Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Imagine you have a star which is on the brink of turning into a black hole. Lets say it is infinitely close to become a black hole, but not there yet.

Since there is no event horizon, but a great gravity pull. Will the speed of light originating from this star be close to 0?

A follow-up question: What will the speed of light be after it travels far enough from the star to escape its gravitational pull?

share|improve this question
    
    
As noted in the other answers: The speed of light is the speed of light. It doesn't slow down it either escapes or it doesn't. –  user6972 Apr 1 at 22:28
    
@user6972: in GR the speed of light is only locally constant. –  John Rennie Apr 4 at 5:59

3 Answers 3

up vote 4 down vote accepted

One of the tricky things with general relativity is that different observers may use different coordinate systems and measure very different things.

The exterior geometry of any static spherically symmetric object is described by the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 d\Omega^2 $$

The only difference between a black hole and your almost black hole is that in your case this metric applies only to some minimum radius $r_{min} > r_s$ while for a black hole it applies everywhere.

The coordinates $t$ and $r$ correspond to times and lengths as measured by an observer at infinity, and as these will not be the same as times and lengths measured by observers nearer the star. If you bear this in mind, you can calculate the speed of light in the Schwarzschild coordinates simply by setting $ds = 0$ and choosing a radial path so $d\Omega = 0$, and you get:

$$ \frac{dr}{dt} = 1 - \frac{r_s}{r} $$

As you reduce $r$ towards $r_s$ the speed of light measured in Schwarzschild coordinates will indeed be reduced to below $c$, and will tend to zero as you approach the event horizon. As I said above, this happens for all static spherically symmetric objects whether or not they are black holes.

But you need to think long and hard about the physical significance of this. An observer hovering just above the surface of your collapsing star will measure the speed of light locally to be $c$, and so will a freely falling observer plunging towards the star's surface. In fact all observers making a local measurement of the speed of light will get the value $c$. The reason the observer at infinity measures a velocity of less than $c$ is that the different sets of observers disagree about the meaurement of times and lengths.

Re your last question, when light from the star's surface reaches the observer at infinity it will be moving at $c$ (as all locally measured light is) but it will have lost energy on the journey from the star and will be red shifted. This is the gravitational red shift and happens in all gravitational fields, even the Earth's. The only special thing about a black hole is that the red shift tends to infinity as you approach the event horizon.

share|improve this answer

The speed of light is a constant regardless of where or when you measure it. The speed of the light as it leaves the star will be $c=299792458\frac{m}{s}$. The speed of the same light far from the star will also be $c$.

Instead of slowing down like newtonian objects, the light will instead lose energy as it attempts to leave the star. This will correspond to a red-shift, or a decrease in the light's frequency. The light will look redder, but it will still be travelling at $c$

share|improve this answer

I assume you don't mean the speed of light, but you are essentially asking: Will light escape that strong gravitational pull?

If this is your question then first:

Direct quote from wikipedia -> "An object whose radius is smaller than its Schwarzschild radius ($r_s = \frac{2GM}{c^2}$) 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)"

Everything has a Schwarzschild radius. Ordinary object's Schwarzschild radius is very very small though (Earth's is only about 9.0 mm). If your star is infinitely close to become a black hole then its radius is equal to the Schwarzschild radius and light will not escape from it.

As for your follow up question, if any light escapes this Schwarzschild radius (because as noted by Jim, light could hypothetically escape if your star is not yet definitely a black hole) it will always travel at the speed of light $c = $ 299 792 458 m / s

share|improve this answer
    
clearly in this hypothetical, light will escape. The OP has established that the star is not currently a black hole. If light from the surface cannot escape, then it would be a black hole. –  Jim Apr 1 at 20:49
    
Well in my explanation I assumed that $r_s$ = the radius of the star. Thats how I comprehend the "infinitely close to becoming a black hole" statement. I guess it makes sense to argue that light will escape though. That's exactly why I added the last paragraph. I shall add it and change the word 'any' to 'if any'. –  PhotonicBoom Apr 1 at 20:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.