A Speed of Light for Bonnor Beam

[diracs-cat]

TL;DR Summary

In general relativity, does light "in a vacuum" actually move at "c" as viewed by a distant observer after accounting for the light beam's gravitational influence on itself?

Background: The "Bonnor beam" is an exact solution to the Einstein Field Equations for an infinitely-long, thin cross-section of electromagnetic radiation. It is an example of a "p-p wave" solution of the EFEs.

The interior metric solution is as follows, where m is the energy density of the electromagnetic wave, and r and $\theta$, u, and v are a cylindrical form of Brinkmann coordinates defined by $u \equiv \frac{1}{\sqrt{2}} \left( z - t \right), v \equiv \frac{1}{\sqrt{2}} \left( z + t \right).$

$$
\left\{
\begin{array}{lr}
ds^2 =
-8 \pi m r^2 \, du^2 - 2 \, du \, dv + dr^2 + r^2 \, d\theta^2,\\-\infty < u,\\
v < \infty,\\
0 < r < r_0,\\
-\pi < \theta < \pi.\\
\end{array}
\right.
$$
My Question: Am I correct in assuming that since this metric represents curved spacetime including gravitational time dilation, this means that the light beam will actually appear to be traveling slower than the speed of light as viewed by a distant observer (assuming the light beam has non-zero frequency and energy density)? In other words, it appears that general relativity actually predicts that light "in a vacuum" doesn't actually travel at "c" as viewed by a distant observer. Is that really correct, or am I missing something? Any insight would be greatly appreciated!

 

[Ibix]

Light always travels at $c$ viewed locally. Speed "viewed from a distance" isn't a uniquely defined quantity and you certainly can get such things that aren't equal to the locally measured speed. This is true in much less exotic circumstances - for example it's one way of interpreting Shapiro delay in radar ranging of Venus.

I haven't actually checked the metric you asked about, but it's more than possible that you can find a way to define velocity from a distance such that the speed so defined is not $c$.

 

[PeroK]

TL;DR Summary: In general relativity, does light "in a vacuum" actually move at "c"

A simpler example is to consider a light pulse travelling backwards and forwards between points above and below each other in a gravitational field. The time between pulses must be greater as measured at the higher point. Owing to gravitational time dilation. If we assume that we can measure a fixed proper distance between the points, then the ratio of "proper distance travelled divided by coordinate time" must be different as measured using local clocks at the two points. In other words, coordinate speed cannot be the same for both.

In curved spacetime, the invariance of the speed of light applies to local measurements. In the above case, the two measurements converge to $c$ as the distance between the points reduces to zero. As in all physics, a local measurement of speed is something of a theoretical construct, made possible by the mathematics of the differential calculus.

 

[Orodruin]

As has already been alluded to: There is no such thing as “velocity viewed at a distance” in general relativity. Only relative velocities for co-located objects at a particular event. The underlying reason for this is that GR spacetime is generally curved, which means velocities (more specifically 4-velocities) at different events cannot be uniquely compared. A Riemannian analogy would be that it does not make much sense to compare velocities along the surface of the Earth at diffferent positions.

 

[PeterDonis]

this metric represents curved spacetime

Yes.

including gravitational time dilation

No. "Gravitational time dilation" is only a meaningful concept in a stationary spacetime. This spacetime is not stationary, so it does not have a meaningful concept of "gravitational time dilation".

[Edit: corrected in post #10 below.]

it appears that general relativity actually predicts that light "in a vacuum" doesn't actually travel at "c" as viewed by a distant observer.

No, that's not what GR says. What GR says is that, in a curved spacetime, the concept of the "speed" of a light beam (or anything else, for that matter) that's at a distant location from an observer is meaningless. The only meaningful measurements of "speed", or more precisely relative speed, in a curved spacetime are local--at the observer's location.

 

[Ibix]

This spacetime is not stationary,

Are you sure? Plugging in the definition of $u$ and $v$ in terms of $z$ and $t$ gets you a form of the metric that's independent of $z$, $t$, and $\theta$, suggesting Killing fields associated with those coordinate fields, and $\partial_t$ appears to be timelike. And that's consistent with an infinitely long beam propagating along its own length, I think. There's spatial cylindrical symmetry and time translation symmetry because nothing really changes.

I haven't bothered to check if it's static, although I wouldn't be surprised if it's not - I think there's directionality to the flow of energy along the beam so I wouldn't expect symmetry under $z\rightarrow -z$.

 

[PeterDonis]

an infinitely long beam propagating along its own length

...would be propagating along a null worldline, not a timelike one, since this is EM radiation. So the "cylinder" of which the beam itself is the "center" would lie along a null direction, not a timelike one.

 

[Ibix]

...would be propagating along a null worldline, not a timelike one, since this is EM radiation. So the "cylinder" of which the beam itself is the "center" would lie along a null direction, not a timelike one.

Yes, but it's infinitely long and "thin", so it's best described as a sheet in the z-t (or u-v) plane. And I don't think there's anything to choose between events in that plane.

 

[Orodruin]

...would be propagating along a null worldline, not a timelike one, since this is EM radiation. So the "cylinder" of which the beam itself is the "center" would lie along a null direction, not a timelike one.

I don’t see how that is an argument for the spacetime itself not being stationary. Nobody is claiming the propagation worldline is a timelike killing field (although it is everywhere except at the beam itself).

Edit: The full beam is a world sheet as @Ibix points out.

 

[PeterDonis]

Killing fields

It's manifest that $\partial_u$ and $\partial_v$ are Killing fields. $\partial_v$ is null everywhere, and $\partial_u$ is null at $r = 0$, which is where the beam is supposed to be.

Since (I think) $\partial_z$ and $\partial_t$ are linear combinations of $\partial_u$ and $\partial_v$ with constant coefficients, they would also be Killing fields. So if $\partial_t$ is indeed timelike, then yes, you're right, the spacetime would be stationary.

 

[PeterDonis]

I don’t see how that is an argument for the spacetime itself not being stationary.

You and @Ibix are correct that it isn't. See my post #10 just now.

Nobody is claiming the propagation worldline is a timelike killing field (although it is everywhere except at the beam itself).

What is "the propagation worldline"? Is $\partial_u$ tangent to it?

 

[PeterDonis]

Yes, but it's infinitely long and "thin", so it's best described as a sheet in the z-t (or u-v) plane. And I don't think there's anything to choose between events in that plane.

So the beam would not be propagating "along its length". The "infinitely long", i.e., "along the length of the beam", defines one direction, and the "direction of propagation" defines another, different direction, and those two directions together span a 2-plane.

 

[Ibix]

So the beam would not be propagating "along its length". The "infinitely long", i.e., "along the length of the beam", defines one direction, and the "direction of propagation" defines another, different direction, and those two directions together span a 2-plane.

I don't think so - that wouldn't be stationary. I think $\partial_u$ is the wave vector, and $\partial_z$ is the "direction in space" of the beam, both the axis of cylindrical symmetry and the flow direction of the energy.

 

[Orodruin]

I don't think so - that wouldn't be stationary. I think $\partial_u$ is the wave vector, and $\partial_z$ is the "direction in space" of the beam, both the axis of cylindrical symmetry and the flow direction of the energy.

This should be relatively easy to check by simply computing the stress-energy tensor of the interior solution.

Edit: From what I gather, this is not a vacuum solution outside an infinitely thin beam. It has an interior and an exterior solution, the interior solution corresponding to the cylindrical propagating beam.

 

[diracs-cat]

Thanks for the interesting discussions and feedback everybody! I'm not sure I understand why non-stationary spacetimes would have "no gravitational time dilation" rather than just having gravitational time-dilation that is time-varying (or perhaps not uniquely defined), but that's not quite my main issue.

My main issue is that I'm seeing arguments above that the speed of light is always "locally c". I understand this is fundamental to general relativity. But does that statement really have meaning anymore once we take into account light's gravitation on itself? Even if we were inside of the Bonnor beam using local interior coordinates, the only way we could measure the "local speed of light" is to shine another light, which then has its own gravitation, etc.

Bottom line: When fully taking into account general relativity (per above), is there any actual experiment we can ever perform in which we would measure the speed of light being exactly "c"? What experiment would this be? Thanks!

 

[Orodruin]

I'm not sure I understand why non-stationary spacetimes would have "no gravitational time dilation" rather than just having gravitational time-dilation that is time-varying (or perhaps not uniquely defined), but that's not quite my main issue.

Gravitational time dilation is an effect between two stationary observers. As such it requires you to have a way of defining what constitutes a stationary observer, which is impossible in a non-stationary spacetime. In a stationary spacetime, it is an observer whose world line is the integral curve of a the timelike Killing field.

My main issue is that I'm seeing arguments above that the speed of light is always "locally c". I understand this is fundamental to general relativity. But does that statement really have meaning anymore once we take into account light's gravitation on itself? Even if we were inside of the Bonnor beam using local interior coordinates, the only way we could measure the "local speed of light" is to shine another light, which then has its own gravitation, etc.

The stress-energy tensor of a beam propagating at the speed of light is going to be traceless. It should be easy to verify that this is the case for the given metric.

 

[PeroK]

Bottom line: When fully taking into account general relativity (per above), is there any actual experiment we can ever perform in which we would measure the speed of light being exactly "c"? What experiment would this be? Thanks!

These days the speed of light is defined to be whatever it is. And the metre is defined in terms of the speed of light and the second, which has its own definition.

In that sense, you can no longer measure the speed of light. Instead, you can use the speed of light to calibrate a metre stick.

 

[Ibix]

In $u,v,r,\theta$ coordinates, $G^{vv}=G_{uu}=16\pi m$. All other components of both upper and lower forms are zero.

If I did the coordinate substitution correctly, in $t,z,r,\theta$ coordinates $G_{tt}=G_{zz}=G^{tt}=G^{zt}=G^{tz}=G^{zz}=8\pi m$ and $G_{tz}=G_{zt}=-8\pi m$. All other components are zero.

This is indeed traceless. It's also independent of $r$, so does need to be patched to an external vacuum spacetime as @Orodruin noted. (And, indeed, as is implied in the OP.)

 

[Orodruin]

The exterior solution is also given on the Wikipedia page: https://en.m.wikipedia.org/wiki/Bonnor_beam

 

[PeterDonis]

I'm seeing arguments above that the speed of light is always "locally c".

That's not a good way of putting it. A much better way of putting it is that spacetime has a geometric structure called a "light cone" structure--at every point in spacetime, there is a local light cone. All timelike objects (ordinary objects with nonzero rest mass) propagate inside the light cone; light itself, and other massless objects, propagate on the light cone. Nothing propagates outside it.

This way of putting it makes it clear that it has nothing to do with "speed"; it's a property of the geometry of spacetime. That's why, as @PeroK pointed out, you actually don't measure "the speed of light"; you use light, which propagates on the light cones and so marks out the geometric light cone structure of spacetime, to calibrate meter sticks.

 

[diracs-cat]

This is really great information, everyone! Below is a summary of my understanding of the above discussion and one clarifying question below. Apologies for any more mistakes. I'm very much still a beginner at GR.

1) The Bonnor beam's spacetime metric is non-stationary, which means that gravitational time dilation is ill-defined (per Orodruin's explanation above). The trace of the stress-energy tensor and Einstein tensor are zero, which I think implies zero Ricci scalar curvature of the spacetime interior to the Bonnor beam.

2) For an "Observer A" at a distance far away from the Bonnor beam, the speed of the Bonnor beam is technically undefined (per Orodruin's analogy to velocity vectors on the earth's surface).

3) For an "Observer B" within the Bonnor beam, the "local speed of light" is c by definition. But this still doesn't necessarily mean that "Observer B" measures the speed of the Bonnor beam itself as being equal to the "local speed of light" does it?

 

[Orodruin]

1) The Bonnor beam's spacetime metric is non-stationary

According to @Ibix post #6, it is stationary. (I have not independently verified this but see no a priori reason to doubt this result.)

2) For an "Observer A" at a distance far away from the Bonnor beam, the speed of the Bonnor beam is technically undefined (per Orodruin's analogy to velocity vectors on the earth's surface).

Yes. This holds for any velocities that are not colocated.

3) For an "Observer B" within the Bonnor beam, the "local speed of light" is c by definition. But this still doesn't necessarily mean that "Observer B" measures the speed of the Bonnor beam itself as being equal to the "local speed of light" does it?

They can measure the stress energy tensor of the beam and conclude that it is traceless.

 

[Ibix]

According to @Ibix post #6, it is stationary. (I have not independently verified this but see no a priori reason to doubt this result.)

Argument extended to the exterior (thanks to you) and neatened up (thanks to Peter): the metric is independent of $u$ and $v$, both internally and externally, so those coordinate basis vector fields are Killing fields. By inspection, the $r-\theta$ plane is purely spacelike and orthogonal to the $u-v$ plane (both $dr^2$ and $d\theta^2$ terms have plus signs, and there are no cross terms involving either), so some linear combination of $u$ and $v$ is a timelike vector field. A linear combination of Killing fields is also a Killing field, so there exists a timelike Killing field.

In fact, since any timelike combination of the $u$ and $v$ basis vectors is a timelike Killing field there are many choices. Presumably this is because there's no natural rest frame parallel to the beam, so all observers at fixed $r,\theta$ see space as unchanging regardless of their velocity parallel to the beam.

 

[PeroK]

2) For an "Observer A" at a distance far away from the Bonnor beam, the speed of the Bonnor beam is technically undefined (per Orodruin's analogy to velocity vectors on the earth's surface).

This isn't a special case. In curved spacetime, vectors are local objects. When you first learn about vectors in "flat" Euclidean space, there are two important concepts: a position vector from the origin to any point in space; and, the ability to compare vectors directly at two different points in space. This is covered in any introductory course on vectors.

In curved spacetime, you lose both of these concepts. This takes some time and effort to get used to. Vectors become local objects, technically defined on a tangent space at any point in spacetime. One way to think about curved spacetime generally is that at each point you have something that looks locally flat (and where Special Relativity applies). Then all these points are patched together according to some global geometry.

In any case, this is fundamental to GR generally. If the spacetime is not static, then the concept of the distance between two points in space becomes ambiguous and not well-defined. For example, in our expanding universe, light may travel from a distant galaxy to Earth. Using comoving time, we can attribute a time in the past "when" the light set out. At that time, the galaxy was $d_0$ light years away. The light takes (comoving) time $\Delta t$ to reach Earth and now the Earth and the galaxy are $d_1$ light years apart. And if you take the quantity $c\Delta t$ as some sort of measure of how far the light has travelled in that time (as measured by a series of local observers), then $c \Delta t$ is equal to neither $d_0$ nor $d_1$, but something in between.

And, in fact, even the statement that the universe is spatially expanding is ambiguous. That is, of course, the popular narrative. But, that is a coordinate dependent statement - i.e. a description of the universe in comoving coordinates. And, although these coordinates are natural and very useful, you don't have to use them.

I think that you are trying to fit the concepts of GR into a fundamentally Newtonian worldview. We all do that when we start. The important thing is to open your mind to the more subtle and flexible mathematics and physics of GR.

 

[Orodruin]

n fact, since any timelike combination of the u and v basis vectors is a timelike Killing field there are many choices. Presumably this is because there's no natural rest frame parallel to the beam, so all observers at fixed r,θ see space as unchanging regardless of their velocity parallel to the beam.

Indeed, as it must. It should transform the metric to one with a different energy density, but otherwise similar structure.

 

[diracs-cat]

That makes sense to me. The Bonnor beam is infinite, so the distribution of energy and momentum within any finite-length "control volume" cylinder surrounding it would be constant over time. So that results in a static spacetime. I assume if the beam were instead finite then the spacetime could end up being dynamic.

 

[PeterDonis]

A linear combination of Killing fields is also a Killing field

Only if the combination has constant coefficients. In this case, as I said before, I believe it does. But not all cases are like that.