# I'm standing on a black hole, is light approaching red or blue?

1. May 6, 2013

### elevin

Little thought experiment. Hypothetically lets say I'm standing on the surface of an imaginary black hole and not completely crushed to death, or better yet, I'm in the center of the thing (never mind the singularity) and looking outward into deep space. Is the light coming toward me red or blue shifted? I'm going with red, that the immense gravity surrounding me will elongate the wavelength of light before it reaches my eye.

2. May 6, 2013

### tom.stoer

You as a physical observer can't "stand on the black hole" if you mean "staying at the event horizon at rest". The event horizon is a light-like surface, so only photons can stay at rest at the horizon, you as a physical observer with mass will pass the horizon with the speed of light (better: the horizon as a light-like surface will pass with speed of light)

Falling towards a black hole you will see infalling light blue-shifted. When you start a rocket engine slowing down your free-fall towards to horizon, the blue-shift will increase.

Last edited: May 6, 2013
3. May 6, 2013

### George Jones

Staff Emeritus
Actually, if you started falling from a great distance, you will see the light red-shifted.

4. May 6, 2013

### tom.stoer

Why?

Radiation created near the horizon and observed far away is redshifted. So in the opposite direction I expect blueshift.

5. May 6, 2013

### George Jones

Staff Emeritus
This is for two static observers. The falling observer is not a static observer.

Suppose observer that A hovers at a great distance from a black hole, and that observer B hovers very close to the event horizon. The light that B receives from A is tremendously blueshifted. Now suppose that observer C falls freely from a great distance. C whizzes by B with great speed, and, just past B, light sent from B to C is tremendously Doppler reshifted. What about light from A to C? The gravitation blueshift from A to B is less that the Doppler redshift from B to C. As C crosses the event horizon, C sees light from distant stars redshifted, not blueshifted.

If observer A, who hovers at great distance from the black hole, radially emits light of wavelength $\lambda$, then observer C, who falls from rest freely and radially from A, receives light that has wavelength

$$\lambda' = \lambda \left( 1+\sqrt{\frac{2M}{R}}\right).$$
The event horizon is at $R = 2M$, and the formula is valid for all $R$, i.e., for $0 < R < \infty$. In particular, it is valid outside, at, and inside the event horizon.

See posts 5 and 7 in

I have since done the calculations using Painleve=Gullstrand coordinates.

Last edited: May 6, 2013
6. May 6, 2013

### tom.stoer

great, thanks!

7. May 6, 2013

### elevin

Totally get that one can't "stand there" but this is a thought experiment :) Can we say for a minute that hypothetically I'm an observing mass at rest at the center of the black hole looking outward (and I can see through all the layers of dense material and outward through the event horizon). I'm seeing the cosmos through a tremendous gravitational lens. It would seem that the wavelength of light coming toward the black hole would be accelerated or stretched towards me. Time dilation aside, and considering that this light would cross an event horizon on its way toward me, wouldn't that stretch this wavelength a considerable amount?

8. May 6, 2013

### DrGreg

You are asking what do the laws of physics predict in a situation where the laws of physics are being broken. There's no answer to that!

The best we can do is tell you about an observer hovering just outside the event horizon, or one inside the event horizon who is falling inwards.

9. May 6, 2013

### elevin

Schwarzchild and Pound-Rebka

The Schwarzchild gets exponentially bigger the more mass you pump into the hole. For example, our universe of 10^55kg (give or take an exponent) has an event horizon roughly approximate to its age, and yet we don't consider the universe a black hole. Seems to be a bit of a gray area on our definition of a black hole.

It would appear that the Schwarzchild is a measurement of mass/density on the path of light (and thus time/matter, etc). Time dilation aside, it would seem that in a theoretical pocket of extreme gravity (lets remove ourselves from the term black hole) of which we were at the center, the observer would experience redshift because of a stretching of a wavelength. The Pound-Rebka experiment would indicate otherwise however so perhaps I'm just talking nonsense. Of course, Pound-Rebka was done on earth. Can anyone clear this up?

10. May 6, 2013

### phinds

Say what ???

11. May 6, 2013

### elevin

HA! If you calculate the Schwarzchild radius of the Mass of the visible universe (roughly 10^80 atoms or 10^55kg) you ironically get its age equivalent. rs=2GM/c^2 (rs=schwarz, G is newtons G constant, M=mass).

Done my own calc, but below is from Wiki.
"The Schwarzschild radius of an object is proportional to the mass. Accordingly, the Sun has a Schwarzschild radius of approximately 3.0 km (1.9 mi) while the Earth's is only about 9.0 mm, the size of a peanut. The observable universe's mass has a Schwarzschild radius of approximately 10 billion light years.[citation needed]
(m) (g/cm3)
Universe 4.46×10^25[citation needed] (~10B ly)
Milky Way 2.08×10^15 (~0.2 ly)
Sun 2.95×103 1.84×10^16
Earth 8.87×10−3 2.04×10^27
An object whose radius is smaller than its Schwarzschild radius is called a black hole. The surface at the Schwarzschild radius acts as an event horizon in a non-rotating body (a rotating black hole operates slightly differently). Neither light nor particles can escape through this surface from the region inside, hence the name "black hole". The Schwarzschild radius of the (currently hypothesized) supermassive black hole at our Galactic Center would be approximately 13.3 million kilometres.

12. May 6, 2013

### Staff: Mentor

No, it doesn't. Plugging numbers for the universe into a formula that doesn't even apply to the universe doesn't prove anything.

No, just in your understanding of what a black hole is. It's a particular static solution to the Einstein Field Equation. Note carefully the "static" part; the universe is not static, so the black hole solution does not apply to it.

13. May 6, 2013

### phinds

Yes, I'm not arguing whether or not you can compute a Schwarzschild radius for anything, but do you really want to call it an "event horizon" ???

14. May 6, 2013

### Staff: Mentor

No, this is not correct. A black hole is a region of spacetime from which light cannot escape. Black holes can be formed by the gravitational collapse of massive objects, and once such a collapsing object's radius is smaller than the Schwarzschild radius for its mass, it is indeed inside the event horizon of the black hole that forms from its collapse. But that does not mean the object is identical to the black hole. The object collapses to a singularity and disappears, but the black hole remains.

15. May 6, 2013

### Staff: Mentor

You should, at least if "compute" means "compute something with physical meaning". Computing the "Schwarzschild radius" of an object that is not described by the Schwarzschild solution to the Einstein Field Equation is computing something that has no physical meaning.

16. May 6, 2013

### elevin

Obviously Schwarzchild metrics get far more complicated such as that of a spinning black hole, however in the most simple sense it seems to say that the mass of an object will affect the path of light. An object is what? A cluster of gravitating constituents. So just as our bodies are made of constituent cells/atoms/subatomic particles (and we indeed have mass), so are stars/galaxies/clusters and our universe, a body of collective constituents. It's only our perspective that puts them into discrete packets. That gravity has a falloff of 1/r^2, would we not find that another solar system in a very weak way would affect ours, that our galaxy, affect another galaxy (gravitationally)? Gravity is everywhere. With respect to this weakly gravitating field interspersed with pinpoints of mass (stars, galaxies), there seems to be an ironic comparison between (my thoughts on) residual gravity, the simple Schwarzchild metric (of which a center of mass would be located at the position of the observer), and the Hubble Constant.

17. May 6, 2013

### Staff: Mentor

The correct general term is "Kerr-Newman metrics"; this covers the whole range of possible black holes with angular momentum and electric charge as well as mass. The Schwarzschild solution is a special case of a Kerr-Newman metric with zero angular momentum and zero charge.

However, all of these metrics share a key property: they describe *isolated* systems surrounded by vacuum. (The technical term is "asymptotically flat".) The universe is not an isolated system, so it can't be described by any Kerr-Newman metric.

An *isolated* object, yes. As above, the universe is not an isolated object.

But the universe is not isolated, as all of your other examples are. Subatomic particles, our bodies, stars, galaxies, and clusters are all isolated objects surrounded by empty space. The universe is not.

18. May 6, 2013

### phinds

Fair enough.

19. May 6, 2013

### laurub

Is it known what the universe IS surrounded by?
If not, any ideas?

20. May 6, 2013

### Staff: Mentor

It's not surrounded by anything. According to our best-fit current model, the universe is spatially infinite; but even in models with a spatially finite universe, the universe is unbounded (its spatial topology is a 3-sphere), so there is nothing surrounding it.