Energy Momentum Tensor

  • #1
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Consider a system that is weakly gravitating. The metric is slightly perturbed from the corresponding flat metric

i.e. [itex]g_{ab}=\eta_{ab}+h_{ab}[/itex] with [itex]|h_{ab}| \ll 1[/itex]

The system is also non relativistic meaning that time derivatives can be taken to be much smaller that spatial derivatives. This implies that the components of the stress energy tensor can be ordered [itex]|T_{00}| \gg |T_{0i}| \gg |T_{ij}|[/itex]

Under such circumstances, I want to show that stress energy conservation reduces to [itex]T^{\mu k}{}_{,k}=0[/itex]

Well we have to start with our standard GR defn [itex]T^{\mu \nu}{}_{; \nu}=0[/itex]
[itex]T^{\mu \nu}{}_{, \nu} + \Gamma^\mu{}_{\sigma \nu} T^{\sigma \nu} + \Gamma^\nu{}_{\sigma \nu} T^{\sigma \mu}=0[/itex]
[itex]T^{\mu \nu}{}_{, \nu} + \frac{1}{2} \eta^{\mu \rho} \left( h_{\rho \sigma , \nu} + h_{\rho \nu , \sigma} - h_{\sigma \nu , \rho} \right) T^{\sigma \nu} + \frac{1}{2} \eta^{\nu \rho} \left( h_{\sigma \rho , \nu} + h_{ \rho \nu , \sigma} - h_{\sigma \nu, \rho} \right) T^{\sigma \mu}=0[/itex]

Now I don't know where to go with this???
 

Answers and Replies

  • #2
dextercioby
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This time I haven't done the calculations, but if I did them, this is what I would do: in your last expression i would use the fact that T_ij can be discarded, as being the smallest, next I would discard the time derivatives of the metric perturbation (because I can do that as per the assumptions) and of the energy-momentum 4-tensor and eventually would use that the flat metric is diag of (+1, -1, -1, -1) or the other conventions.

Use all things and post your work.
 
  • #3
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This time I haven't done the calculations, but if I did them, this is what I would do: in your last expression i would use the fact that T_ij can be discarded, as being the smallest, next I would discard the time derivatives of the metric perturbation (because I can do that as per the assumptions) and of the energy-momentum 4-tensor and eventually would use that the flat metric is diag of (+1, -1, -1, -1) or the other conventions.

Use all things and post your work.

Thanks for the reply.

So if we discard the [itex]T^{ij}[/itex] terms then that means that at least one of the indices on [itex]T^{\mu \nu}[/itex] must be a zero . So we are left with

[itex]T^{\mu \nu}{}_{, \nu} + \frac{1}{2} \eta^{\mu \rho} \left( h_{\rho 0 , \nu} + h_{\rho \nu , 0 } - h_{0 \nu , \rho} \right) T^{0 \nu} + \frac{1}{2} \eta^{\nu \rho} \left( h_{0 \rho, \nu} + h_{\rho \nu , 0 } - h_{0 \nu , \rho} \right) T^{0 \mu} = 0[/itex]

Then, discarding time derivatives we get
[itex]T^{\mu \nu}{}_{, \nu} + \frac{1}{2} \eta^{\mu \rho} \left( h_{\rho 0 , \nu} - h_{0 \nu , \rho} \right) T^{0 \nu} + \frac{1}{2} \eta^{\nu \rho} \left( h_{0 \rho, \nu} - h_{0 \nu , \rho} \right) T^{0 \mu} = 0[/itex]

Then I rewrote it as (using symmetry on h indices)

[itex]T^{\mu \nu}{}_{, \nu} + \eta^{\mu \rho} h_{0 [ \rho , \nu]}T^{0 \nu} + \eta^{\nu \rho} h_{0 [ \rho , \nu]} T^{0 \mu}=0[/itex]

Then I am confused as to what to do next...
 
  • #4
dextercioby
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The 3rd term of your last equation is 0, right ?
 
  • #5
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The 3rd term of your last equation is 0, right ?

Do we know that [itex]h_{\mu \nu}[/itex] is diagonal?

If so, then [itex]\eta^{\nu \rho} h_{[0 \rho , \nu ]} T^{0 \mu}[/itex]

would require [itex]\rho = 0[/itex] so that we are talking about [itex]h_{00}[/itex]

but then if [itex]\rho=0[/itex], the [itex]\eta^{\nu \rho}[/itex] will force [itex]\eta=0[/itex]

but then [itex]h_{0 [ \rho , \nu ]}[/itex] will be [itex]h_{0 [0,0]}[/itex] which is a time derivative and hence negligable.

Is that logic correct?

So now we are down to just [itex]T^{\mu \nu}{}_{, \nu} + \eta^{\mu \rho} h_{ 0 [ \rho , \nu ] } T^{0 \nu}[/itex]
 
  • #6
dextercioby
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The 3rd term above is not [itex]\eta^{\nu \rho} h_{[0 \rho , \nu ]} T^{0 \mu}[/itex]
, but [itex]\eta^{\nu \rho} h_{0 [\rho , \nu ]} T^{0 \mu}[/itex] which is zero because it's a contraction b/w an antisymmetric tensor (the one with brackets) and the metric of flat spacetime.

For this term, [itex] \eta^{\mu \rho} h_{ 0 [ \rho , \nu ] } T^{0 \nu}[/itex], i would say to keep only the term containing [itex] T^{00}[/itex], because the other can be assumed negligible.

What do you get then ?
 
  • #7
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The 3rd term above is not [itex]\eta^{\nu \rho} h_{[0 \rho , \nu ]} T^{0 \mu}[/itex]
, but [itex]\eta^{\nu \rho} h_{0 [\rho , \nu ]} T^{0 \mu}[/itex] which is zero because it's a contraction b/w an antisymmetric tensor (the one with brackets) and the metric of flat spacetime.

For this term, [itex] \eta^{\mu \rho} h_{ 0 [ \rho , \nu ] } T^{0 \nu}[/itex], i would say to keep only the term containing [itex] T^{00}[/itex], because the other can be assumed negligible.

What do you get then ?

Surely we want to get rid of the [itex]T^{00}[/itex] term as this will set [itex]\nu=0[/itex] and result in a time derivative on the h term - which would be negligable, no?

However, if I follow it through
[itex]\frac{1}{2} \eta^{\mu \rho} ( h_{0 \rho , 0} - h_{00,\rho} ) T^{00}[/itex]
Now the first term has a time derivative which means we ignore it. This leaves
[itex]-\frac{1}{2} \eta^{\mu \rho} h_{00, \rho} T^{00}[/itex]
Now we want [itex]\rho \neq 0[/itex] to avoid a time derivative so pick [itex]\rho = i [/itex] but this then forces [itex]\mu=i[/itex] and we get
[itex]-\frac{1}{2} \eta^{ii} h_{00,i}[/itex]
 
  • #8
dextercioby
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Yes, that term apprently remains. Something must be wrong then. What could we have missed ?
 

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