How Does a Cosmic Ray Measure Temperature Changes in the Sun?

[Living_Dog]

(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

$$\frac{dT}{d\tau} = \partial_u T = <dT,u>$$.​

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

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

Why is this result reasonable?"

--

The temperature gradient inside the sun would be $$\frac{dT}{dt}$$. But wrt the cosmic ray, the temperature gradient is $$\frac{dT}{d\tau}$$, which is the above equation. No? I guess I am asking what am I supposed to show?

 

[hamster143]

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".

 

[Altabeh]

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 $$v=0$$. Other than this, I think there wouldn't be any "reasonable" answer to their question.

AB

 

[Living_Dog]

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".

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.

 

[dodelson]

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.

-Matthew

 

[Living_Dog]

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.

-Matthew

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.)

 

[Living_Dog]

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$$\nabla T(Q)$$,​

namely, the directional derivative.

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

$$\gamma \frac{\partial T}{\partial t} + \gamma v^j \frac{\partial T}{\partial x^j}$$,​

which equals $$u^\alpha\frac{\partial T}{\partial x^\alpha} \equiv \frac{dT}{d\tau}$$.

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,

$$<dT,u> \equiv \partial_u T$$.​

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

$$<dT,u> \equiv \partial_u T = \frac{dT}{d\tau}$$.​

Q.E.D. (I believe)