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Spacetime intervals again - still figuring out the formula

  1. Apr 20, 2005 #1
    ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2

    I'm still trying to figure it out. What I was thinking, you're always moving through spacetime at c, right? Through time, but that will be less when traveling in spatial demensions. (If any of this is wrong, let me know, I read this somewhere but can't remember where and it's some time ago so my memory might've messed it up or I may simply have read something that isn't true)

    So wouldn't that make the distance traveled in spacetime only dependant on the amount of time? d = ct? I can't think of anything else that would follow the above so the spacetime formula must've come from somewhere else. Where?
     
  2. jcsd
  3. Apr 20, 2005 #2

    pervect

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    I'm not sure what you are "trying to figure out". The formula, just written down by itself, seems to me to be a lot clearer than your explanation of it.
     
  4. Apr 20, 2005 #3

    jtbell

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    In relativity we deal with events that occur at a particular point in space (x, y, z) at a particular time (t). The spacetime interval between two events is

    [tex]\Delta s = \sqrt {c^2 (t_2 - t_1)^2 - (x_2 - x_1)^2 - (y_2 - y_1)^2 - (z_2 - z_1)^2}[/tex]

    That's basically all your equation means. It has nothing to do with motion, specifically. It's written with ds, dt, dx, etc. because we often work with infinitesimal differences using calculus.

    The spacetime interval is important because it has the same numeric value in any reference frame, for a given pair of events. We say it is invariant under Lorentz transformations between reference frames.
     
  5. Apr 20, 2005 #4

    JesseM

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    I'd say the "everything always moves through spacetime at the speed of light" is just a bad explanation of relativity which has been made popular by Brian Greene, but which is more confusing than helpful (the math works out, but the way he defines 'speed through spacetime' is very counter-intuitive). If you want more details this was discussed at length on this thread.
     
  6. Apr 20, 2005 #5
    JesseM,

    Amen!
     
  7. Apr 20, 2005 #6

    quasar987

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    I don't get where the formula using differentials comes from. If I calulate the total differential

    [tex]ds = \frac{\partial{s}}{\partial{t}}dt + \frac{\partial{s}}{\partial{x}}dx [/tex]

    of the function

    [tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

    and square it, I get the ugly

    [tex]ds^2=\frac{c^4t^2dt^2-x^2dx^2-2c^2txdtdx}{c^2t^2-x^2}[/tex]

    So what am I missing?
     
    Last edited: Apr 21, 2005
  8. Apr 20, 2005 #7

    dextercioby

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    He meant the very famous

    [tex] ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu} [/tex]


    Daniel.
     
  9. Apr 20, 2005 #8

    jtbell

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    Where did you get the function

    [tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

    from?
     
  10. Apr 20, 2005 #9

    quasar987

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    I figured

    [tex]s(x,t) = \sqrt{(ct)^2-x^2}[/tex]

    is kinf of

    [tex]\Delta s = \sqrt{(c\Delta t)^2 - (\Delta x)^2}[/tex]

    where t1 = x1 = 0...

    How do you get the expression

    [tex](ds)^2 = (cdt)^2 - (dx)^2 - (dy)^2 - (dz)^2[/tex]

    anyway?
     
  11. Apr 20, 2005 #10

    dextercioby

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    By definition.That's the definition of the relativistic interval between 2 inifnitesimally close points in a flat Minkowski space...


    Daniel.
     
  12. Apr 20, 2005 #11

    JesseM

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    Another thing to notice is that the spacetime interval defined above is just c times the proper time (time as measured on the clock of someone moving along that worldline), [tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex]

    Since all frames must agree on how much time ticks on a clock between two points on its worldline, this shows why it's necessary that [tex]d\tau[/tex], and hence [tex]ds = cd\tau[/tex], must be a frame-invariant quantity.
     
    Last edited: Apr 20, 2005
  13. Apr 21, 2005 #12
    I'm trying to figure out why the formula looks like that. I don't understand. Why are we multiplying dt^2 by c^2? Why are we substracting dx through z^2?
     
  14. Apr 21, 2005 #13

    Ich

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    It´s like pythagoras in a weird space.
    time can be considered with some right to be very closely related to space. Some would say that you could measure time in meters and space in seconds - that meter and second are merely different units for basically the same thing. The conversion factor between those units would then be: 1m = 300000000 m/s * 1s.
    And like pythagoras gives you a distance in normal space, ds gives you a distance in spacetime. Mostly you express this distance in units of meters, so when you have time given in seconds you first have to transform it to meters -> c*t = time in meters.
    The only difference to normal space is that the time component counts as a negative space component, that´s why it´s called Minkowski space instead of euclidean space.
    However this distance has a physical meaning: it equals the time between two events in the frame where they happen at the same place, or the spatial distance in the frame where they happen simultaneous, whichever is appropriate.
     
  15. Apr 21, 2005 #14

    JesseM

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    Like I said, what you're really calculating is the proper time. If you see a clock moving past you at velocity v, you know that if the time elapsed on your own clock between two points is [tex]dt[/tex], and if the time elapsed on that clock between those same two points is [tex]d\tau[/tex], then the two time intervals are related by the time dilation formula:

    [tex]d\tau = dt/\gamma[/tex]
    where [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex]

    And in your coordinate system, v is equal to the distance the clock covered during the time interval [tex]dt[/tex], giving

    [tex]v = \sqrt{dx^2 + dy^2 + dz^2}/dt[/tex]

    So, plugging that into the time dilation formula:

    [tex]d\tau = dt \sqrt{1 - v^2/c^2}[/tex]
    [tex]d\tau = dt \sqrt{1 - (dx^2 + dy^2 + dz^2)/(c^2dt^2)}[/tex]
    [tex]d\tau = dt \sqrt{(1/dt^2)(dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2))}[/tex]
    [tex]d\tau = \sqrt{dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)}[/tex]

    which gives the proper time formula I posted above:

    [tex]d\tau^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2)[/tex]

    And as I said, if you define [tex]ds = cd\tau[/tex], you get the formula for the spacetime interval that you posted, they are pretty much interchangeable (although proper time [tex]d\tau[/tex] has a more obvious physical meaning than [tex]ds[/tex], I'm not sure what the motivation is for ever using [tex]ds[/tex]).
     
    Last edited: Apr 21, 2005
  16. Apr 21, 2005 #15
    quasar,

    In your post #6, I think there are some mistakes in your expressin for ds^2. Doesn't the first term needs a factor of t, the second term a factor of x and the cross term an x and a t? And didn't you forget to square the denominator?

    Or am I missing something?
     
  17. Apr 21, 2005 #16

    quasar987

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    No, that's true.

    I'll edit it.
     
  18. Apr 22, 2005 #17
    Thanks, that helps. The first part of the formula (c^2dt^2) is immediately more clear. I still don't understand, however, why we're substracting the spatial parts. In Pythagoras, you use addition.
    I'm just looking for a geometrical (if one's allowed to call it that) explenation to the formula. Such complex (well, to me, anyway) maths like that of JesseM is no doubt correct and might lead me to better understanding, but it would take me a lot of time. I'm not especially gifted at physics and maths and I haven't studiedthem extensively (I'm still in high school, or whatever it's called in your part of the world).
     
  19. Apr 22, 2005 #18

    quasar987

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    Since c is the speed of light, and dt is the time interval between the events, then c*dt is the distance travelled by light between the time the first event hapened and the time the second event hapened.

    And dx²+dy²+dz² = D², according to pythagoras, is just the spacial distance (squared) separating the position of hapening of the two events.

    We conclude from the substraction of D² from (c*dt)² that if the result is negative (ds² < 0), then it means that (c*dt) < D. That is to say, the distance traveled by light during the event is smaller than the actual distance separating the event. So not even light could have gone from A to B between their hapening. And since nothing can travel faster than light, then nothing in event A could have caused event B.

    If the result is positive (ds² > 0), it means that (c*dt) > D. Then A might be the cause of B.

    If, finally, the result is 0 (ds² = 0), then (c*dt) = D and the only thing in A that could have caused B is the light that was emited from A.
     
  20. Apr 22, 2005 #19
    quasar,

    I just saw the corrections you made to your expression for ds^2 in post #6. They look right.

    Now remember you defined your spatial and time intervals as being from 0 to x and from 0 to t, respectively. But when you calculate the differential ds^2, those intervals become differentials as well. That is x -> dx and t -> dt. if you make those substitutions in your expression, you'll see that your numerator is just the square of your denominator. So your whole expression is equal to what you now have in the denominator. And that's the right answer.

    This one really had me wrapped around the axle for a while; the solution finally hit me while I was jogging yesterday! The way you got into trouble was by using the variables x and t to represent not only VALUES of the domain of S but also INTERVALS in the domain of S. That works fine in algebra and analytic geometry, but in calculus you need different variables for x and dx or you'll always be calculating derivatives at x=0, because the calculation automatically sends dx->0.

    In this case, since there's nothing physically significant about assigning the coordinates (0,0) to your first event, you don't lose any physics as long as you remember that once the derivative is taken x = dx and t = dt.

    Did that help at all, or was it confusing?
     
    Last edited: Apr 22, 2005
  21. Apr 22, 2005 #20

    quasar987

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    Thanks jdavel. I will look seriously into this after my exams, but I do see your point, I just need to really see it for myself. Thanks again!
     
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