# Shapiro delay

1. Mar 6, 2014

### jartsa

Why is there a large Shapiro delay when a light beam passes a black hole?

Coordinate velocity of the beam is slowed down in a gravity field, except for the straight down component of the coordinate velocity?

Some part of the trip takes a long time anyway.

2. Mar 6, 2014

### Staff: Mentor

No. The Shapiro delay is caused by the distance between source and emitter being longer when a substantial (no need for it to be black hole) mass is nearby. Light still travels at speed c, but it has longer to go when the mass curves the space between source and emitter, so it takes longer to make the same journey.

(A fair case could be made this question belongs in a different thread)

3. Mar 7, 2014

### jartsa

The radar distance between black hole photon sphere and black hole event horizon is infinite, according to an observer at photon sphere.

The radar distance between black hole photon sphere and black hole event horizon is zero, according to an observer at rest at event horizon.

A radar beam sent towards the event horizon reaches the event horizon in finite time, according an outside observer.

A radar beam sent outwards from the event horizon will not reach an outside observer in finite time.

Right?

It seems like the space around a large mass is such kind of space that the distance from A to B is larger than the distance from B to A, if A is located below B.

Last edited: Mar 7, 2014
4. Mar 7, 2014

### Staff: Mentor

Yes, or any other radius r>R.

There is no such observer.

What do you mean by "time according to an outside observer" as applied to events on the horizon?

Yes.

It depends how you define the distances.

5. Mar 7, 2014

### Staff: Mentor

No, what your example shows is that under the conditions of that example (which is only vaguely related to the Shapiro delay) those distances are not easily defined.

6. Mar 7, 2014

### A.T.

It's not just the spatial distortion (more distance), but also gravitational time dilation (slower light according to distant clock).

7. Mar 7, 2014

### PAllen

No, it is sort of either/or, take your pick. The invariant observation is the delay. Pick one set of coordinates, and you have pure distance distortion; pick another and you have pure coordinate lightspeed change; or pick a mixture with funky coordinates.

8. Mar 7, 2014

9. Mar 7, 2014

### A.T.

What about Schwarzschild coordinates?

10. Mar 7, 2014

### pervect

Staff Emeritus
MTW works out the shapiro delay on pg 1107 in PPN coordinates. These are basically the same as isotropic coordinates (not the usual schwarzschild) - though the surfaces of time simultaneity will be the same for isotropic and schwarzschild coordinates.

They use the cartesian form of the isotropic / ppn coordinates (dt, dx, dy, dz).

MTW's derivation is approximate - they spend some time explaining why assuming light travels in a straight line in PPN coordinates (rather than the actual curved and not time-invariant path it actually follows) gives small errors, due to Fermat's principle.

The basic answer they get with these approximations is:

$$t_{TR} = \int_{-a_t}^{a_r} \left( \frac{g_{xx}}{-g_{tt}} \right) ^ {\frac{1}{2}} \, dx$$

where $-a_t$ and $a_r$ are the x-coordinate positions in the ppn system. (There is a parameter b in the result which is the constant y-coordinate in the ppn system as well). These are hard to observe, a later formula eliminates them by considering the rate of change of the Shapiro delay.

I would say that the answer that bests fits the above derivation is "both". Certainly we see $g_{tt}$ there, we also see the spatial terms $g_{xx}$. It's a bit hard to summarize this in the simplistic way the thread is being discussed :-(.

[add]
Probably the best simple answer is not an answer, but another question, i.e. to ask "what do you really mean by distance? in GR? How are you measuring it?".

There is certainly no conflict with the notion that the notion of proper distance is based on a constant speed of light though, which seems to be the thrust of the question. But the details of carrying out the calcuation (even with an approximation) typically involve setting up a coordinate system, using changes in coordinates (dx - which isn't a distance), and calcuating the proper time by using the coordinate change dx and the non-constant coordinate speed of light (the number in the square root).

MTW makes a few more approximations, and subtract out a Newtonian part, and get an answer that depends only on the ppn parameter $\gamma$

There may or may not be other ways of calculating it, this is what my textbook does though.

Last edited: Mar 7, 2014
11. Mar 9, 2014

### A.T.

That was my point.

12. Mar 9, 2014

### PAllen

In SC coords, coordinate light speed plays a role. But in radar coordinates, the same prediction is purely from spatial distortion.

13. Mar 9, 2014

### pervect

Staff Emeritus
Personally, I think of distance informally as the length of the shortest curve connecting two points - more formally this mutates into "the greatest lower bound" or infimum of the set whose members are the lengths of all curves connecting two points. (We replace the concept of minimum for a finite set with the concept of the infimum which applies to finite and infinite sets, the two concept are equivalent for finite sets). In the context of physics, we also require that the curves be in a hypersurface of constant time, this construction then gives "distance as a function of time".

When one computes this distance by means of coordinates, we can see from the above example that the coordinate expression of gravitational time dilation and the coordinate expression of spatial curvature both play a role in computing the distance in that coordinate system. However, this should not be interpreted as the local speed of light varying - the local speed of light is always equal to c. Furthermore we note that the distance (by this definition) really does increase when the connecting curve passes near a massive object - again the point is not to "throw away" the fundamental notion that the speed of light when measured locally is a constant.

14. Mar 9, 2014

### yuiop

The horizontal coordinate velocity of light (travelling at constant radius from the gravitational centre) is slower than the local horizontal speed of light, by a factor of sqrt(1-2m/r) in the Schwarzschild metric, which is purely due to gravitational time dilation. The vertical component of the coordinate speed of light that is slower by a factor of (1-2m/r), which you can think of as a slow down due to gravitational time dilation by a factor of sqrt(1-2m/r) and an additional factor of sqrt(1-2m/r) due to spatial distortion.

15. Mar 10, 2014

### A.T.

Yes, according to a local clock. But the whole point of the Shapiro delay is what duration a distant observer measures with his clock.

16. Mar 10, 2014

### jartsa

Let me think ... I mean we have clocks with gravitational time dilation elimination mechanism, these clocks, when lowered slowly to various altitudes, can be used to measure times.

As an example of such clocks: light clocks that shrink in such a way that gravitational time dilation is cancelled.

Last edited: Mar 10, 2014
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