Solving Dimensional Analysis Problems in Stress Energy Momentum Tensor

[space-time]

Yes, it's correct, and it's equivalent to mine (though you forgot to take the square root of sin²(Θ)). In metrics with these signatures, the definition of 'orthonormal' is with a -1 for the inner product of the e₀ vector with itself (this is because this vector is timelike). Also, you don't need that -1 in the components, it's just a +1 (the vector is future directed). Second, be careful with the notation: the metric is a purely dual tensor, i.e., you contract it with tangent vectors, not dual vectors like dΘ. If you want to use the dual basis given by the dΘs, etc., you need to contract with the inverse metric then. I actually used the dual basis, and that's why the result looks different. But the dual basis to the tangent vector basis you got, is the basis (of the dual space) I got previously.

If you want to use this to calculate curvature, you will have to study the method (the so called 'tetrad method'). You can find it in Wald's or Carroll's textbooks.

As for the units, I have not followed the discussion here. In my case, I use the normal practice of taking G= 1 and c=1. After I finish the calculation, I insert the adecuate factors in order to get the right units. You can find the details of this in appendix F of Wald.

Thank you. Do you know where I can find Wald's textbook or have a link that you can post by any chance?

 

[PeterDonis]

keep your basis vector positive, or you will flip the direction of time

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.

$(e_0) \cdot (e_0) = -1$ ...

...because $(e_0)$, which we would write in abstract index tensor notation as $(e_0)_a$ or $(e_0)^a$ depending on whether it was covariant or contravariant, is timelike.

If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

 

[pervect]

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.
If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

Good point, I stand corrected.

add.

Though I saw a recent example in Wald that did make the convector positive (pg 121, eq 6.1.6a sets $(e_0)_a = f^{1/2} (dt)_a$. I think your way has a lot to recommend it though.

 

[space-time]

If you're using a (-+++) signature, you can't do this with both the vector and the covector. See below.
If $(e_0)^a (e_0)_a = -1$, then both of them can't possibly be positive; one has to be negative, which means either the vector or the covector has to have a negative 0 component. I've always seen it be the covector that is negative; i.e., if we stipulate that $(e_0)^a > 0$ means time is moving forward, then we must have $(e_0)_a < 0$ for time moving forward if we are using a (-+++) signature.

Hey Peterdonis, do you know where I can find Wald's textbook or where I can find the method to use orthonormal bases to help me derive the Einstein tensor without having to worry about units?