Yes, for escape from a spherical mass M, with a slightly different interpretation of what ##V_e## and ##r## mean. In Newtonian physics, ##V_e## would be ##\frac{dr}{dt}##, while in General Relativity, it would be ##\frac{dr}{d\tau}##, where ##\tau## is proper time (time on a clock moving at speed ##V_e##). The interpretation of ##r## is slightly different, also.Hi,
Do we obtain the same escape velocity equation: V_{e} = sqrt(2GM/r) using both Newtonian and Relativistic approach?
The most obvious difference is at ##r=2GM/c^2##, the Schwarzschild radius. At and beyond this point, GR doesn't really have the concept of an escape velocity, since you can't escape. Newtonian physics (if it were an accurate description of such strong fields) would continue to permit escape from below this point.The interpretation of ##r## is slightly different, also.
This is probably getting into the weeds, a bit, but what the mathematics tells us is that if ##r < \frac{2GM}{c^2}##, then it is impossible for ##\frac{dr}{d\tau}## to change sign. But technically, the sign could be either positive or negative. A positive value would correspond to a particle arising from a white hole, which is described by the same metric as a black hole. (I think)The most obvious difference is at ##r=2GM/c^2##, the Schwarzschild radius. At and beyond this point, GR doesn't really have the concept of an escape velocity, since you can't escape. Newtonian physics (if it were an accurate description of such strong fields) would continue to permit escape from below this point.
This looks a lot like dr/dr, but I see by the latex that it's dr/d tau.##\frac{dr}{d\tau}##
I don't know if there is a way to increase the size of Latex displays. They always seem too small to me, especially fractions.This looks a lot like dr/dr, but I see by the latex that it's dr/d tau.
I think that it is important to point out that the black and white hole regions are disjoint regions of spacetime even if they look just the same up to time reversal. Also, the white hole is unlikely to correspond to anything physical and even if it did it would have no meaningful concept of escape velocity (everything goes out of it regardless of state of motion).This is probably getting into the weeds, a bit, but what the mathematics tells us is that if ##r < \frac{2GM}{c^2}##, then it is impossible for ##\frac{dr}{d\tau}## to change sign. But technically, the sign could be either positive or negative. A positive value would correspond to a particle arising from a white hole, which is described by the same metric as a black hole. (I think)
Inline maths mode tries not to mess up the line spacing, so it likes to shrink symbols. Either use paragraph mode (dollar signs) or write ##dr/d\tau##. Or make it ##\huge{\\huge}## (https://texblog.org/2012/08/29/changing-the-font-size-in-latex/).I don't know if there is a way to increase the size of Latex displays. They always seem too small to me, especially fractions.
Do the white hole and "other" exterior region actually correspond to anything in a coordinate chart covering Newtonian flat spacetime?I think that it is important to point out that the black and white hole regions are disjoint regions of spacetime even if they look just the same up to time reversal. Also, the white hole is unlikely to correspond to anything physical and even if it did it would have no meaningful concept of escape velocity (everything goes out of it regardless of state of motion).
I am not sure I understand the question. Nothing in the Schwarzschild spacetime corresponds to flat spacetime.Do the white hole and "other" exterior region actually correspond to anything in a coordinate chart covering Newtonian flat spacetime?
For inlines you can always use \dfrac instead of \frac to make your fractions larger, looks awful though ... ##\dfrac{dr}{d\tau}##, ##\frac{dr}{d\tau}##.I don't know if there is a way to increase the size of Latex displays. They always seem too small to me, especially fractions.
Sorry. Newtonian physics envisages a flat space on which one can define spherical polars with ##r## in the range ##0\rightarrow\infty##. The OP is implicitly identifying Newton's ##r## with Schwarzschild's ##r##, and we're exploring the limits of that.I am not sure I understand the question. Nothing in the Schwarzschild spacetime corresponds to flat spacetime.
It's a little confusing to me to say that a white hole is an unphysical time-reversal of the black hole, because they are actually both described by the same metric. So as far as the structure of space-time is concerned, they are the same thing. The distinction is not about the holes themselves, but about the behavior of the test particles going in or out.I think that it is important to point out that the black and white hole regions are disjoint regions of spacetime even if they look just the same up to time reversal. Also, the white hole is unlikely to correspond to anything physical and even if it did it would have no meaningful concept of escape velocity (everything goes out of it regardless of state of motion).
An even more pressing issue is that there is no stationary observer to have an escape velocity relative to in both cases.
I did not say it was unphysical, I said that it does not correspond to anything physical. The reason for this is that real black holes formed from some matter distribution in the past, meaning that the Schwarzschild solution (which is a vacuum solution) does not extend infinitely into the past. Instead, you need to paste on some metric (and matter distribution) that describes the past.It's a little confusing to me to say that a white hole is an unphysical time-reversal of the black hole, because they are actually both described by the same metric.
As illustrations of @Orodruin 's point compare Fig. 6.9 and Fig. 6.11 in Wald, or the figures on pages 188 and 191 in Carroll's online notes,So I'm a little confused about the claim that black holes are physical and white holes are not.
Thanks.I can use Carroll's diagram to discuss what escape velocity means for radially-directed geodesics.As illustrations of @Orodruin 's point compare Fig. 6.9 and Fig. 6.11 in Wald, or the figures on pages 188 and 191 in Carroll's online notes,
https://arxiv.org/pdf/gr-qc/9712019.pdf
As I said in my post, in the derivation of escape velocity for a black hole, the relevant quantity is not ##\frac{dr}{dt}## butAnother question. In the Newtonian approach we use the classical kinetic energy expression mv^{2}/2 but why not to use the expression from the special relativity m - m_{0} ?