Understanding Tensors: Is Misner Thorne and Wheeler Enough?

[Demystifier]

E.g., the explanation, why the phase-space distribution function in statistical physics is a relativistic scalar, makes this issue (which confuses even practitioners in the field sometimes) very clear.

Thanks for the tip! But I looked into it and found their derivation, in terms of Lorentz contraction, rather clumsy. Do you know a reference with more elegant derivation? Perhaps a derivation that works even in curved coordinates?

 

[vanhees71]

It works in curved spacetime, because it's a local concept. I don't know, whether this is the same as in MTW, my approach is this: There's the "lab frame" with spacetime coordinates $x=(x^{\mu})$ and on-shell momenta $p=p^{\mu}=(E_p,\vec{p})$. Now we consider the particles with momentum $\vec{p}$ and define $\mathrm{d} N^*$ as the number of particles at time $t^*$ with $\Sigma^*$ the reference frame, where these particles are at rest and write
$$\mathrm{d} N^*=f^*(x^*,\vec{p}^*=0) \mathrm{d}^3 x^* \mathrm{d}^3 p^*.$$
Now we express this very same particles in terms of the quantities in our observational frame $\Sigma$. Because we measure our volume element at fixed time $t$ we have length contraction and thus
$$\mathrm{d}^3 x=\frac{1}{\gamma} \mathrm{d}^3 x=\frac{E}{m} \mathrm{d}^3 x.$$
For the on-shell momenta volume elements you have
$$\mathrm{d}^3 p^*/E^*=\mathrm{d}^3 p^*/m=\mathrm{d}^3 p/E$$
and thus
$$\mathrm{d}N^*=\mathrm{d} N = f(x,\vec{p}) \mathrm{d}^3 x \mathrm{d}^3 p = f(x,\vec{p}) \frac{m}{E} \mathrm{d}^3 x^* \frac{E}{m} \mathrm{d}^3 p^*=f(x,\vec{p}) \mathrm{d}^3 x^* \mathrm{d} p^* \' \Rightarrow \; f^*(x^*,\vec{p}^*=0)=f(x,\vec{p}).$$
In this way it's clear that $f(x,\vec{p})$ is a Lorentz-scalar one-particle phase-space distribution function.

Another argument is that the particle-number four-current,
$$N^{\mu}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 p \frac{p^{\mu}}{E} f(x,\vec{p})$$
must be a vector field, and thus since for on-shell particles $\mathrm{d}^3 p/E$ is a Lorentz scalar and $p^{\mu}$ is a four-vector $f$ must be a scalar.

For GR everything goes through analogously. You only have to write everything in a general covariant way with the corresponding tensor densities.

 

[andresB]

Possibly interesting:
http://www.oobject.com/john-wheeler-and-his-elaborate-blackboard-presentations/
View attachment 289118

That's a dedicated teacher.

 

[robphy]

https://webofstories.com/play/john.wheeler/96 [part of a series]

John Wheeler - Origins of the book 'Gravitation' with Thorne and Misner (96/130)

 

[robphy]

That's a dedicated teacher.

Possibly enlightening:

https://physicstoday.scitation.org/doi/pdf/10.1063/1.3120897
John Wheeler’s mentorship: An enduring legacy
Terry M. Christensen

https://ir.library.oregonstate.edu/concern/graduate_thesis_or_dissertations/8w32r7988
Theoretical physics takes root in America : John Archibald Wheeler as student and mentor
Christensen, Terry M.

[update]
https://arxiv.org/abs/1901.06623
John Archibald Wheeler: A Biographical Memoir
Kip S. Thorne

 

[ohwilleke]

That's a dedicated teacher.

FWIW, what makes sense for a course taught by one of the textbook's authors in a classroom setting and what makes sense for someone studying independently without the backup of lectures, office hours, TAs and study groups from people in the same class, can be quite different.

 

[yucheng]

Why is MTW a pun though? I seriously do not get it.

 

[Frabjous]

Why is MTW a pun though? I seriously do not get it.

It’s so massive that it exerts its own gravitational field or it demonstrates the topic by its weight.

 

[PeterDonis]

It’s so massive that it exerts its own gravitational field or it demonstrates the topic by its weight.

The old joke was that Wheeler wanted the book to undergo gravitational collapse and form a black hole.

 

[Frabjous]

The old joke was that Wheeler wanted the book to undergo gravitational collapse and form a black hole.

I hadn’t heard that one.

 

[PeterDonis]

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