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Show that ds^2 = ds'^2

  1. Mar 14, 2008 #1
    Every time I try to learn special relativity the statement is made that ds^2 = ds'^2. This time I am trying to derive it. And so here is where I get stuck. (I use 'D' as the Greek capital "delta".)

    Accepting Postulate 2 ('c' is constant in all inertial frames), and that the front of a flash of light travels with speed v = c in the original S frame, then:

    vx2 + vy2 + vz2 = c2

    To an inertial observer in the S' frame, then they write the equation as:

    v'x2 + v'y2 + v'z2 = c2

    or, in terms of the coordinates:

    (Dx/Dt)2 + (Dy/Dt)2 + (Dz/Dt)2 = c2


    (Dx'/Dt')2 + (Dy'/Dt')2 + (Dz'/Dt')2 = c2

    Since Dt is the same interval for all three components of speed in S,
    and Dt' is the same interval for all three components of speed in S',
    then we can write:

    (Dx)2 + (Dy)2 + (Dz)2 = c2Dt2


    (Dx')2 + (Dy')2 + (Dz')2 = c2Dt'2

    Subtracting the "cdt" terms from their respective equations this yields:

    c2Dt2 - (Dx)2 - (Dy)2 - (Dz)2 = 0​


    c2Dt'2 - (Dx')2 - (Dy')2 - (Dz')2 = 0​

    Here is the exact place where I get stuck. How can I equate these? (Yes, they are both zero, but is that a valid argument? Are any two things, which are equal to zero, themselves equal? E.g. does: 1 - cos2 - sin2 = Ds2??)

    Thanks in advance for any help you may offer.
  2. jcsd
  3. Mar 14, 2008 #2


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    Of course. But this equation is only valid if the intervals are zero, you still have to show the identity for arbitrary intervals. I don't know how to do that without using the transformation equations.
  4. Mar 14, 2008 #3


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    Hi Living_Dog! :smile:

    What you've (succcesfully) proved is only:
    The separation of two events on the path of a ray of light is always 0.​

    You need to consider the separation of a general pair of events.

    Start with a ray of light going from event A to a mirror B, and then to event C.

    Can you prove that s(A,C) and s´(A,C) are the same? :smile:
  5. Mar 14, 2008 #4
    The transformation equations ... meaning the Lorentz transformation? Shouldn't it be simpler than that??

    Thanks Ich

    hey tiny_tim!

    When you say two general events, do you mean e.g. a flash of light (event A) and a balloon popping (event B)? What's the point of the mirror?

    BTW, the text says that the wave front moving makes and interval " ds " and someone in an inertial frame sees this same interval as " ds' ". Then they say "it follows" that the two are equal. So that is why I approached the problem this way.

    I'll think about general intervals.

    Thanks once again for the help.
  6. Mar 15, 2008 #5


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    Hi Living_Dog! :smile:

    Yes, basically, saying "light always moves at c" is the same as saying "along a ray of light, [tex]x^2 + y^2 + z^2[/tex] always equals [tex]ct^2[/tex]" - and therefore along a ray of light [tex]ds^2\,=\,0[/tex] - and so [tex]ds^2\,=\,ds'^2\,=\,0[/tex]
    Suppose a ray goes from a stationary flashlight at (0,0,0,0) to a mirror at (c,0,0,1) and then back to the flashlight at (0,0,0,2).

    The separation between (0,0,0,0) and (0,0,0,2) is 4c^2: but is it obvious that it's also 4c^2 to another observer who doesn't regard the flashlight as stationary?

    I agree it's mathematically necessary - I just don't think it's obvious.
  7. Mar 15, 2008 #6


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    In fact, if the flashlight were moving with a velocity v, you'd expect the speed of light to range anywhere between c - v and c + v (depending on the angle under which you look at it). This is also what we would expect, based on our (low-velocity) intuition - in fact it is what people did expect for a long time. The entire point of special relativity is that this is not so, and the light always moves with speed c. If this were obvious, special relativity wouldn't have been such a turning point in our understanding of physics.

    If you want to show that the spacetime interval is the same for all observers in all generality, you should consider an event taking place at a position (t, x, y, z) for a "stationary observer" and calculate the ds2 for this observer. Then do a(n arbitrary) Lorentz transformation to a moving frame, in which the event takes place at (t', x', y', z') and show that the corresponding interval (ds')2 is equal to ds2. The case ds2 = 0 (events on the light cone) is just a special case of this, then.
  8. Mar 15, 2008 #7


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    Yes … but Living_Dog wants a proof without Lorentz transformations, based directly on the constancy of the speed of light! :smile:
  9. Mar 15, 2008 #8
    Yes, tiny_tim is right - that is what I am trying to show. (Although I see your point.) But I want to show it sans the "fancy" math - as if I have to help an undergraduate in a modern physics course who hasn't any linear algebra.

    Starting from Postulate 2: 'c' is constant in all inertial reference frames,
    show that:
    ds2 = ds'2.​

    I will try your suggestion now that I understand it: (0,0,0,0) -> (0,1,0,0) -> (0,0,0,0) = 4c2. What is it in a frame moving at a speed v?

    Last edited: Mar 15, 2008
  10. Mar 15, 2008 #9
    Concerning your last question: Can you equate two quantities which are both equal to zero ?

    In mathematics, yes. In physics, that depends. In your example, equating the two quantities would give dimensional inconsistencies (a pure number = (length)² ???).
  11. Mar 15, 2008 #10
    The unit of the terms on the left are m2. :)

    IOW, I can make the 1, 1 m2, and the same for the amplitudes of the trig terms.
  12. Mar 16, 2008 #11

    Ken G

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    I think the problem here is that a simple "equation" by itself is not enough to carry the full logic of a situation, so it should not be expected to do so. That's why students who only memorize the equations always fail the test. So I think you need to step back from the equations and ask yourself what you are really trying to do, what is the logic. It sounds like, you are trying to use the fact that the speed of light must be the same in two frames to derive a general expression that takes on the same value in both frames. That's very much the spirit of relativity-- we assume we know some results as seen by one observer, but we want to know what another observer will see, without asking them. To do this, you need some kind of "rock" that will be true in both frames. So you've done that, you've shown that a certain general form will give zero (for light) in either frame. If you now equate them, you are not making some philosphical remark on all things that are zero, you are simply saying that you have shown those two forms give the same answer in different frames, in one special case.

    But using only the logic thus far, you haven't shown that it works for everything, it just works for light. Indeed, if you multiply the primed equation by 2, it still gives zero, and can be equated to the unprimed zero-- but the result will then only be true for light (i.e., "null" geodesics), and you will have violated the "same form" constraint. So there's additional logic there-- you seek a truth for light, that is written in the same form in all frames, and hope that this guides you to an invariant for things other than light. I think that's what is bothering you-- your approach is a necessary part of identifying the invariant, but the logic is not sufficient to establish the invariant other than for light, you also need to assert that you are looking for physics that takes on the same form in all frames.
    Last edited: Mar 16, 2008
  13. Mar 16, 2008 #12
    I think it is almost a definition:
    Lorentz transformations are just rotations in 4D, to see it use the immaginary time formulation----> you pass from M^4 to R^4.... and it is easy to show that a rotation leave the modulus of a vector unchanged. In other words it is an Invariant.

    To prove it you nedd only the formal properties of the inner product....

  14. Mar 16, 2008 #13


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    … Noether's theorem …

    Ah, but does he (apart from a multiplicative factor)?

    Surely the laws of physics - well, the conservation laws at least - are inevitable consequences of the symmetries of the space-time metric (from Noether's theorem)?

    Can't Living_Dog derive the space-time metric directly from the constancy of the speed of light - as he wants to - and then get the laws of physics from that? :smile:
  15. Mar 16, 2008 #14

    Ken G

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    No, he doesn't seem to, and that's the incompleteness he is puzzled about.
    But one could also think of it the other way-- the symmetries of the spacetime metric, which we seek, are derivable from the laws of physics. In other words, relativity doesn't mean Newton's laws were wrong, it means Galilean invariance was wrong. From the point of view of Noether's theorem, we never needed Galilean invariance to find the conserved quantities, it was always a kind of "orphan" to the theory that depended on what symmetries were actually in place, it seems to me.
    As I understand it, the constancy of the speed of light is not sufficient to get the laws of physics, one must also assume additional symmetries that can be expressed in the expectation that the form of the metric look the same in all frames. I think that was what was missing in Living_Dog's analysis, though I'm not exactly sure what the minimum requirements are.
  16. Mar 16, 2008 #15
    Hello Living Dog.

    See pages 10-14 of Schutz, A First Course in General Relativity for a proof of the invariance of the spacetime interval. Such proofs appear in many textbooks on SR eg. Rindler's Special Relativity etc.. It needs, of necessity, a certain amount of mathematics but really no more than a basic knowledge of linear transformations. It can probably be given in simple outline without the mathematics ?

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