How to Prove the Energy-Momentum Tensor Identity?

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Homework Help Overview

The discussion revolves around proving an identity involving the energy-momentum tensor, specifically relating an integral of the second time derivative of a volume integral of a product of coordinates and density to an integral of the energy-momentum tensor itself.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy-momentum and the use of Stokes' Theorem in their attempts to manipulate the integrals involved. There are questions about the correctness of applying Stokes' Theorem and integration by parts.

Discussion Status

Some participants have provided alternative approaches, such as suggesting integration by parts to simplify the expressions. Others express uncertainty about their current methods and seek clarification on specific steps.

Contextual Notes

There is mention of boundary conditions affecting the application of Stokes' Theorem, with assumptions about the behavior of the energy-momentum tensor at the boundaries of the volume integral being discussed.

WannabeNewton
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Homework Statement


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

Homework Equations


The Attempt at a Solution


[itex]\partial _{t}T^{t\nu } = -\partial _{i}T^{i\nu }[/itex] from conservation of energy - momentum. [itex]\partial_{t}\partial_{t}(T^{tt}x^{j}x^{k}) =(\partial_{t}\partial_{t}T^{tt})x^{j}x^{k}[/itex] 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 [itex]\partial_{t}\partial_{t}(T^{tt}x^{j}x^{k}) =(\partial _{i}\partial _{m}T^{im})x^{j}x^{k}[/itex] and by using the product rule on [itex]\partial _{i}\partial _{m}(T^{im}x^{j}x^{k})[/itex] to solve for the right hand side of the previous equation I get [tex]\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}[/tex] and this is where I am stuck. I don't know if what I am doing after this is exactly correct. For instance, [tex]-2\int_{V}\partial _{i}(T^{ij}x^{k})dV = -2\int_{\partial V}(T^{ij}x^{k})dS_{i} = 0[/tex] as per Stoke's Theorem and because [itex]T^{ij}[/itex] 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 [itex]T^{ik}[/itex] also in the parentheses and for the first expression I did [tex]\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[/tex] for the same reason as before so that [itex]\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[/itex]. 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:
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try using the fact that

[tex]\int d^3 x T^{ij} = \int d^3 x [ \partial_k (T^{ik} x^j ) - ( \partial_k T^{ik} ) x^j ][/tex]
 
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.
 

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