# A Purely Hypothetical Question regarding Special Relativity Theory.

Discussion in 'Physics & Math' started by geistkiesel, Jan 29, 2005.

1. ### geistkieselValued Senior Member

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2,471
SRT Theorists, Can you handle a purely hypothetical question?

Before we get to any subjects regarding the effect of expeimental results this question is designed and intyended as purely hypothetical.

Let’s discuss the following gedanken. Two photons, l and r (<|>), are emitted simultaneously from the midpoint M of photon sensitive clocks L and R mounted at the front and rear of an inertial frame, Vf. The experiments can be conducted with Vf velocity v = 0, with respect to the embankment, or where Vf moves to the right with v > 0.

The diagram defines the condition just as l, <, and r,>, are emitted:

Code:
L |_________________<|>_________________| R     v ->
l M r
L and R are synchronized photon sensitive clocks that record the time of arrival of the photons l and r. tl and tr are travel times of the simultaneously emitted photons from M -> L and from M -> R.

Hypothetical question: Is the measured tl < tr when v = 0 and when v > 0?

Geistkiesel?

3. ### YuriyRegistered Senior Member

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1,080
geistkiesel,

1. LM=MR=S
2. Both photons are emmited at the same moment of time.
1.Case v=0.
Tlm = S/c = Tmr
2.Case v>0.
Tmr = S/(c-v)
Tlr = S/(c+v)

What is now?

5. ### superluminal.Registered Senior Member

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10,717
Yuriy:

1)
Let S = 1m
Let v = 0m/s
Tlm = 1/(300e6) = 3.33e-9s
Tlr = 1/(300e6) = 3.33e-9s

This seems correct.

2)
Let S = 1m
Let v = 200e6m/s
Tlm = 1/(300e6-200e6) = 1e-8s
Tlr = 1/(300e6+200e6) = 2e-9s

If I am on the spacecraft travelling at 200e6m/s, sipping tea in the lab, and I do this experiment, will I not get the results in 1)? How can two observers, who always agree on the fact that the clocks are synchronized (although at different observed tick rates) disagree on the photon travel times?

Last edited: Jan 30, 2005

7. ### YuriyRegistered Senior Member

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1,080
superluminal,
you said in thread on pulsars?
Now here you have the same problem: the answer on your pathetic question
lies exactly ... in content of my statement marked as "2. Case v>0": it shows how and why...

8. ### superluminal.Registered Senior Member

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Oh Great One:
That was a super-nice explanation to a question.

9. ### superluminal.Registered Senior Member

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Oh Great One:
I am not a physicist. I am trying to get clarification to questions of interest here. You are an arrogant human being, and a horrible teacher.

10. ### superluminal.Registered Senior Member

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10,717
Maybe if you boned up on you english, you wouldn't have so many problems with your explanations.

11. ### YuriyRegistered Senior Member

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superluminal,
So, a simple reminder of people's wisdom: " do not spit in a well where from you get water to drink" caused such a nervous reaction of you?
Calm down and think a little about what just happened...

12. ### XgenRegistered Senior Member

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315
I started an entire thread conserning this question called Absolute Space. I had even propossed a device based on exactly the same scheme that you had draw, called PASD.

The answer is we cant know. If that is everything we know about the light source then I suppose that what Yuriu had calculated is correct:

But if you realize this here on Earth you cant be sure because except velocity v there would come into play and Earth velocity around Sun, The Sun velocity around galaxy center and so on....

So in the above equations v is not just some kind of velocity but the absolute velocity of the light source WRT the abosolute space.

If the observer is placed at the center of the light-source he will observe that both clocks are running exactly in the same rate in the case 2). That is because the photons should go back to the source and both times will becomes equal. If you had observer somewhere else he will notice the same - time is running in the same rate according to all clocks and that is noyt in contradiction to the above formulaes. But if we have two observers at the two clocks, they will observe different photon arrival times.

As I had sayed and before 'observe' is a very slicky word. You should be very carefull with it. Whenever be have movement of light back and forth observation times are equal. The PASD device however is designed so that only uni-directional movement of light gets into acount.

13. ### superluminal.Registered Senior Member

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Xgen,
Not sure I get it yet, but thanks for the response.

14. ### YuriyRegistered Senior Member

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1,080
geistkiesel,
I have to emphasize for the record:
in my answers on your questions v meant exactly what you have defined in your initial post - WE see once v =0, and in another case the same WE see v>0. So v is velocity of your device in respect to WE (I hope, you meant WE as a simple Laboratory Reference Frame.) So in my post there was no trace of any mysterious Absolute Space, Absolute Time or Absolute Motion...

Last edited: Jan 30, 2005
15. ### XgenRegistered Senior Member

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315
superluminal,

People do errors all the time. It is bad when they dont realizes them. If I am wrong with something I had write I will inform you and the others and will apologize.

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10,717
Xgen:
Thank you.

17. ### superluminal.Registered Senior Member

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A simple question, presented with the utmost humbleness:

If I do the experiment above, sitting here in my house sipping an espresso, I measure the same travel times for each photon. Both clocks read the same.

If I do the experiment above, sitting in my spaceship, sipping tea this time, travelling at 0.8c toward Rigel, I measure the same travel times for each photon. Both clocks read the same.

18. ### geistkieselValued Senior Member

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Superluminal, I am jumping in here ahead of you, please excuse the barging in and answer as you will.
Geisntkiesel.

Consider the following. If we let the experiment go to its natural conclusion we see that the observers on the moving platform will always observe that the photons arrive at L first, then R. Does ths not tell the moving observer that he can always determine that he is moving with respect to the (a)stationary frame?

This pathetic post is the first half of a popular Einstein gedanken. AE had the photons reflected back in the direction of the source of the photons where the photons arrived simultaneously (in both the stationary and moving cases), thereby giving the moving observer, persuant to SRT of course, his "right" to claim he is at rest.

How does the moving observer explain the difference between the stationary frame case 1 and case 2 without concluding he is moving wrt the embankment in case 2? I am referring to the "difference" of arrival times of the photons at L and R.

geistkiesel

19. ### geistkieselValued Senior Member

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2,471
Would not an observer always determine he is moving in case 2?
Geistkiesel

20. ### superluminal.Registered Senior Member

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geistkiesel:

You see. This is my problem. I have always understood that any experiment performed in any valid frame should always yeild the same results to the observer in that frame, no matter what the frame's state of motion. Especially that the speed of light is always measured to be c.

If the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. In the ship, nothing appears odd to me.

This makes no sense. Help me.

Last edited: Jan 30, 2005
21. ### superluminal.Registered Senior Member

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Note: As long as the frame above is not accelerating.

22. ### geistkieselValued Senior Member

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Yuri,
It is apparrent you haven't grasped the scope of the sffect of your answer.

If the arrival times of the photons at L and R are different, as you concluded, then how can the observer ever conclude he is at rest when the arrival times of the photons are dofferent? Does this not tell the moving observer that he is moving wrt the stationary case? Even if the obsrve on the moving frame was born there and was never on a "staationary frame" where the experiment would result in simultaneous arrival times, she would conclude that she is moving. Further, she is using the velocity of light as a standard of velocity measurement. When she uses the c-v and c+ v terms she is comparing her absolute velocity with the velocity of the speed of light.

One must be carefull and always distinguish how much "slower" a platform is moving wrt light velocity as well as how much "faster" is a closing (collision trajectory) speed of inertial frames relative to the speed of light.

In all of the discussions on the subject in the last year or so no one has ever stated that a massive object ever moved at a velocity greater than the speed of light. An object's motion and "relative velocity" of inertial frames are two different physical conditions and should not be confused.

Geistkiesel

23. ### geistkieselValued Senior Member

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2,471
Read the question again Yuri. I am only discussing the fact that the light always arrives at different times on the frame when movinmg, and the photons arrive simultaneously when stationary. From whatever perspective you want to view the problem the moving observer and the statioanry observer will see the identical result when looking a the printout of the arrival times from the clocks at L and R. If you want to impose some superficial frame contraction or time dilation, go right ahead, but you must still explain to all observers why the photons arrive at different times on the moving frame.

Yes, there is a trace of what you call "mysterious" as the difference in arrival times on the printouts from the clocks will attest. Absolute motion is detected and I would like to share with you my condolences for your admission to this seemingly trivial physical reality. We all know what passion you place in your posts when expressing your views which is very refreshing and stimulating.

Some days are just sadder than others Yuri.

Geistkiesel