# Does the speed of light depend on the I and w of the light?

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1. Jul 16, 2015

### ORF

Hello.

Let's consider a beam of monochromatic light (just one frequency).

1. Light creates gravity field.
2. Gravity affects the speed of light.

Q1: Are 1. and 2. true?
Q2: Does the speed of light depend on the intensity and frequency of the light?

Thank you for you time :)

Greetings!

2. Jul 16, 2015

### Staff: Mentor

Yes.
No. More specifically, the locally-measured speed of light (in a vacuum of course) is always c. That is, the speed of a light pulse as it whizzes past you, measured by instruments that are sufficiently near your location that the effects of spacetime curvature are negligible, is always c.

3. Jul 16, 2015

### ORF

Hello

Thanks for your answer. There is one thing that I don't understand clearly: when you says "the effects of spacetime curvature are negligible", are you referring to the curvature caused by light?

Greetings!

4. Jul 16, 2015

### Staff: Mentor

I was referring to spacetime curvature in general. I don't know whether the spacetime curvature produced by an object affects the object itself, assuming of course that the object is sufficiently small.

5. Jul 16, 2015

### PAllen

That, or any other curvature. The locally measured speed of light in vaccuum being c is universally true in GR. Statements about light being slowed by gravity are about inferred speed of light 'somewhere else' using some natural coordinate system.

6. Jul 16, 2015

### PAllen

I guess I should add that as an intense, high frequency, light is passing you by such that you are inside it, is a case not covered by jtbell's statement (or mine). In this case, the region around you is not even close to vacuum in the GR sense (assuming the intensity is such as to cause detectable gravity). Note that electromagnetic fields are not vacuum. The universal statements given previously would apply if you are measuring such a pulse going by a little away from you, but not so intense as to distort your instruments or make it problematic to define a sufficiently extended local inertial frame in which to make the measurement.

Also, note that routine statements about light traveling c in vacuum apply to plane waves. By manipulating light pulses to substantially differ from a plane wave approximation, labs have measured light speeds slightly slower than c in vacuum, and this has nothing to do with gravity.

Dealing accurately with gravitationally significant light intensities in GR is complicated business and gets into areas where it is not clear it is valid to use pure classical treatment. For example, classically, you can get things like colliding EM waves producing singularities. Many physicists would suggest that simply means the classical treatment is no longer applicable.

7. Jul 16, 2015

### bahamagreen

When light's path is curved, is c the magnitude of the vector tangent to the curve or the vector sum of that tangent vector and the lateral vector pointing into the curve... which is the one assumed referencing light at c in vacuum?

8. Jul 16, 2015

### PAllen

The magnitude of a tangent vector to a null world line is identically zero everywhere per the mathematical structure of GR. There are no exception to this - it is really the definition of a null world line. The local speed of any null or timelike world line can only be defined relative to some orthonormal frame at an event on it. Introducing such a frame (there are infinitely many, corresponding the the group of all rotations and boosts), you then can take the scalar product of the the tangent 4-vector with each basis vector. Then, the Euclidean norm of the spatial scalar products divided by the timelike product gives you the mathematically defined local speed. Note, the arbitrary scaling of a null vector cancel out from the division. The result of this exercise is that you get c, identically, with no exceptions, in GR.

However, all of this is geometrical optics. The OP seemed to be asking about classical EM waves where the amplitude and frequency matter, rather than the geometrical optics approximation (which is what using null geodesics gives you).

9. Jul 16, 2015

### Staff: Mentor

If you choose a small enough region around yourself, you can always treat a curved spacetime as if it is flat. This somewhat analogous to the way that we can treat the surface of the earth as flat when we're using stretched strings to lay out the foundation of a building, but not when we're laying out an aircraft flight plan from Capetown to Buenos Aires. The smaller the region you consider, the better the "it's flat" approximation gets.

In a curved spacetime, the notion of speed (meaning change in spatial position with time) is only meaningful in a region small enough that it can be treated as flat, so when you're talking about the speed of light being $c$, we're talking about its speed in the locally flat region around us.

10. Jul 17, 2015

### bahamagreen

I'm thinking of scenarios like the bent path of light passing the Sun...

For example, the Earth has a velocity tangent to its orbit in the orbital plane, but also has an acceleration directed toward the Sun; in the case of light passing the Sun, is c the tangential velocity or does it comprise also the lateral motion?

11. Jul 17, 2015

### PAllen

You seem to be confusing Newtonian physics and GR. By definition, any geodesic (null for light, or timelike for a massive body) has zero proper acceleration, always. That is a direct consequence of the definition of geodesic, and physically, of the principle of equivalence.

As for locally measured velocity of light, I already explained that at every point in light's path, all the infinite possible observers in different states of relative motion measure the speed of the light as exactly c. The direction will vary by the observer.

In GR, velocity of something at a distance from you (whether a material body or light) has no invariant meaning. Curvature makes it fundamentally ambiguous. Depending on coordinates chosen, it can be whatever you want.For a static GR solution, you can introduce coordinates such that the speed of light varies, and the path of light is determined by Fermat's principle according to varying speed expressed in such coordinates. However, this feature is very specialized (only static solutions) and totally coordinate dependent. Einstein used this approach in his original derivation of light bending, but essentially all modern textbooks do not use this approach - instead they analyze using null geodesics in Schwarzschild geometry (because this approach generalizes, while Fermat's principle does not rigorously apply to our universe anywhere; there are no static regions in our universe).

12. Jul 17, 2015

### Staff: Mentor

Neither. The curvature effects are not negligible in this case, so you can't calculate the speed of the flash of light by dividing the distance it travels by the time it took to cover that distance - that only works when you and both endpoints of the distance are located in the same approximately flat region of spacetime. The problem is defining the the time it took to cover the distance - that requires assigning times to distant events, which cannot be done unambiguously in curved spacetime.

13. Jul 17, 2015

### Staff: Mentor

It's worth noting that this issue is not purely a coordinate issue. The coordinate-independent way of describing the difference is that, in a curved spacetime, parallel transport of vectors is path-dependent, whereas in flat spacetime, it isn't. In flat spacetime, we could still choose wacky coordinates that make the coordinate "speed" of light something other than $c$; but since parallel transport is not path-dependent, there is always a unique way of comparing two vectors at different spacetime points, and if one of the vectors is null and the other is timelike, this comparison will always yield $c$ as the relative velocity. In curved spacetime, this no longer works, since parallel transport is path-dependent; even if we choose the "nicest" possible coordinates, we can't get a unique answer if we try to compare vectors at different spacetime points. Vectors can only be compared locally, i.e., at the same spacetime point, so the concept of "relative velocity" is only well-defined for two objects at events where they are co-located.

14. Jul 17, 2015

### PAllen

True, those two sentences were coupled in a way not intended.

A further invariant statement that can still be made is that parallel transport, though path dependent, never changes the classification of a vector (timelike, null, spacelike) because it is norm preserving. Thus, while comparing the speed of material bodies at a distance is fundamentally ambiguous, the relative speed between a timelike and null vector is actually path independent - you get c for the relative speed for every possible path.

15. Jul 17, 2015

### Staff: Mentor

Hm, I see what you mean; in this case the "frequency/wavelength" of the light would be path dependent, but the "relative speed" would not be.