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What do I see wrong here ? T2 = T1 . (1 - V/C)

  1. Oct 5, 2011 #1
    SRT = the theory of special relativity

    I can't find for myself what is wrong with this ?

    A <---------------------- C x T1 ----------------------------------------> B : system S1 observer 1

    A <--- V x T1 ---><----------------------------- C x T2 -----------------> B : system S2 observer 2

    At observer 1 a light wave starts in A and arrives 1 seond later over a distance of 300.000 km in B.

    Observer 2 travels with speed V from A with the same light wave in direction B.

    Observer 2 uses the light wave to meassure the distance between A and B (when the light wave arrives in B, the meassurement ends), so counts the periods, the counted periods

    will be lesser (than the measurement for observer 1 in A) because of his higher speed, so the expectation is like in the SRT, the distance and time will be shorter for observer

    2, his time goes slower.

    So C . T1 = V . T1 + C . T2, so T2 = T1 . (1 - V/C). It fits with the expectation of the SRT, when V=C than T2 = 0, the time stands still. When V= 0, T2 = T1.
    In both situations S1 and S2 is the speed of light C, the same value.

    But this formula is not the same as in the SRT.

    What do I see here wrong (maybe my view of light waves), but if fits in the SRT, seems it (it is clearly visible that distances and times are shorter as the SRT says) ?

    I think that a light wave is always the same (independed for speeds of the observers, because of mass = 0) anywhere in the universe ...
     
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  3. Oct 5, 2011 #2

    ghwellsjr

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    When you use a light wave to measure a distance, you have to wait for the light to reflect off your target and come back to you. But if you are moving with respect to the target, what distance are you actually measuring, the one where you were when the light left you, or the one where you were when it returned to you or something in between?
     
    Last edited: Oct 5, 2011
  4. Oct 5, 2011 #3
    Thanks for the answer, purely measurement, this is not correct.

    I mean the distance where observer 2 is when the light wave arrives in B (one way only).

    Then it looks shorter for him than for the observer 1 in A (observer 2 moves).

    B is exactly the same end point for both observers.

    I was looking for an example to get/present a view for shorter times/distances in the SRT.

    Maybe this is not the right one, do you know a better one ?

    Strange but in some way my example gives a feeling ... purely using the characteristics of a light wave ...
     
  5. Oct 5, 2011 #4

    ghwellsjr

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    You are thinking that some measurement of time (T2) correlates to some distance (from Observer 2 at then end to B), correct? I would agree, if Observer 2 were stationary at his end location and he used light to measure his distance from there to B, that he would get a meaningful result. But when the light starts at A and goes to B, why should that provide information about a different distance?

    But in addition to that, how do the observers know when the light arrives at B unless they wait for a reflection or some other signal to get back to them to let them know when to stop their timers?

    The normal example that people use to demonstrate the correct timing relationship between two observers in relative motion is the "light clock". Why don't you look that up in wikipedia?
     
  6. Oct 5, 2011 #5
    I know now better what I want to demonstrate with this example in fact, not the distance measure.

    I want to show that somebody with a higher speed (compared to the other) goes slower in time.

    I see than just a light wave from 1 second long between A and B so the distance is 300.000km. This one second illustrates a time frame in the universe, so illustrates time (light speed is always the same everywhere, so illustrates time on a light clock). A and B are for both observers exact the same locations.

    When observer 2 has a higher speed passing A the light wave goes slower when he/she "looks" to it than for observer 1 in A.

    When the light wave arrives in B, for oberserver 1 is it 1 second later, but for observer 2 is it I thought 1 - V/C seconds passed.

    Imagine there is a light clock since the beginning of the universe and this 1 second is a part of it, but observer 2 was traveling in the direction of the light wave, so his time is shorter (and the distance traveled in that period shorter compared with the light speed.

    Do you agree, but the time difference is not the same as calculated in the SRT, what is wrong in my thinking ?

    But in my thinkings I see maybe a wrong view of light, so I think that the crossing of two light waves (send at the same location on the same time) will always be that location, if somebody is in that location and still there after a while he/she stands completely still, if he/she is not there anymore he/she moves with a speed away. So A and B could be such a locations. Maybe I go wrong here, and that is maybe the reason I go wrong in my formula (but it illustrates maybe something like slower going through time ...).
     
  7. Oct 6, 2011 #6
    I try to describe it more clearly now. This is just theoretically (experiment in thought).

    See picture in first answer.

    E.g. one mile before location A we let start a long light wave passing A and B and further.

    Observer 1 stands in A, starts a special clock (which can count periods of the lightwave) and counts the periods which are passing and stops the clock after 1 second. The starting point of the light wave on the moment of starting the clock has arrived in B, 300.000km further. If observer 1 has a speed too is not important, this is always possible in his system S1.

    Observer 2 has a speed V (> observer 1 and goes direction B) and starts his ingenious clock too at the same time with observer 1 when passing A, his clock counts lesser periods until the starting point of the lightwave from observer 1 has arrived in B. Exactly what Einstein tells us, his time (observer 2) goes slower. Near the shorter time, observer 2 could measure the one-way distance to B too with the counted periods and of course is shorter too (Einstein tells us too).

    We are using here just the lightspeed with his fixed periods for our clock (and one long light wave), and the clock for observer 2 goes slower.

    Once again (read other anwers too), what do I wrong in my thinkings ?
     
    Last edited: Oct 6, 2011
  8. Oct 6, 2011 #7

    ghwellsjr

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    There are several things wrong with your thinking.

    First off, a device that counts the cycles of light can be used as a clock, but only as long as the light source is stationary with respect to the counting device. This would work for observer 1 but since observer 2 is moving away from the light source, he will count fewer cycles than he should for it to be a legitimate clock so the count of the wave cycles won't be correlated with time. This is not the time dilation that Einstein described. It is instead Relativistic Doppler which would result in your "clock" running even slower than it should. But if observer 2 carried an identical light source and a second period counter, it will count at the correct rate predicted by Einstein's time dilation. Then observer 2 could compare the rates of the two devices and that will demonstrate the Relativistic Doppler Factor.

    A second problem is what I asked you about in post #4:

    "how do the observers know when the light arrives at B unless they wait for a reflection or some other signal to get back to them to let them know when to stop their timers?"

    Of course, observe 1 could stop his counter after one second but this presupposes that he has another clock. If instead, he is simply waiting until his counter reaches the frequency of the light source (the number of cycle in one second), then he is simply stating that the front of the light wave has reached one-light second away and he is not measuring anything.

    Even worse for observer 2, what is he supposed to count to "measure" his length contracted distance to point B?
     
  9. Oct 7, 2011 #8
    Thanks for your answer.

    But this is an experiment in thought, not a question how do I measure.

    So I have changed quickly my ingenious clock (was not easy), my clock can measure the length of the passing light wave, for observer 1 is 300.000 km wave passing.

    For observer 2 (1 - V/C).C light wave. So my clock eliminates the doppler effect (the light speed is always the same).

    Everything has to do with the constant lightspeed in Einsteins theory (and so with light), I want to visualize this (if possible) in details ... and may use 300.000km as 1 second (like they always do with a light clock, but I want to avoid this because very confusing, so I use the same light wave for both observers to be sure and to learn of course where I am wrong in my thinkings).
     
  10. Oct 7, 2011 #9

    ghwellsjr

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    If you want to use a rest frame for observer 1, then in your thought experiment you can talk about the time it takes for the light to propagate from A to B and you can call that one second and you can say the distance is 1 light second or 300.000 km. No problem there. You can also say that observer 2 travels at speed V and you can identify where he will be after 1 second according to your specified rest frame. Let's call that position X. No problem there. Then you can also talk about how long it will take for light to go from X to B according to this same rest frame. No problem there. Now you can create your formula as you did in your first post and it will be correct.

    But it has nothing to do with what observer 2 experiences. At the time that observer 2 reaches point X, the light has already reached point B. So why do you want to make an equation involving a fictitious time interval for how long it took for your light wave to go from point X, long before observer 2 arrived there, to point B?
     
  11. Oct 7, 2011 #10

    ghwellsjr

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    After thinking about your scenario some more, I think I understand what you are talking about. You are treating the light wave as a sort of tape measure with your ingenious device keeping track of how many waves, and therefore how far the light has traveled past each observer, correct?

    Let's say the light source has one wave per meter (way too few for real light, but this is a thought experiment so we can pretend). Then for observer 1, at point A, after he counts out 300.000.000 waves, he knows that the beginning of the light wave will be one light second away from him, at point B.

    But since observer 2 is traveling from A toward B at speed V, he will count the waves at a lesser rate than observer 1 and when the front of the light wave reaches point B, he will be at point X and will have counted out the correct number of waves corresponding to the distance he is from point B.

    So we could demonstrate that your equation from post #1 would fit the data as just described.

    But here's the problem: we can talk about where observer 2 is when the front of the light wave reaches point B, because we can see everything that is going on as defined according to a frame of reference in which observer 1 is at rest but observer 2 cannot know any of this. In fact, he will not consider the waves to be one meter apart. Since his rulers are length contracted by one over gamma, he will think the waves are too long by a factor of gamma. So when he determines how far away the front of the light wave is when he reaches point X, he will think it is past point B. Furthermore, he will then think point X is closer to point A by a factor of one over gamma. So from the point of view of observer 2, your equation is no longer valid.
     
  12. Oct 8, 2011 #11
    Thanks for your answers, we come closer to my errors probably.

    Forget this situation and consider my space-time-diagram (attached).

    So my originally thought was, if two persons A and B are "looking" to the same light wave and person B with a higher speed than A (and in the direction of the light wave), person B goes slower in time because the light wave is passing him slower. I see this light wave as time (it represents time, if your speed is equal to the lightspeed, times stand still for you.

    So I try to make Einstein visible to show to other people.

    My diagram:

    - one/same system S
    - light wave starts in Xoc
    - B has speed V compared with A
    - when B meets A, we start our enginous clocks for measure the passing light wave length
    - on any moment time ΔTb = ΔTa . (1 - Vb/C) so slower (see attachment)
     

    Attached Files:

  13. Oct 9, 2011 #12
    Yes you are right, Ghwellsjr, thank you very much for your answer.

    I know my error now I think.

    I calculate the situation for B, being A (seen from A). how it would be for B probably.

    And you see than already that time goes slower for B.

    But I have to calculate it too for being B (seen from B) and that will be like Einstein calculates it (a refinement).

    But my idea to explain it to others fits I think (compare the lightspeed with time, when you go faster the light waves are going slower passing you, so time will go slower).

    If you agree, how do you think about the crossing of two light waves resolving in a point standing still, other objects around are moving ?
     
    Last edited: Oct 9, 2011
  14. Oct 10, 2011 #13

    ghwellsjr

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    Digi99, instead of using your diagrams and analyses to describe what happens when two observers count the waves of light as they go by them, consider a similar scenario on the surface of water where two observers are counting the waves of water going by them from a distant wave maker. Won't your equation apply in this situation as well? And yet I don't think you will conclude that time is going slower for an observer moving on the water, just because he counts the waves passing him at a slower rate than for the stationary observer, would you?
     
  15. Oct 10, 2011 #14
    Attached the complete diagram, so you can see ΔTb as well.

    I find this very teachable, what you can do wrong in mathematics compared with physics.

    Ghw, the lightspeed has always the same value in the universe, so it represents time.

    My clock is not counting waves but the length of the passing lightwave (divided by time difference always C). You can say the lightspeed represents time.

    This all is not true for other moving objects like water waves.

    I go to work this out with 2 rest systems and will come back with other diagrams (but takes a few weeks) ...
     

    Attached Files:

  16. Oct 10, 2011 #15

    ghwellsjr

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    Digi99, I'm trying to get you to see that your explanation is no different than normal Doppler which is true for water waves but not true for light waves. It does not teach time dilation. Light waves are like water waves when you view them from a single reference frame and only consider what an observer at rest in that reference frame analyzes using that reference frame and only pay attention to what that one observer sees and measures.

    But if you take into account what the moving observer sees and measures, then your explanation, like water waves, does not fit. You have mistakenly interpreted Doppler to be the same as time dilation. You can easily see that this is wrong if you were to do your same thought experiment with the moving observer approaching the stationary light source. Then his count of the waves would be higher than the stationary observer and you would wrongly conclude that time was going faster for the moving observer than for the stationary observer.

    You need to learn Special Relativity before you try to teach it to others. And this is the place to learn it if you are willing.
     
    Last edited: Oct 10, 2011
  17. Oct 11, 2011 #16
    Thanks Gwh, also for the other answer (crossing lightwaves), that answer helped me a lot in the way as confirmation how it see it.

    Still reading the SRT and I will come back later (after weeks) with this subject (my wrong diagram). For now it is difficult (for me) to agree with you .. I have to find out more ... but "time" in daily life is a problem for me generally ..
     
  18. Oct 12, 2011 #17
    Gwh, thinking about you say here, is it not true that every object B with a higher speed as object A is always going slower in time (is that a misunderstanding from me) ? (Δt' = γ . Δt in Einstein, A and B in their own rest frames)

    So this is also true for water waves except that the speed of a water wave seen from A and B is not constant as it is for lightspeed.
     
  19. Oct 13, 2011 #18
    In my attachments I have proofed my statement.

    With and without Lorenz transformations.

    In timediagram 1 seen by oberserver 1 + seen by observer 1 for observer 2, in timediagram 2 seen by observer 2.
    Both in/from their own rest frames.

    So if an observer 2 (B) has more speed than an observer 1 (A), observer 2 sees the light wave slower passing and his/her time goes slower.

    So you can explain to somebody, sees time as an passing light wave in thought, if you go faster, the light wave in thought will be slower passing you, so time will go slower. If
    the light wave in thougt is passing you with maximum time, you are standing still in the universe, if the light wave stands still for you, time stands still for you. I think
    that lightspeed represents our understanding of time.

    In my second attachment I have something strange in a Lorenz equation, I get V = -V, anybody an explanation ?
     

    Attached Files:

  20. Oct 15, 2011 #19
    In time_e2.jpg I had changed t1 and t2 by mistake, that was the reason.

    In time_e1.jpg :

    it seems that space time diagram 1 was correct, for diagram 2 I was thinking for B (in his own rest frame) to be sure but gives the same result (dividing by γ, the length contraction).
    So without Lorenz is the time dialation already visible in diagram 1.

    I hope that I am thinking completely wrong, than I learn at most.

    Conclusion:

    If you are standing still somewhere in the universe, time flies at maximum speed, if you move times flies too but slower.
     
  21. Oct 16, 2011 #20

    ghwellsjr

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    I've been trying to tell you that your ideas only work for one observer. It is true that if B is traveling with respect to A, then A will determine that B is experiencing time at a lower rate than A. But the same thing is true in the opposite direction: as far as B is concerned, A is the one that is traveling (in the opposite direction) and B will determine that A is experiencing time at a lower rate than B. How does your diagram illustrate this point?
     
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