# Temperature during black hole collapse

1. Oct 29, 2009

### nickyrtr

Let's say you are an observer on the surface of a massive object that is collapsing to form an eventual black hole. As the object's surface approaches its own Schwarzschild radius, light from the object is increasingly redshifted, as seen by distant observers, approaching infinite redshift as the event horizon forms.

Correspondingly, you on the object surface see that light from distant sources is blueshifted. Now let us assume that the top layer of the object surface remains in thermal equilibrium with the incident radiation, consisting of the cosmic microwave background plus light from other stars and galaxies. As the object shrinks closer to its Schwarzschild radius, the blueshift raises the effective temperature of the cosmic background (and other sources), so it would seem that the object's surface temperature becomes infinite!

Surely this infinite temperature is unphysical, so what is the mistake in this picture I have described? Is the assumption of thermal equilibrium false? Do I misunderstand what blueshift means or how it works? Any explanation is appreciated.

2. Oct 29, 2009

### George Jones

Staff Emeritus
The blueshift isn't infinite, and, where blueshift might be expected, there can even be a redshift.

Consider an observer that freely falls along a radial worldline feet-first into a black hole. The stars directly above the observer are redshifted for the observer. Roughly, for these stars, Doppler redshift dominates gravitational blueshift.

Also, as you note, the time taken for stellar collapse might be too short to reach (either way) thermal equilibrium.

3. Oct 29, 2009

### nickyrtr

Well I was picturing an observer who is not falling freely, but who is supported by the surface of the collapsing object. This is the (non-inertial) reference frame of the matter that makes up the surface of the object itself.

For this surface-bound observer is the blueshift still always finite?

4. Oct 29, 2009

### George Jones

Staff Emeritus
Yes, as long the surface observer doesn't experience unbounded 4-acceleration (infinite g-force) during the collapse.

5. Oct 29, 2009

### nickyrtr

Interesting. Will this surface-bound observer also reach the event horizon in finite proper time?

6. Oct 29, 2009

### George Jones

Staff Emeritus
I wrote this without doing a calculation, and without looking at any references, i.e., I used

Wheeler's First Moral Principal: Never make a calculation until you know the answer. ... Courage: no one needs to know what the guess is. Therefore make it quickly, by instinct. A right guess reinforces this instinct. A wrong guess brings the refreshment of surprise.

I'll take a closer look.
Yes. (Without using Wheeler's First Moral Principal)

7. Oct 29, 2009

### Ich

I'm sure you're right with the acceleration. You don't need maths, just the comparison with the local free falling frame, like Rindler-Minkowski.

8. Oct 29, 2009

### nickyrtr

Perhaps my mistaken assumption was that light moving toward the collapsing star and light moving away from the star are affected equally by gravity. For example I found this formula on Wikipedia for the gravitational redshift of light emitted outward by the star:

$$z = \frac{1}{\sqrt{1-\frac{r_s}{r}}}-1$$

Of course this diverges as the star's radius approaches rs, but perhaps this formula does not apply to light propagating inward, toward the star. Is there a different formula for the blueshift of inward-propagating light?

9. Oct 30, 2009

### George Jones

Staff Emeritus
No, but this expression is valid only for an observer who has a fixed $r$ value in Schwarzschild coordinates; I call this type of observer a hovering observer.

Consider another "falling in" observer B in the neighbourhood of a hovering observer A, but with very slightly smaller $r$ than A. What wavelength shift for distant stars does B see? Due to relative motion between A and B, there is a Doppler redshift between A and B, so B sees the gravitational blueshift between the stars and A combined with the Doppler redshift between A and B.

Now, let A and B both approach the event horizon. A has unbounded 4-acceleration, and, if B has bounded 4-acceleration, the relative speed between A and B approaches the speed of light. Consequently, for B there is an infinite Doppler redshift combined with and infinite gravitational blueshift. The net result depends on the rates at which these two infinities are approached, and could be zero, finite redshift or blueshift, or infinite redshift or blueshift.

I maintain that there is finite redshift or blueshift for B with bounded 4-acceleration,

10. Oct 30, 2009

### Ich

The relative velocity of an observer infalling from infinity is $$\sqrt{\frac{1}{r}$$, IIRC. Combine that with your formula, you'll find z = 1 at the horizon.

11. Oct 30, 2009

### George Jones

Staff Emeritus
This is not what I get for the speed. Typo? For a derivation of the combined effect (which gives Ich's z result), see

https://www.physicsforums.com/showthread.php?p=861282#post861282.

I'm not sure if the last mathematical line is showing up correctly. The final result should be

$$f' = \frac{f}{1+\sqrt{\frac{2M}{r}}}.$$

Last edited: Oct 30, 2009
12. Oct 30, 2009

### Ich

Yes, 1=2M, of course. But I see that you've calculated and posted all this long before.

13. Oct 30, 2009

### nickyrtr

OK, I think I understand now, thank you. Only a "hovering" observer ever sees infinite blueshift, and a surface-bound observer cannot possibly be hovering if the surface is contracting. Therefore, the contracting surface never sees an infinite effective temperature from the infalling radiation.

EDIT: Furthermore, it is impossible to hover long enough to even see the infinite blueshift, because you need an infinitely strong force to maintain the hovering!

Last edited: Oct 30, 2009
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