Recent content by craigthone

We know in Lorentzian signature spacetime, in the case of timelike or spacelike hypersurfaces ##\Sigma## with \begin{align} n^\alpha n_\alpha=\epsilon=\pm1 \end{align} where ##\epsilon=1## for timelike and ##-1## for spacelike. We can define a tensor ## h_{\alpha\beta}## on ##\Sigma## by...

For the 2nd one, $$\frac{d^2x^\mu}{d\tau^2}=\frac{dU^\mu}{d\tau}$$ Generally ##dU^\mu## does not give the relative velocity, so we need something can be directly subtracted, i.e. ##\theta## For the uniformly accelerating object, the proper acceleration is defined as $$\alpha=\frac{d\theta}{d\tau}$$

I guess you missed some terms in ##\delta R## $$\delta R=R_{\alpha\beta} \delta g^{\alpha\beta}+g_{\alpha\beta} \delta R^{\alpha\beta}=R_{\alpha\beta} \delta g^{\alpha\beta}+g_{\alpha\beta}\nabla^2 \delta g^{\alpha\beta}-\nabla_\alpha \nabla_\beta \delta g^{\alpha\beta}$$

Thank all of you for discussions, especially PeterDonis's clear physical picture.

From the discussion above, there seem to be another possible solution. (1) Suppose at each position there is an observer with fixed ##(r,\theta,\phi)##, (called Schwarzschild observer), carrying two clocks. One is the standard clock which record his proper time. One Schwarzscild clock records...

Indeed I had a possible solution to the formula (1.4) which I think is messy. Consider the out-going radial light ray from event ##(t_E,r_E)## to ##(t_R,r_R)## where signal received position is at infity ##r_R \rightarrow \infty##. The light ray equation can be written is...

I should say that the relation between ##dt_{H}## and ##dt_{\infty}##. What we need is just the relation $$ dt_\infty = \left( 1 - \frac{2GM}{r} \right)^{1/2} dt_{ H} $$ This relation is from the argument

You mean this relation is not for infalling observer? For infalling observer, we have $$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2=\frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2\approx...

I mean the approach here using the information of the free falling observer rather than using the out going light ray equation.

Start with the above formula $$ d\tau^2 \approx 2 \left( 1 - \frac{2GM}{r} \right)^2 dt_{ H}^2 $$ This is the relation for the infalling observer near the horizon. Then we need to relate the coordinate time ##t_H## to the coordinate time, and also the proper time of distant observer...

You are right again. Why does it not work? This is strange to me.

$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2$$ $$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1}\left(\frac{r-2GM}{2GM}\right)^2dt^2\approx 2\left(1-\frac{2GM}{r}\right)dt^2$$ $$d\tau\approx...

Yes, you are right and you give the answer. Though ##dr^2## terms can not be ignored, it has the same form as $dt^2$ term. In the end we get the relation between ##d\tau## and ##dt##. Thanks for all your posts.

Thanks for your corrections again. How about the following argument, for the infalling observer near the horizon we have $$\frac{dt}{dr}\approx -\frac{2GM}{r-2GM}=-\frac{2GM}{\varepsilon}$$ This is the formula from Dirac (page 33). Then the proper time for the infalling observer...

The metic ##g_{\mu\nu}## can be written as $$ \left[ \begin{array}{ccc} -A+r^2\Omega^2 & 0 &0\\ 0&\frac{B}{A} & r^2\Omega\\ 0&r^2\Omega&r^2 \end{array} \right] $$ Am I right?