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I've seen that Vaidya metric does not use the Schwarzschild metric, but instead to get it you simply go to Eddington-Finkelstein coordinates and there put mass as a function of retarded time. From

$ds^2=\left(1-\dfrac{2M}{r}\right)du^2+2dudr-r^2\left(d\theta^2+\sin^2{\theta}\right)$

We can go to

$ds^2=\left(1-\dfrac{2M\left(u\right)}{r}\right)du^2+2dudr-r^2\left(d\theta^2+\sin^2{\theta}\right)$

Is it because the black hole is not the same after time translations, and therefore the $g_{0i}$ components of the metric would not be zero in Schwarzschild coordinates or is there another motivation?

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  • $\begingroup$ You can have it in Schwarzschild-like coordinates, but the function for the mass becomes a rather complicated function of time and radius in that case, see here. For an approximate metric of a Hawking radiating black hole where the source of the radiation is not exactly localized you can still use an approximation where the mass is only a function of time though and neglect the r-dependency, see here, although I wouldn't use it for the last stage of evaporation before it disintegrates. $\endgroup$
    – Yukterez
    Commented Jun 30, 2023 at 19:11
  • $\begingroup$ If there is a component (mass) dependent on time, wouldn't $g_{0i}=g_{i0}$ components of the metric ($i\in\{1,2,3\}$) not necessarily be zero? See en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution. On the Wikipedia page, they consider static time in which after $t\rightarrow -t$ positive and negative components must be equal and therefore $e_te_r=\left(-e_t\right)e_r=0$. If the body is not time-static, wouldn't these components be non-zero? In what situations? (See: physics.stackexchange.com/questions/770183/…) $\endgroup$
    – Antoniou
    Commented Jun 30, 2023 at 22:55
  • $\begingroup$ No, the axes can still be orthogonal so the cross terms can be 0. If you want your local observers not hovering at fixed spatial coordinates but instead have them for example free falling with v in the r direction like in Gullstrand Painlevé style coordinates set √(gₜᵣgᵗʳ)=v. For evaporating black holes we have the mass function dependend on t, not u, and the source of the radiation isn't localized anyway, so for that case the diagonalized form in the 2nd link of my 1st comment might be the best fit. $\endgroup$
    – Yukterez
    Commented Jul 3, 2023 at 1:26

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The motivation is that the metric becomes simple if the mass M(u) is supposed to be a function of only u, but with Hawking radiation we have mass M(t)=³√(M₀-tћc⁴/5120/G²/π) as a function of asymptotic coordinate time t.

Since u depends on r and t, the real M(t) becomes M(u,r) if you transform that back, and the metric would become less elegant. Therefore the original form on the Vaidya metric is only good for qualitative, but not for quantitative studies of evaporating black holes.

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