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Single equation for space-time curvature?

  1. Dec 16, 2013 #1
    Is there a single equation that can model both spacial and temporal metric contraction simultaneously? And also what's that equation that can model the actual degree of curvature n space-time that uses trig functions and how do you use that in combination with the two previously mentioned equations to model the dimensional metric contraction caused by curvature?
     
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  3. Dec 16, 2013 #2

    PeterDonis

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    Can you give a specific example of a situation in which "spatial and temporal metric contraction" occurs?
     
  4. Dec 16, 2013 #3
    Uh...ok...gravity? Traveling over 1% the speed of light (observably, anyway)? You put a ruler stick in a higher gravity field and from an outside frame of reference it looks smaller while simultaneously clocks run slower. You travel near the speed of light you are in a 4-dimension-ally rotated frame such that your ruler appears smaller and your clock is slower to an outside observer that measures you traveling near the speed of light.
     
    Last edited: Dec 16, 2013
  5. Dec 16, 2013 #4

    K^2

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    The thing about General Relativity is that you can do a general coordinate system transformation, and still be able to get a metric out of it. There isn't really a formula for a specific transform you can do, because it's just whatever you need. You do have to figure out the new metric, of course. But the way you do that is by making sure that all possible scalar products are preserved.
     
  6. Dec 16, 2013 #5
    Well I mean you rotate a space-time coordinate system a certain amount to observe a certain specific metric contraction though right? Let's say both axis rotate 10 degrees from 20 miles away. What's are the new metric contractions from an observer 20 miles away?
     
  7. Dec 16, 2013 #6

    Nugatory

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    You are describing two different effects governed by two different sets of equations.

    The gravitational case is described by the Einstein field equations, which specify how spacetime is curved and how this curvature is represented in the metric. This is general relativity.

    The length contraction and time dilation effect has no curvature effects and the metric doesn't change, and is described by the Lorentz transformations of special relativity.

    Neither can be described in a single equation in the way that (I think) you're asking for.

    The Einstein field equations can be written as ##G=8\pi{T}##, but neither ##G## nor ##T## are simple variables. They're multi-component tensors and that innocent-looking equality is actually a compact representation of a collection of shriekingly difficult non-linear differential equations.

    The Lorentz transforms can be written as a single matrix equation, but that matrix equation is a compact representation of the four transformation equations, one for each coordinate, that you see more often in introductory texts. By choosing the relative velocity in the direction of one of the axes, you can get it down to two equations, but that's about it.
     
  8. Dec 16, 2013 #7

    Nugatory

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    Interesting that you're talking in terms of "rotation".... The Lorentz transformations are not rotations.

    Long ago SR was sometimes taught using a formalism in which the ##t## coordinate was imaginary, an ##ict## term appeared in various equations, and the Lorentz transforms in Minkowski space had the same formal appearance as a rotation in Euclidean space. This approach has pretty much fallen out of favor because spacetime is not Euclidean and trying to understand SR in terms of Euclidean rotations is a dead end, something you have to unlearn, before you can tackle GR.
     
  9. Dec 16, 2013 #8
    Well I mean I was looking up stuff like this http://www.phy.olemiss.edu/HEP/QuarkNet/time.html
    http://www.phy.olemiss.edu/HEP/QuarkNet/length.html
    and while I was looking something up I ound more complicated equations to model the angle o curvature, literally the actual "angle", theta, modeled using a sine function. And so, I was wondering how exactly can you combine the two different views to say "length and time metrics contract because space-time 4 dimensiontally rotates. Rotating space-time at a certain angle will give you a certain metric contraction".


    Yeah it's not a Euclidean rotation, but it's a 4-dimensional rotation...you can't just use polar coordinates because both axis of the coordinate system are rotating...and those axis represent a total of 4 dimensions...
    And I think that's how Einstein described it as one point, he said gravity was a fictitious force and that actually a gravitational field is an accelerated frame of reference caused by the rotation of space-time.
     
  10. Dec 16, 2013 #9

    PeterDonis

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    These are two different kinds of phenomena; traveling fast is not the same as being in a gravity field. See below.

    Does it? How would you measure this? There is a difference between this and the situation with clocks--see next comment.

    This can be directly measured: have two observers exchange round-trip light signals, one far out in empty space and another deep in a gravity well. The second observer will see fewer ticks of his clock elapse between successive round-trip light signals than the first observer.

    This is not the same as being in a gravity well. First, with respect to clocks, there is no way to directly measure the clock effect the way I described it in the gravity case above. (If you disagree, please describe explicitly how you would measure it.) But that's not as important as the next point.

    The more important point is, unlike the case with gravity, both effects, even if we can settle on how they are to be physically measured, are symmetric: if you and I are in relative motion, you see my clocks run slow and my rulers contracted, and I see your clocks run slow and your rulers contracted. But the two observers I described above in the gravity case are *not* symmetric with respect to their clocks: the observer far out in empty space sees the clock of the observer in the gravity well run slow, but the observer in the gravity well sees the clock of the observer far out in empty space run *fast*. So the analogy you are trying to draw between the two kinds of phenomena fails.
     
  11. Dec 16, 2013 #10

    K^2

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    Boosts are hyper-rotations. A lot of the algebra is similar, but distinction is important.
     
  12. Dec 16, 2013 #11

    PeterDonis

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    Yes.

    No. If nothing else, a real gravitational field is not uniform: for example, the "acceleration due to gravity" of the Earth points in different directions at different points. But a "rotation of spacetime", at least as you're using the term, must be in the same direction everywhere.

    What you *can* say is that, in a sufficiently small region of spacetime (small enough that tidal effects--the effects of changes in the magnitude or direction of the gravity field--are negligible), being at rest in an accelerated frame is equivalent to being at rest in a gravity field. That's what Einstein meant when he said that gravity was a fictitious force. But this equivalence only holds locally, not globally.
     
  13. Dec 16, 2013 #12

    pervect

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    There are 10 simultaneous partial differential equations involving the 10 metric coefficients g_ij, and their first and second partial derviatves with respect to time and space. This is the "left side" of Einstein's equation,

    G_ij = 8 pi T_ij

    The capital G_ij is not the same as the lower case g_ij G_ij is the "Einstein tensor". The equation above becomes a partial differential tensor equation when one expands each of the G_ij in some basis in terms of the g_ij and their first and second partial derivatives.

    In a vacuum, the right hand side of the equation, i.e. all the T_ij are zero. More generally the T_ij represent the matter distribution.

    By some counts there would be 16 g_ij, as i and j can each take on the values 0,1,2,3. But because of the symmetry condition g_ij = g_ji, there are only 10 "different" metric coefficients.

    While there are 10 equations, they're not independent, there is one "constraint equation", so there are actually 9 indepenent equations and 9 variables. Similar remarks can be made about the right hand side, T_ij.

    The 10 g_ij are the metric coefficients - I'm not sure why you refer to them as "metric contractions", that's not a standard term. Possibly you are thinking of something else when you ask about "metric contraction" (perhaps related to time dilation?), but it's not clear to me what you might be thinking of. I'm concerned that you may have intended to asked a different question than the one I answered due to semantic issues.

    The metric coefficients are a quadratic form that gives the square of the "distance" (more precisely the Lorentz interval) between two points.

    Thus, if you have (t,x,y,z) as coordinates, there will be some expression for the square of the lorentz interval, ds^2

    ds^2 = g_00 dt^2 + 2 g_01 dt dx + 2 g_02 dt dy + 2 g_03 dt dz + g_11 dx^2 + 2 g_12 dx dy + 2 g_13 dx dz + g_22 dy^2 + 2 g_23 dy dz + g_33 dz^2

    the 10 numbers (g_00, g_01, g_02, g_03, g_11, g_12, g_13, g_22, g_23, g_33) are the metric coefficients.

    ds^2 could be the square of a distance, or the square of a proper time, depending on your sign conventions.
     
  14. Dec 16, 2013 #13

    K^2

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    Local equivalence is all that matters for there to be a fictitious force, though. That might be what you meant, but I wanted to clarify.

    It also just occurred to me that the confusion with rotation and gravity might have come from a centrifuge analogy. If you write Minkowski metric in polar coordinates, and then go to a rotating frame of reference, you will get "gravity" terms that are equivalent to Centrifugal and Coriolis forces.
     
  15. Dec 16, 2013 #14
    it's very strange you say that because I thought you're suppose to be able to treat acceleration and gravitational field strength the same because of the whole gravity being fictitious force thing...
    http://en.wikipedia.org/wiki/Equivalence_principle

    (Inertial mass) \ (Acceleration)) = (Intensity of the gravitational field) \ (Gravitational mass))


    Uh yeah I'm pretty sure that's how it works, I've had more than one person tell me that. You measure it simply by seeing that a ruler is smaller than it should be and that a clock is slower than it should be. They've measured that clocks go faster with very precise clocks in above the Earth...


    But the point is, if I accelerate near the speed of light, an outside observer would observe the same time dilation as if I were in an intense gravitational field...

    Ok, I do disgree, http://en.wikipedia.org/wiki/Time_dilation#Experimental_confirmation
    read the "experimental confirmation" section.

    But...if both people are moving at the same speed, their clocks and rulers will agree with each other, but disagree with an outside frame of reference, and the same is true of them being in a gravitational field...

    Alright do you have a link that can more precisely and in a more organized way display the equations you are trying to show? When I say metric contraction I am trying to generalize length contraction and time dilation simultaneously, since you would measure time passing slower via a meter of time appearing to shrink.
     
    Last edited: Dec 16, 2013
  16. Dec 16, 2013 #15

    WannabeNewton

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    There is some merit to what you're trying to say but what you have actually stated is inaccurate. Allow me to make it more rigorous. In Minkowski space-time, an arbitrary boost can be represented by the vector field ##\xi^{\mu} = 2E^{[\nu}u^{\mu]}x_{\nu}## where ##E^{\mu}## is a space-like vector field, ##u^{\mu}## is the 4-velocity of an inertial observer, and ##x^{\mu}## is the position vector field relative to this inertial observer. If ##\eta^{\mu}## is a constant vector field on Minkowski space-time then it can be shown (with little effort) that ##\mathcal{L}_ {\xi} \eta^{\mu} = 0## if and only if ##\eta^{\mu}## is orthogonal to both ##u^{\mu}## and ##E^{\mu}##, where ##\mathcal{L}_{\xi}## is the Lie derivative along ##\xi^{\mu}##. What this means is that boosts generated by ##\xi^{\mu}## rotate all codimension 2 submanifolds to which ##u^{\mu}## and ##E^{\mu}## are tangent. In this sense the boost is within the plane determined by ##u^{\mu}## and ##E^{\mu}## and rotates this plane. If we had a simple boost along the ##x##-axis of this inertial observer's frame then this plane would correspond to the ##t##-##x## plane.

    In contrast, an arbitrary rotation can be represented by ##\xi^{\nu} = \epsilon^{\mu\nu\alpha\beta}x_{\mu}u_{\alpha}B_{\beta}## where ##B^{\mu}## is also a space-like vector field. It can similarly be shown with little effort that ##\mathcal{L}_{\xi}\eta^{\mu} = 0## if and only if ##\eta^{\mu} = \alpha u^{\mu} + \beta B^{\mu}## hence the rotations generated by ##\xi^{\mu}## rotate all codimension 2 submanifolds to which ##u^{\mu}## and ##B^{\mu}## are orthogonal. And in this sense ##B^{\mu}## determines the axis of rotation.

    This is a nonsensical statement.
     
  17. Dec 16, 2013 #16

    Nugatory

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    Traveling fast at a constant velocity is a very different thing than acceleration, so the equivalence principle between gravitation and acceleration doesn't contradict PeterDonis.
     
  18. Dec 16, 2013 #17
    Ok so there you use the word rotation, a different type of rotation though. How do I use this to described metric contraction in an x,y,z,t coordinate system?



    No it's completely sensical you just obviously don't agree that it's true.

    So what if you indefinitely "accelerated" to approach the speed of light? And what about the equivalence between the time dilation and length contraction? Does it mean nothing? Just pure coincidence?
     
  19. Dec 16, 2013 #18

    WannabeNewton

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    Yeah, a textbook on general relativity would be a logical place to start. It would also clear up the numerous other confusions you seem to be having about general relativity.
     
  20. Dec 16, 2013 #19

    WannabeNewton

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    Yes, you're clearly right. I know nothing about general relativity, sorry for trying to correct you.
     
  21. Dec 16, 2013 #20
    Like...measuring time in meters? Did Einstein not postulate that time was a dimension? You think time has never been measured in meters?

    No idea why you're getting sarcastically defensive, just take a break. Maybe it's because Einstein's models were more accurate than Newton's.
     
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