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

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

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

    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.

    AB
     
  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:

    [tex]
    \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
    [/tex]

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

    -Matthew
     
  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}
    [/tex].​


    Q.E.D. (I believe)
     
    Last edited by a moderator: Aug 3, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: GRAVITAION, by MTW, Exer. 2.6
  1. MTW wrong ? (Replies: 2)

  2. MTW questions (Replies: 12)

Loading...