect. In frame M', both the light path lengths and the travel times are different.

[vuquta]

This is incorrect. In frame M', both the light path lengths and the travel times are different.

 

I cannot see how the times will be different for M'.

 

When the strikes hit M, M and M' are co-located.

 

I do realize LT claims they will be different. But, this method is testing the validity of the relationship t = d/c.

[Sisyphus]

I don't understand where the difficulty is. Why do you think the travel times are the same? The flashes are not simultaneous, and they meet at M'. Hence, different travel times from source to M'.

[vuquta]

I don't understand where the difficulty is. Why do you think the travel times are the same? The flashes are not simultaneous, and they meet at M'. Hence, different travel times from source to M'.

 

 

That is the reasoning that must be used to get it to work. But, it is wrong.

 

Correct me if I am wrong, here is the reasoning.

 

Both M' and M are hit by both the light of A and B simultaneously.

 

Since they were equidistant, by design, they were also simultaneous by design, then this logic is in place for M without question.

 

Now, for M' as stationary, M' is struck by both at the same time.

 

Then, one says, well since the light paths are different, it must be the case that the lights emitted at different times in the frame of M'.

 

Correct?

 

Well, instead of saying is must be the case, I am looking if it is.

 

Let A and B correspond to positions in frame M'.

 

At each point of M' there is a synchronized clock. Also A and B are synched in M.

 

Given any time interval, A and B are at some positions in M' and M' calculates that interval as ∆t and views that interval as ∆t/λ elapsed in M. M' is not concluding M has different times at different positions of M unless light receiving events are taking place. This is GPS. The earth frame does not conclude two satellites in the same orbit have different times based on positions.

 

So, when the interval is at ∆t1/λ in M, both A and B emit.

 

That corresponds to one interval ∆t1 in M'. Therefore, in spite of the fact that SR says they must emit at different times in M' they do not.

Edited January 25, 2010 by vuquta

[swansont]

Both M' and M are hit by both the light of A and B simultaneously.

 

No, you cannot assume this. In one frame they are simultaneous, but in the other frame they are not. This assumption causes your contradiction.

[vuquta]

Originally Posted by vuquta

Both M' and M are hit by both the light of A and B simultaneously.

 

 

 

No, you cannot assume this. In one frame they are simultaneous, but in the other frame they are not. This assumption causes your contradiction.

 

M' is co-located with M when both light beams meet at M.

 

There is no choice that M' is also met by both light beams simultaneously also.

[Sisyphus]

 

Given any time interval, A and B are at some positions in M' and M' calculates that interval as ∆t and views that interval as ∆t/λ elapsed in M.

 

And vice versa. You can't just set up a one to one correspondence between reference frames. Each frame views the other as under time compression.

 

M' is not concluding M has different times at different positions of M unless light receiving events are taking place. This is GPS. The earth frame does not conclude two satellites in the same orbit have different times based on positions.

 

I have no idea what this section means.

 

So, when the interval is at ∆t1/λ in M, both A and B emit.

 

In M? As viewed by who?

 

That corresponds to one interval ∆t1 in M'.

 

No, it doesn't. There isn't a direct correspondence like that.