If $g_{00}=\pm c^2$, and $g$ is diagonal, and $T^{00}=\rho$ (a mass density, not an energy density), then $T_{00}=ρc^4=\rho_E c^2$. This is a good example of why trying to track factors of $c$ through calculations in GR is not a very good idea....

(Iff) we use a very unusual and unconventional definition of $T_{00}$ with dimensions of mass density, then really we should write $\frac{1}{c^2}T_{00} =\rho_m$. Still, in such a case, say we wish to evaluate the tensor in its correct dimensions ( which is why I said this is unconventional because we are dealing with the tensor with time time components, which is the energy density) we still have $T_{00} = \rho_m c^2$. In no circumstances is $T_{00} = \rho c^4$ correct because the dimensions are grossly wrong.

The only time the mass density with a coefficient of $c^4$ can be entertained as a relation to the stress energy $T_{00}$, is if we had an extra factor of $c^2$ as a coefficient on the tensor as $T_{00}c^2 = \rho_m c^4$.... but I've never ran into such an expression in all my days.

And I don't generally agree that we should work in a system where c=1. By removing constants, we make ourselves intentionally forget where they should be in the first place. I don't know what your background is, but it's a pain in the backside writing a theory when you've set an array of constants to 1, when dimensional analysis is such an important aspect of physics and is a vital tool in constructing equations.

If $T_{00}=ρc^2$, then $T^{00}=ρc^{-2}$. Generally I agree re units, but in GR there is nowhere that factors of $c$ "should" be, since space and time are unified (as another answer to this question said).

Em no. Again, you seem to be chosing what dimensions you want for $T_{00}$. You're defining the tensor as a mass density, and a symbol for mass density almost universally as an energy density hence why in my reply to the OP, we are careful to write $\rho_E$ to express we are talking about the energy density, not just simply $\rho$. And anyhow, who goes about calling $T_{00}$ an object with units of mass density? That is so weird.

Do you understand the difference between $T^{00}$ and $T_{00}$?

So your using covariance and contravariance to change the units?

I've worked in upper and lower indices but not once used them to define (different dimensions) to an observable. Maybe there is an area of math I'm unaware of... but i was sure you cannot change an objects dimensional meaning like this.

If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review

It's not a new question, it's a reply to a statement in which I was asking justification for changing the units based on using upper or lower indices.

If the metric is $ds^2=c^2dt^2-dx^2-dy^2-dz^2$, then $g=\text{diag}(c^2,-1,-1,-1)$, and $T_{00} = T^{00} g_{00} g_{00} = T^{00} c^4$. You can give either $T_{00}$ or $T^{00}$ your preferred units, but the other one can't have the same units.

I've never seen this as a way to transport one observables dimensions to another. In the Einstein field equations, where I have used upper and lower indices, I've never seen a justification of changing the dimensions of something based on what you wrote above. Take $T^{\mu \nu} = g^{\mu \alpha} T^{\nu}_{\alpha}$ as an example. It is a stress energy no matter if the index is lowered like $T_{\mu \nu} = g_{\mu \alpha} T^{\alpha}_{\nu}$ - do you have a reference I can read?

I'm sorry I have to ask for a reference, but you were also the same poster who was trying to tell me that $\ddot{a}$ was not a second derivative when it clearly was. So I'm not going to take your answer on the face of it.

I'll tell you what, if you can't provide me with a reference by tomorrow, I'll just assume you can't and I'll leave it at that. What I will say is that dimensional analysis is my forte in physics. I'm acquainted with constructing equations based on dimensional analysis. As the OP wrote their question, I didn't see any problems with the article they referenced. I am also going to point out that it isn't just about convention, but about being consistent. Plus, I've never seen an object like $T_{00} = \rho c^{-2}$ so convention aside, the consistency for physics is wrong.

If you don't put @benrg in your comments, I'm not notified of them. I told you that $\dot a^2$ isn't a second derivative. Then I asked if you understand the difference between $\dot a^2$ and $\ddot a$ because you seemed not to. You were offended and said you do. Now you claim I told you that $\ddot a$ isn't a second derivative, which I didn't. Are you sure you understand the difference? I don't think I can help you any more.

@Benrg Yeah, yet it is a second derivative. I gave you two examples, then you equated the two, that's not my fault. I'm seriously worried that you don't understand derivatives.

You should know that $\frac{d^2a}{dt^2} = \ddot{a}$ is a second derivative. I also said $a \frac{d^2a}{dt^2} =\dot{a}^2$ (is also) an example of a second derivative on the scale factor... but not according to you. You then questioned me on saying, "you do realize that $\frac{d^2a}{dt^2} \ne a \frac{d^2a}{dt^2}$ in a style as if I said that, when I didn't. @benrg

@benrg in effect, you're not just spreading misinformation, you are actively trolling. And the fact you don't know what a second time derivative is, you shouldn't be in any position to be speaking about physics or math until you understand why you were wrong.

By the way, I am a physicist. I write physics all the time, and I'm well aquiainted with second derivatives. That's how I knew you didn't know what you were talking about by denying it was a second derivative.

@benrg and if you had read what I wrote earlier, you would have seen I knew what I was talking about. I asked what is the third derivative of $\dot{a}^2$ and I gave you the answer which was $2\ddot{a}\dot{a}$... do you know why that is the answer, or why it is called a third derivative? I doubt it if you don't understand why the former was a second derivative.