# Relativistic escape velocity

1. Dec 28, 2008

### stevebd1

Curved space induces velocity, according to-

$$\gamma\equiv\frac{1}{\sqrt{1-(v/c)^2}}=\frac{1}{\sqrt{1-(r_s/r)}}$$

which results in

$$v_{rel}=-\sqrt{\frac{r_s}{r}}\ c$$

The velocity induced by the curvature at the surface of the Earth is 11.185 km/s, for the Sun 617.746 km/s which, in each case, matches the escape velocity for each object. These velocity's aren't relativistic and you would, with enough power, be able to hover above Earth without feeling any adverse effects. Around a black hole, the velocity induced by curvature increases relativistically, reaching the speed of light at the horizon. Curvature also increases distance within a confined radial distance (inverse of the time dilation), increasing to infinity at the event horizon. If you travel at the velocity induced by the curvature then dr will always equal dr (i.e. not increasing as suggested by Schwarzschild metric). My question is, if you were able to stop yourself deep within the gravitational field of a black hole (outside the EH), would you 'feel' the ultra-relativistic streaming of space around you, trying to make you pick up speed? I'm not talking about tidal forces but the actual curving of space itself, the gradient of which increases the closer you got to the EH. On one or two occasions, I've seen curvature described as a form of pressure which might be a fair comparison.

For example, the stress-energy tensor at the event horizon of a 10^12 sol mass black hole, if such a black hole could exist, would be negligible but the metric tensor (curvature) would induce a free-fall velocity that approached c at the EH. If you were to stop just outside the EH, what would be the resistance you would feel, it wouldn't be from the stress-energy tensor (I understand that gravity is a result of both stress-energy and metric tensors but in most cases, the stress-energy tensor is more significant, is there a difference to what would be experienced when the metric tensor is more significant?)

A comparison might be that in order to remain static within the gravity field, you would have to travel equal to the escape velocity for that specific radius, which might be compared to traveling at ultra-relativistic speed through flat space (hence the inclusion of the Lorentz factor above).

On a slightly different note regarding rotating black holes and frame dragging, the Killing surface gravity equation suggests that gravity reduces with higher spin-

$$\kappa_\pm=c^2\frac{r_+-r_-}{2(r_\pm^2+a^2)}$$

is this in some way related to centripetal acceleration induced by frame-dragging cancelling out the gravitational pull? Also, if you were an observer with zero angular momentum that fell from infinity radially towards a rotating black hole, due to frame dragging, you would orbit the BH before crossing the event horizon. Would you 'feel' the rotation or would the fact that it is the frame that was rotating and that technically, you are relatively static within the frame mean you would not feel any centripetal/centrifugal force? I imagine that some centripetal acceleration is induced by frame dragging but the question is how much.

Last edited: Dec 29, 2008
2. Dec 29, 2008

### Mentz114

I'm puzzled by this. In GR, the value of the metric tensor depends on the stress-energy tensor. The field equations provide the means of getting the metric for a given SE tensor. When the metric is known, the SE tensor is no longer required, and does not 'act' in any way independently of the metric.

If I've misunderstood what you're saying, I'm sure you'll tell me.

Last edited: Dec 29, 2008
3. Dec 29, 2008

### stevebd1

Thanks for your response, Mentz114. I think I may have my semantics wrong. I was looking for a term for simple Newtonian gravity $(Gm/r^2)$. Maybe I should have referred to gravity stress tensor (T?) rather than stress-energy tensor $(T_{\mu \nu})$ which I now realise is the sum of the stress tensor and metric tensor $(T_{\mu \nu}=T\,g_{\mu \nu})$ or Newtonian gravity multiplied by coordinate acceleration.

To add a bit more info to my question, the Killing surface gravity for a static black hole is-

$$\kappa=\frac{c^4}{4Gm}$$

which is basically equivalent to $g=Gm/r_s^2$ which tells us that Newtonian gravity (or the stress tensor?) is still 'at work' even when you cross the event horizon at the speed of light. This is backed up to some extent by the fact that the tidal force equation remains unchanged once past the EH, $dg=2Gm/r^3$

I think I may have answered my own question by understanding the term stress-energy tensor a bit better but in most cases, the gravity stress tensor (or basic Newtonian gravity) contributes significantly to the gravitational field outside a black hole but the larger the black hole gets, the less it appears to contribute and the gravitational 'pull' comes about almost purely from the curvature of space (the metric tensor). Using the 10^12 sol mass black hole as an example, the Killing surface gravity is about 1.5 Earth g at the event horizon but you would still have to accelerate close to c away from the BH in order to remain static outside the EH, regardless of how low the gravity stress energy tensor is. I suppose this is one of the consequences of gravity in GR. My point is, would you be aware of the curved space around you trying to induce a velocity close to c as you accelerated against it? I suppose the only clues would be that anything visable outside the gravitation field would be highly blueshifted and anything that fell into the black hole radially close by would whiz by at close to the speed of light. I imagine the event horizon would appear further away too due to the increase in radial length due to curvature (which you wouldn't experience if you were falling in at $v=\sqrt{r_s/r}\,c$) but would you 'feel' the ultra-relativistic streaming of space around you, trying to make you pick up speed?

Last edited: Dec 29, 2008
4. Dec 29, 2008

### Mentz114

I'm even more puzzled. I re-read your first post. When you say 'curvature induces velocity' I presume you mean that gravity attracts. Your terminology is unfamiliar and I can't follow what you're trying to say. Separating gravity into 'curvature' and 'stress tensor' doesn't make sense to me. Newtonian gravity is not a different thing from the gravity of GR. In weak fields the two theories converge in their predictions. But gravity is gravity.

The behaviour of falling bodies, rockets etc is well understood from studies of the Shwarzschild metric.

5. Dec 30, 2008

### Ich

I dindn't understand either, but I want to clarify a few points: The "surface gravity" is not the acceleration an object has to endure to stay put at the EH. The former is a rather theoretical value which cannot be measured, the latter diverges as it should. And the distance to the EH is not infinite for a static observer, only ds/dr diverges, but not the integral. And the equivalence principle is still valid at or near the EH, which means that a free falling object feels nothing unusual, irrespective of his/her velocity (which might be zero wrt stationary observers for a time).

6. Dec 31, 2008

### stevebd1

Hi Ich, you say the (Killing) surface gravity is a theoretical value. I've seen it stated that 'the surface gravity κ of a static Killing horizon is the acceleration, as exerted at infinity, needed to keep an object at the horizon' which implies it might have no local effect but if this is the case, why does it play a significant role in calculating Hawking radiation which suggests it has local effect-

$$T_H=\frac{\kappa \hbar}{2\pi k_bc}$$

where $\kappa$ is the Killing surface gravity of the black hole, $\hbar[/tex] is the reduced Planck constant and $k_b$ is the Boltzmann constant. Also, the equation for tidal forces (the gravity gradient) remains unchanged once past the event horizon and can still provide information. While gravitation acceleration becomes infinite at the event horizon, the tidal force equation implies that there is still some form of stress still at work- $$\Delta g=\frac{2Gm}{r^3}$$ If, as you say, the Killing surface gravity is a theoretical quantity that has no effect locally, how is it the gravity gradient equation can still apply inside the black hole when gravitational acceleration has reached infinity at the event horizon? On a side note, I'm assuming it's correct to say that if an object at rest at infinity was to fall to Earth, not taking into account atmosphere, it would hit the Earth at 11.185 km/s. I was trying to imagine the relativistic effects that would occur as you remained static in highly curved space (I suppose the statement 'curved spacetime induces velocity' would have also been a more appropriate statement). Incidentally, how would you calculate the thrust/power required to remain static just outside the event horizon and does anyone have any comments regarding the question about centripetal acceleration induced by frame dragging? Last edited: Dec 31, 2008 7. Jan 1, 2009 ### Ich AFAIK Hawking radiation is not a local effect, but I'm no expert. It may sound funny, but "gravitational acceleration" is not a locally defined quantity either. It is defined wrt static observers, and "static" is a global attribute. Maybe visualize it this way: the local light cones tilt smoothly as you approach the singularity, following the "gravity gradient equation". The EH is where it is tilted 45° wrt the original one, not where the tilting becomes pathological in any way. I don't like both wordings very much, the "induced velocity" is nothing but the escape velocity, a global quantity, while one unsually means a local quantity when speaking of curvature. Especially the Riemann scalar is locally defined. I would solve the equation of motion for d²x/dt²=0. 8. Jan 1, 2009 ### Mentz114 9. Jan 5, 2009 ### stevebd1 I thought it was worthwhile posting some extracts from the webpage that Mentz114 recommended (M is the gravitational radius, M=Gm/c^2, 2M is the Schwarzschild radius)- '..relative to the frame of a particle falling in from infinity, a hovering observer must be moving outward at near light velocity. Consequently his axial distances are tremendously contracted, to the extent that, if the value of Dr is normalized to his frame of reference, he is actually a great distance (perhaps even light-years) from the r=2M boundary, even though he is just 1 inch above r=2M in terms of the Schwarzschild coordinate r. Also, the closer he tries to hover, the more radial boost he needs to hold that value of r, and the more contracted his radial distances become. Thus he is living in a thinner and thinner shell of Dr, but from his own perspective there's a world of room. Assuming he brought enough rocket fuel to accelerate himself up to this "hovering frame" at that radius 2M+Dr (or actually to slow himself down to a hovering frame), he would thereafter just need to resist the local acceleration of gravity to maintain that frame of reference.. Quantitatively, for an observer hovering at a small Schwarzschild distance Dr above the horizon of a black hole, the radial distance Dr' to the event horizon with respect to the observer's local coordinates would be- $$\Delta r'=\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}$ Local gravitation acceleration for the Schwarzschild coordinate (r) and proper time (τ)- [tex]\frac{d^2 r}{d\tau^2}=\frac{M}{r^2}\,c^2$$

when r=2M-

$$\frac{d^2 r}{d\tau^2}=-\frac{1}{4M}\,c^2\equiv\kappa$$

where $\kappa$ is the Killing surface gravity.

However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor, i.e., we have dr'=dr/(1-2m/r)^1/2. Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer solving for proper distance (r') and proper time (τ)-

$$\frac{d^2 r'}{d\tau^2}=-\frac{1/\Delta r'}{1+\sqrt{1+(\Delta r'/M)^2}}\,c^2$$

which also equals-

$$\frac{d^2 r'}{d\tau^2}=-\frac{M}{r^2\sqrt{1-2M/r}}$$

which is basically-

$$a=-\frac{Gm}{r^2}\left(1-\frac{2Gm}{rc^2}\right)^{-1/2}$$

where G is the gravitational constant and m is mass

What's interesting about the first equation is that it shows that coordinate-wise an observer could 'hover' within 100 mm of the event horizon of a 3.7e+6 sol static black hole but from the observers perspective, the distance would appear to be ~33 km to the EH.

Incidently, why is it d2r/dτ2 as opposed to dr2/dτ2 and what's the implication of squaring d?

Last edited: Jan 5, 2009
10. Jan 5, 2009

### Mentz114

By convention, second derivatives are written

$$\frac{d^2y}{dx^2}$$.

So

$$\frac{d^2y}{dt^2}$$

is an acceleration, and

$$\frac{dy}{dt}$$

is a velocity.

11. Jan 5, 2009

### Ich

...and $$dr^2/d\tau^2$$ is an unusual way of representing a squared velocity.