# The shape of black holes and their surroundings

1. Sep 5, 2011

### motorman

Hi Guys,

Just a thought, if a job of a black hole is to suck everything in it's surrounding to nothing-ness.

Then why are the there examples of black holes with jet streams flowing away from their axial poles?

Apparently the earths gravity field is weaker at the equator, so why does a black hole appear to suck more mass (stars, dust & whatever) at it's equator and potentially spew out jet streams from it's poles?

What goin on here?

I look forward to your replies.

2. Sep 5, 2011

### DaleSwanson

As matter falls into a black hole it ends up in a spiraling orbit around it. This is called an accretion disc.
http://en.wikipedia.org/wiki/Accretion_disc" [Broken]

Due to the tremendous amount of matter and energy near a black hole, the accretion disc becomes a very energetic place. The result is magnetic fields that funnel matter towards the poles. The exact nature of these jets is still unknown. A google search for 'black hole jets' will return a lot of recent stories about these jets, like this one:

The force of gravity on Earth's surface is weaker at the equator because of the spin of Earth. This effect is only felt by objects that are spinning with Earth (like us). An object coming from space wouldn't experience it.

Last edited by a moderator: May 5, 2017
3. Sep 6, 2011

### stevebd1

'Why do Saturn's rings all lie in the same plane? Why are planetary rings always found in their equatorial planes and not sometimes crossing their poles?'

http://saturn.jpl.nasa.gov/faq/FAQSaturn/#q14 [Broken]

Last edited by a moderator: May 5, 2017
4. Sep 6, 2011

### motorman

Hi chaps,

The accretion disc theory has a number of assumptions in it, that had to have a number of tweaks made to model in or for it to work on computers.

I was under the impression that the black hole sucks stuf in from "all" directions, no just one plane.

The images suggest that black hole may have spin. I think that a point on the the surface of the equator of the black hole spins at the same rate as a point on the surface at one of its' 'poles'. However it implies that spatially, the equator is moving at a higher RPM than the poles, and using extreme masses, this may be having an effect on the strength of gravity in the equatorial regions.

The gravity in the poles is may be weaker than at the equator, hence the point of no return is further away(or weaker) from the black hole, allowing jets of stuff to escape. My assumption.

This may imply:speed of rotation of mass is proportional to strength of grativational(or otherwise) pull.

It's what the evidence in sat. images suggests. Any counter arguements?

All are invited to shoot this theory down, with facts tho'.

stevebd1:
That's an excellent observation. Anyone care to explain that?

5. Sep 6, 2011

### Cosmo Novice

A black holes gravitational pull is symmetrical pretty much - remember we are not dealing with Newtonian Gravity but extreme curvature under GR. This is abvious subject to some change by certain factors - such Black Hole composition.

I think a crucial failing point in your understanding is understanding the mechanism of the accretion disk. The matter that comprises of the accretion disk is not hidden behind the Event Horizon - the orbital of the accretion disc is further out than the outer Event Horizon so it does not require "more" energy to release jets.

Hope this helps

6. Sep 6, 2011

### motorman

Hi Cosmo Novice,

I do accept that the disc is beyond the event horizon, otherwise we wouldn't see the stuff.

The essence of the 2 questions are:

1) Why shape the matter around the BH as a disc? (esp around what may appear to be an equator); Why not suck matter in from all directions?

2) Why have jets? esp axially opposite to each other?

So far I've not read anything convincing on the web that explains the above two points, without needing special mods to algorithms, special exceptions to theories, or just, "we don't know yet"

7. Sep 6, 2011

### Cosmo Novice

There is no specific mechanism that is any different to saturns rings - the accretion discs are 3dimensional and they just orbit based on gravity. Remember until we get inside the EH a BH exerts the same gravitational curvature as any other object with mass. The explanation of Saturns rings can be applied here.

The truth is we are not entirely sure, please see the following article outlining mechanisms for relativistic jets.
http://en.wikipedia.org/wiki/Relativistic_jet
It is worth remembering that Black Holes with relativistic jets are the exception and not the common.

I hope this information helps.

8. Sep 6, 2011

### WannabeNewton

First of all, a black hole doesn't suck in anything. The gravitational field outside of the event horizon has the same nature as that of any other spherically symmetric object's. Secondly, it isn't on one plane only, the dust is distributed throughout many different planes around the black hole but note that it is around the black hole. Because a black hole is spherically symmetric, dust that settles on a certain plane will remain on that plane.
It isn't a simple proportion for a kerr black hole. The metric that determines the geometry due to a kerr black hole contains many terms other than angular momentum.

9. Sep 6, 2011

### tvscientist

Apparently all gravity concentrations produce accretions disks. I am unaware of any that do not. Accordingly, as mass approaches a Black Hole two things happen. First, any mass that approaches the poles of the spinning BH will collide and produce polar jets. The extreme version is gamma ray bursts.

Second, conservation of angular momentum produces relativistic speeds as matter approaches the BH. Accordingly, time comes to a near stop, and the mass simply accumulates in a spiral onion layer. The event horizon can expand beyond these layers, but that just means electomagnetic radiation goes into orbit within the BH.

My question is what happens as electormagnetic radiation passes the event horizion. My supposition is that space/time becomes so dense the speed of light also approaches zero. Accordingly, a BH is not all that mysterious. It is simply an area of the universe where time slows to near zero. This is not a big deal IMHO.

10. Sep 6, 2011

### WannabeNewton

It takes an infinite coordinate time to get past the event horizon. Lorentz boost to the frame of the infalling object and you will see that it falls in finite proper time.

11. Sep 6, 2011

### tvscientist

WN

What is coordinated time? It seems to me the speed of light is the same in all directions to all observers. Accordingly, as a BH absorbs an accretion disk of matter, the matter approaches C and time approaches zero. Accordingly, the speed of light seems to bifurcate: Specifically, the observer approaching the BH experiences C at near zero, from our perspective.

Further, we can not see any photon that passes the event horizon, and accordingly can not measure its speed. Even though it should be C. I suspect the photon, from out perspective, DOES NOT equal C. We assume the speed of light in a vacuum is C. I suspect the speed of light varies in the presence of very high gravitational forces when those forces approach infinity.

But I am just quessing....

from the point of view of the observer who is approaching the BH, as observed from an outside observer, and time approaches zero, the speed of light, from the outside perspective, also approches zero.

12. Sep 6, 2011

### tvscientist

Dale - You wrote: "The force of gravity on Earth's surface is weaker at the equator because of the spin of Earth." I believe the force of gravity is an independent variable. Accordingly, if the earth were exactly spherical, the force of gravity at the equator would be exactly the same as any other place.

However, the earth is spinning, and at the equator there will be outward centrifugal force applied against gravity. Still, the equator bulges as a result of this centrifugal force, and accordingly the surface of the earth at the equator is further then in other places. The measured force of gravity will be accordingly less.

In summary, the WEIGHT of any given mass, as measured at the equator, will vary according to 1) the increased distance from the center of mass; and 2) the centrifugal force of the spinning earth.

13. Sep 6, 2011

### WannabeNewton

A radially moving photon starting from rest at r = infinity will have $$ds^{2} = 0 = -(1 - \frac{2GM}{r})dt^{2} + (1 - \frac{2GM}{r})^{-1}dr^{2}$$ so $$\frac{dr}{dt} =(1 - \frac{2GM}{r})$$ which, at r = 2GM, becomes $\frac{dr}{dt} = 0$ so yes the photon stops at r = 2GM but this is coordinate velocity. This is what far observers will conclude. Again, time only approaches a stop from the perspective of a far observer as a test particle falls closer to r = 2GM. In the reference frame of the test particle, it falls in finite proper time. These problems that occur at the event horizon are due to problems in the schwarzchild coordinate chart. If you switch to, say, eddington coordinates then the issues don't occur anymore.