The measurement of the relative velocity of frame and photon.


A. Velocity of objects moving with respect to inertial frames in motion.

First, we see how to determine the relative velocity of objects moving with respect to a inertial frame moving with respect to the embankment.

A bullet fired from a gun at a target 1 km down range on the stationary frame will arrive in time t = 1/Vbe, where Vbe is the velocity of the bullet with respect to the embankment. A train with an identical set up measured when the train is at rest with respect to the embankment moves along side the embankment arrangement. The two guns fire when adjacent to each other and the bullets race to their respective targets. Observers on both frames can determine the bullet moving with respect to the train is moving faster than the bullet moving with respect to the embankment; or from Vbe = Vte + Vbt the velocity of the bullet with respect to the train is Vbt – Vte = Vbt as we set the Vte at rest in the bullet train inertial system.

As we all know, the velocity of the bullet with respect to the embankment is the sum of the velocity of the train with respect to the embankment plus the velocity of the bullet with respect to the train assumed at rest. This physical fact is verified from experiment.

B. The velocity of light with respect to the embankment.

The same form of measurenment as above is used here.

The physical fact that Vce = Vct is verified from experiment which can only occur if the velocity of light with respect to the embankment is the sum of the velocity of the frame Vte, plus the velocity of light with respect to the inertial frame Vct, minus the velocity of the inertial frame Vte, or

Vce = Vte + Vct – Vte,

or,

Vce = Vct,

with the observation that Vct – Vte > 0 in general. We do this knowing the velocity of the light is independent of the velocity of the source of the light, or Vte. To start off with Vce = Vct hides the fact that the velocity of the light with respect to the embankment is the velocity of the train with respect to the embankment plus the velocity of light with respect to the train minus the velocity of the train.

C.What is the velocity of light with respect to the train?

We have a train moving at Vte. The train and the embankment have 1 km ranges established on each frame.
A |---------------------------------------|B
A’|-----A’’|------------------------------|B’------|B’’
The distance AB = A’B’ and A’A’’ = vt = B’B’’.
When the emitters are adjacent, both sources emit light. The light from both will arrive at the 1 km target located on the embankment (at B and B’)as observed from both frames of reference, and both lights are a distance vt from the target/clock on the train now at B’’. This is also verified by observers on both frames of reference. From here the light will arrive at the clock in time t’ after moving an additional vt plus the distance the frame moves in the interim vt’ or,
ct’ = vt + vt’, where

v = ct’/(t’ + t)

or,

c = V(t' + t)/t'.

Even without clocks the observers on both frames can observe the simultaneous arrival of the light at B' and B''

If the measured t’ is zero the inertial frame is at rest, otherwise the inertial frame is moving absolutely with velocity Vte > 0. If we set t’ = xt then Ve = cxt/t(x + 1) = cx/(x + 1). As, x goes from zero to infinity in the expression c*x/(x + 1), Ve -> c.

The second expression says by measuring the velocity of the frame Vte and then determining C, where C = Vte(x + 1)/x or C = Vte(t + t')/t' which means we cannot measure the velocity of light thus unless we have a t' > 0.


SRT grossly underestimates the effect of relative motion of frame and photon.

Here is a link to a unique POV. (http://www.aquestionoftime.com/).

It should be clear that the velocity of light measured from any inertial frame is always C and that the relative velocity of light with respect to a moving inertial frame is Vct - Vte. The expression Vct - Vte is merely the algebraic sum indicating the difference in absolute speeds of frame and photon and in no way diminishes the speed of light. The expression is an algebraic sum as distinguished from a physical operator. This of course drives us to a universe of absoute time and space.

Conclusion
Treating light motion directly analogous to physical objects, bullets for instance, we arrive at Vce = Vte + Vct, which is obviously erroneous as light speed is independent of the motion of the source of light, here Vte, the train. So, Vce = Vbt.
The speed of bullets is functionally determined by the speed of the frame on which the bullet is emitted.

Geistkiesel :cool:

geistkiesel:

A. Velocity of objects moving with respect to inertial frames in motion.

First, we see how to determine the relative velocity of objects moving with respect to a inertial frame moving with respect to the embankment.

A bullet fired from a gun at a target 1 km down range on the stationary frame will arrive in time t = 1/Vbe, where Vbe is the velocity of the bullet with respect to the embankment. A train with an identical set up measured when the train is at rest with respect to the embankment moves along side the embankment arrangement. The two guns fire when adjacent to each other and the bullets race to their respective targets. Observers on both frames can determine the bullet moving with respect to the train is moving faster than the bullet moving with respect to the embankment; or from Vbe = Vte + Vbt the velocity of the bullet with respect to the train is Vbt – Vte = Vbt as we set the Vte at rest in the bullet train inertial system.

I assume you meant to write Vbe - Vte = Vbt.

This is correct in a Newtonian world, but only an approximation in a special relativistic world. At the normal speed of trains, though, it isn't a bad approximation.

As we all know, the velocity of the bullet with respect to the embankment is the sum of the velocity of the train with respect to the embankment plus the velocity of the bullet with respect to the train assumed at rest. This physical fact is verified from experiment.

Yes.

B. The velocity of light with respect to the embankment.

The same form of measurenment as above is used here.

The physical fact that Vce = Vct is verified from experiment which can only occur if the velocity of light with respect to the embankment is the sum of the velocity of the frame Vte, plus the velocity of light with respect to the inertial frame Vct, minus the velocity of the inertial frame Vte, or

Vce = Vte + Vct – Vte

That is an incorrect conclusion to draw, since it mixes Newtonian physics with relativistic physics. In special relativity we actually have:

Vce = (Vte + Vct)/(1 + Vte Vct/c^2) = c = Vct

Velocities don't add in relativity in the same way that they add in Newtonian physics. In a Newtonian world, we would expect Vce = Vte + Vct, rather than the above (correct relativistic) formula.

To start off with Vce = Vct hides the fact that the velocity of the light with respect to the embankment is the velocity of the train with respect to the embankment plus the velocity of light with respect to the train minus the velocity of the train.

You are working entirely in a Newtonian world. But experiments tell us that our actual world is non-Newtonian. The starting point for the special theory of relativity is the constancy of the speed of light in ALL inertial reference frames, which is amply supported by experimental evidence.

Since you have made a fundamental error at this point, there's no point in considering the rest of your post.

Conclusion
Treating light motion directly analogous to physical objects, bullets for instance, we arrive at Vce = Vte + Vct, which is obviously erroneous as light speed is independent of the motion of the source of light, here Vte, the train.

Yes. The equation Vce = Vte + Vct is erroneous, and should be replaced by the correct relativistic expression, which I gave above.

The speed of bullets is functionally determined by the speed of the frame on which the bullet is emitted.

Yes, it is, but you need to realise that the expression Vbe = Vbt + Vte is only a Newtonian approximation to the correct relativistic expression, which works reasonably well provided that both Vbt and Vte are both much less than the speed of light.

geistkiesel:

Originally Posted by geistkiesel
A. Velocity of objects moving with respect to inertial frames in motion.

First, we see how to determine the relative velocity of objects moving with respect to a inertial frame moving with respect to the embankment.

A bullet fired from a gun at a target 1 km down range on the stationary frame will arrive in time t = 1/Vbe, where Vbe is the velocity of the bullet with respect to the embankment. A train with an identical set up measured when the train is at rest with respect to the embankment moves along side the embankment arrangement. The two guns fire when adjacent to each other and the bullets race to their respective targets. Observers on both frames can determine the bullet moving with respect to the train is moving faster than the bullet moving with respect to the embankment; or from Vbe = Vte + Vbt the velocity of the bullet with respect to the train is Vbt – Vte = Vbt as we set the Vte at rest in the bullet train inertial system. ”



I assume you meant to write Vbe - Vte = Vbt.

This is correct in a Newtonian world, but only an approximation in a special relativistic world. At the normal speed of trains, though, it isn't a bad approximation.

I meant to write what I wrote. As you can see I am starting out where AE started in his book "Relativity". You are jumping ahead by claiming that Newtonian and relativistic mechanics were "mixed", your error here.

Geistkiesel
“ As we all know, the velocity of the bullet with respect to the embankment is the sum of the velocity of the train with respect to the embankment plus the velocity of the bullet with respect to the train assumed at rest. This physical fact is verified from experiment. ”



Yes.


Geistkiesel
“ B. The velocity of light with respect to the embankment.

The same form of measurenment as above is used here.

The physical fact that Vce = Vct is verified from experiment which can only occur if the velocity of light with respect to the embankment is the sum of the velocity of the frame Vte, plus the velocity of light with respect to the inertial frame Vct, minus the velocity of the inertial frame Vte, or

Vce = Vte + Vct – Vte ”



That is an incorrect conclusion to draw, since it mixes Newtonian physics with relativistic physics. In special relativity we actually have:

Vce = (Vte + Vct)/(1 + Vte Vct/c^2) = c = Vct

Velocities don't add in relativity in the same way that they add in Newtonian physics. In a Newtonian world, we would expect Vce = Vte + Vct, rather than the above (correct relativistic) formula.


No, I made no error. I got the same result you did without mixing anything and using "Newtonian" physics as you call it.
Geistkiesel
“ To start off with Vce = Vct hides the fact that the velocity of the light with respect to the embankment is the velocity of the train with respect to the embankment plus the velocity of light with respect to the train minus the velocity of the train. ”



You are working entirely in a Newtonian world. But experiments tell us that our actual world is non-Newtonian. The starting point for the special theory of relativity is the constancy of the speed of light in ALL inertial reference frames, which is amply supported by experimental evidence.
First you say I am working entirely in a mixed newtonian and relativistic world, then you say I am working in entirely in a newtonian world, You are being confusing here, to yourself, not to me.

No experiments tell you nothing when assuming relativitic mechanics. What you have missed here is that when AE came to the conlusion that Vce - Vte = Vct he asserted that Vct should have been = to Vce, which is a true statement. The only problem here is he treated the light as the bullet, where we set the velocitty of the Vte = 0 when determining the velocity of the bullet with respect to the train, In the case of light, due to light's constant speed nature the velocity of the train with respect to the embankment should have been subtracted from the velocity of the light with espect to the train. Then AE would have come up with the correct conclusion that Vce = Vct and we would have avoided special realtivity which as we can see now was due to AE miss-using the dynamics of inertial frame representation.

Since you have made a fundamental error at this point, there's no point in considering the rest of your post.
It is your mistake, your fundamental error so do as you please.


“ Conclusion

Treating light motion directly analogous to physical objects (as did AE), bullets for instance, we arrive at Vce = Vte + Vct, which is obviously erroneous as light speed is independent of the motion of the source of light, here Vte, the train. Vct should have bee subtracted from the right hand side and all would be well.


Geistkiesel :cool:

This Answer is directed to the all those adhering to any aspect of Special Relativity Theory as determined by AE in 1905. The following is a direct replay of the analysis used by AE in arriving at the postulate that the speed of light is constant as measured from any inertial frame of reference. The reader is urged to see the difference in the measurement of the speed of light, which is a constant C with respect to the relative speed of frame and photon, which in general is less than C.

This post is similar to the original post of this thread. I mean to emphasize here that using realtivity theory to counter the conclusions here is premature as special relativity was derived from an erroneous analysis of the following matter as outlined by AE in his 1916 book "Relativity", chapters 6 and 7.

I do this like I have to counter in a coherent manner James R's arguments using SRT and such vague references "as I told you before". To defeat the conclusions here one must necssarily defeat the conclusions as established within the four corners of this post.

The use of SRT to object to the following (as did James R in his response to the opening post of this thread) is premature as SRT follows this analysis as performed by AE. I refer to chapters 6 and 7 of Einsteins book “Relativity” (published in 1916) where he shows us the underlying basis for relativity theory published 11 years earlier.

The notation used here helps to keep tabs on where we are in regards to inertial frames and which frame is being referred to.

Notation:
In general, Vxy means the velocity of x with respect to the inertial frame y.

Velocity of a man moving on the moving train.

AE starts with a man walking on the train and establishes the velocity of the man with respect to the train Vmt = Vm + Vte where Vte is set to zero. Now the Velocity of the man with respect to the embankment Vme = Vte + Vmt, which is the result we see experimentally.

The velocity of light with respect to the moving train.

Here the same logic used by AE is duplicated here. We say the velocity of light with respect to the embankment Vce is equal to the velocity of the train with respect to the embankment plus the velocity of light with respect to the train. Here seen as, Vce = Vte + Vct. AE was using the same physics and logic as used in determining the velocity of the man with respect to the embankment. Here, AE showed us that Vce – Vte = Vct, saying this cannot be as the velocity of the light with respect to the train should not be less than Vce (or C). He is correct of course, as the velocity of the light does not depend on the velocity of the source of the light. But, what is Vct? This is easy enough, as Vct = Vce – Vte and this merely says how much faster is the light than the train as measured with respect to the embankment.

So returning to the expression Vce – Vte = Vct = Vce – Vte and we arrive at the conclusion that Vce = Vce where we must remember that the right hand side is the light emitted on the train. Clearly, the speed of light is independent of the speed of the source of the light.

Here is a simple experiment to unravel any ambiguities lingering in anyone’s mind.

There is a 30 km light range on the embankment and the train with targets able to measure the arrival of the light. As the train moves with some velocity Vte lights emit simultaneously in the train and the embankment when the sources are adjacent to each other. As the light beams race to their targets we see that the lights both arrive at the target on the embankment simultaneously. Later both lights arrive at the target on the train which has moved a distance Vt during the time the lights moved t = 30/C during the “race”. The only explanation is that the light moving with respect to the train is the same velocity as the light moving on the embankment as observers on both frames can easily determine.

So why does the light strike the target on the train later?

Because the distance the light has to travel to reach that target is longer than the distance the light has to travel on the embankment. There is no mystery here and surely there is no need to invoke a frame contraction and or time dilation to explain the difference. Certainly, the observer on the train need not even scratch her head. See the simple experiment below describing the above.

A|----------------------------------------------|B ;30 km embankment range.
C|-----C’|-------------------------------------|D-----|D’ ; 30 km train range with motion shown as Vt that CC’ = Vt = DD’.

AS both observers agree on the results, there is no conflict and certainly, no special attributes attach to what the observer on the train sees. The only way around this is to blind the observer on the train to the extent she is not aware of any motion of the train.

However, she has been involved in the experiments shown above as observer on the embankment and observer on the train. Now if she is not aware of her motion what will she observe? If she is located at the point on the train where the embankment target is located she will know immediately she is moving as the light will not have arrived at the target on the train. Simply said, if she is actually at rest, with respect to the embankment, the light arrives at her target in time t = 30/c. If the train is in motion with respect to the embankment, she will see an extended time of arrival of the light at the train target.

How much extended time?

When the light arrives at the 30 km point with respect to the embankment, the light is Vt from the target on the train. The light will arrive from this point in a time t’ after the light moves the distance Vt plus a small amount Vt’ that the train moves in the interim. This added distance [of motion] then is, Ct’ = Vt + Vt’ or,

t’ = Vt/(C - V).

Some might claim a “time dilation” as the explanation for the difference, but then it would be necessary to explain why the lights traveled farther to reach the moving target?

The time exptressed here should replace the so-called "tiem dilation" of SRT.

Would not this mean then a frame contraction to bring the targets to a position of co-location? If so then, the experimental results will contradict that line of reasoning. Everyone in the universe knows the time for the light to reach the moving target is longer than the time to reach the light on the embankment by the amount t'. The universe watching from the embankment all verifies this. Are we to allow the moving observer to warp this result into some relativistic form to agree with her erroneous “perception”? And how could she possibly have such a perception that differes from observation and her experience as an observer on the embankment earlier?

Is the relativistic time the same as the classic time observed above? In other words we have found that t + t’ = 30 + v/(1 – v) for unit speed of light or (30 – 30v + v)/(1 – v) = (30 – 29v)/(1 – v). Using relativistic models will show you that relativity “kicks in" later than the effect shown here. This is left as an exercise for those proficient in SRT calculations.


Conclusion:
Relativity theory is based on the assumption that the relative velocity of frame and photon is C. This result comes from the erroneous reading that Vce = Vte + Vct, or Vce – Vte = Vct where Vct the velocity of the light with respect to the train was deemed “Less than the speed of light in vacua”. This is not so of course as Vct = Vce – Vte and hence Vce = Vct + Vce – Vct and Vce = Vce, with the understanding that the right hand side of the expression is measured where the light is emitted on the moving train.

What AE misunderstood, rather misapplied, is that the speed of light is independent of the speed of the motion of the source of light. Had he used this postulate properly he would have arrived at the same result above. He was aware of the independence postulate as in the first paragraph of chapter 7 AE says,

”...De Sitter was able to show that the velocity of propagation of light cannot depend on the velocity of motion of the body emitting the light”
AE misused this postulate in arriving at his conclusions.

Geistkiesel :cool: