Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spacetime separation and dilation/contraction

Tags:
  1. Sep 21, 2014 #1
    I'm trying to understand the relationship between the spacetime metric [itex]\Delta s^2 = \Delta x^2 - c^2\Delta t^2[/itex] and the simple formulas for time dilation and length contraction in special relativity [itex]x = \frac{1}{\gamma} \bar{x}[/itex] and [itex]t = \gamma \cdot \bar{t}.[/itex]

    Suppose from an inertial reference frame, I observe an object passing through two points [itex](x_1, ct_1)[/itex] and [itex](x_2, ct_2)[/itex], and I want to consider how these two points appear from a different inertial reference frame. I suppose that a change in reference frame amounts to a particular structure-preserving transformation of Minkowski space, and while I'm not sure exactly which structure ought to be preserved, at least spacetime separation ought to be preserved. (So if the original points were separated by [itex]\Delta s^2[/itex] they should remain so in any reference frame.)

    Depending on the sign of [itex]\Delta s^2[/itex], there is either a reference frame in which the two events occur at the same time, or one in which they occur at the same place. Proper distance is the spatial separation between two events from a reference frame in which they are simultaneous, and proper time is the temporal separation between two events from a reference frame in which they occur in the same place.

    From what I understand, length contraction implies that the spatial separation between two events is never more than the proper distance, and similarly time dilation implies that temporal separation never less than the proper time.

    I have trouble reconciling this with the following: if in one reference frame I measure the separation between the two events as [itex](\Delta x_A, c\Delta t_A)[/itex] and in another as [itex](\Delta x_B, c\Delta t_B)[/itex], then

    $$\Delta x_A^2 - (c\Delta t_A)^2 = \Delta s^2 = \Delta x_B^2 - (c\Delta t_B)^2.$$

    But if the measurements are simulatenous in reference frame A, so that [itex]\Delta t _A = 0[/itex] and[itex]\Delta x_A [/itex] is the proper length, it seems like we get

    $$\Delta x_A^2 = \Delta x_B^2 - (c\Delta t_B)^2$$

    which suggests that the spatial separation [itex]\Delta x_B [/itex] measured in any other reference frame is always longer than the proper length. Something's wrong --- please advise.
     
  2. jcsd
  3. Sep 21, 2014 #2

    Nugatory

    User Avatar

    Staff: Mentor

    Because you're trying to relate the spacetime interval to length contraction, let's consider the two events to be the turning on of two lights, one at either end of a given object, such that the two events are simultaneous in the A frame, in which the object is at rest. You are quite right that ##\Delta x_A^2## will give you the (square of the) proper length in the A frame.

    However, when you go to calculate the contracted length in the B frame you need two flashes that are simultaneous in the B frame, so you can't use these two events. In fact, the distance between them in the B frame is the sum of the contracted length in the B frame and the distance that the other end of the object moves in the B frame during the time interval between the first flash and the second flash.
     
  4. Sep 21, 2014 #3
    That makes some sense. You're telling me that because the (easiest?) way to measure length in a given reference frame is to use two simultaneous events located at the endpoints of the object, different reference frames will in general require different pairs of events for measuring the length of the same object. To put it another way, the length of an object in your reference frame is operationally defined as the proper distance between two simultaneous events located at its endpoints.

    Proper distance is still an invariant: all observers will agree on its value. But what they will disagree on, if they are in different reference frames, is whether the proper distance between those two events is an accurate measure of the object's length (after all, I can only call it an accurate measure of length if the events are simultaneous in my reference frame---and simultaneity is observer-dependent.)

    So the meaning of my earlier formula [itex]\Delta x_A^2 = \Delta x_B^2 - (c\Delta t_B)^2[/itex] is this: if you set up two events for the length-measuring protocol, then from your own vantage point (in A), the events are simultaneous and the distance between them is proper. For someone in another reference frame B, however, the events you set up are non-simultaneous (they are separated by an interval of [itex]\Delta t_B[/itex]) and so the distance between the two events as seen from B ([itex]\Delta x_B[/itex]) is too large to be the proper length of the object.

    If all that is right, consider the following scenario: I set up an experiment to measure the length of an object using simultaneous flashes in my reference frame. I obtain a measurement of [itex]\ell_0[/itex]. A friend in a different reference frame watches my experiment and concludes that I have erroneously measured the length to be [itex]\ell[/itex], which is an overestimate of its proper length in her frame; in her frame, after setting up a similar experiment, she concludes that its length is [itex]\ell^\prime[/itex]. Finally, the velocity difference between our two frames is [itex]\beta[/itex].

    Regarding the four quantities [itex](\ell_0, \ell, \ell^\prime, \beta)[/itex], is it possible to solve for any two quantities given the other two? I know, for example, that the length contraction gives [itex]\frac{1}{\sqrt{1-\beta^2}} = \frac{\ell^\prime}{\ell_0}[/itex], but I haven't been able to figure out how the others are related.

    In particular, can you solve for the velocity difference [itex]\beta[/itex] between our two frames if all you know is the result of the experiment in my frame ([itex]\ell_0[/itex]) and how that same result appears to my friend ([itex]\ell[/itex])?
     
  5. Sep 21, 2014 #4

    Nugatory

    User Avatar

    Staff: Mentor

    Easiest and only way - the length of an object is by definition the distance between its ends at the same time. You can calculate it from other measurements in other frames, especially if you know ##\beta##, but if you want a direct measurement of the length, that distance is what you have to be measuring.

    Actually, it's not a proper length at all. It's a spacetime interval. Those spacetime intervals for which ##\Delta{t}## is zero are called "proper lengths", those for which ##\Delta{x}## is zero are called "proper times", and both are special cases of "spacetime interval".

    You're doing fine. Everything I've said above is minor terminology stuff.

    If I'm understanding your question properly, you need one more piece of information. Any one of your velocity relative to your friend; your velocity relative to the object; your friend's velocity relative to the object; or the object's proper length will suffice. In most realistic problems, one of us will be at rest relative to the object or have a pretty good idea what the proper length is, so we'll have this additional information.
     
  6. Sep 21, 2014 #5
    Thanks for your help.

    To clarify: suppose you ignore the object for now; all you have is a spacetime interval being viewed from two different inertial reference frames. (Maybe there are two lights in particular places being switched on at particular times, for example.) The length of the spacetime interval is invariant, but the spatial/temporal separations will be different in each reference frame: $$\Delta x^2 - (c\Delta t)^2 = \Delta s^2 = \Delta \bar{x}^2 - (c\Delta \bar{t})^2$$

    1. A (tangential) question about inertial reference frames: is it possible, by varying the choice of inertial reference frame, to make the temporal separation between the events appear arbitrarily large in that frame? Or is there some constraint (like light speed, or the kind of allowed transformation) that somehow keeps that from being possible? (Also is the answer different if the two events are constrained to lie on the worldline of some object?)
    2. A clarification of my original question: knowing only ##\Delta x##, ##\Delta \bar{x}##, ##\Delta s^2##, and the fact that the reference frames are inertial, can you compute the difference ##\beta## between the velocities of the two reference frames? If not, what other information would you need?
     
  7. Sep 21, 2014 #6

    Nugatory

    User Avatar

    Staff: Mentor

    If the square of the interval is positive in one frame then it will have the same positive value in all frames, so although ##\Delta{X}## and ##\Delta{t}## will be vary from frame to frame ##\Delta{x}^2## will always be greater than ##\Delta{t}^2##. We refer to the intervals as "space-like"; there exists no inertial frame in which the two events are happen at the same place and one inertial frame in which they happen at the same time (##\Delta{t}=0##).

    If the square of the interval is negative in one frame then it will have the same negative value in all frames, so although ##\Delta{X}## and ##\Delta{t}## will be vary from frame to frame ##\Delta{t}^2## will always be greater than ##\Delta{x}^2##. We refer to the intervals as "time-like"; there exists no inertial frame in which the two events are simultaneous and one inertial frame in which they happen at the same place (##\Delta{x}=0##).

    That's sufficent, you don't need anything more.
     
  8. Sep 21, 2014 #7
    Regarding (1), my question is: by choosing the appropriate intertial reference frame, is it possible to make a spacetime interval seem like it spans an arbitrarily large interval of time (it doesn't matter to me what happens to the spatial separation)?

    Regarding (2), that's surprising! What's an equation relating ##\beta## to ##\Delta x##, ##\Delta \bar{x}##, and ##\Delta s^2##? It doesn't seem obvious to me.
     
  9. Sep 21, 2014 #8

    Nugatory

    User Avatar

    Staff: Mentor

    I'm sorry, I didn't quite answer that question. Yes, if the two events are time-like separated, you can find a frame in which ##\Delta{t}^2## is arbitrarily large. It'll be a frame that is moving very rapidly relative to the frame in which the two events happened at the same place, so also has a very large ##\Delta{x}##.

    Try it. Because ##\Delta{s}^2## is the same in both frames, knowing it and one of ##\Delta{x}## or ##\Delta{t}## for each frame is enough to compute the missing ##\Delta{x}## and ##\Delta{t}## values. That plus the Lorentz transformations and some algebra will see you home. (You will save yourself much grief by choosing to make the coordinates of the first event be (0,0) in both coordinate systems before you start in on the algebra).[/quote][/QUOTE]
     
  10. Sep 21, 2014 #9
    Regarding (1), it shouldn't matter whether the events are timelike-separated, should it, when you're making the temporal interval arbitrarily large rather than arbitrarily small?

    As for (2), I've been trying something similar and hitting a dead end. Starting from
    $$\begin{eqnarray}
    x^2 - (ct)^2 &=& \bar{x}^2 - (c\bar{t})^2\\
    \bar{x} &=& \gamma(x - vt)\\
    c\bar{t} &=& \gamma\left(ct - \frac{vx}{c}\right)
    \end{eqnarray}$$

    I squared the third and fourth equations to get expressions for ##\bar{x}^2## and ##(c\bar{t})^2##.
    I plugged those expressions into the first equation to eliminate ##\bar{x}## and ##\bar{t}##, yielding an equation in terms of ##x##, ##t##, and ##\beta## only:

    $$x^2 - (ct)^2 = \gamma^2 (x - vt)^2 - \gamma^2 \left(ct - \frac{vx}{c}\right)^2$$

    Each side is the difference of two squares and can be factored ##a^2 - b^2 = (a+b)(a-b)##:


    $$(x + ct)(x-ct) = \gamma^2 \left[ \left(x - vt + ct - \frac{vx}{c}\right)\left(x - vt - ct + \frac{vx}{c}\right) \right]$$

    On the right hand side, rewrite ##x = \frac{cx}{c}## and reorganize terms to yield:


    $$(x + ct)(x-ct) = \gamma^2 \left[ \left(ct - vt + \frac{cx - vx}{c}\right)\left(\frac{cx+vx}{c} - (c+v)t \right) \right]$$

    Factor out ##(c+v)## or ##(c-v)## where appropriate

    $$(x + ct)(x-ct) = \gamma^2 \left[ (c-v)(c+v)\left(t + \frac{x}{c}\right)\left(\frac{x}{c} - t \right) \right]$$

    Factor out ##c^2## on the right hand side:


    $$(x + ct)(x-ct) = \frac{\gamma^2}{c^2} (c-v)(c+v)(x + ct)(x - ct).$$

    If the interval is lightlike/null, both sides vanish because of the ##(x-ct)(x+ct)## term (which represents the interval length) and so the equation isn't informative. Otherwise, divide by the interval length ##(x+ct)(x-ct)## to get

    $$1 = \frac{\gamma^2}{c^2}(c-v)(c+v)$$

    which is almost a function of ##\beta##. Expanding out ##\gamma(\beta) = (1-\beta^2)^{-1/2}## and ##v(\beta) = \beta c## to get

    $$\begin{eqnarray}
    1 &=& \frac{\gamma^2}{c^2}(c-v)(c+v)\\
    &=& \frac{1}{c^2(1-\beta^2)}(c-\beta c)(c+\beta c)\\
    1 &=& \frac{1}{c^2(1-\beta^2)}(c-\beta c)(c+\beta c)\\
    1 - \beta^2 &=& (1-\beta)(1+\beta)
    \end{eqnarray}$$

    which is always true, and so isn't useful. Where have I gone wrong?
     
  11. Sep 21, 2014 #10

    Nugatory

    User Avatar

    Staff: Mentor

    Ah, yes, you're right. I'm still not reading your questions carefully enough :-(


    by listening to me, of course. If one of the other science advisors doesn't step in first, I'll check my math tomorrow...
     
  12. Sep 21, 2014 #11
    No worries. I've been trying a number of different ways to derive a function ##\beta = \cal{F}(\Delta x, \Delta \bar{x}, \Delta s^2)##; using the Lorentz transform along with the spacetime metric was just the first thing I tried. I thought about starting from other equations like the length contraction/time dilation formulas, but (a) based on the discussion above, I believe those equations relate the proper lengths of two different spacetime intervals, rather than the same interval viewed from two different inertial frames, which is what we want here, and (b) those contraction/dilation equations ought to follow from the Lorentz transform and so they probably don't lead to any different results than just using the Lorentz transform.

    I'll keep trying different approaches though!
     
  13. Sep 21, 2014 #12

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Where you went wrong is not noticing that delta t can be expressed in terms of delta s and delta x. Similarly, for delta t prime. The delta s is the same. Plugging these into one (either) of the Lorentz transforms gives you a relation of delta x, delta x prime, delta s, beta and gamma. Of course, gamma can be expressed in terms of beta. Then solve for beta.
     
  14. Sep 22, 2014 #13
    Naturally I noticed, but there are several dead ends when deciding which equations to substitute into which others.

    Here is one of my more successful attempts: Start by expressing the Lorentz equation ##\bar{x} = \gamma(x-vt)## in terms of ##\beta##, then re-arrange:

    $$\begin{eqnarray}
    \bar{x} &=& \frac{1}{\sqrt{1-\beta^2}}(x - \beta c t)\\
    \bar{x} \sqrt{1-\beta^2} &=& x- \beta c t \\
    \end{eqnarray}$$

    From the spacetime interval formula, we know that ##ct = \sqrt{x^2 - S^2}##.

    $$\bar{x} \sqrt{1-\beta^2} = x - \beta \sqrt{x^2 - S^2}$$

    Square both sides (this looks problematic if ##\beta=0## or ##\beta=1## or ##x = vt##).

    $$\bar{x}^2 (1-\beta^2) = x^2 - 2\beta x \sqrt{x^2 - S^2} + \beta^2 (x^2-S^2)$$

    This is quadratic in ##\beta##. Reorganizing terms a little gives:

    $$\begin{eqnarray}
    %%\bar{x}^2 - \beta^2 \bar{x}^2 &=& x^2 - (2x \sqrt{x^2 - S^2})\beta + (x^2 -S^2) \beta^2\\
    (\bar{x}^2 + x^2 - S^2)\beta^2 - (2x \sqrt{x^2 - S^2})\beta + (x^2 - \bar{x}^2) &=& 0\\
    \end{eqnarray}$$

    I don't see where, but I may have botched the algebraic manipulations somewhat. One troubling result is that this equation seems to lack symmetry: why are the two solutions of beta not centered about zero (I would have expected a solution of the form ##\pm \beta_0##)? Why isn't the equation obviously (anti)symmetric with respect to interchange of ##x## and ##\bar{x}## --- and can it be manipulated into such a form (I don't immediately see how)? And finally, it's surprising that only two of the equations were needed (although in retrospect, using four produced a trivial result.)


    In any case, the quadratic formula explicitly yields:

    $$\begin{eqnarray}\beta &=& \frac{2x\sqrt{x^2-S^2} \pm \sqrt{4x^2(x^2-S^2) - 4(\bar{x}^2+x^2 -S^2)}}{2(\bar{x}^2 + x^2 -S^2)}\\
    &=&\frac{x\sqrt{x^2-S^2} \pm \sqrt{x^2(x^2-S^2) - (\bar{x}^2+x^2 -S^2)}}{\bar{x}^2 + x^2 -S^2}\\

    \end{eqnarray}$$

    Is that about right?
     
  15. Sep 22, 2014 #14

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    abs(beta) < 1 is required. Note, you can see from your form right after substitution, that beta=0 corresponds to equal delta x, as expected. Beta= 1 is prohibited. As for x=vt, that simply gives you that delta x prime = 0 which means the events correspond to a stationary world line per primed frame and that delta t prime = delta s. You should find that there is only one beta meeting physical constraints for any scenario. [ Actually, if both solutions have abs < 1, then both are possible. An example is delta x = delta S. Then +/- beta produce any valid delta x prime.]

    Your first formula after "quadratic formula explicitly yields" is incorrect. Your quadratic equation prior to this is correct. You should be able to spot the error. After fixing this and simplifying, the result will have a lot more symmetry.
     
    Last edited: Sep 22, 2014
  16. Sep 22, 2014 #15
    Thanks. To clarify: my question is why the result seems asymmetric with respect to ##x## and ##\bar{x}##. Since the problem didn't specify which is the primed frame and which is the unprimed frame, I would expect a result that would have some (anti)symmetry with respect to an exchange of ##x## and ##\bar{x}##, or one which could be put into such a form.
     
  17. Sep 22, 2014 #16

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    See my edited comment. You have an error, which I point out. The correct formula is symmetric, after simplification. I don't want to give you the answer explicitly because you are doing a good job working things out yourself (and I am lazy transcribing from my paper to latex). But it is quite symmetric.
     
  18. Sep 22, 2014 #17
    Gotcha.

    $$(\bar{x}^2 + x^2 - S^2)\beta^2 - (2x\sqrt{x^2 - S^2})\beta + (x^2 - \bar{x}^2) = 0$$

    The quadratic formula explicitly yields (if I've done my arithmetic correctly):
    $$\begin{eqnarray}\beta &=& \frac{ 2x\sqrt{x^2 - S^2} \pm \sqrt{ 4x^2(x^2 -S^2) - 4 (\bar{x}^2 + x^2 - S^2) (x^2 - \bar{x}^2)}}{2(\bar{x}^2 + x^2 - S^2)}\\
    &=& \frac{ x\sqrt{x^2 - S^2} \pm \sqrt{ x^2(x^2 -S^2) - (\bar{x}^2 + x^2 - S^2) (x^2 - \bar{x}^2)}}{\bar{x}^2 + x^2 - S^2} \\
    &=& \frac{ x\sqrt{x^2 - S^2} \pm \sqrt{ x^2(x^2 -S^2) - \bar{x}^2 (x^2 - \bar{x}^2)- (x^2 - S^2) (x^2 - \bar{x}^2)}}{\bar{x}^2 + x^2 - S^2} \\
    &=& \frac{ x\sqrt{x^2 - S^2} \pm \sqrt{ \bar{x}^2(x^2 -S^2) - \bar{x}^2 (x^2 - \bar{x}^2)}}{\bar{x}^2 + x^2 - S^2} \\
    &=& \frac{ x\sqrt{x^2 - S^2} \pm \sqrt{ \bar{x}^2(\bar{x}^2 -S^2)}}{\bar{x}^2 + x^2 - S^2} \\
    &=& \frac{ x\sqrt{x^2 - S^2} \pm \bar{x}\sqrt{ \bar{x}^2 -S^2}}{\bar{x}^2 + x^2 - S^2} \\
    \end{eqnarray}
    $$
    which is as symmetric as I imagined! Very cool.
     
  19. Sep 22, 2014 #18

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    You got it!!! Much better than if I just copied my paper for you.
     
  20. Sep 23, 2014 #19
    Perhaps.

    So earlier we thought that some of the analytic solutions for ##\beta## were physically excluded. If that's still the case now that I've fixed the solution, what physical constraint would exclude them (e.g. if I plug in ##\bar{x}=\frac{x}{2}##, I get two possible answers for ##\beta##, neither of which seems obviously implausible)?
     
  21. Sep 23, 2014 #20

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    As I corrected in post #14, I think multiple solutions for beta are valid unless abs >= 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook