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GR: body at rest

  1. Jul 9, 2009 #1
    Consider a 4-coordinate system [tex]x=(x^0,x^1,x^2,x^3)[/tex], [tex]x^0[/tex] plays a role of time, [tex]x^1, x^2, x^3[/tex] are some kind of space coordinates. In what follows greek letters will be 1,2 or 3, latin will be 0,1,2,3, [tex]c = 1[/tex], proper time squared [tex] ds^2 = g_{ik}(x) dx^{i} dx^{k}[/tex].

    Now, if the body is "at rest" we have
    [tex]x^{\alpha}={\rm const}[/tex],
    [tex] dx^{\alpha} = 0 [/tex],
    so space components of the 4-velocity are zeroes
    [tex] u^k = (dx^0/ds, 0, 0, 0)[/tex]
    and space components of the contravariant 4-momentum [tex]p^k = m u^k[/tex] are zeroes as well.
    But [tex]p^{\alpha}=0[/tex] is not equivalent to [tex]p_{\alpha}=0[/tex]:
    [tex]p_{\alpha}=g_{\alpha k}p^k = g_{\alpha 0}p^0=g_{\alpha 0} g^{00} p_0 + g_{\alpha 0} g^{0 \beta} p_{\beta}[/tex],
    so we have a system of the form (with non-trivial solutions in general)
    [tex] (\delta^{\beta}_{\alpha} - g_{\alpha 0} g^{0 \beta}) p_{\beta} = g_{\alpha 0} g^{00} p_0[/tex].
    On the other hand, momentum may be considered as a gradient of the action [tex]S(x)[/tex]:
    [tex]\vec{p}=\nabla S(x)[/tex],
    energy as minus time derivative of [tex] S(x) [/tex]:
    [tex] E = -\partial S/\partial x^0 [/tex],
    in other words [tex]p_k = -\partial S/\partial x^k = (E, -\vec{p})[/tex].
    So we are having a very interesting situation when body is at rest ([tex] dx^{\alpha} = 0[/tex]), but has a non-zero momentum (that is not constant!).

    Are my thoughts correct?
     
  2. jcsd
  3. Jul 11, 2009 #2
    Maybe I'm wrong, but [tex]-\vec{p}[/tex] in this case should be zero. It's the 3-d momentum that for a body at rest is zero. In fact you should have only the energy of the body at rest E=m.
    I remember that we have the 4-momentum conservation (only the covariant vector) if the metric doesn't depend on (or from I don't kown, I'm not a native english speaker...) some coordinates. If the metric doesn't depend on (from) all the coordinates (flat) we have that also the controvariant 4-momentum is conserved.
    If the body is at rest you are in a local inertial frame of reference, where the space-time is flat (only in one point but it doesn't matter now), hence you should find the 4-momentum conservation.
    I repeat, maybe I'm wrong.
     
  4. Jul 12, 2009 #3

    DrGreg

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    For what it's worth, you might like to look at a question I posted on a closely related topic, Energy & momentum in non-Minkowski coordinates. I never got a satisfactory answer, but that was entirely my fault for not replying to the answers I did get. Follow some of the other links in that thread, and see if you can make better sense than I did at the time!

    I think the problem may be related to whether spacetime is "static" or not. And if you are taking the gradient of an action, should this be the coordinate derivative or the covariant derivative?
     
  5. Jul 12, 2009 #4
    I was not hoping to get a reply... Thanx guys :)

    The question comes when we are not in local inertial reference frame. But here is what i thought... If take a simple case of stationary metric - metric on a rotating disk with [tex]x = (t,r,\varphi,z)[/tex] ([tex]t,r,z[/tex] have the same sense as to inertial observer with cylindrical coordinates) then
    [tex]ds^2 = (1-(\Omega r)^2) dt^2 - dr^2 - 2 \Omega r^2 dt d\varphi -r^2 d\varphi^2 - dz^2[/tex]
    if I didn't screw up [tex]p_2 = -\Omega r^2 p_0/(1-(\Omega r)^2)[/tex]. As you said, in local inertial reference frame energy is just mass so [tex]m = p_{k}u^{k}=p_0 u^0 = p_0/\sqrt{g_{00}}[/tex] hence [tex]p_0 = m\sqrt{1-(\Omega r)^2}[/tex], and at last
    [tex]p_2 = -\Omega r^2 m/\sqrt{1-(\Omega r)^2}[/tex].
    If we consider [tex]p_2[/tex] as a component of [tex]-\vec{p}[/tex] it has just the same value as angular momentum component of the rotating body in inertial reference frame. And still [tex]p^2=0[/tex]. Very interesting coincidence =)

    It doesn't matter because action is a scalar.
    I briefly looked through your links and didn't find the answer too =) I'll keep looking deeper...
     
  6. Jul 13, 2009 #5
    Surely you are more qualified than me.
    But let me understand...
    If you want a body at rest I don't understand why you choose a rotating disk. Let's take the simplest case, when the rotation is on a plane. Hence dz=0 but the body rotates so d(phi) isn't zero and so the body isn't at rest (when you put the center of the frame of reference, for example, in the orbit's center).
    If we want a body at rest I think we should choose the simplest case: to put the center of the frame of reference on the body. In this case r=0 and so you should find p_0=m and p_2=0.
    I hope this makes sense.

    Bye
     
  7. Jul 13, 2009 #6
    The body is at rest with respect to the rotating disk. [tex]d\varphi = 0[/tex] because [tex]\varphi[/tex] has sense of angle with respect to the rotating observer (perhaps =)), I didn't write it clearly, my fault. For an inertial observer the angle would be [tex]\varphi + \Omega t[/tex].

    Of course, what you say to be the simplest way is quite true, but in GR we can choose any reference frame :)
     
  8. Jul 22, 2009 #7
    So you put your center on the disk... ok
    In this case you are moving at a distance r from the body, that is constant because both the center and the body are moving on the disk.
    If also omega is constant you should find that p_2 (and obviously p_0) is constant...
    So... what is omega?
     
  9. Jul 24, 2009 #8
    [itex]\Omega[/itex] is angular frequency of the disk =) with respect to inertial observer. [itex] p[/itex] is constant.
     
  10. Jul 24, 2009 #9
    Ok, so for a constant motion all is constant...
    Forgive me but I have forgotten the point, what is the problem? :)
     
  11. Jul 25, 2009 #10
    The problem is that the body is at rest but has a non zero momentum %)
     
  12. Jul 27, 2009 #11
    Let me ask you this:
    how can you justify the non-rotational motion with your choices? In this particular case you put the origin on the disk and, if I'm not wrong, you shouldn't find a rotational motion, but, considering there isn't relative velocity between the body and the observer (that is the field acts in the same way in all circumference's points), you should find a body at rest seen by a "translated" observer.
    I ask you this because I'm in trouble finding a good sense interpretation of angular frequency in this case.

    Bye
     
  13. Jul 28, 2009 #12
    I don't understand you? sorry %/
     
  14. Jul 29, 2009 #13
    Oh I'm sorry, I'm not a native speaker and I wrote the last post without dictionary... sorry...
    I'm in trouble finding the meaning of omega in this case.
    In fact your method is obviously correct when the center of the frame is the center of the disk (or circumference if you prefer) that is rotational motion.
    But you put the center on the disk and, if the fields acts in the same way for all the points of the disk (that is there isn't relative velocity between the body and the observer), you should find not a rotational motion, but a body seen at rest by a "translated" (the distance is constant) observer.
    I hope you understand now...
    Bye
     
  15. Jul 29, 2009 #14
    Omega has a meaning of angular frequency of the rotating frame with respect to inertial frame. It's a "property" of the non-inertial frame, not the body.

    The center of the rotating frame is the center of the disk. The body we are looking at is somewhere at the edge of this carousel :) It is at rest with respect to the rotating frame. However the frame itself is rotating at angular frequency Omega with respect to inertial observer (someone standing and looking at kids on the carousel).
     
  16. Aug 7, 2009 #15
    Ok mate, so there is no problem :)
    This problem looks like this:
    does an electron in free-fall in a gravitational field emit?
     
  17. Aug 7, 2009 #16
    Looks like your electron is in pure free-fall, so There's no absorption but there is emission; but as you said in your next post, it does not emit with respect to the observer in free fall. How is this possible?
     
    Last edited: Aug 7, 2009
  18. Aug 7, 2009 #17
    Yes, if the observer is in free-fall he doesn't see emission (although the particle is accelerated)
     
  19. Aug 7, 2009 #18
    This is classical mechanics, the body itself is at rest but not the masses that make up the body. The world line of the body would pass through the time axis but not the world lines of the rotating masses
    Reframe your equations for dx^(alpha)=0 to get the four momentum of the body (E,0)
     
  20. Aug 8, 2009 #19
    I got this interesting and strikingly good pagehttp://www.mathpages.com/HOME/kmath528/kmath528.htm" [Broken]
    I couldn't understand the Wheeler-Fenyman absorber theory
    There's a mistake in rep after eq(2)
    w^4sin^2(wt)
    From the equtions it is clear that a purely accelerating charge does not radiate, there needs a change in acc, but acc is said to do the work
    as has been elegantly described from the math
    P= -2e^2/3c^2(dx/dt) (d^3x/dt^3)
    Here there's no acc
    But it is shown that acc does most of the work from this:
    [d/dt(dx/dt)(d^2/dt^2)]=(d^2x/dt^2)^2 + (dx/dt)(d^3x/dt^3)
    and considering a rotating charged particle
     
    Last edited by a moderator: May 4, 2017
  21. Aug 9, 2009 #20
    There's a well-disguised mistake,the rotational consideration needs w not to be constant for a derivative of acc to exist
    The coefficients of all sines and cosines with powers of w have to be replaced by corresponding powers of tdw/dt +w
     
    Last edited: Aug 9, 2009
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