Matterwave said:
I am actually curious what the result is myself.
I ran it through Maxima and obtained the following Einstein tensor:
$$
G^t{}_r = \frac{2 \frac{dm}{dt}}{\left( r - 2m \right)^2}
$$
$$
G^r{}_t = \frac{2 \frac{dm}{dt}}{r^2}
$$
$$
G^{\theta}{}_{\theta} = G^{\phi}{}_{\phi} = \frac{r^2 \frac{d^2 m}{dt^2} + r \left[ \left( \frac{dm}{dt} \right)^2 - 2 m \frac{d^2 m}{dt^2} \right]}{\left( r - 2m \right)^3}
$$
Matterwave said:
Since the Bianchi identities are a geometric identity, it would seem that they can not be violated by any valid metric to any manifold, not just the solutions to the EFE's.
Actually, since computing the Einstein tensor is just a matter of taking derivatives and multiplying,
any metric with continuous first and second derivatives will give a valid "solution" to the EFE, mathematically speaking. You just take the Einstein tensor, multiply it by ##8 \pi##, and call the result the stress-energy tensor. Whether the solution makes sense physically is another matter.
Matterwave said:
the EFE's will use the Bianchi identities to force conservation of the stress-energy tensor.
Yes. But again, that's just a mathematical fact. Whether the stress-energy tensor makes sense physically is another matter.
Matterwave said:
my guess is that the Bianchi identities will still hold, but there is no way to make the Einstein tensor equal to any valid stress-energy tensor (which would imply that the Einstein tensor is non-zero in this case...).
No, it implies that the stress-energy tensor is nonzero, i.e., that the spacetime is not vacuum. You can see from the results I gave above that indeed this is the case: the "Schwarzschild metric" with ##m## as a function of ##t## gives a non-zero stress-energy tensor, therefore the spacetime is not vacuum. The Bianchi identities still hold (as they must, since, as you say, they are geometric identities that don't depend on the solution being physically reasonable).
Mathematically, this non-vacuum stress-energy tensor is perfectly "valid"; after all, I just wrote it down, and there aren't any expressions in it that are problematic, mathematically speaking. However, there is an obvious issue physically: ##G^r{}_t## is singular at ##r = 0## unless we set ##dm/ dt = 0## at ##r = 0##. However, we specified this metric with the assumption that ##m## was a function only of ##t##, not ##r##. That means that, if ##dm / dt = 0## at ##r = 0##, we must have ##dm / dt = 0## at all values of ##r##. And, of course, if this is true, then the Einstein tensor vanishes and we just have the ordinary Schwarzschild metric.
In other words, what we have done here is not derive a metric that is physically different from the Schwarzschild metric. What we have done is to prove that,
if the metric takes the form of the Schwarzschild metric, then ##m## cannot be a function of ##t##, physically speaking. This is why I said, at the end of post #17, that the metric for the case of a mass that is "radiating away" is
not the Schwarzschild metric with a time-dependent ##m##; it can't be, as we have just shown.
Matterwave said:
The other possibility that I can think of is that there is a defect with this metric itself, and that it can't be a valid metric to a Lorentzian manifold.
No, the metric is perfectly valid; it just doesn't mean what ChrisVer thinks it means. See above.
I am not trying to prove that what you say is wrong, but why it is correct.
I eventually did exactly that by the time I wrote m=m(r,t).
