# The vacuum light speed: Is it really constant, i.e., independent of location in space-time?

I am by no means an expert in this field, however something puzzles me about the speed of light and the relativity of time and space (space-time).

Is is universally acknowledged that the speed of light (299,792,458 m/s) is the universal speed limit, and that nothing can travel faster than light? That is a measurement based on a man-made interpretation of time (hours, minutes and seconds etc. are man made...there is nothing natural dictating how long a second should be).

For instance, according to Einstein, time and space bend around the physical matter of the universe, so for example, time near or on the surface of a "super massive black hole" should be drastically slower, relative to that of earth. Lets say for example that for every second that passes on the black hole, 10 seconds pass on earth, so essentially time on the surface of the black hole is 10 times slower than the time on earth.

Given the example above, is the speed of light at the surface of the black hole still 299,792,458 m/s, or is it 299,792,458,0 m/s?

• Lightspeed is measured as constant in local frames, i.e. when you measure it where it is. In your example with the black hole, if time is slower there as seen from a more distant observer, that observer will of course see the speed of light there as slower (not faster, as in your numerical figures...). – BjornW Sep 6 '12 at 9:13
• – user10001 Sep 6 '12 at 14:01

There are more ways to understand this, and the answer is the same: yes, the speed of light is constant. I will try to explain this in the way I consider simpler, but I am sure that others have their preferred explanation. Anyway, to receive a fair explanation, I suggest you to read something more detailed about special relativity, and then about general relativity. My explanation is geometric. Being mathematical, this means that you try first to understand what I mean in an imagined world of mathematics, to see that the ideas are consistent. Think at it as a sci-fi movie, in which you are only concerned with logical possibility. I suggest you that only after you are happy with the self-consistency of the model, you try to judge this image and compare it with what you know about physical world.

Spacetime is a space with four dimensions. Near each point, the spacetime is almost flat, but as we depart from the point, it becomes curved. On very short (infinitesimal) distances, the spacetime being almost flat, we can write a Pythagoras' theorem. In four dimension it is like

$$d s^2= - c^2d t^2+d x^2 + d y^2 + d z^2.$$

We are interested to make it work also in frames which are not normalized, and have different scales (denoted here by $g_{aa}$), and hence different measurement units:

$$d s^2= g_{00}d x_0^2+g_{11}d x_1^2 + g_{22}d x_2^2 + g_{33}d x_3^2.$$

Here I replaced $ct,x,y,z$ with $x_0,x_1,x_2,x_3$. But we also want to write this in coordinates whose axes are not necessarily orthogonal, so we have to add some cosines between the axies $a$ and $b$, which are written as $g_{ab}$:

$$d s^2=\sum_{a,b}g_{ab}d x_a d x_b.$$

This also works for curvilinear coordinates (we allow the metric coefficients $g_{ab}$ to vary from point to point), which are the suitable ones for curved spacetime.

The length is given by Pythagoras' theorem. Some infinitesimal distances are $d s^2>0$, and they separate points which can be in the same space. Some are $d s^2<0$, and they measure time intervals. Some are $d s^2=0$, and such directions are called light-like direction. So, if you measure the length of a curve described by a photon in spacetime, you measure it with this theorem, and you always obtain it to be $=0$. If you choose the reference frame so that $d x_2=d x_3=0$, and go back to $x,y,z,t$ notation, you obtain that

$$g_{11}d x^2+g_{00}c^2d t^2=0,$$

and the light speed is apparently

$$\frac{d x}{d t}=c\sqrt{-\frac{g_{00}}{g_{11}}},$$

which is not necessarily $=c$. Can we conclude that it is not constant? Well, not, because this formula doesn't show the speed of light in tehe units in which $c$ is expressed, but in some other units, which are scaled. To find the correct answer, we either choose the frame to be orthonormal, which gives $g_{11}=-g_{00}=1$, or we make sure to divide each infinitesimal distance with the unit "conversion factor" along that direction (they are just $\sqrt{-g_{00}}$ and $\sqrt{g_{11}}$). Hence, the speed of light is always $c$, although in rescaled coordinates may appear not to be. You cannot make it different, no matter how you try, unless you rescale it (i.e change the units).

Now, please see that this is not a proof that the speed of light is constant. It is constant by the very construction of the spacetime. It would be circular to claim that this shows the speed of light is constant. I showed you this construction to explain how it is consistent to have constant speed of light, even if time and space intervals change in different frames. Now you can compare this model with the physical data.

Now, the "true" speed of light is that present in the wave equation describing the light, in vacuum. And this is still $c$. Can it vary from point to point? It may be possible, in principle, but it is consistent with the observations that it remains constant. If it would vary, Maxwell's equations would not be covariant. This would not be a big deal, one can imagine worlds in which they are not covariant. But the theory of relativity originated from the study of their invariance.

• The mathematical aspects of this went way over my head, however given enough space-time...I might understand ;-) Thanks!, +1 – series0ne Sep 6 '12 at 10:21

The speed of light is always locally constant. If you do a local experiment to measure the speed of light the result will always be $c$. In this context local means a small enough area that spacetime curvature is insignificant.

However the speed of light is not constant on a larger scale. There's an easy way to see this. The Schwartzschild metric for a black hole is:

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

The co-ordinates used here are the ones a distant observer would measure i.e. if you and I are sitting on a space station far from the black hole $t$ is the time we measure on our clocks and $r$ is the radial distance we measure from the centre of the black hole.

Now suppose we consider a ray of light moving radially inwards (or outwards). Because it's radial $d\Omega$ = 0, and because it's a light ray it moves on a null geodesic so $ds$ = 0. The metric simplifies to:

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

and a quick rearrangement gives:

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

Remember that these co-ordinates are what we measure, so what we've ended up with is the equation for the speed of light that distant observers measure (in units where the speed of light is 1). Far from the black hole $dr/dt \approx 1$ as you'd expect, but as $2M/r$ approaches unity the speed slows and at $2M/r = 1$, i.e. at the event horizon, the speed of light is zero.

• The mathematical aspects of this went way over my head, however given enough space-time...I might understand ;-) Thanks!, +1 – series0ne Sep 6 '12 at 10:20
• The last equation should have a square on the left hand side or a $\pm$ instead of a square on the right hand side. I also don't like saying that $r$ is the raidal distance to the black hole wwhen you are far away, sicne you can't be far away and measure all the way to the event horizon. Saying it is the circumference of a spherical shell divided by $2\pi$ is fine, or even that it looks like a normal radial coordinate when you stay out there far from the black hole. – Timaeus Jun 15 '15 at 20:32
• @Timaeus: thanks. Please feel free to correct obvious typos in any of my posts. Re the $r$ coordinate, it's always a judgement call how much to explain when dealing with non-GR heads. In this case I think I struck about the right balance, though obviously not everyone will agree. – John Rennie Jun 16 '15 at 5:30

According to some theory space (vacuum) has a refractory index determined by the presence of mass. In optical dense medium light is retarded according to the value of the refractory index "n". The bending of light near the sun (Einstein-Eddington) has its ground through the polarization of the "vacuum" (zero point energy) having near the mass of the sun "n" larger than 1 (1.00001).

• "Some theory"? What theory, if you don't mind? – Deer Hunter Aug 6 '13 at 16:06

Well, so far, all experiments, that people have performed, have shown, that the speed of light in vacuum is constant. Whether on earth or in space, whether in the direction of the earth's orbit around the sun or in the opposite direction, the speed was always the same.

This led the famous physicist Einstein to the declaration, that the speed of light is ALWAYS constant, even at extreme situations (like close to a black hole), that are (as of today) completely inaccessible to our measurements. Einstein then did the maths to calculate the consequences of the constancy of the speed of light, and out came the theories of special and general relativity.

So far, both special and general relativity theory have passed ALL tests, that we've put them to. This is an extreme success. On the other hand, we DO know, that relativity theory and quantum theory come to completely different and contradicting result for example with respect to black holes. In the end, Einstein's relativity theory might be wrong at real tiny scales (like in the order of the so called Planck length, which is 1.6 * 10^-35 meters), or Einstein might be right, but quantum theory needs corrections in the presence of strong gravitational fields (alias space curvature).

We do know from experiments (like Quantum Cryptography), that quantum states are non-local, and that the state collapse, that happens due to measurements, affects the entire wave function of a quantum state immediately. That even applies to systems of macroscopic scale: For example, for two entangled photon, if the spin of one of them is measured, the spin of the other photon is affected immediately. So the "information" about the measurement somehow travels faster than light. Unfortunately (for all time-travel enthusiasts) or fortunately (for relativity theory), no "classical" information can be transferred that way. For example, if one of the two entangled photons was measured as spin-up, the other has to be spin-down. But if the first photon is measured as spin-left, the other becomes spin-right. Nonetheless, we cannot know, if that other measurement was already taken or not, and how it was taken: If the first measurement for example is for spin up or down, and the second for left or right, the second measurement yields arbitrary results independent of the first measurement.

Basically, the entanglement of both photons can only be verified, if both spin measurements are transmitted (back) to one place, where a judge then verifies, whether both measurements were equal. But that brings us back to Schroedinger's cat: Could it be, that both devices, that measure those two far-apart photons, actually send a superposition of signals to the judge, and only the judge then decides, if it was spin-up-down or spin-left-right?

To be honest: Even though quantum theory is used extensively, the whole thing about state vector reductions is actually still not understood till today. So the likelyhood, that relativity (and thus the constancy of the speed of light) holds true, and the quantum theory still needs some correction, is there. But it could easily be the other way round, that the speed of light is not constant at miniscule scales.

As pointed out by Michael Duff here, the speed of light in vacuum is a mere conversion constant. So, just as Michael Duff puts it in his article, you can just as well ask if the number of liters to the gallon is independent of location in space-time.