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Why is invariant interval invariant?

  1. Nov 11, 2009 #1
    The invariant interval is defined to be
    [tex]\Delta {s^2} = \Delta {x^2} + \Delta {y^2} + \Delta {z^2} - {c^2}\Delta {t^2}[/tex]
    and despite which inertial frame we are in, [tex]\Delta s[/tex] for two particular events would be the same.
    If I use Lorentz transformation, this can be proved easily. But is there any more "intuitive" way to verify the invariance? Like when (x,y,z,t) describes propagation of light, it'll be trivially true,e.g. [tex]\Delta s=0[/tex] because of the principle of constancy of light velocity.
    But what about other cases, when two events don't lie on the same light cone? Of course [tex]\Delta s[/tex] is not 0 but still remains invariant, but how to convince myself it's true without doing the arithmetic manipulation of Lorentz transformation?
     
  2. jcsd
  3. Nov 11, 2009 #2

    DrGreg

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    It's not exactly intuitive, but you might find this post relevant.
     
  4. Nov 11, 2009 #3
    It's easy. A space-like interval is just a length, distance in the frame of reference where delta_t=0. I.e., it is a tangible physical thing. In other RFs you can obtain its value too if you make the neccesary combination of delta_t' and delta_x'.
    The same is valid for a time-like interval. It is just one hour. It is normal that this result can be obtained in other RFs - they also have means to recalculate their observations into your system. It would be highly frustrating not to be able to have a rule to get such an information exchange between physical RFs.
     
    Last edited: Nov 11, 2009
  5. Nov 11, 2009 #4

    clem

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    Think of the Lorentz T as a rotation in Minkowski space. Then the length \Delta S doesn't change.
     
  6. Nov 11, 2009 #5

    Dale

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    That is my preferred visualization also. Then it comes down to realizing that in Minkowski space "lengths" are defined by hyperboloids instead of circles. That realization also makes it clear why you cannot rotate a timelike interval into a spacelike interval and also why a future-directed timelike interval cannot be rotated into a past-directed timelike interval.
     
  7. Nov 12, 2009 #6
    Thank you guys, but I think the rotation point of view is also derived from LT. Actually I was asking if we can skip the LT and verify the invariance just by two postulates.
     
  8. Nov 12, 2009 #7

    Dale

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    No, that is certainly not possible. The term "invariant" means that it does not change under the Lorentz transform. So it is certainly not possible to say that something does not change under a transform without defining the transform (other than by tautology).
     
  9. Nov 12, 2009 #8
    We can look at it without LT. Let us see a stick moving fast along us. It bears a label "WalMart". We can make two measurements of its "length". One measurement will give a contracted length and another can give the length in our RF if the stick were still (its proper length = interval). The latter case needs special experiment but no LT. Just two values: delta_x and delta_t. The numerical value of the interval answers the question about the stick length at rest in our RF if we would buy it at WalMart. So you may think of invariance of the interval as of invariance of the stick length from WalMart however fast you move after buying it.
     
    Last edited: Nov 12, 2009
  10. Nov 12, 2009 #9

    Meir Achuz

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    You have it backwards. The physical principle is that going to a moving coordinate system corresponds to a generalized rotation in space-time. The LT can be derived from that.
     
  11. Nov 13, 2009 #10
    But still, is it possible to skip the explicit formula of LT, and verify the invariance.
    I mean, certainly invariance only makes sense under some transformation. But now we know the real transformation is not Galilean, but something can make the speed of light constant. So I want to ask, is it possible to verify the invariance under this "real transformation" pretending we don't know such thing as Lorentz formula?
     
  12. Nov 13, 2009 #11
    That is a good way to think of it, space-like region gives the proper length and time-like gives proper time, but still I can't see why if I don't apply the Lorentz T explicitly
     
  13. Nov 13, 2009 #12

    Dale

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    Not that I know of (except tautologically).
     
  14. Nov 13, 2009 #13
    We can understand it from an ordinary relativity and ambiguity of measurements.

    For example, if a direct measurement of the proper length can be done by arranging a ruler along the stick, in a distant but still RF you with your ruler will obtain a different value because of the perspective law. You have to transform your data with help of a certain rule to get the proper length (which is, by the way, is visible on the first ruler).

    Similarly with the interval - you should not think of its invariance at all but of the proper length solely. Think of proper length as of certainty, not as of invariance.

    delta_x and delta_t (your data) squared in the interval are not the LT. The interval formula is just a rule of transforming your measurement data to learn the proper length: there is no relative velocity involved in the interval!

    The only "postulate" needed to assure you is that it is always possible to do.
     
    Last edited: Nov 13, 2009
  15. Nov 13, 2009 #14
    Hi Bob, I'm totally lost by your post 13, can you explain a bit more? And what do you mean by " a distant but still RF"
     
  16. Nov 13, 2009 #15
    A distant but still RF is a reference frame resting at some distance from the stick with its own ruler attached. No relative velocity. You just use your eye, your ruler and your hand (the stick is out of reach). So it looks shorter, isn't it? Knowing the distances - from the stick to you, the ruler readings, and your hand length, you calculate the proper stick length.
     
  17. Nov 13, 2009 #16

    Dale

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    In SR reference frames have infinite spatial extent, so there is no such thing as a reference frame being located some distance from a specific object. I think you simply mean a reference frame where the object is at rest at some location other than the origin.

    Having said that can you show how your idea of measuring lengths using similar triangles results in the formula for the invariant interval without using the Lorentz transform? It is certainly not clear to me that it is possible to derive the interval formula that way.
     
  18. Nov 13, 2009 #17
    It does not result in the formula for the interval. I never said that! I said usual (daily) relativity, not special relativity. I meant to say that measuring something (in any physical situation) in a RF different from the proper RF needs recalculation or transformation of the data to get the proper values. The proper values are certain. Any proper value is unique. Interval is just a proper value of time or distance. It is invariant because it is unique, certain, tangible.
     
    Last edited: Nov 13, 2009
  19. Nov 13, 2009 #18

    Dale

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    I don't think that answers the OP's question.
     
  20. Nov 13, 2009 #19

    DrGreg

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    I disagree...
     
  21. Nov 13, 2009 #20
    kof9595995,

    You are right that light rays in inertial frames obey the differential equation ds²=0.
    Therefore ds²=0 in one inertial frame R should imply ds'²=0 in any other inertial frame R'.
    This is only possible if there is this kind of transformation of the element from frame R to frame R':

    [1] . . . ds'² = a(V) ds²

    where V is the relative velocity of R' with respect to R.
    More complicated relations should be excluded based on the uniformity of space-time.

    Consider now a frame R" with the opposite relative velocity with respect to R.
    We should have:

    [2] . . . ds"² = a(-V) ds² .

    We expect to have ds'² = ds"² because of the symmetry and therefore

    [3] . . . a(V) = a(-V)

    In addition, the transformation from R' to R should be based on the opposite velocity (-V) as compared to [1]:

    [4] . . . ds² = a(-V) ds²

    Therefore by combining [1] and [4]:

    [5] . . . a(V)a(-V) = 1

    Combining [3] and [5] leads to a(V) = 1.

    The relation [1] defines a group.
    The symmetry conditions lead to a very trivial group!

    Michel
     
    Last edited: Nov 13, 2009
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