I'm on Earth. Joel is on a planet orbiting near a black-hole's horizon. Every year on Joel's planet is equivalent to 10000 years on Earth due to the gravity of the BH. The two planets are 10000 light-years apart. I send a message to Joel via light. 10000 years later Joel got the message, but only 1 year for Joel, because of the time difference. The distance between the two planets and maximum speed of light is the same for both of us. Therefore it should take 10000 years for both of us. Why Joel got the message so fast?

  • $\begingroup$ The speed of light is not constant when there is gravitational field present $\endgroup$ Apr 4, 2015 at 6:58
  • $\begingroup$ @Rogelio, does it get faster than 299,792,458 m/s? $\endgroup$ Apr 4, 2015 at 7:02
  • 1
    $\begingroup$ The speed of light in a vacuum is. $\endgroup$ Apr 4, 2015 at 7:05
  • $\begingroup$ Sorry for the wrong title, updated. $\endgroup$ Apr 4, 2015 at 7:07
  • $\begingroup$ I believe wormhole is a solution to Einstein's equation and you may need to collect some hypothetical exotic matters for it to work. Speed of light is the same no matter the observers, see relativity written by same guy that give you wormhole and I wouldn't recommend you to have a vacation at P which is close to B. $\endgroup$
    – user6760
    Apr 4, 2015 at 7:11

2 Answers 2


While it's often stated that the speed of light is a constant, this is an oversimplification. It's more precise to say that light follows a path called a null geodesic.

Suppose we are using some coordinate system $(t, x, y, z)$ to measure the positions of spacetime points. If we observe the light to move an infinitesimal distance in space $dx, dy, dz$ in an infinitesimal time $dt$. Then the proper distance moved by the light is given by:

$$ ds^2 = g_{\alpha\beta}x^\alpha x^\beta \tag{1} $$

where $g$ is the metric tensor. When we say light follows a null geodesic we mean that equation (1) will always give the result $ds = 0$. The proper distance $ds$ is an invarient i.e. all observers will agree on its value.

In special relativity the metric tensor takes a simple form, and equation (1) becomes:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{2} $$

Remember that for light $ds = 0$, so if we put this into equation (2) and rearrange it we get:

$$ c^2 = \frac{dx^2 + dy^2 + dz^2}{dt^2} $$

But if you recall Pythagoras' theorem, $dx^2 + dy^2 + dz^2$ is just equal to the total (coordinate) distance moved squared, $dr^2$, so our equation becomes:

$$ c = \frac{dr}{dt} $$

And this just defines a velocity $dr/dt$ for the light, and it tells us that velocity is always equal to $c$. This is the origin of the claim that the velocity of light is the constant $c$.

However in general relativity the metric $g$ is more complicated. For example, for a static black hole the metric (in polar coordinates this time) is the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{2GM}{r c^2}\right)c^2dt^2 + \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 + r^2 d\Omega^2 \tag{3} $$

Even though this looks a lot more complicated, we can calculate the speed of light in the same way by setting $ds = 0$. We'll make the additional simplifaction that the light is moving in a radial direction so $d\Omega = 0$. In that case we get:

$$ 0 = -\left(1-\frac{2GM}{r c^2}\right)c^2dt^2 + \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 $$

And this rearranges to:

$$ \frac{dr}{dt} = \left(1-\frac{2GM}{r c^2}\right)c $$

So in this case the speed of light is not $c$. In fact it's less than $c$ for any value of $r$ less then infinity.

The speed of light I've calculated here is the speed as observed by an observer far from the black hole, and it's always lower than $c$. However if you were hovering just outside the event hole you'd observe the speed of light to be less than $c$ everywhere nearer the black hole than you, but faster than $c$ everywhere farther from the black hole than you. That's why in the situation you describe light can travel faster than $c$.

However it's important to state that if you make a local measurement of the speed of light i.e. measure it at your position then the speed you measure will always be $c$. This is an important principle in general relativity. No matter how curved the spacetime, in your immediate vicinity it is always (approximately) flat and special relativity applies.

  • $\begingroup$ Oh, I really appreciate the effort. Never thought someone give me such a long answer, and sorry that I'm not able to understand it solely. When you say faster than $c$, do you mean faster than the local $c$? $\endgroup$ Apr 4, 2015 at 9:32
  • $\begingroup$ The local speed is always $c$, i.e. 299,792,458 m/s, so I mean faster than 299,792,458 m/s. $\endgroup$ Apr 4, 2015 at 9:35
  • $\begingroup$ How much local is local? $\endgroup$
    – Paul
    Apr 4, 2015 at 11:24
  • $\begingroup$ @Paul: that would need to be a new question. $\endgroup$ Apr 4, 2015 at 11:38
  • $\begingroup$ Yeh,okey.But I have another question,light doesnt really travel at faster than the speed of light ,we just observe it to be going faster than 'c' otherwise that would mean information travelling faster?right? $\endgroup$
    – Paul
    Apr 4, 2015 at 11:59

Common misconception: the speed of light in a vacuum is constant. It's not proper to say "speed of light"- because, the speed of light can be slowed in a gravitational field. Heck, traveling at v=0.8c, there could be a light beam that gets slowed to 0.75c=v because of gravity- voila! You're traveling faster than light!!!! I always like to think that since c is a peculiarly chosen max speed, it is more like the information loading speed of spacetime. It 'predicts' one planck length each planck time...

So, yes, in response to what you most likely meant, the speed of light in a vacuum is a constant, as far as we know. 299792458 m/s. My favorite number.


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