- #1
erogard
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Hi, I am trying to show explicitly the isotropy of the stress energy tensor for a scalar field Phi.
By varying the corresponding action with respect to a metric g, I obtain:
[tex]
T_{\mu \nu} = \frac{1}{2} g_{\mu \nu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\nu \Phi
[/tex]
Assuming further that Phi has not spatial dependence (only function of time) and using the Friedmann Robertson Walker metric, which is diagonal, I end up with a diagonal energy tensor:
[tex]
T_{\mu \mu} = \frac{1}{2} g_{\mu \mu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\mu \Phi
[/tex]
where
[tex]
g_{\mu \mu} = Diag \left( 1 , -R(t)^2, -s^2 R(t)^2, -\sin^2(\theta) s^2 R(t)^2 \right)
[/tex]
R(t) being the scale factor.
I then use
[tex]
T_{\mu \nu}' = R_{\mu i} R_{\nu j} T_ij
[/tex]
where R is a 4x4 rotation matrix, with the (1,1) entry being equal to one, and the remaining 3x3 matrix being the usual rotation matrix (here about the z axis).
However, I am unable to recover the original tensor upon performing the above product. For example, I get
[tex]
T_{22}' = \cos^2(\theta')T_{22} + \sin^2(\theta')T_{33}
[/tex]
where I use theta prime (the angle of rotation) to differentiate between theta in the FRW metric, and since T_22 is not equal to T_33 (unless sin theta is one), I don't quite get the same result. Haven't worked out the other diagonal entries yet.
I suspect that I am using a wrong form for the RW metric (maybe wrong coordinate?). Any help would be appreciated. Thanks.
By varying the corresponding action with respect to a metric g, I obtain:
[tex]
T_{\mu \nu} = \frac{1}{2} g_{\mu \nu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\nu \Phi
[/tex]
Assuming further that Phi has not spatial dependence (only function of time) and using the Friedmann Robertson Walker metric, which is diagonal, I end up with a diagonal energy tensor:
[tex]
T_{\mu \mu} = \frac{1}{2} g_{\mu \mu} \left( \partial_\alpha \Phi g^{\alpha \beta} \partial_\beta \Phi + m^2 \Phi^2 \right) - \partial_\mu \Phi \partial_\mu \Phi
[/tex]
where
[tex]
g_{\mu \mu} = Diag \left( 1 , -R(t)^2, -s^2 R(t)^2, -\sin^2(\theta) s^2 R(t)^2 \right)
[/tex]
R(t) being the scale factor.
I then use
[tex]
T_{\mu \nu}' = R_{\mu i} R_{\nu j} T_ij
[/tex]
where R is a 4x4 rotation matrix, with the (1,1) entry being equal to one, and the remaining 3x3 matrix being the usual rotation matrix (here about the z axis).
However, I am unable to recover the original tensor upon performing the above product. For example, I get
[tex]
T_{22}' = \cos^2(\theta')T_{22} + \sin^2(\theta')T_{33}
[/tex]
where I use theta prime (the angle of rotation) to differentiate between theta in the FRW metric, and since T_22 is not equal to T_33 (unless sin theta is one), I don't quite get the same result. Haven't worked out the other diagonal entries yet.
I suspect that I am using a wrong form for the RW metric (maybe wrong coordinate?). Any help would be appreciated. Thanks.
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