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As I understand it, the proper length of an object is defined as the length of the object in its rest frame. In terms of the metric it is defined as the length of the spacetime interval between two space-like separated events, i.e. $$dl^{2}=\sqrt{ds^{2}}$$ (with the "mostly plus" signature).

Now, suppose that an observer is at rest in an inertial frame that is itself at rest with respect to a given object that the observer wishes to measure. Why is it the case that one considers the "simultaneous" length of the object, i.e. when $dt=0$, such that $$dl^{2}=\sqrt{dx^{2}+dy^{2}+dz^{2}}$$ Is it simply so that it agrees with the definition of spatial distance in Euclidean geometry or are there other intuitive reasons for why it must be the case (analogous to the definition of proper time in which the proper time of an object is equal to the coordinate time of an observer who is at rest [i.e. $dx=dy=dz=0$] with respect to the object, such that $d\tau=\frac{1}{c}\sqrt{-ds^{2}}=dt$)?!

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    $\begingroup$ You either need to replace $dl^2$ with $dl$, or get rid of the square roots. $\endgroup$
    – user4552
    Apr 4, 2019 at 21:10

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When two guys are in the same reference frame then events for both of you occur at the same time coordinate - ie they are simultaneous. What I mean by this is that both guys will ascribe the same time coordinate to when an event A occurred. Of course depending on how far they are from A means they have to use different values for their measurements but they should both agree.

Now imagine a meter stick. In the meter sticks frame of reference let's suppose there are two events. Event A is one end of the meter stick being at x = 1m at t = 0 and event B is the other at x = 2m at t = 0. According to your formula of the space time interval we can find the distance between these two events. Since t = 0 for both A and B in the sticks frame, we then get your formula for the length. (Edit: dt = 0 because the time for both events is t = 0)

Now if I am in the meter sticks frame, according to my first paragraph, A and B should be simultaneous events for me, because these are simultaneous events for the meter stick - and we are in the same frame.

I guess it can be kind of confusing.

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    $\begingroup$ So is it simply that in the rest frame of the stick, its end points can be viewed as two events that occur simultaneously and hence if the observer is at rest with respect to the stick then these two events occur simultaneously for the observer too? $\endgroup$
    – user35305
    Oct 25, 2016 at 7:11
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    $\begingroup$ Yea that is pretty much it $\endgroup$
    – user45815
    Oct 25, 2016 at 10:23
  • $\begingroup$ Ok cool. Can one also approach the problem in terms of and objects rest length, $L_{0}$ defined by Pythagoras's theorem, $dL_{0}=\sqrt{dx^{2}+dy^{2}+dz^{2}}$, and the proper length between two space-like separated events, labelling the endpoints of the object, defined as $dL=\sqrt{ds^{2}}$. Then, the two quantities coincide in the case where the space-like separated points are simultaneous (i.e. dt=0), such that $dL=\sqrt{ds^{2}}=\sqrt{dx^{2}+dy^{2}+dz^{2}}=dL_{0}$?! $\endgroup$
    – user35305
    Oct 25, 2016 at 10:33
  • $\begingroup$ Well Pythagoras's Theorem also works when you aren't in the rest frame of the stick. Both ends of the stick still simultaneously exist no matter which frame you are in. Even if I am in a moving frame S' I can denote event A' as one end of the stick in S' and B' as the other. Then the t' for A' and B' is the same and you can use Pythagoras (with coordinates x',y' and z'). So again this is the simultaneous length in frame S'. However you cannot do this for determining the distance between A' and B as in most cases t' - t isn't 0. Then you need to determine the space-time interval s. $\endgroup$
    – user45815
    Oct 25, 2016 at 16:07
  • $\begingroup$ Just read your comment again. I guess what you are saying is true, but I wanted to clarify that it doesn't have to be the rest length. $\endgroup$
    – user45815
    Oct 25, 2016 at 16:09

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