# I Escape velocity - General relativity

#### sha1000

Hi,

Do we obtain the same escape velocity equation: Ve = sqrt(2GM/r) using both Newtonian and Relativistic approach?

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#### stevendaryl

Staff Emeritus
Hi,

Do we obtain the same escape velocity equation: Ve = sqrt(2GM/r) using both Newtonian and Relativistic approach?
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.

#### Ibix

The interpretation of $r$ is slightly different, also.
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.

#### stevendaryl

Staff Emeritus
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 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)

#### DaveC426913

Gold Member
$\frac{dr}{d\tau}$
This looks a lot like dr/dr, but I see by the latex that it's dr/d tau.

#### stevendaryl

Staff Emeritus
This looks a lot like dr/dr, but I see by the latex that it's 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.

#### Orodruin

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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)
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.

#### Ibix

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.
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 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).
Do the white hole and "other" exterior region actually correspond to anything in a coordinate chart covering Newtonian flat spacetime?

#### Orodruin

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Do the white hole and "other" exterior region actually correspond to anything in a coordinate chart covering Newtonian flat spacetime?
I am not sure I understand the question. Nothing in the Schwarzschild spacetime corresponds to flat spacetime.

#### Orodruin

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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.
For inlines you can always use \dfrac instead of \frac to make your fractions larger, looks awful though ... $\dfrac{dr}{d\tau}$, $\frac{dr}{d\tau}$.

#### Ibix

I am not sure I understand the question. Nothing in the Schwarzschild spacetime corresponds to flat spacetime.
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.

Schwarzschild coordinates cover the exterior of a black hole, and interior coordinates with a similar functional form for the metric can be defined. But that seems to me to exhaust the range of $r$ values that have any analogue (no matter how shaky) in the Newtonian $r$. Regions III and IV on a Kruskal diagram don't correspond to a Schwarzschild $r$ value that could be (vaguely) associated with a Newtonian spatial coordinate - do they?

Edit: e.g. this Kruskal diagram from Wikipedia only labels $r$ in regions I and II. I'm asking if that means that the r coordinate can't cover regions III and IV, or if it's merely an editorial choice for that diagram. I thought the former, but may be wrong

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#### Orodruin

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The white hole can get a coordinate function $r$ just as much as the black hole region can. Technically, the Schwarzschild coordinate patch you cover with the outside coordinates is just the region outside of the event horizon. You can then extend this region to the black hole region or white hole region (easiest to see by introducing Eddington-Finklestein coordinates based on the ingoing/outgoing light-like geodesics. You can introduce coordinates r and t on the white hole region and the mirror universe region that will also give you exactly the same functional form for the metric. I see no real reason to favour the black-hole region over the white-hole region when considering the Schwarzschild spacetime (a real black hole is something different). I would rather not draw any analogy too far.

#### stevendaryl

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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.
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.

A black hole corresponds to the case where all particles have

$$\frac{dr}{d\tau} < 0$$

whenever

$$r < \frac{2GM}{c^2}$$

A white hole corresponds to the case where all particles have

$$\frac{dr}{d\tau} > 0$$

whenever

$$r < \frac{2GM}{c^2}$$

But actually, the sign of $\tau$ is just conventional, so there really isn't any difference, as long as you're talking about elementary particles.

So I'm a little confused about the claim that black holes are physical and white holes are not.

#### Orodruin

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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.
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.

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#### sha1000

Thank you for your kind replies. I got the full answer and even more.

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#### stevendaryl

Staff Emeritus
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
Thanks.I can use Carroll's diagram to discuss what escape velocity means for radially-directed geodesics.

So here is Carroll's diagram of the full Schwarzchild spacetime for an eternal black hole. The regions are:
• Region I is outside the black hole event horizon.
• Region II is inside the event horizon. The curved line at the top of Region II is the singularity.
• Region III is inside the "white hole". The curved line at the bottom is the white hole singularity.
• Region IV is a second external region. It is ordinary space, like Region I, but it can't be reached from Region I
The two diagonal lines are the event horizons. In terms of Schwarzschild coordinates, which are singular at the event horizon, these lines correspond to $r = \frac{2GM}{c^2}$ and $t = \pm \infty$.

I've drawn in three types of timelike geodesics:
• Type A: A test particle emerges in Region III from the white hole in the past,crosses the white hole event horizon into Region I, then escapes to spatial infinity. As described using Schwarzschild coordinates, the test particle seems to arise from the event horizon at time $t \rightarrow -\infty$, and its radial velocity, $dr/d\tau$ is greater than escape velocity. As $t$ goes to $+\infty$, the particle escapes to spatial infinity.
• Type B: A test particle emerges in Region III from the white hole region in the past, crosses the white hole event horizon into Region I, then crosses the black hole event horizon into Region II and finally reaches the black hole singularity. As described using Schwarzschild coordinates, the particle arises from the event horizon at time $t \rightarrow -\infty$ but it has a radial velocity $dr/d\tau$ that is less than escape velocity. It falls back below the event horizon as $t \rightarrow +\infty$.
• Type C: A test particle comes in from spatial infinity in Region I and falls through the black hole event horizon into Region II and eventually hits the black hole singularity.
(There are three corresponding geodesic types involving the second external region, IV. There are also orbits, where the geodesic never crosses the event horizon and never goes to spatial infinity. Those are boring. More specifically, they are not radially-directed.)

So for the full Schwarzschild spacetime, the notion of "escape velocity" has the following meaning in different regions:
• In Region I, a particle with positive radial velocity greater than escape velocity will escape the black hole.
• In Region II, all particles have negative radial velocity. But those with velocities of magnitude greater than escape velocity came from spatial infinity (assuming that they are following geodesics)
• In Region III, all particles have positive radial velocity. Those with radial velocities greater than escape velocity will enter Region I and escape to spatial infinity.

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#### sha1000

Another question. In the Newtonian approach we use the classical kinetic energy expression mv2/2 but why not to use the expression from the special relativity m - m0 ?

#### stevendaryl

Staff Emeritus
Another question. In the Newtonian approach we use the classical kinetic energy expression mv2/2 but why not to use the expression from the special relativity m - m0 ?
As I said in my post, in the derivation of escape velocity for a black hole, the relevant quantity is not $\frac{dr}{dt}$ but

$$\frac{dr}{d\tau}$$

where $\tau$ is proper time. So that expression is relativistic.It's not the relativistic energy, though.
It's sort of a coincidence that the relativistic equation involving $\frac{dr}{d\tau}$ looks the same as the nonrelativistic equation involving $\frac{dr}{dt}$.

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