# Almost black holes

1. Jun 9, 2012

### mrspeedybob

If there were a star which was almost, but not quite massive enough to become a black hole it seems as though gravitational time dilation should make it appear very dim. If time dilation resulted in 1/1000 of the time passing within the star as passes for us, we should see a star emitting 1/1000 the energy that it's mass would otherwise suggest. Do we see such stars in the universe?

2. Jun 9, 2012

### Bobbywhy

Just an off-the-wall question from a non-astronomer: Would these "almost black holes" exhibit an intrinsic red-shift? I think I remember that Halton C. Arp proposed that for explaining the red-shift of quasars.

3. Jun 9, 2012

### Drakkith

Staff Emeritus
No, such objects can only be Neutron Stars, which are extremely tiny in size, yet comparable to the mass of the Sun. There are no stars that are still fusing elements in their cores that are anywhere close to being a black hole.

Your question should apply to a neutron star though. I would expect to see significant redshift on radiation coming from a neutron star.

4. Jun 10, 2012

### ImaLooser

A star that is not quite a black hole is a neutron star. The energy that comes off of a neutron star is red shifted, but I don't know how much.

Last I looked only one neutron star had been directly observed. (Usually we see the glow of the surrounding matter.) But that should enough to get a real figure.

5. Jun 10, 2012

### stevebd1

The smallest a neutron star can get before run away collapse occurs is r0=9M/4 or 2.25M in order to have positive tangential pressures (according to the Schwarzschild interior metric, this is the point that the time dilation at the very interior of the star becomes zero). If we consider the gravitational redshift-

$$z=\frac{1}{\sqrt{1-\frac{2M}{r}}}-1=\frac{\lambda_o-\lambda_e}{\lambda_e}$$

where z is the redshift, $\lambda_o$ is the wavelength observed and $\lambda_e$ is the wavelength emitted. The above can be rewritten-

$$\lambda_o=(z\cdot\lambda_e)+\lambda_e$$

if we consider a 3 sol mass as an absolute maximum, then M=4430.55, then you can pretty much work out what z would be and what the emitted wavelength of light would be shifted to.

Regarding gravitational time dilation, this would be expressed as-

$$d\tau=dt\sqrt{1-\frac{2M}{r}}$$

so the greatest time dilation due to gravity at the surface of a stable neutron star at the very boundary of collapse would be 0.333 or ~1/3.

Last edited: Jun 10, 2012
6. Jun 10, 2012

### Chronos

7. Jun 11, 2012

### ImaLooser

Duh.

What's M? What's z?

8. Jun 11, 2012

### Drakkith

Staff Emeritus
I believe M = mass, and Z = the measure of redshift.

9. Jun 12, 2012

### Chronos

Yes, but, just to clarify, M is expressed in units of solar mass.

10. Jun 12, 2012

### Rorkster2

I don't think it would appear dim to us. Light would travel very slowly away from it when released because of massive gravity pulling backwards, but as soon as the light moves far enough away it would resume its 'speed of light' pase. And reach us just like a normal star. Unless gravity pulled back some photons and not others it would have the same luminosity

11. Jun 12, 2012

### Bobbywhy

It appears you have a mistaken view of "gravitational redshift". It is true when we say photons "climb out of the gravitational potential" of a massive object. But this does not mean their velocity is changed...they still travel at c. What happens is just what's been mentioned above: redshift. This term comes from the optical part of the EM spectrum where red is a longer wavelength than, say, blue. This means the WAVELENGTH of the light has been stretched out and its "color" will be different (redder). Light never "slows down" or "resumes its pace", at least in this kind of example.

12. Jun 12, 2012

### Algr

Does this mean that one second passes on the nutron star for every 3 seconds outside of it? If so, then I'd think that that would indeed dim the star by a factor of 3. Regardless of red shift, the star would produce one third the number of photons in 1 second that it would in 3. And an observer on the neuron star would see the rest of the universe as sped up, and three times as bright.

13. Jun 12, 2012

### Rorkster2

Sorry to say but i believe your wrong. First off, light can indeed change its speed. For example, in near absolute zero it is possible to slow photons to a crawl, and also 'speed of light' refers to its speed when traveling threw a vaccuum, not its constant speed. Next, you are right in saying that the redshift and blueshift waves are a derived from its frequnecy, however in this example you also have to include the fact that light DOES slow down, because black holes change light momentum from going 'forward' to 'backwards', and a near blackhole object would be in the very close inbetween where it slows down but does not quite go 'backward'

14. Jun 12, 2012

### Drakkith

Staff Emeritus
It is a well known fact that light does indeed change speeds in a material, but when anyone says that the speed of light doesn't change or is invariant they are referring to the velocity in a vacuum, which is c always.

This is incorrect. The change in momentum in no way affects the speed of light. As for the forward and backward stuff, I'm not sure what you are getting at.

15. Jun 12, 2012

### Chronos

The time dilation factor goes by (1 + z), so a neutron star with a gravitational redshift of .33 would be time dilated by 1/1.33, or basically it would take 1.33 seconds to emit the same amount radiation as it would radiate in 1 second if not redshifted.

16. Jun 12, 2012

### ImaLooser

Aha. Well, I'm duh again because I don't know what it means to express radius of an object in units of mass.

17. Jun 13, 2012

### Drakkith

Staff Emeritus
The radius isn't being expressed in units of mass. You are dividing the mass by the radius because more mass packed into a smaller radius means more gravity and more time dilation.

18. Jun 14, 2012

### stevebd1

In the paper posted by Chronos, M refers to mass, in the post I made, M is the geometric unit of mass, sometimes referred to as the gravitational radius where M=Gm/c2 where G is the gravitational constant, m is the mass in SI units (i.e. kg) and c is the speed of light in m/s.

19. Jun 18, 2012

### grantwilliams

If a photon is considered to be without mass how can light have momentum?

note: my only exposure to physics has been in an AP mechanics class and in the books i've read over this summer. (i won't begin more formal study until i begin college next fall)

20. Jun 18, 2012

### Drakkith

Staff Emeritus
The momentum and energy of a photon is determined by the frequency. Higher frequency photons have more momentum and energy. You can read a bit more here: http://en.wikipedia.org/wiki/Photon#Physical_properties

If that isn't detailed enough or doesn't answer your question just say so.