How to Prove the Energy-Momentum Tensor Identity?

Click For Summary
SUMMARY

The discussion focuses on proving the energy-momentum tensor identity, specifically the equation \(\frac{1}{2}\frac{\mathrm{d} ^{2}}{\mathrm{d} t^{2}}\int_{V}\rho x^{j}x^{k}dV = \int_{V}T^{jk}dV\). Participants explore the application of conservation of energy-momentum and Stokes' Theorem in their attempts to manipulate the tensor equations. Key insights include the correct application of integration by parts to eliminate surface terms and the importance of boundary conditions for the tensor \(T^{ij}\). The conversation highlights common pitfalls in applying Stokes' Theorem and emphasizes the need for clarity in the manipulation of tensor calculus.

PREREQUISITES
  • Understanding of tensor calculus, specifically energy-momentum tensors
  • Familiarity with Stokes' Theorem and its applications in physics
  • Knowledge of integration techniques, including integration by parts
  • Basic concepts of conservation laws in physics, particularly energy and momentum conservation
NEXT STEPS
  • Study the application of Stokes' Theorem in tensor calculus contexts
  • Learn about integration by parts in the context of vector fields and tensor analysis
  • Review the derivation and implications of the energy-momentum tensor in general relativity
  • Explore advanced topics in tensor calculus, including divergence and curl operations
USEFUL FOR

This discussion is beneficial for graduate students in physics, researchers working on general relativity, and anyone involved in theoretical physics focusing on tensor analysis and conservation laws.

WannabeNewton
Science Advisor
Gold Member
Messages
5,848
Reaction score
552

Homework Statement


Show that \frac{1}{2}\frac{\mathrm{d} ^{2}}{\mathrm{d} t^{2}}\int_{V}\rho x^{j}x^{k}dV = \int_{V}T^{jk}dV.

Homework Equations


The Attempt at a Solution


\partial _{t}T^{t\nu } = -\partial _{i}T^{i\nu } from conservation of energy - momentum. \partial_{t}\partial_{t}(T^{tt}x^{j}x^{k}) =(\partial_{t}\partial_{t}T^{tt})x^{j}x^{k} since the x's are fixed coordinates of their respective volume element inside the source. So using the equality from conservation of energy - momentum I get \partial_{t}\partial_{t}(T^{tt}x^{j}x^{k}) =(\partial _{i}\partial _{m}T^{im})x^{j}x^{k} and by using the product rule on \partial _{i}\partial _{m}(T^{im}x^{j}x^{k}) to solve for the right hand side of the previous equation I get \partial_{t} \partial_{t}(T^{tt}x^{j}x^{k}) = \partial _{i}\partial _{m}(T^{im}x^{j}x^{k}) - 2\partial _{i}(T^{ij}x^{k} + T^{ik}x^{j}) + 2T^{jk} and this is where I am stuck. I don't know if what I am doing after this is exactly correct. For instance, -2\int_{V}\partial _{i}(T^{ij}x^{k})dV = -2\int_{\partial V}(T^{ij}x^{k})dS_{i} = 0 as per Stoke's Theorem and because T^{ij} has to vanish at the boundary of the source so that the pressure differs smoothly from the source to the outside but I don't think I applied Stoke's Theorem correctly here. I did the same with the T^{ik} also in the parentheses and for the first expression I did \int_{V}\partial_{m} \partial _{i}(T^{im}x^{j}x^{k})dV = \frac{\mathrm{d} }{\mathrm{d} x^{m}}\int_{V}\partial _{i}(T^{im}x^{j}x^{k})dV = \frac{\mathrm{d} }{\mathrm{d} x^{m}}\int_{\partial V}(T^{im}x^{j}x^{k})dS_{i} = 0 for the same reason as before so that \int_{V}\partial _{t}\partial _{t}(T^{tt}x^{j}x^{k})dV = \frac{\mathrm{d} ^{2}}{\mathrm{d} t^{2}}\int_{V}\rho x^{j}x^{k} = 2\int_{V}T^{jk}dV. Could anyone tell me where and how I used Stoke's Theorem wrongly here and how I am supposed to correctly use it in the context of this problem? Thanks in advance.
 
Last edited:
Physics news on Phys.org
try using the fact that

\int d^3 x T^{ij} = \int d^3 x [ \partial_k (T^{ik} x^j ) - ( \partial_k T^{ik} ) x^j ]
 
I tried that but I couldn't find any to use it to get to the answer? Could you show me how you would use it for one of the expressions or something along the lines? Thanks for the reply.
 
I have attached the section of my cambridge notes covering this formula
 

Attachments

Oh thank you very much. I didn't even think of integration by parts lol. Talk about a much simpler way of getting rid of surface terms.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K