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GRAVITAION, by MTW, Exer. 2.6

  1. Jul 27, 2010 #1
    (This is not homework. I am reading MTW on my own.) I have worked on this and worked on this and cannot see how to solve it. It is Exercise #2.6, on page 65 of MTW. I quote verbatim:

    "To each event Q inside the sun one attributes a temperature T(Q), the temperature measured by a thermometer at rest in the hot sun there. Then T(Q) is a function; no coordinates are required for its definition and discussion. A cosmic ray from outer space flies through the sun with 4-velocity u. Show that, as measured by the cosmic ray's clock, the time derivative in its vicinity is

    [tex]\frac{dT}{d\tau} = \partial_u T = <dT,u>[/tex].​

    In a local Lorentz frame inside the sun, this question can be written

    [tex]\frac{dT}{d\tau} = u^\alpha \frac{\partial T}{\partial x^\alpha} = \frac{1}{\sqrt(1 - v^2)} \frac{\partial T}{\partial t} + \frac{v^j}{\sqrt(1 - v^2)} \frac{\partial T}{\partial x^j}[/tex].​

    Why is this result reasonable?"


    The temperature gradient inside the sun would be [tex]\frac{dT}{dt}[/tex]. But wrt the cosmic ray, the temperature gradient is [tex]\frac{dT}{d\tau}[/tex], which is the above equation. No? I guess I am asking what am I supposed to show?
  2. jcsd
  3. Jul 27, 2010 #2
    You're asked to show from coordinate-free viewpoint (and it is quite simple to show) that

    \frac{dT}{d\tau} = <dT, u>

    I don't know what they mean by "reasonable".
  4. Jul 27, 2010 #3
    I think this is reasonable because the expression for the time derivative of T turns out to largely depend on the first term at the lower speeds and actually reduces to the non-relativistic case by putting [tex]v=0[/tex]. Other than this, I think there wouldn't be any "reasonable" answer to their question.

  5. Jul 28, 2010 #4
    Well, the reasonable part is more of a conceptual answer. It's basically testing you if you understand the equation.

    But my concern was the first bit - where you say it is "simple." My problem is that it seems tautological!

    If I write out the derivative I get the bottom expression. So how does one show the top expression without using the bottom expression... in "coordinate free" language??

    Am I being clear - I see both expressions, top and bottom, as being identical.
  6. Aug 1, 2010 #5
    I think what they want is this:

    \frac{dT}{d\tau}=\frac{dT}{dx^\alpha}\frac{dx^\alpha}{d\tau}=\frac{dT}{dx^\alpha}u^\alpha=\partial_{\boldsymbol{u}}T=\langle \boldsymbol{d}T,\boldsymbol{u}\rangle

    The reason I love MTW is that you can never tell if a problem is infuriatingly easy or impossible.

  7. Aug 1, 2010 #6
    I did that but these expressions are supposed to be from two different points of view - one from the rest frame of the sun and the other from the cosmic ray's moving frame. But what you did, and I did this also thus why I still don't get it, is put the two equations together.

    We have different ideas about books. I think they should explain - and so teach. (I mean to say, that's how I see what the above equation is - and thus my ignorance.)
    Last edited: Aug 2, 2010
  8. Aug 3, 2010 #7
    Re: GRAVITAION, by MTW, Exer. 2.6 SOLUTION

    I think I have a solution for this problem. It came to me this early AM.

    In the sun's frame:

    Suppose the particle was moving with non-rel. speed, v, then the temperature gradient along that direction would simply be:

    v o[tex]\nabla T(Q)[/tex],​

    namely, the directional derivative.

    But in this case the speed is relativistic, |v| <~ c. So u = [tex](\gamma_v,\gamma_vv)[/tex] and so the directional derivative becomes Eq. (2.37):

    [tex]\gamma \frac{\partial T}{\partial t} + \gamma v^j \frac{\partial T}{\partial x^j}[/tex],​

    which equals [tex]u^\alpha\frac{\partial T}{\partial x^\alpha} \equiv \frac{dT}{d\tau}[/tex].

    In the particle's frame:

    The particle "sees" a number of (hyper-)planes of temperature levels called dT (the temp. grad. 1-form). So, for the particle moving with 4-velocity u, the directional derivative is the number of piercings of u with the 1-form dT, or <dT, u>. Thus,

    [tex]<dT,u> \equiv \partial_u T[/tex].​

    Finally, since [tex]\partial_u f \equiv \frac{df}{d\lambda}[/tex] with [tex]f \rightarrow T[/tex], and [tex]\lambda \rightarrow \tau[/tex] we have Eq. (2.36):

    [tex]<dT,u> \equiv \partial_u T = \frac{dT}{d\tau}

    Q.E.D. (I believe)
    Last edited by a moderator: Aug 3, 2010
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