Gravity for a stellar black hole

1. Feb 27, 2008

stevebd1

I'm currently looking at the various gravity gradients (tidal stresses) for black holes and I've realised that in respect of small stellar black holes, as you approach the event horizon, not only does the gravity increase but it (appears) to exceed c. For example, for a 3 sol black hole, the gravity at the event horizon, using GM/r^2, works out at 5.0845x10^12 m/s^2. With the notion that this is the rate of acceleration in every second, is it acceptable to use a figure like this to express something which is considered to travel as the same as c? I also noticed this only occurs with neutron stars and small black holes (and any other compact star that might exist in between). Doing some research, it seems acceptable to state that the surface gravity for neutron stars can range from 2x10^11 to 3x10^12 Earth g (which are in the same ball park). Also, working backwards, you can work out at what point the gravity exceeds 299,792,500 m/s^2, r = (GM/c)^0.5, for a 3 sol black hole this would be at 1152.2 km radius. Does this radius denote something significant? I'd appreciate any feedback.

regards
Steve

P.S. the gravity gradient at the event horizon for a 3 sol mass black hole increases at 1.1490x10^9 m/s^2 per metre which in itself appears to be superluminal (I'm aware that any object within a gravity field of this magnitude would not exceed c but be dragged by the gravity field to speeds very close to c). I'm also aware there are theories that propose that the speed of gravity exceeds c but this seems to be met with considerable skepticism.

Last edited: Feb 27, 2008
2. Mar 3, 2008

stevebd1

Hopefully I'm not being bullish as I understand that not all questions are answered straight away but I'm only looking for a simple response (or maybe there isn't one) to the above, such as 1) I've missed something out in the calculations, 2) This is nothing to be concerned about, it's just the way it is, 3) There's something theoretical I'm missing. I'd appreciate any feedback.

Steve

Last edited: Mar 3, 2008
3. Mar 3, 2008

George Jones

Staff Emeritus
I'm not sure how you got GM/r^2. Are you using Newtonian gravity? Or are you using coordinate acceleration according to general relativity of a hovering observer?

Are you saying that acceleration exceeds c? The magnitude of acceleration can be any non-negative real number.

4. Mar 3, 2008

stevebd1

Hi George and thanks for the reply. I'm using Newtonian gravity which seems to be an acceptable solution according to most sources I look at, correctly it would be expressed as

$$g = G\frac{m}{r^{2}}$$

For earth, this (obviously) means a gravity acceleration of 9.8207 m/s^2 at sea level (based on a mean earth radius of 6,371,000 m and a mass of 5.9736x10^24 kg), an acceleration of 9.79 m/s^2 at 8850 m, the top of Everest (r would equal 6,379,850 m) and 8.85 m/s^2 at 340 km above sea level, the orbit of the ISS (r would equal 6,711,000 m).

Using this same equation and principle for a black hole with a mass of 3 suns (5.9673x10^30 kg) and a Schwarzschild radius of 8861 m, based on the Schwarzschild equation-

$$R_{s}=\frac{2GM}{c^{2}}$$

Using the Newtonian equation to get an idea of the gravity at the event horizon, I get 5.0715x10^12 m/s^2, which tells me for every second an object is falling, it should increase in velocity by 5.0715x10^12 m/s, which is obviously superluminal. I'm not expecting that matter would accelerate at this rate but I just seem to be having a problem with the fact that it is acceptable to produce a figure for gravity that suggests acceleration in excess of the speed of light when it's common knowledge that gravity itself travels at the speed of light.

Steve

Last edited: Mar 3, 2008
5. Mar 3, 2008

George Jones

Staff Emeritus
Could you give a reference either to a URL or a book?

I don't think that Newtonian gravity is acceptable here. The GR (Schwarzschild) coordinate acceleration has the same expression as Newtonian acceleration, but an accelerometer carried by a hovering observer would actually read

$$\left(1 - \frac{2Gm}{c^2 r} \right)^{-\frac{1}{2}} G\frac{m}{r^2}$$

This makes things worse, and diverges at the event horizon!

In relativity, velocities no longer add in the usual way. For example, if you zip off to my left at a speed of 3c/4 with respect to me, and Bob zips off to my left at a speed of 3c/4 with respect to me, you see bob zipping off to your right with a speed 24c/25, not 3c/4 + 3c/4 = 3c/2.

If, with respect to me, you start from rest, and then you undergo acceleration $a$ for a time $\tau$ as measured by your watch, then you observe our relative speed to be

$$v = c \tanh \left(\frac{a \tau}{c} \right).$$

This never exceeds $c$ no matter how large $\tau$ is. For $x \ll 1$, $\tanh x \doteq x$, which gives that $v \doteq a \tau$ when $a \tau \ll c$.

Last edited: Mar 3, 2008
6. Mar 3, 2008

stevebd1

Thanks George, I see the addition of the second part to the Newtonian equation is the way to go. When you say 'accelerometer carried by a hovering observer would actually read' are you implying that this is the gravity as it would be read and experienced real time by an observer? (as apposed to say redshifted light which has been stretched and isn't an accurate representation of itself as it was when it started its journey). The gravity read by the accelerometer is the gravity as it is (as the equation) and not dilated in some way.

One thing I did find while looking on the web was a slightly different expression of the equation you've provided-

http://astro.pas.rochester.edu/~aquillen/ast142/Lecture/Lecture9.pdf pages 18

$$a=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

This implies that gravity blows up towards infinity at the event horizon (as also stated in the paper). This is quite a difference from what the equation you've posted suggests which is that the gravity drops to 0 and inverts at the event horizon. Are there 2 different theories as to what happens at the event horizon in respect of gravity.

Would you also apply this addition to the Newtonian gravity equation for neutron stars and other compact stars also.

cheers
Steve

Last edited: Mar 3, 2008
7. Mar 3, 2008

George Jones

Staff Emeritus
Yes, this is the "g-force" that the hovering observer would feel.

The two formulas are exactly the same.

8. Mar 4, 2008

stevebd1

I see now how the negative power works, thanks. I've noticed that this works perfectly with a Schwarzschild black hole but when applied to a rotating Kerr black hole, the infinite gravity occurs at the edge of the ergosphere (or the static limit) in the equatorial plane (where the Schwarzschild radius would be if it were a static black hole) and the event horizon is further back. Also, because the ergosphere is a flattened sphere, the infinite gravity would occur somewhere outside the event horizon/ergosphere at the poles (see attached picture). Is this the norm or is there a seperate equation or a modification required for a Kerr black hole?

Steve

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Last edited: Mar 5, 2008
9. Mar 5, 2008

stevebd1

I was looking for surface gravity for a Kerr black hole on the web and happened upon a section on wiki which made reference to something called the Killing horizon (named after Wilhelm Killing) that had the following equation-

$$k = \frac{\sqrt{M^{2}-Q^{2}-J^{2}/M^{2}}}{2M^{2}-Q^{2}+2M\sqrt{M^{2}-Q^{2}-J^{2}/M^{2}}}$$

Based on a 3 sol mass black hole with a rotation factor of 0.8, M = 5.9673x10^30 and J = 6.3408x10^42 and considering the charge (Q) as zero, the answer comes out at 4.196276x10^-32. What significance does this figure have?

Oddly enough, when I calculate k using the more simple Schwarzschild solution, I get a very similar answer of 4.1897x10^-32

$$k = \frac{1}{4M}$$

Again, what is the significance of this number (from what I understand, it appears to be a geometric number) and how might it relate to surface gravity of a black hole. I'd appreciate any feedback.

http://en.wikipedia.org/wiki/Surface_gravity

Steve

Last edited: Mar 6, 2008
10. Mar 7, 2008

stevebd1

Attached is an image I put together comparing a 3 sol mass static black hole and a 3 sol mass rotating black hole (a = 0.95). I've used the modified Newtonian equation to calculate gravity for both black holes which results in gravity reaching infinite at the Schwarzschild radius for both types (the Schwarzschild radius is only apparent on the rotating black hole but the point of infinite gravity is the same for both models). I'd appreciate any feedback regarding the properties of gravity on the rotating black hole between the Schwarzschild radius and the outer event horizon (areas marked with G?).

Equations for rotating black hole-

Event horizon-

$$R+=(GM/c^{2}) (1+\sqrt{(1-a^{2})})$$

Cauchy horizon-

$$R+=(GM/c^{2}) (1-\sqrt{(1-a^{2})})$$

Static limit-

$$Re=(GM/c^{2}) (1+\sqrt{(1-a^{2}\,sin^{2}\theta)})\,\,\,\,\,\,\,\,\theta=latitude$$

(equatorial edge equal to Schwarzschild radius, polar edge equal to event horizon)

Photon sphere-

$$Rps=3GM/c^{2}$$

cheers
Steve

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11. Mar 8, 2008

pseudo

according to ur calculations any body travelling near the blackhole vil tare apart due to its tidal forces and the remaining debris vil revolve around it.if its gravitational accleration is more than c then its tidal force vil tremendously increase therefore increasing its roche limit.so it would bcome dangerous as a blackhole vil show a vicinity of 10 to 9 mpc.therefore v should get explotions near from a blackhole which v dont get.

12. Mar 8, 2008

pseudo

according to stevebd1 a body vil fall inside a blackhole vit a velocity more than c.then it vil have increasing mass and finally its mass vil reach infinity. how can it b possible that a body having infinite mass cannot escape from a blackhole ?

13. Mar 8, 2008

stevebd1

I've realised that the equation that modifies the Newtonian equation for gravity is based on the Schwarzschild radius which is for a static black hole, if I modify it to incorporate the the Kerr equation for the outer event horizon I get-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-\frac{GM(1+\sqrt{(1-a^{2})})}{rc^{2}}}}$$

Basically I replaced the 2 with (1+(1-a^2)^1/2), the equation that dictates the outer event horizon for a rotating black hole (while the 2 dictated the event horizon for a static black hole). The equation works for calculating gravity in close proximity to rotating black holes, taking into account the reduced event horizon and keeping the infinite gravity at the event horizon.

The source for the original rotating black hole equations-
http://www.engr.mun.ca/~ggeorge/astron/blackholes.html

If anyone sees anything at odds with the equation, I'd appreciate it if you could let me know.

regards
Steve

Last edited: Mar 9, 2008
14. Mar 10, 2008

stevebd1

It appears common practice for models of black holes that gravity diverges (or reaches infinity) at the event horizon and the tidal forces (or gravity gradients) continue to the singularity where they reach infinity. I've looked extensively on the web and these forums and haven't been able to answer the following questions-

How is it possible to have a gravity gradient within the event horizon when gravity already reached infinity at the event horizon?

How crucial is it that gravity diverges (reaches infinity) at the event horizon? Is it related in some way to the way light is effected by gravity? Would it not be a reasonable proposal to establish infinity gravity (or divergence) at the singularity?

I'd appreciate any comments or feedback.

regards
Steve

Last edited: Mar 10, 2008
15. Mar 10, 2008

pseudo

i still dont get it dat how can gravity reach infinity at the event horizon. i accept that gravity and the tidal forces reach infinity at the singularity but how can it be infinite at the event horizon. if the tidal forces are infinite at the event horizon then any matter even near to the horizon would tare apart into small moleculesbefore falling inside . and if v consider gravity to be infinite the the mass of the blakhole(event horizon) can reach infinity and would tare the fabric of the universe bending it to an infinite point.

16. Mar 12, 2008

stevebd1

I was looking at the GR (Schwarzschild) coordinate acceleration equation while considering how gravity might be calculated (or estimated to some degree) beyond the event horizon. Keeping in mind that space and time are suppose to flip (or invert) at the event horizon, I looked at the option of inverting the 2GM/rc^2 part of the equation in order to maintain the flow of gravity and this produced results shown plotted in the attached document. While this isn't strictly speaking gravity as we know it, it does seem to collaborate with the curve outside the event horizon and the tidal forces.

equation for gravitational acceleration-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-2GM/rc^{2}}}$$

Based on the principle that the original equation diverges (or inverts) at the event horizon, for gravity beyond the event horizon the following might apply-

$$\frac{2GM}{rc^{2}}$$ inverts to become $$\frac{1}{2GM/rc^{2}}$$ or $$\frac{rc^{2}}{2GM}$$

This provides the following equation-

$$g=\frac{GM}{r^{2}}\frac{1}{\sqrt{1-rc^{2}/2GM}}$$

Using the revised equation beyond the event horizon, gravity appears to pull back from infinity, picks up more or less on the same curve it had previously been following, then increases to infinity again as it approaches the singularity.

The equation may be more of a geometric curiosity but it does appear to provide worthwhile information.

regards
Steve

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Last edited: Mar 13, 2008
17. Mar 12, 2008

pseudo

i can help u in it with an example. consider the blackhole is a centrifuge.any object falling in a centrifuge acclerates and finally comes to the centre. consider the same for the blackhole where u will get that the accleration gradient of a body keeps on increasing once it gets into the event horizon. so simply get the relation between the accleration gradient and the gravitational accleration considering that gravity is not infinite. i got a relation that mass is directly proportional to event horizon, while it is also directly proportional to the gravitational accleration of the blackhole. using the law of proportionality we say that event horizon is directly proportional to gravitational accleration. therefore g$$\proptor.$$

18. Mar 12, 2008

pseudo

19. Mar 12, 2008

pseudo

20. Mar 12, 2008