SR and c measurement

[Klaynos]

Thanks. I agree 1 is not consistent with 2, but that is SR.

 

Here is the light postulate.

Any ray of light moves in the ``stationary'' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

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

 

Now, since it is "any ray of light", that means in any direction also.

 

Thus, a light sphere proceeds spherically from the emission point in the frame at c in all directions.

 

Also, it says, "whether the ray be emitted by a stationary or by a moving body.". Thus, even if the the light source were moving, then light would still proceed spherically from the emission point in the frame.

 

Yes, but one of your frames is moving relative to the other, so it wont be a sphere in both.

[vuquta]

Just a real quick glance here, but this popped out as being wrong

 

The last line has a sign error: when you have a - on the outside of a parentheses with both a positive and a negative sign inside them, both terms cannot achieve the same sign on the inside by multiplying on the outside.

 

Also (λ - (1+λ))/(1+λ) becomes λ/(1+λ) - 1, NOT (1+λ). You have a division error, too.

 

I haven't checked any of the rest of the math, but to be frank, with these basic of errors, the rest of the math has to be considered suspect until it is verified.

 

You also may want to consider using the LaTeX capabilities of this forum, to improve the ease of reading.

 

x' = vtλ( (λ - (1+λ))/(1+λ) ) becomes

 

[math]x' = vt\lambda \left( \frac{\left( \lambda - (1+\lambda) \right)}{(1+\lambda)} \right) [/math] is significantly easier to read (I hope that I got all the ()'s correct, because when it is all in one line, it is very hard to read.)

 

No, I only forget to type in the / symbol at the end.

I make mistakes like that.

 

But, x' = vtλ( (λ - (1+λ))/(1+λ) ) = -(vtλ) / (1+λ) = -x since

 

(λ - (1+λ) = (λ - 1 - λ) = -1

 

and no the math is not suspect for leaving out a symbol.

 

My link has easy to read word math and did not contain this typing error.

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Just a real quick glance here, but this popped out as being wrong

 

The last line has a sign error: when you have a - on the outside of a parentheses with both a positive and a negative sign inside them, both terms cannot achieve the same sign on the inside by multiplying on the outside.

 

Also (λ - (1+λ))/(1+λ) becomes λ/(1+λ) - 1, NOT (1+λ). You have a division error, too.

 

I haven't checked any of the rest of the math, but to be frank, with these basic of errors, the rest of the math has to be considered suspect until it is verified.

 

You also may want to consider using the LaTeX capabilities of this forum, to improve the ease of reading.

 

x' = vtλ( (λ - (1+λ))/(1+λ) ) becomes

 

[math]x' = vt\lambda \left( \frac{\left( \lambda - (1+\lambda) \right)}{(1+\lambda)} \right) [/math] is significantly easier to read (I hope that I got all the ()'s correct, because when it is all in one line, it is very hard to read.)

 

Here it is in that format. Thanks for the suggestion. It is easy to make typing mistakes with all those parens.

 

[math]

x' = \left( x - vt \right)\lambda

[/math]

and

 

[math]

x = \frac{vt\lambda}{ \left( 1 + \lambda \right)}

[/math]

 

So,

[math]

x' = \left( \frac{vt\lambda}{ \left( 1 + \lambda \right)} - vt \right)\lambda

[/math]

 

 

[math]

x' = vt\lambda \left( \frac{\lambda}{ \left( 1 + \lambda \right)} - 1 \right)

[/math]

 

 

[math]

x' = vt\lambda \left( \frac{\lambda}{ \left( 1 + \lambda \right)} - \frac{\left( 1 + \lambda \right)}{ \left( 1 + \lambda \right)} \right)

[/math]

 

 

[math]x' = vt\lambda \left( \frac{\left( \lambda - (1+\lambda) \right)}{(1+\lambda)} \right) = vt\lambda \left( \frac{ \lambda - 1 -\lambda}{(1+\lambda)} \right) = vt\lambda \left( \frac{-1}{(1+\lambda)} \right)= \frac{-vt\lambda}{(1+\lambda)} = -x

 

 

[/math]

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I don't know what this means. What is simultaneous? You've already stated that the strikes will not be seen as simultaneous in the other frame.

 

Yes, I said only that all the points will not hit simultaneously from the POV of the stationary frame to the moving frame.

 

However, this point is struck in each frame at the same time.

 

This is something completely different.

 

SR claims both spheres will see the light sphere emerge spherically in their frames and strike all points at the same time.

 

Now, since the moving sphere is centered at (vt,0,0) in the coords of the stationary frame, and the stationary sphere is centered at (0,0,0), then it is natural to ask how this happens and further what will a clock at the origin of stationary frame read when this occurs.

 

That is the purpose of this point.

Both light spheres hit this point at the same time aoocrding to LT and this point in on both rigid body sphere at any time t.

 

This allows the question to be answered by the clock in the stationary of when the simultaneity occurs in the moving frame.

[mooeypoo]

No, I only forget to type in the / symbol at the end.

I make mistakes like that.

Please try not to. Math is a concise endeavor. It is exactly what it's written, not what it's meant. Errors like these produce absolutely different results, and turn the entire exercise 'suspect', because it doesn't follow.

 

At the very least, don't be surprised we don't understand what you're doing when you are being careless with your mathematics.

 

But, x' = vtλ( (λ - (1+λ))/(1+λ) ) = -(vtλ) / (1+λ) = -x since

 

(λ - (1+λ) = (λ - 1 - λ) = -1

Just a side comment here -- this is another example of 'so-so' math exercise. I know that it's relatively clear to understand that you meant to close your missing brackets before the 'equal' sign, but it's only clear because I *ASSUME* that is what you meant. If you closed it anywhere else, it would produce an absolutely different result.

 

This (assuming you forgot a symbol as well as your closed brackets?):

(λ - )(1+λ)

 

is not the same as this:

(λ - (1+λ))

 

So while this error was relatively easy to 'solve' by assumption, you shouldn't make us assume *anything*, because our assumptions may well be wrong, and your exercise may well have lots of errors in it that you (and us) will not notice if you don't keep your math in order.

 

With a bit more concentration on your part, we can move forward in this discussion without getting stuck on the nitty-gritty of stuff that might be less problematic after all.

 

and no the math is not suspect for leaving out a symbol.

Yes, it is, it's showing either bad math or wrong result (which overturn your hypothesis). Try not to forget symbols. At the very least, try not to scold us for noticing it and asking for clarification.

 

Here it is in that format. Thanks for the suggestion. It is easy to make typing mistakes with all those parens.

Yes, it's understandable, but just take into account that parens' can *completely* change an equation.

 

Yes, I said only that all the points will not hit simultaneously from the POV of the stationary frame to the moving frame.

So in the moving frame there's no symultaneity.

 

However, this point is struck in each frame at the same time.

Relative to which frame? a separate one from the mentioned two? without pointing out the frame of reference the "same time" means nothing.

 

This is something completely different.

 

SR claims both spheres will see the light sphere emerge spherically in their frames and strike all points at the same time.

Wait, sorry, I'm not sure I understand you, so here's my attempt to be clearler:

 

SR states that each frame of reference will see the light emanating at their individual frame of reference as spherical.

 

Each frame of reference will NOT NECESSARILY (or, likely not!) see the light emanating from the *OTHER* frame of reference as spherical.

 

Your sentence was less clear, I'm not sure I understand which frame you mean sees what as spherical.

 

Now, since the moving sphere is centered at (vt,0,0) in the coords of the stationary frame, and the stationary sphere is centered at (0,0,0), then it is natural to ask how this happens and further what will a clock at the origin of stationary frame read when this occurs.

I don't understand what you're saying here, sorry.. can you describe your hypothetical clock situation again? Where is the clock, in which frame of reference, and what event is it timing?

 

That is the purpose of this point.

Both light spheres hit this point at the same time aoocrding to LT and this point in on both rigid body sphere at any time t.

There's no meaning to "same time" unless you say in which reference those two events happen simultaneously. You just said that neither frame will see both simultaneously, so are you talking about a third frame?

 

This allows the question to be answered by the clock in the stationary of when the simultaneity occurs in the moving frame.

I'm completely confused. Describe your hypothetical clock and what event you are measuring with it, because I don't know if I understand what reference frame you're using with it..

 

~moo

Edited January 6, 2010 by mooeypoo

[vuquta]

Yes, but one of your frames is moving relative to the other, so it wont be a sphere in both.

 

Agreed but only when viewed from one frame to another.

 

Inside each sphere, the light is a sphere in its own frame.

 

So, I am not really comparing frames with SR.

 

I am concluding that each light sphere within each frame proceeds spherically from the emission point in the frame just by the requirements of SR.

 

Intellectually, once the theory is put in this light, it is clear there are two different light spheres one in each frame.

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Please try not to. Math is a concise endeavor. It is exactly what it's written, not what it's meant. Errors like these produce absolutely different results, and turn the entire exercise 'suspect', because it doesn't follow.

 

At the very least, don't be surprised we don't understand what you're doing when you are being careless with your mathematics.

 

 

Just a side comment here -- this is another example of 'so-so' math exercise. I know that it's relatively clear to understand that you meant to close your missing brackets before the 'equal' sign, but it's only clear because I *ASSUME* that is what you meant. If you closed it anywhere else, it would produce an absolutely different result.

 

This (assuming you forgot a symbol as well as your closed brackets?):

(λ - )(1+λ)

 

is not the same as this:

(λ - (1+λ))

 

So while this error was relatively easy to 'solve' by assumption, you shouldn't make us assume *anything*, because our assumptions may well be wrong, and your exercise may well have lots of errors in it that you (and us) will not notice if you don't keep your math in order.

 

With a bit more concentration on your part, we can move forward in this discussion without getting stuck on the nitty-gritty of stuff that might be less problematic after all.

 

 

Yes, it is, it's showing either bad math or wrong result (which overturn your hypothesis). Try not to forget symbols. At the very least, try not to scold us for noticing it and asking for clarification.

 

 

Yes, it's understandable, but just take into account that parens' can *completely* change an equation.

 

 

So in the moving frame there's no symultaneity.

 

 

Relative to which frame? a separate one from the mentioned two? without pointing out the frame of reference the "same time" means nothing.

 

 

Wait, sorry, I'm not sure I understand you, so here's my attempt to be clearler:

 

SR states that each frame of reference will see the light emanating at their individual frame of reference as spherical.

 

Each frame of reference will NOT NECESSARILY (or, likely not!) see the light emanating from the *OTHER* frame of reference as spherical.

 

Your sentence was less clear, I'm not sure I understand which frame you mean sees what as spherical.

 

 

I don't understand what you're saying here, sorry.. can you describe your hypothetical clock situation again? Where is the clock, in which frame of reference, and what event is it timing?

 

 

There's no meaning to "same time" unless you say in which reference those two events happen simultaneously. You just said that neither frame will see both simultaneously, so are you talking about a third frame?

 

 

I'm completely confused. Describe your hypothetical clock and what event you are measuring with it, because I don't know if I understand what reference frame you're using with it..

 

~moo

 

 

My apologies. I will use the math language here though it sure does take time, but is more readable for all and also avoids silly mistakes on my part.

[Klaynos]

Agreed but only when viewed from one frame to another.

 

Inside each sphere, the light is a sphere in its own frame.

 

So, I am not really comparing frames with SR.

 

I am concluding that each light sphere within each frame proceeds spherically from the emission point in the frame just by the requirements of SR.

 

Intellectually, once the theory is put in this light, it is clear there are two different light spheres one in each frame.

 

WARNING WARNING FRAME MIXING DETECTED!!!

 

You need to all the transforms when moving from one frame to the other, else you will find inconsistencies.

[mooeypoo]

I agree with Klaynos here - other than the math inconsistencies which hopefully will be clearer, you seem to be confusing frames. Read what I wrote about and my questions about your framess -- in a few cases you seem to relate to a *THIRD* frame of reference to declare your "simultaneity", which would require you to define it and make sure you use the transformations for it to make sense.

[vuquta]

Originally Posted by vuquta

Agreed but only when viewed from one frame to another.

 

Inside each sphere, the light is a sphere in its own frame.

 

So, I am not really comparing frames with SR.

 

I am concluding that each light sphere within each frame proceeds spherically from the emission point in the frame just by the requirements of SR.

 

Intellectually, once the theory is put in this light, it is clear there are two different light spheres one in each frame.

 

 

 

 

WARNING WARNING FRAME MIXING DETECTED!!!

 

You need to all the transforms when moving from one frame to the other, else you will find inconsistencies.

 

 

No, I did not mix frames. I am only stating a fact that within each frame, the light sphere emerges spherically. I am not doing a frame comparison.

 

If I did a frame comparison, I would find the light sphere in the moving frame to be an ellipsoid. But, I would also find its center located at (vt,0,0).

 

That is the basic problem, the light sphere in the moving frame emerges at vt at any time t in the coords of the stationary frame and thus, that light sphere is moving because its center is moving.

 

But, the one in the stationary frame emerges at (0,0,0) and does not move.

[vuquta]

I agree with Klaynos here - other than the math inconsistencies which hopefully will be clearer, you seem to be confusing frames. Read what I wrote about and my questions about your framess -- in a few cases you seem to relate to a *THIRD* frame of reference to declare your "simultaneity", which would require you to define it and make sure you use the transformations for it to make sense.

 

Yes, I am going through your stuff.

 

It will take a little while.

[Klaynos]

No, I did not mix frames. I am only stating a fact that within each frame, the light sphere emerges spherically. I am not doing a frame comparison.

 

If I did a frame comparison, I would find the light sphere in the moving frame to be an ellipsoid. But, I would also find its center located at (vt,0,0).

 

That is the basic problem, the light sphere in the moving frame emerges at vt at any time t in the coords of the stationary frame and thus, that light sphere is moving because its center is moving.

 

But, the one in the stationary frame emerges at (0,0,0) and does not move.

 

If you define it as spherical in Frame A, and Frame B is moving relative to Frame A then it cannot be spherical in Frame B, your assumption that it is is flawed.

[vuquta]

Yes, I said only that all the points will not hit simultaneously from the POV of the stationary frame to the moving frame.

 

So in the moving frame there's no symultaneity.

 

Well, the sphere points of the moving frame will not be struck simultaneously from the context of the stationary frame.

 

But, the moving frame concludes its sphere points are struck simultaneously.

 

 

 

However, this point is struck in each frame at the same time.

Relative to which frame? a separate one from the mentioned two? without pointing out the frame of reference the "same time" means nothing.

 

Relative to each. There are only two frames. LT calculates that t' = t when this point is struck in each frame. So, by the SR light postulate, t = r/c for all points on the stationary rigid body sphere. Also, LT calculates this point is hit in O', the moving frame, at t' = t.

 

 

This is something completely different.

 

SR claims both spheres will see the light sphere emerge spherically in their frames and strike all points at the same time.

 

Wait, sorry, I'm not sure I understand you, so here's my attempt to be clearler:

 

SR states that each frame of reference will see the light emanating at their individual frame of reference as spherical.

 

Each frame of reference will NOT NECESSARILY (or, likely not!) see the light emanating from the *OTHER* frame of reference as spherical.

 

Your sentence was less clear, I'm not sure I understand which frame you mean sees what as spherical.

 

Individually, each frame sees its own light sphere as spherical and then sees the other's as an ellipsoid.

 

 

Now, since the moving sphere is centered at (vt,0,0) in the coords of the stationary frame, and the stationary sphere is centered at (0,0,0), then it is natural to ask how this happens and further what will a clock at the origin of stationary frame read when this occurs.

 

I don't understand what you're saying here, sorry.. can you describe your hypothetical clock situation again? Where is the clock, in which frame of reference, and what event is it timing?

 

Oh, there is a clock at the origin of each frame or anywhere for that matter. Inside the frame, that clock will read t = r/c when the sphere points are struck. But, it will read various different times as to when it calculates the moving sphere points are struck by the stationary light sphere.

 

But, SR tells us that the moving frame, even though it cannot be seen for some unknown reason, will experience simultaneity for its sphere points also.

 

So, here is the question. What will the clock in the stationary read when this simultaneity occurs in the moving frame? That is what I am attempting to answer.

 

 

 

 

That is the purpose of this point.

Both light spheres hit this point at the same time aoocrding to LT and this point in on both rigid body sphere at any time t.

 

There's no meaning to "same time" unless you say in which reference those two events happen simultaneously. You just said that neither frame will see both simultaneously, so are you talking about a third frame?

 

It just means LT calculates t' = t for that point. So, with the point, you are allowed to talk about the "same time". You cannot with any other, but you can with this one.

 

Let me be more specific. Assume a light timer. This device is synched to 0 when the light is flashed in a frame and the clock stops when it is hit by light.

 

If I install one of the in the stationary frame at that x point and install one in the moving frame at x' = -x, then each are synched in their respective frames when the light is emitted.

 

Now run the experiment and bring the clocks back together at some later time. They will read the exact same time.

[mooeypoo]

If I did a frame comparison, I would find the light sphere in the moving frame to be an ellipsoid. But, I would also find its center located at (vt,0,0).

In which frame!

 

 

vuquta, you have to start paying attention to which frames you're talking about and you have to start being more concise in telling us which frame you mean. We won't be able to discuss any of it without it, because without a frame the above statement is meaningless.

 

Not in all frames the center will be at (vt,0,0). Be clear.

 

 

That is the basic problem, the light sphere in the moving frame emerges at vt at any time t in the coords of the stationary frame and thus, that light sphere is moving because its center is moving.

It seems to only be a problem because you mix frames. It will be much easier if you start specifying the frames you mean in your calculations. But I don't want to jump the gun here -- I mentioned all this in my previous reply, and I don't want to rush you for nothing.

 

Answer my previous reply, we'll continue from there when things are a bit clearer.

[vuquta]

If you define it as spherical in Frame A, and Frame B is moving relative to Frame A then it cannot be spherical in Frame B, your assumption that it is is flawed.

 

I completely agree with you.

 

But, the observer in each frame concludes the light is spherical from its emission point in the frame.

 

That is what I am saying. I am talking only about what each individual frame concludes about itself.

 

Now, if a frame were to look at another frame, then it would see an ellipsoid.

 

I actually think on these particular issues, we are all in agreement.

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In which frame!

 

 

vuquta, you have to start paying attention to which frames you're talking about and you have to start being more concise in telling us which frame you mean. We won't be able to discuss any of it without it, because without a frame the above statement is meaningless.

 

Not in all frames the center will be at (vt,0,0). Be clear.

 

OK, this is in the coords of the stationary frame. The moving sphere will be origined (vt,0,0) in the coords of the stationary frame after any time t.

 

 

It seems to only be a problem because you mix frames. It will be much easier if you start specifying the frames you mean in your calculations. But I don't want to jump the gun here -- I mentioned all this in my previous reply, and I don't want to rush you for nothing.

 

Answer my previous reply, we'll continue from there when things are a bit clearer.

 

I think I have answered them, but please, if I did not seem to, let me know.

Thanks

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In which frame!

 

 

vuquta, you have to start paying attention to which frames you're talking about and you have to start being more concise in telling us which frame you mean. We won't be able to discuss any of it without it, because without a frame the above statement is meaningless.

 

Not in all frames the center will be at (vt,0,0). Be clear.

 

 

 

It seems to only be a problem because you mix frames. It will be much easier if you start specifying the frames you mean in your calculations. But I don't want to jump the gun here -- I mentioned all this in my previous reply, and I don't want to rush you for nothing.

 

Answer my previous reply, we'll continue from there when things are a bit clearer.

 

Please look at figure 2 in the pdf file.

 

pdf file

 

It shows the context of the stationary frame, the light spheres and also this point.

[Klaynos]

How many emitters have you got? I thought you had one, but you appear to have one per frame?

[vuquta]

How many emitters have you got? I thought you had one, but you appear to have one per frame?

 

I just have one emitter for the spheres. It is located in the moving frame.

 

But, it emits when the origins of the rigid body sphere are coincident.

 

Then, it each frame, I am using the emission point in the frame as the origin of the light sphere for the frame since this is SR. That would make it appear like I have two emitters then.

[Klaynos]

I just have one emitter for the spheres. It is located in the moving frame.

 

Then in the "moving frame" it'll be spherical but not in any frame that is moving relative to the "moving frame"

 

But, it emits when the origins of the rigid body sphere are coincident.

 

Coordinate systems are arbitrary, this is therefore completely superfluous.

 

Then, it each frame, I am using the emission point in the frame as the origin of the light sphere for the frame since this is SR. That would make it appear like I have two emitters then.

 

In fact it means you have two emitters, not one.

[vuquta]

Then in the "moving frame" it'll be spherical but not in any frame that is moving relative to the "moving frame"

Agreed. When each of the frames looks at its own light sphere, it is spherical.

 

Yet, the moving frame will have one that is an ellipsoid.

 

 

Coordinate systems are arbitrary, this is therefore completely superfluous.

Agreed.

 

In fact it means you have two emitters, not one.

 

No, there is one emitter. The light postulate is as follows,

Any ray of light moves in the ``stationary'' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

 

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

 

It says any ray of light, so that means spherical in the stationary coordinates.

 

It also says, whether the ray be emitted by a stationary or by a moving body.

 

Therefore, whether the light source is moving, or not, the light will proceed spherically from the emission point in the frame.

 

So, when the two rigid body spheres are coincident, the moving sphere emits.

 

In the at rest frame, the light sphere proceeds from its origin since the motion of the light source has no effect.

 

Also, in the moving frame, since the light souce is at rest with it, the light proceeds spherically from the emission point inside that frame as well.

[mooeypoo]

Agreed. When each of the frames looks at its own light sphere, it is spherical.

 

Yet, the moving frame will have one that is an ellipsoid.

This is getting a bit annoying. You agree, and proceed to explain how your claim totally disagrees with what Klaynos said.

 

Either stop saying "Agreed" (because you're confusing us) or stop confusing us with a followup non-agreed-upon sentence.

 

We're talking different languages here. Is have a feeling this is a literal problem, vuquta. Are you using a translator to answer us, by any chance?

 

It might explain some of the problems we have here in talking in circles.

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Agreed. When each of the frames looks at its own light sphere, it is spherical.

 

Yet, the moving frame will have one that is an ellipsoid.

Let me try and make this simpler:

 

You have 2 frames of reference. We will stop calling it 'moving' and 'stationary' because these terms are only valid to a yet ANOTHER external reference. To the "moving frame" the "stationary" is moving and itself is stationary, so those terms are MOOT.

 

We will call it Frame A and Frame B.

Both Frame A and Frame B has a light emitter in each. Light is emitted at time=0 in BOTH frames. Initial conditions are equal.

 

When the observer is standing at Frame A:

The observer at Frame A will see the light at Frame A as a sphere.

Frame B is moving in respect to Frame A, so the observer (at Frame A) will see the light from Frame B as an ellipsoid.

 

When the observer is standing at Frame B, you have the EXACT OPPOSITE:

The observer (now at Frame B) will see the light at Frame B as a sphere.

Frame A is moving in respect to Frame B, so the observer (at Frame B) will see the light from Frame A as an ellipsoid.

 

Without adding or protesting *any other information*, please just tell me if the case above is what you mean. Obviously, the drawings you supplied weren't enough, we each have a different case in mind and we keep talking in circles.

 

Before we can discuss what does and doesn't happen we need to make sure we're all talking about the SAME CASE.

 

 

No, there is one emitter. The light postulate is as follows,

Okay, see, this is confusing AGAIN. You kept saying "both" frames with spherical light, seeing the other frame as ellipsoid light, and you semed to have agreed with us when we asked if there are two light sources. And yet, now it seems there's only one.

 

I'm confused.

Any ray of light moves in the ``stationary'' system of co-ordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body.

 

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

The term "stationary" and "moving" are confusing, and I have a feeling they're partly the source of half the confusion in this thread. The stationary frame is only stationary if you're the observer on it, and the moving frame is only moving if you're observing it through another frame. Since half of our discussion is what happens when you switch between frames *AND* move to a completely separate third frame (when you can 'watch' both frames), we might as well just call them "Frame A" and "Frame B" and define the location of the observer in each case.

 

Let's first try to get rid of the confusion and then we can move on to the more particular part of your claim.

 

~moo

[vuquta]

This is getting a bit annoying. You agree, and proceed to explain how your claim totally disagrees with what Klaynos said.

 

Either stop saying "Agreed" (because you're confusing us) or stop confusing us with a followup non-agreed-upon sentence.

 

We're talking different languages here. Is have a feeling this is a literal problem, vuquta. Are you using a translator to answer us, by any chance?

 

I said exactly what I wanted to say here and was completely accuracte.

 

I just think the model I am talking about is non-standard and perhaps not readily seen.

 

But, we will communicate.

 

Let me try and make this simpler:

 

You have 2 frames of reference. We will stop calling it 'moving' and 'stationary' because these terms are only valid to a yet ANOTHER external reference. To the "moving frame" the "stationary" is moving and itself is stationary, so those terms are MOOT.

 

We will call it Frame A and Frame B.

Both Frame A and Frame B has a light emitter in each. Light is emitted at time=0 in BOTH frames. Initial conditions are equal.

 

When the observer is standing at Frame A:

The observer at Frame A will see the light at Frame A as a sphere.

Frame B is moving in respect to Frame A, so the observer (at Frame A) will see the light from Frame B as an ellipsoid.

 

When the observer is standing at Frame B, you have the EXACT OPPOSITE:

The observer (now at Frame B) will see the light at Frame B as a sphere.

Frame A is moving in respect to Frame B, so the observer (at Frame B) will see the light from Frame A as an ellipsoid.

 

Without adding or protesting *any other information*, please just tell me if the case above is what you mean. Obviously, the drawings you supplied weren't enough, we each have a different case in mind and we keep talking in circles.

 

 

The only thing I change above is that there is only one light emitter and only one is needed under the light postulate.

 

Otherwise, this is exactly the case.

 

 

Okay, see, this is confusing AGAIN. You kept saying "both" frames with spherical light, seeing the other frame as ellipsoid light, and you semed to have agreed with us when we asked if there are two light sources. And yet, now it seems there's only one.

 

I'm confused.

 

 

I think you are confused because you believe two light sources are necessary. By the light postulate, only one is needed to exhibit this behavior described.

 

If you need two of them to think about it, I am OK with that because it makes no difference with the outcome according to the light postulate.

 

I can show this if you want me to.

 

The term "stationary" and "moving" are confusing, and I have a feeling they're partly the source of half the confusion in this thread. The stationary frame is only stationary if you're the observer on it, and the moving frame is only moving if you're observing it through another frame. Since half of our discussion is what happens when you switch between frames *AND* move to a completely separate third frame (when you can 'watch' both frames), we might as well just call them "Frame A" and "Frame B" and define the location of the observer in each case.

 

Stationary and moving are terms I use when there exists two frames and I am taking the view that one is stationary.

 

I am OK and able to communicate using frame A and B though.

You tell the me the terminology and I will use that.

[mooeypoo]

Alright, that's great, just take into account that it seems we're all talking in circles here, vuquta.

 

Somewhere, obviously, we've lost communication. That's why I'm trying to restate things here.

 

~moo

[vuquta]

Alright, that's great, just take into account that it seems we're all talking in circles here, vuquta.

 

Somewhere, obviously, we've lost communication. That's why I'm trying to restate things here.

 

~moo

 

I am OK with anything in logic.

 

I just want to make sure communication is established also.

 

I think you do have the model down.

 

If not, I will continue until it is established.

 

It is two rigid body spheres that are coincident when light is emitted at their common origins.

 

One is in relative motion.

[vuquta]

This will show two different calculations under SR for one way light transfer and will support one of them.

 

Link

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Consecutive posts merged

Here is an alternative.

High orbit GPS satellites are programmed to beat with the ground based clocks by changing their their clock frequency using the λ factor.

Here is the setup.

 

 GPS1       ->v                                                          GPS2
   |                                      
 Ground

 

 

The ground clock, earth based station, and GPS satellite 1 will be located at the same vertical position at say time t0 for both. There is a second satellite GPS2 located a distance d from GPS1. At that time t0 when the ground clock and the GPS1 are vertical, GPS1 fires a laser at GPS2. GPS2 records the time the light is received. After any time t1, the ground knows the position of the GPS2 relative to the earth. So, GPS2 beams the recorded time, t1, to the earth based station. The earth will be able to determine the exact position of GPS2 at that time.

 

So, GPS1 is located at v(t1-t0) from the original position when the laser was shot and GPS2 is located at v(t1-t0) from its original position when the light strikes GPS2 at time t1.

 

Since the clocks are all in sync, from the earth frame, the light moved from the station to v(t1-t0) + d in time t1-t0.

 

The GPS frame concludes light moved a distance d in time t1-t0.

 

As such, they cannot measure the same one-way light speed

[swansont]

GPS analysis requires GR, not just SR. You have to worry about the Sagnac effect.

 

But … GPS works. You can't use it as an example of the failure of relativity. It only demonstrates your lack of understanding of relativity.

[vuquta]

GPS analysis requires GR, not just SR. You have to worry about the Sagnac effect.

 

But … GPS works. You can't use it as an example of the failure of relativity. It only demonstrates your lack of understanding of relativity.

 

Yes, GPS contains Sagnac, GR and time dilation.

 

This paper is the leading one showing relativity and GPS.

 

At low orbits, time dilation is dominant, at high orbits, GR gravity is.

 

 

Now, what is clear is that the simultaneity shift term of LT is missing from GPS.

t' = ( t - vx/c^2)λ

 

t' = λt - λvx/c^2

 

λvx/c^2 is the simultaneity shift term.

 

GPS contains GR and the time dilation term λt.

 

So, if I shoot light and time it as in the example, the experiment turns out as I said. Then, you can apply GR and time dilation, that what is applied to the satellite prior to launch such that the satellite clocks would be in sync with the earth clocks. In a reverse way to discover what the clocks would have been due to GR and SR time dilation, you could apply the inverse operations but you will not get the simultaneity shift term λvx/c^2 as part of the equation.

 

Therefore, the simultaneity shift term λvx/c^2 is not valid.

 

And no, this does not mean I do not understand SR. It just means using GPS and ground based clocks as an actual laboratory, the logic of LT does not hold up.

 

In particular, given two GPS clocks and one ground based clock, the further a gps satellite is away, the worse you would expect it to be out of sync with the ground based clock using LT.

 

But, that is not the case. They all remain in sync.

 

Therefore, if a GPS satellite and ground clock are vertical and a laser is shot freom GPS1 to GPS2, you know for a fact, a corresponding ground clock vertical with GPS2 will read the same exact time as GPS2 when light strikes the GPS2.

Edited January 14, 2010 by vuquta

[vuquta]

Originally Posted by swansont

Are you not following the discussion?

 

He is claiming that the orbit about the sun will add a much larger Sagnac phase shift that the rotation of the earth.

 

 

 

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No, this is not correct. Einstein discusses a "stationary" frame — he even uses quotes when first discussing this, and at various times afterward. He also discusses another frame, moving at an arbitrary velocity v with respect to the first.

 

Further, he states

 

 

light (as required by the principle of the constancy of the velocity of light, in combination with the principle of relativity) is also propagated with velocity c when measured in the moving system.

all in On the Electrodynamics of Moving Bodies

Yes, well he also claimed that needed to be proven later and not simply assumed as true.

 

Einstein:

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.

 

At the time t = τ = 0, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

 

x² +y² +z² =c² t² .

 

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

 

ξ² + η² + ς² = c² τ²

 

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible.5

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

 

He claimed the above is the proof of logical consistency. He stated the requirement for the logical consistency of the light postulate and the relativity postulate is the spherical light wave in the stationary frame must be spherical when translated by LT in the moving frame. He made the statement, "The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system."

 

Here is what he did.

He started with a point (x, y, z) that was struck by the spherical light in stationary K.

 

He performed LT on that point.

 

He then claimed in moving k, ξ² + η² + ς² = c² τ² . This is in fact true for any arbitrary light beam.

 

Thus, he claimed he proved the spherical wave in K is also spherical in moving k, based on the equation above.

 

But, for the spherical light wave in K to be spherical in k, c² τ² would need to be constant for all (x, y, z ) attained by the stationary light sphere after translation by LT.*This is easily proven false by considering (r,0,0,r/c) and (-r,0,0,r/c) in the stationary frame and translating them using LT.

 

Hence, he committed the logical fallacy, dicto simpliciter.

Therefore, the light sphere in K is not spherical in k when translated by LT

Hence, the light postulate and the relativity postulate are logically inconsistent.

 

Here is a picture of the LT calculated light sphere from the view of the moving frame.

 

http://www.freeimagehosting.net/uploads/04e0a878a2.gif

 

The yellow circle is the stationary light sphere.

 

The yellow rays are the LT calculated light beams.

 

The yellow rays do not constitute a sphere. Therefore, his compatability proof is false.

[swansont]

This is easily proven false by considering (r,0,0,r/c) and (-r,0,0,r/c) in the stationary frame and translating them using LT.

 

I'm not sure what you're trying to show here. r/c is the time it takes light to travel a distance r. The transforms won't work, and don't purport to work, on a frame traveling at c.