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Question about Special Relativity and matching results

  1. Apr 5, 2015 #1
    Assume a light was emitted from the origins of two frames in relative motion.

    Assume the light acquired some position (x,y,z,t) in an F frame. Then map that point using the lorentz transformations to (x',y',z',t') in a F' frame.

    Can anyone prove under all conditions that the the light postulate in the F' frame agrees this lorentz transformations to (x',y',z',t') is the correct answer in its own frame?
     
  2. jcsd
  3. Apr 5, 2015 #2

    mfb

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    I don't understand what exactly your question is, but there is a very general result that gives "yes" for all those questions: the Lorentz transformations form a group (called Lorentz group). No matter how you transform things, you will always get correct physics in all reference frames.
     
  4. Apr 5, 2015 #3
    Yes, can you prove mathematically that the other frames agree with the any answer in the Lorentz group. That is what I am asking.
     
  5. Apr 5, 2015 #4

    pervect

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    If by the light postulate you mean that the speed of light is constant, yes. On any lightcone originating at the coordinate origin in F, a constant speed of light implies that distance^2 = c^2 * time^2, or x^2 + y^2 + z^2 - c^2 t^2 = 0.

    Now, x^2 + y^2 + z^2 - c^2t^2 is known as the Lorentz interval, and algebra will confirm that if you use a generalized boost, the Lorentz interval will not change as a result of applying the transform - it's invariant. This implies that if x^2 + y^2 + z^2 - c^2 t^2 = 0, then x'^2 + y'^2 + z'^2 - c^2 t'^2 = 0 in frame F'. Thus the Lorentz interval is independent of the frame of reference, and so is the speed of light.

    One particularly easy to write proof. consider a standard Lorentz boost in the x direction. Using the concept of rapidity and hyperbolic functions http://en.wikipedia.org/wiki/Rapidity, we can write a Lorentz boost in the x direction as:

    x' = x ##\cosh \theta## + c t ##\sinh \theta##
    c t' = c t ##\cosh \theta## + x ##\sinh \theta##
    (see the wiki article for details)

    Then by direct substitution of the above, x'^2 - c^2 t'^2 = (cosh^2 ##\theta## - sinh^2 ##\theta##) x^2 - c^2 t^2 (cosh^2 ##\theta## - sinh^2 ##\theta##) , using the identity cosh^2 - sinh^2 = 1 we can simplify this to x^2 - c^2 t^2

    It's not necessary to use rapidities if they are unfamiliar or intimidating, but using them makes the formal similarity between rotations leaving distances unchanged in 3d Euclidean space and Lorentz transforms leaving Lorentz intervals unchanged in 4d Minkowskii space.
     
  6. Apr 5, 2015 #5

    mfb

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    That follows from the group structure. Showing that the group is well-formed is a typical homework question for students, I'm sure you can find the derivation in many textbooks.
    Showing that electrodynamics and so on are Lorentz invariant is done in textbooks as well. And I guess you can find websites deriving it, too.
     
  7. Apr 5, 2015 #6
    Yes, but prove the primed frame agrees for all possible circumstances that the Lorentz transforms are correct.

    If it is so simple prove it with the math.
     
  8. Apr 5, 2015 #7

    PeterDonis

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    pervect showed you the math in post #3. You should be able to work from there on your own. We don't mind helping you to understand, but we are not going to post detailed proofs of results that are basic facts of relativity. That's what textbooks are for.
     
  9. Apr 6, 2015 #8

    Orodruin

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    Apart from what has been said already, I think it should be mentioned that leaving the speed of light invariant is essentially the definition of the Lorentz group and you use this to derive the form of the transformations, not the other way around.
     
  10. Apr 6, 2015 #9
    Thanks for your advice on textbooks.

    Anyway, after we have this mapped point using the math as I posted above, we get say (x',y',z',t') from (x,y,z,t) using the lorentz transformations which you call the math.

    And I am not asking for detailed proofs of the lorentz transformations. That is not at all what I asked.

    I asked once the point (x,y,z,t) maps to (x',y',z',t'), does the prime origin for example using the light postulate in its own frame agree (x',y',z',t') is the correct answer for the circumstances.

    This is a simple enough request. This is not requesting a detailed proof. It is simple verification that (x',y',z',t') is the correct answer in the view of the primed frame. It is sort of like checking your work using long division by multiplying the result by the denominator to get to the dividend. Then you know the answer is correct.

    That is what I am asking for here.
     
  11. Apr 6, 2015 #10
    Yes, the speed of light is invariant.

    But, that is not what I am asking.

    And, I know exactly how to derive x'^2 + y'2 + z'^2 = (ct')^2 from x^2 + y^2 + z^2 = (ct)^2 and vice-versa. It is trivial algebra.
     
  12. Apr 6, 2015 #11

    PeterDonis

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    The correct answer to what? If you mean the correct answer to where the light will end up in the primed frame, the answer is yes.
     
  13. Apr 6, 2015 #12
    Once you have the answer (x',y',z',t') from (x,y,z,t) have you ever proved the primed origin agrees with this answer yes or no? I mean, if you are going to make claims about logic in the primed frame, then certainly it is science to request that the primed frame origin for example agrees this is the correct answer.

    Can you demonstrate that (x',y',z',t') is the correct answer for the primed origin that applies the light postulate? As you know, whatever the calculation in the unprimed frame, the primed frame sees a spherical light wave. That calculation must simply agree with the primed origin's light postulate. It is a scientific double check of the lorentz answer.

    The goal of the lorentz transformations is to perfectly match the view of the other frame no?
     
  14. Apr 6, 2015 #13

    PeterDonis

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    I'm not understanding what sort of additional proof you need. Could you give more detail about what exactly you are looking for? For example, an outline of what a proof you would find acceptable would look like?

    The light wave is spherical in both frames.
     
  15. Apr 6, 2015 #14

    Dale

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    Yes. This was done for you already in pervect's post 4 paragraph 2. What more do you want, or what did you not understand?
     
  16. Apr 16, 2015 #15

    Sure, I would like to see a proof that the primed frame origin using the light postulate agrees with the unprimed frame's calculation using the lorentz transformations on the location of the light. So, the unprimed frame origin claims the light is located at a transformed (x'y'z't') under some conditions.

    Now, given the conditions that made the unprimed origin conclude this, under those same conditions, does the primed origin agree that the light is located at (x'y'z't') using its light postulate under those same conditions?

    Does such a proof exist?

    You know, one set of circumstances should lead to (x',y',z',t') using the lorentz transformations for the unprimed frame and also (x',y',z',t') using the light postulate in the primed frame.

    So, condensed

    Unprimed origin
    circumstances->light postulate (x,y,z,t)->LT (x',y',z',t')
    Primed frame origin
    circumstances->light postulate (x',y',z',t')
     
  17. Apr 16, 2015 #16

    mfb

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    Yes it exists, and you'll find it in textbooks. See post #5 for example:
     
  18. Apr 16, 2015 #17

    Ibix

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    Define two frames S and S', where S' moves at velocity v in the +x direction. A pulse of light is emitted from the origin, moving at an angle θ to the x-axis, at the time when the origins of coordinates in the two frames coincide - t=t'=0.

    Working in S, the coordinates of the light pulse at time t=T are ##x=cT\cos\theta## and ##y=cT \sin \theta##. You can substitute those in to the Lorentz transforms to get x', y' and t'. It's simple enough to then show that ##x'^2+y'^2=ct'^2## - in other words, that the transformed coordinates of the light pulse are exactly as far from the origin of the transformed coordinates as light could have travelled in the transformed time. I'll leave the algebra to you.

    Was this what you were looking for? If so, you can get to it much more directly: you can derive the Lorentz transforms from the assumption that the speed of light is constant in all frames; therefore the Lorentz transforms must keep the speed of light constant.
     
  19. Apr 16, 2015 #18
    Really.

    The group structure is based on the Minkowski metric that says, a light beam measures c in one frame iff that same light beam measures c in the other frame.

    The metric does not prove that the transformation produces the same light beam that measures c in the other frame given a set of circumstances and its light postulate. It only produces a light beam that measure c period.

    I am fully capable of proving the Minkowski metric, so that is not at all what I am asking. I assume you know how to do this no?

    Let me simplify it.

    Assume a light beam is emitted up the positive x-axis of both frames when their origins are common in the standard configuration.

    Now, let's set the circumstances.
    Assume the primed coordinate d' is now common with the unprimed origin.
    Where does the unprimed origin conclude the light beam is in the coordinates of the primed frame.
    Given d' and the unprimed origin are at the same place, where does the primed frame conclude the light beam is?

    Are the the same. Note they must be.
     
  20. Apr 16, 2015 #19
    Many thnks.
    But, no that is not what I was looking for.

    Please see my post below yours.
     
  21. Apr 16, 2015 #20

    Ibix

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    So the emission event is (x,t)=(-X,0) - i.e., to he left of the origin (in either coordinate system), with the beam moving to the right. The event you are calling "now" is (x',t')=(0,T') - i.e., at the origin of the primed frame some time after it was emitted.

    Is this a correct description of your set up?
     
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