What is the volume of the system when inserting the 4-velocity of the observer?

[StenEdeback]

Homework Statement

I do not understand the mathematical formula under item 1. at all (and not the other items either).

Relevant Equations

The text says: "1. If we insert the 4-velocity u of the observer into one of the slots...". I do not understand at all. What does it mean that a tensor has "vector slots"? What does "dp/dV" mean? I would like to see the operation step by step.

I have attempted but with no result.

 

[anuttarasammyak]

I do not have read the book but as for vector slots I assume that a rank 2 tensor can be regarded as a 4X4 matrix and one of 4 columns or 4 rows is 4-velocity inserted.
As for dp/dV, what are p and V ? Is p momentum? Is V volume of something ?

 

[strangerep]

After Goldsteins eq(13.72) he immediately gives (13.73) as the explicit-component version of the same equation. "Inserting" a vector into a tensor's "slot" means contracting the tensor with the vector. In (13.73), the "contraction" is the implicit summation over the index $\beta$.

But note that there's a typo in (13.73), at least in my copy of the book. He writes $$T^\alpha_{~\beta} u^\alpha ~=~ T_\beta^{~\alpha} u^\beta ~=~ -\left( \frac{dp^\alpha}{dV} \right) ~.$$ The 1st term should have $u^\beta$, not $u^\alpha$.

 

[StenEdeback]

I do not have read the book but as for vector slots I assume that a rank 2 tensor can be regarded as a 4X4 matrix and one of 4 columns or 4 rows is 4-velocity inserted.
As for dp/dV, what are p and V ? Is p momentum? Is V volume of something ?

Thank you!

 

[StenEdeback]

After Goldsteins eq(13.72) he immediately gives (13.73) as the explicit-component version of the same equation. "Inserting" a vector into a tensor's "slot" means contracting the tensor with the vector. In (13.73), the "contraction" is the implicit summation over the index $\beta$.

But note that there's a typo in (13.73), at least in my copy of the book. He writes $$T^\alpha_{~\beta} u^\alpha ~=~ T_\beta^{~\alpha} u^\beta ~=~ -\left( \frac{dp^\alpha}{dV} \right) ~.$$ The 1st term should have $u^\beta$, not $u^\alpha$.

Thank you! I thought so too, though I was a little bewildered by the typo. I will need to do the contraction to see the result. I have tried to do so in my head but not succeeded, probably because I was not sure what "inserting a vector into the slot of a tensor" really meant. Now I can proceed with a little more confidence, using paper and pen when doing the tensor contraction. I do private studies on my own, and sometimes I get stuck, at times for no good reason. Then Physics Forums is my last resort and very valuable to me. Thanks again!

 

[anuttarasammyak]

He writes T βαuα = Tβ αuβ = −(dpαdV) . The 1st term should have uβ, not uα.

Thanks to your quotes. So V has physical dimension of p/Tu, volume if c=1. What volume is it ?

 

[strangerep]

Thanks to your quotes. So V has physical dimension of p/Tu, volume if c=1. What volume is it ?

IIUC, $V$ by itself means the volume of the system, and $dV$ is an infinitesimal volume element, as one would find in an integral over some total volume. E.g., in eq(13.34) on p568, in a section about the energy-stress-momentum tensor and conservation laws, he illustrates with: $$R_\mu ~=~ \int T_\mu^{~0} dV$$as an example of an integral quantity which is conserved by virtue of a continuity equation.

In the exercise mentioned in the OP, we've switched to the relativistic case, so $T_\mu^{~0}$ is generalized to $T^\alpha_{~\beta} u^\beta$. The idea is that an integral over the volume of the system gives the total energy-momentum (IIUC).