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I Is there any way to derive the time dilation formula?

  1. Sep 27, 2016 #1
    I've already read the derivation in which we use the light pulse clock kept in a spacecraft such that the light pulse follows a zig-zag motion due to motion of the spacecraft being perpendicular to motion of the light pulse. Then, we apply Pythagoras theorem to derive the formula.
    BUT this seems like derivation in a special case, not a universal derivation. I mean there won't be any zig-zag motion of light pulse if it is sent in the same direction as the motion of the spacecraft, so we can't apply Pythagoras theorem in that case to derive the same formula.
    I just need a derivation which doesn't involve any zig-zag motion of light pulse or any other special case. A derivation which just derives the formula for the relationship between proper time and dilated time interval between two events due to relative motion of reference frames.
     
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  3. Sep 27, 2016 #2

    vanhees71

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    That's simple. The proper time of a moving particle is defined by (using natural units with ##c=1##):
    $$\mathrm{d} \tau=\sqrt{\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}}=\mathrm{d} t \sqrt{\eta_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} t}}=\mathrm{d} t \sqrt{1-\vec{v}^2}.$$
    Thus you have
    $$\frac{\mathrm{d} t}{\mathrm{d} \tau}=\frac{1}{\sqrt{1-\vec{v}^2}}.$$
     
  4. Sep 27, 2016 #3

    PeroK

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    It's a special case of a clock. But, you only need to show time dilation applies in the case of one reliable clock and the conclusion is inevitable: that time itself is dilated. The only alternative is that there is something wrong with your clock. The reason a light clock is used is that one of the postulates of Special Relativity is that the speed of light is constant in all inertial reference frames. So, measuring the time light takes to travel a known distance is perhaps the only reliable clock at this stage of the theoretical development of relativity.

    The problem if the light is shining in the direction of motion is that you also have length contraction to take into account. Until you've worked out length contraction, you don't know the relative lengths and you cannot immediately deduce time dilation alone. There is, therefore, an important point about the light clock: how do you know that the height of the clock (i.e. distance perpendicular to motion) is the same for both observers?

    For example, if length were contracted perpendicular to motion for one observer, then that might equally explain the constant speed of light without time dilation.

    You might like to think about this question: why can there be no length contraction for distances perpendicular to the motion?
     
  5. Sep 27, 2016 #4

    stevendaryl

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    It can be derived using Doppler shifts plus the relativity principle (equivalence of inertial frames) plus the fact that the speed of light is constant.

    First, the nonrelativistic Doppler formulas: If you are sending light signals between two observers moving away from each other at relative speed [itex]v[/itex], then the frequency is shifted lower by a factor of [itex]\frac{1}{1+\frac{v}{c}}[/itex], in the case where the sender is moving and the receiver is at rest. If the sender is at rest and the receiver is moving, the frequency is shifted lower by a factor of [itex]1-\frac{v}{c}[/itex]. So nonrelativistically, there is a distinction in the Doppler formula for the two cases. Relativistically, there can't be a difference, since motion is relative; you can't say which one of the sender or receiver is moving. This is the clue to developing the time dilation factor.

    Suppose that we are in a frame where Alice is at rest, and Bob is moving away from Alice at velocity [itex]v[/itex]. Every [itex]T[/itex] seconds (according to Alice's clock), she sends a light signal toward Bob, and every [itex]T[/itex] seconds (according to Bob's clock), he sends a light signal toward Alice. Let [itex]r[/itex] be the rate of Bob's clock, as measured by Alice. We can show that [itex]r[/itex] cannot be equal to 1, and in fact, has to be given by the time dilation factor.

    If Alice sends a signal once every [itex]T[/itex] seconds, according to her clock. Since Bob is moving away from Alice, every signal has farther to travel than the last, so the time between signals as the arrive at Bob, is greater than [itex]T[/itex]. Alice computes that they arrive at Bob every [itex]\frac{T}{1-\frac{v}{c}}[/itex] seconds. However, if Bob's clock is running slow by a factor of [itex]r[/itex], then he will measure the time between signals to be shorter: every [itex]\frac{r T}{1-\frac{v}{c}}[/itex] seconds.

    Now, let's consider the signals Alice receives from Bob. Bob sends the signals once every [itex]T[/itex] seconds, according to his clock. But since his clock is running slow by a factor of [itex]r[/itex], then according to Alice, he is sending the signals once every [itex]\frac{T}{r}[/itex] seconds. Since Bob is moving away from Alice, the signals take longer and longer to arrive at Alice, so (again skipping the derivation), the time between signals as they arrive at Alice is not [itex]\frac{T}{r}[/itex], but [itex]\frac{T (1+\frac{v}{c})}{r}[/itex]

    So, Alice measures signals coming from Bob once every [itex]\frac{T (1+\frac{v}{c})}{r}[/itex] seconds, while Bob measures signals coming from Alice once every [itex]\frac{r T}{1-\frac{v}{c}}[/itex]. By the relativity principle, these numbers should be the same (because their situations are exactly analogous). So we must have:

    [itex]\frac{T (1+\frac{v}{c})}{r} = \frac{r T}{1-\frac{v}{c}}[/itex]

    Solving for [itex]r[/itex] gives:

    [itex]r = \sqrt{1-\frac{v^2}{c^2}}[/itex]
     
  6. Sep 27, 2016 #5
     
  7. Sep 27, 2016 #6

    PeroK

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    You're fundamentally misunderstanding time dilation. It's time dilation, not clock dilation. If time dilates, it dilates. Full stop. Shining a light in a different direction is not going to change the laws of physics.

    You only have to prove it one way. One proof is enough. It doesn't mean that relativity is not proven until someone comes up with a thought experiment that proves time dilation using an hourglass!
     
  8. Sep 27, 2016 #7
    I completely understand the fact that if time dilation is proved for a light clock, then it is proved for any other clock or any other thing which is affected by time. I'm saying that the way it is proved for the light pulse clock seems like a special case to me. I posted this thread to find a proof which does not assume the motion of light pulse relative to the spacecraft in some suitable direction, like assuming it is perpendicular to the motion of the spacecraft. And, if for light pulse traveling in the same direction as spacecraft, the expression for length contraction is required, then I want a proof for length contraction which does not involve any assumptions.
     
  9. Sep 27, 2016 #8

    vanhees71

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  10. Sep 27, 2016 #9

    PeroK

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    Perhaps someone else can do that for you. But, you know, for me it would be a monumental waste of time. In the direction of the ship is also a special case. What you might ask for next is to prove it for light with an arbitrary velocity vector in the x-y-z directions. Or, perhaps, when the light clock is rotating within the spacecraft?

    The fact that the light clock is stationary with respect to the ship is also a special case.

    One solution is to learn SR and prove it for yourself!

    Anyway, you never considered the question I asked in post #3 about length contraction in the perpendicular direction. Explaining that is a valid step in confirming the normal derivation time dilation.
     
  11. Sep 27, 2016 #10

    Ibix

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    If you are happy with the derivation for the transverse light clock and you are happy with the notion that what is true for one clock is true for all then you have a proof for a light clock in a general orientation.
     
  12. Sep 27, 2016 #11
    Thanks, man. This does seem like the real proof.
     
  13. Sep 27, 2016 #12
     
  14. Sep 27, 2016 #13
    I think that makes sense. So, if the people on spacecraft have two clocks: one light clock in perpendicular orientation and one in any other orientation. And, if they do not agree, then the first principle is violated. Thanks, man. Now, I'm considering the question raised by PeroK: How do we know that the time dilation on the light clock was actually due to time slowing down and not due to perpendicular length contraction?
     
  15. Sep 27, 2016 #14

    PeroK

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    @Prem1998 I'm asking you to prove it! It's not as hard as you think. Consider two people moving towards each other. If they are not the same height, then ...?
     
  16. Sep 27, 2016 #15
    I don't have a deeper understanding of relativity. I'm guessing that if they're not of the same height then, is it violating principles of relativity? Can you give me more hints?
     
  17. Sep 27, 2016 #16

    PeroK

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    If they are the same height. Or, better still, each has a vertical metre stick. Then, when they collide, which metre stick is shorter?
     
  18. Sep 27, 2016 #17
    If you're assuming they're the same height, just moving towards each other, then why're you asking which one is shorter on collision?
     
  19. Sep 27, 2016 #18

    PeroK

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    Are you saying that both metre sticks must be the same height in both reference frames? Why? Why can there be no length contraction in this direction?

    Perhaps they both measure the other's metre stick as shorter than their own?
     
  20. Sep 27, 2016 #19
    So, does perpendicular length contraction happen?
     
  21. Sep 27, 2016 #20

    PeroK

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    It can't. The logic is as follows:

    Assume the bottom ends of the metre sticks are lined up and collide. Let's take the stick on the left. Where does the top of that stick collide with the other. If it collides lower than the top of the one on the right, then it is shorter. But, that would be a valid measurement in both frames. Likewise, if it was above the top of the stick on the right, then it would be longer in both frames.

    Any length contraction in the perpendicular direction would have to be absolute: one would be shorter in both frames. But, with the assumption than neither is absolutely moving, there's no reason that either must be shorter than the other.

    The only conclusion is that they must be the same length of ##1m## as measured in both reference frames.

    That then opens the door to use Pythagoras in the light clock experiment. You know the perpendicular length of the light clock is not changed by the relative motion.
     
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