Kinnersley’s “photon rocket” and gravitational radiation

[Povel]

TL;DR Summary

Meaning of "radiative terms" in the connection of Kinnersley's "photon rocket" solution and their relation with usual gravitational radiation?

In this paper by Carlip, a comparison is made between electromagnetic and gravitational aberration.

For the latter case, he takes as a study subject the Kinnersley’s “photon rocket”, an exact solution which is known to have the strange property of not producing any gravitational waves, even though it models a point-particle accelerating due to the anisotropic emission of a photon flux.

The metric has the following linearized form (eq. 2.1):$$g_{\mu\nu} = \eta_{\mu\nu} - \frac{2Gm(s_R)}{r^3}\sigma_\mu\sigma_\nu$$

Where $m$ is the time-varying mass and $\sigma_\mu\sigma_\nu$ is proportional to the stress-energy tensor and represents radiation/null dust streaming out from the world line.

Carlip then writes the connection coefficients, and in particular $-\Gamma^i_{00}$, since in the Newtonian limit this corresponds to the "acceleration" that a test particle undergoes in the metric 2.1

Schematically:$$\Gamma^i_{00} =nonradiative\;terms + radiative\;terms $$

In the previous chapter on electromagnetic aberration he decomposed the electric field in a similar way, and he specified that the radiative terms "depend explicitly on acceleration and fall off as $\frac{1}{R}$"

If the notation is consistent, then, it would seem that the radiative terms in the gravitational case must depend on the same quantities.
The equation for gravitational radiation amplitude also depends on acceleration and falls of as $\frac{1}{R}$ (I'm not talking about the radiated power formula, I mean the equation 4 at page 8 https://www.ego-gw.it/public/events/vesf/2010/Presentations/Quadrupole-Ferrari.pdf)What is the relation between these radiative terms in the connection and the usual gravitational radiation (which should be null in this case)?

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[Povel]

I am going to add some further elements. Hope it will help anyone kind enough to try answering my question. Of course, for getting the full picture is better to read the paper(s).

In Carlip's paper he obtains from the Lienard-Wiechert potential the following expression for the electric field (eq. 1.8):

$$E^i=F^{i0}=\frac{\textit{e}}{\gamma^2_RR^2(1-\textbf{n}\cdot\textbf{v}_R)^3}(\textit{n}^\textit{i}-\textit{v}^\textit{i}_R) + \textit{radiative terms}$$

He then points out how the first member on rhs is retarded, and there is no dependece whatsoever on the instantaneous position of the source. Even so, though, for a charge in uniform motion the direction of the first term, $(\textit{n}^\textit{i}-\textit{v}^\textit{i}_R)$, points toward the instantaneous position of the source thanks to a linear extrapolation term that is present inside the expression for the direction of the force/field. The end result is that the electric field of a charge undergoing uniform motion has no aberration.
In case of non-uniform motion, as Carlip points out:

if a uniformly moving charge suddenly stops at position $\textit{z}(\textit{s}_\textit{0})$, the field at a distant location $\textit{x}$ will continue to point toward its “extrapolated” position—even though the charge never actually reaches that position—until the time $\textit{t−}\textit{z}(\textit{s}_\textit{0})$ that it takes for light to travel from $\textit{z}(\textit{s}_\textit{0})$ to $\textit{x}$. At that time, the field will abruptly switch direction to point toward the true position of the source. This sudden change in the field, propagating outward from $\textit{z}(\textit{s}_\textit{0})$ at the speed of light, is what we mean by the electromagnetic radiation of an accelerated charge.

I think that it is pretty clear (even obvious) that the omitted "radiative terms" in the above equation are exactly those associated with electromagnetic radiation. These terms describe a retarded field that points toward (it is centered around) the retarded position of the source, that is proportional to the acceleration of the charge and to the inverse of the distance. See this simulation and the following image for reference:

So far so good.

Now, Carlip analyzes the case of gravitational aberration. There are two stumbling block:

- There is no preferred time-slicing in general relativity, and thus no unique definition of an “instantaneous” direction. For weak fields, we can use the nearly flat background to define a nearly Minkowski coordinate system
-Second, we cannot simply require by fiat that a massive source accelerate

For solving this second issue, Carlip studies the Kinnersley's photon rocket, which I already described in the first post. He uses the flat metric (simply obtained from the Kinnersley's one by putting $m = 0$) for defining Minkowski coordinates. In Newtonian language, then, the gravitational acceleration is determined by the connection, and in particular by the $\Gamma^i_{00}$ coefficient.

Arguably, other choices are possible for the coordinates, and $\Gamma$ is frame dependent, but I don't think that these facts change anything or make less physical the possibility of dividing $\Gamma$ in radiative and non-radiative part. Afterall, with the Kinnersley's rocket, the spacetime curvature is not zero, so even if one can locally annihilate a component of $\Gamma$, I presume its contribution will reappear in its derivatives.

I found another paper (https://arxiv.org/abs/1006.1583) that has the entire expression for $\Gamma^i_{00}$ written down. In 4 dimensions it is (equation 106):

$$\Gamma^i_{00}=\frac{\textit{m}}{R^2\gamma^3(1-\textbf{n}\cdot\textbf{v})^5}[ (1-2\textbf{n}\cdot\textbf{v} -2(\textbf{n}\cdot\textbf{v})^2+3\textbf{v}\cdot\textbf{v})\textit{n}^\textit{i}-(1-\textbf{n}\cdot\textbf{v})\textit{v}^\textit{i}-\frac{2\textit{m}}{R\gamma^3(1-\textbf{n}\cdot\textbf{v})}\textit{n}^\textit{i}] - \frac{3\textit{m}\textit{k}_\sigma\ddot z^\sigma + \dot m}{R\gamma^4(1-\textbf{n}\cdot\textbf{v})^4}\textit{n}^\textit{i}$$

The very last term is the "radiative term". The paper I linked above describes it like this:

The last term, proportional to the acceleration $\ddot z^\sigma$ and mass decrease ̇$\dot m$ of the rocket, represents the radiative part of the gravitational field which behaves as $∝ R^-1$. Close to the photon rocket this can be neglected with respect to the first “Newtonian force”-term $∝R^-2$.

The paper then goes on pointing out how the Newtonian-like (non-radiative part?) gravitational acceleration points toward the extrapolated instantaneous position of the rocket. Therefore, there is no aberration.

From the look of the last term of this last equation, it appears I was right and that in the other paper Carlip was consistent: the radiative term is indeed proportional to the inverse of the distance and to acceleration, like in the electromagnetic case. Moreover, if I'm reading the last term correctly, it seems like the direction of this field is $\textit{n}^\textit{i}$, which is the purely retarded position of the rocket, without any linear, quadratic or higher order extrapolation toward the instantaneous one, as one would expect for an actual radiation field.

Carlip ends this section of the paper by writing:

Every term in the connection $\Gamma^\rho_{\mu\nu}$ depends only on the retarded position, velocity, and acceleration of the source; [..] there is no dependence, implicit or explicit, on the “instantaneous” direction to the source. [..]. In particular, as in Maxwell’s theory, if a source abruptly stops moving at a point $\textit{z}(\textit{s}_\textit{0})$, a test particle at position $\textit{x}$ will continue to accelerate toward the extrapolated position of the source until the time it takes for a signal to propagate from $\textit{z}(\textit{s}_\textit{0})$ to $\textit{x}$ at light speed.

This paragraph is almost a replica of the one he wrote previously about electromagnetism I quoted. It seems to me that a signal that communicates to a far away particle about the new state of motion of the rocket and changes its path cannot be interpreted in any other way than gravitational radiation.

However, as I mentioned in the first post, the Kinnersley's solution is non-radiative, at least according to common definitions in literature: the Weyl tensor is of type D far away from the rocket, so there are no gravitational waves. This is something Carlip doesn't mention. What is all this talking about "radiative part of the gravitational field" or "radiative terms" then?The physical picture described above seems to imply there is radiation in this case, but in the connection coefficients, and not in the invariant Riemann curvature tensor.

 

[PeterDonis]

the Kinnersley's solution is non-radiative, at least according to common definitions in literature: the Weyl tensor is of type D far away from the rocket, so there are no gravitational waves.

Yes, this is correct according to the definition of "gravitational waves" that is usually used in the literature.

What is all this talking about "radiative part of the gravitational field" or "radiative terms" then?

Because those terms are more general than the term "gravitational waves", at least as those terms are usually used in the literature.

You're right that Carlip doesn't mention the fact that the Kinnersley solution has no gravitational waves as that term is normally used; but he does mention something that I think is a clue to the difference between "gravitational waves" and the more general term "radiative part of the gravitational field". Consider Carlip's description of the EM scenario that illustrates what we mean by "the radiation from an accelerated charge": the charge changes its motion in a way that causes a change in the field to propagate outward at the speed of light.

If we carry this viewpoint over to the gravitational case, we encounter a difference that Carlip also remarks upon. In the EM case, we could just declare by fiat that the charge accelerates. But we can't do that with a source of gravity in GR: as Carlip notes, we can only consistently represent an accelerating mass if we include the energy responsible for its acceleration. In the Kinnersley case, that means including the photons ejected out the back of the rocket and the corresponding decrease in the mass of the source.

But doing this also changes the definition of a "change in the motion of the source" as far as our concept of "gravitational waves" is concerned. The Kinnersley solution does not "change its motion" in the sense required for gravitational radiation, because its acceleration is constant--or, to put it another way, its rate of photon emission and mass loss is constant. To put it still another way, the "gravitational field", since it depends on the acceleration of the source, only "changes" if the acceleration changes.

In other words, the analogue of Carlip's description of radiation due to an accelerated charge propagating a change in the EM field would be a photon rocket that changes its photon emission and mass loss rate; this change in the field would then get propagated as gravitational waves, in the usual sense of that term. For example, suppose the rocket just shut itself off at some particular point on its worldline; distant observers could only detect that shutoff when the change in the "gravitational field" propagated to them at the speed of light. And they would detect the change as a gravitational wave front exactly analogous to the EM wave front Carlip describes for the EM case.

But if the photon rocket never changes its acceleration/photon emission/mass loss rate, there is no "change in the field" to propagate. The "radiative terms" in the field don't inform distant observers of any change, because there is no change to inform them of. The nomenclature "radiative terms" only refers to the fact that these terms depend on $1 / R$ instead of on $1 / R^2$; it does not mean they are "carrying information about a change in the field".

 

[Povel]

Thank you for the reply Peter.
Your explanation makes a lot of sense, but there are still a number of things that are unclear to me concerning the difference between a "radiative term" in the connection/gravitational field and gravitational waves/radiation:-As you point out, for having a change of the field be trasmitted the field should be changing in the first place.
In general, though, an accelerating object does produce changes in the field. The Kinnersley's solution is, in this regard quite special. Damour, for example, has shown in this paper that in a Kinnersley's rocket both the photon flux and the mass accelerating due to recoil produce gravitational waves, but their coherent addition cancel each other contribution out; the gravitational field is actually time-dependent, but non-radiative due to symmetry.
If the photon flux has anisotropies of higher multipole order than monopolar and dipolar, then there will be gravitational radiation.
There are also other solutions that can be interpreted as an object undergoing a constant acceleration due to the action of an external force, like the C-metric, and these are radiative in general. It seems then that, generally speaking, an object uniformly accelerating in a physically reasonable way does produce changes in the field that could propagate as radiation.-The "shutted down" Kinnersley's rocket is simply the Schwarzschild's solution. Of course this is also non-radiative. Therefore, in turning off the rocket the solution doesn't change its non-radiative nature. If, hypothetically, instead of shutting down the rocket I could change the amount of thrust by adjusting the intensity of the flux streaming out from the wordline without introducing anisotropies of higher order than monopolar and dipolar (something that doesn't look impossible to me, unless this automatically introduces higher order anisotropies in the flux somehow), there should be no gravitational waves emitted at any point during the procedure, thanks to the mechanism Damour describes above.
Clearly, though, the gravitational field far away would not be able to change instantly, to account for the different acceleration of the rocket, wouldn't it?- In connection with the previous points, Carlip states that:

A source with a constant second derivative of Q can therefore radiate no angular momentum or energy, and any nonconservation of mechanical angular momentum and energy must again be canceled by additional terms in the interaction. The second derivative $\frac{d\textbf{Q}^2}{dt^2}$ involves terms proportional to acceleration and to the square of the velocity, so the cancellation must occur at a higher order than it did for electromagnetism

From this quote it seems like for having radiation either a time-dependent acceleration or a non null product of velocity and acceleration (the time derivative of the square of the velocity) is needed in general. This would seem to support the idea that constant acceleration too can produce gravitational radiation.-Finally, the direction of the "radiative term" in the last equation I wrote above is toward the retarded position of the rocket $\textit{n}^\textit{i}$ (correct?), while the other non-radiative terms are pointing, effectively, toward the "instantaneous position". How is then correct to say that there is no aberration? If one is considering a point far away enough from the rocket, only the last term will be relevant, and this term does not track the rocket position as the other terms do.
But if there is no compensation for this aberration, how is it possible there is no "radiation" in some form?

 

[Vanadium 50]

an exact solution which is known to have the strange property of not producing any gravitational waves

Why is it strange? The quadrupole moment isn't changing, so why should it radiate?

 

[Povel]

Why is it strange? The quadrupole moment isn't changing, so why should it radiate?

I guess that naively one would expect the quadrupole moment to change. Afterall the total system of photon flux + mass accelerating due to recoil is changing shape. That the system nonetheless does not emit gravitational waves has surprised some people, at least Bonnor and Damour.

The quadrupole moment of the flux and mass, taken separately, are changing, and can be considered to emit radiation. However their contribution cancel out due to symmetry. I think this is what Damour is saying here:

We find that the amount of gravitational radiation at infinity depends on the anisotropies of multipole order$\textit{ℓ}≥2$ in the photon flux, there being a cancellation between the gravitational wave amplitude emitted by the energy-momentum distribution of the monopolar and dipolar photon flux, and the gravitational wave amplitude emitted by the accelerated massive particle

 

[PeterDonis]

Damour, for example, has shown in this paper that in a Kinnersley's rocket both the photon flux and the mass accelerating due to recoil produce gravitational waves, but their coherent addition cancel each other contribution out; the gravitational field is actually time-dependent, but non-radiative due to symmetry.

The radiative term in the Kinnersley solution is indeed time-dependent, yes, since it includes the mass itself as well as the acceleration and the rate of change of the mass.

Damour's overall treatment seems reasonable to me as far as the math goes. What, exactly, deserves the term "gravitational waves" as opposed to just "radiative terms" is a matter of terminology, not physics. I don't think there is disagreement about the physics.

The "shutted down" Kinnersley's rocket is simply the Schwarzschild's solution.

A rocket that is always shut down is the Schwarzschild solution. But a rocket that starts out running and then shuts down means the spacetime has two regions: a Kinnersley region and a Schwarzschild region. The boundary between those two regions will, I think, contain gravitational waves (but see the comment on terminology below), which, heuristically, will propagate the information about the rocket shutting down.

If, hypothetically, instead of shutting down the rocket I could change the amount of thrust by adjusting the intensity of the flux streaming out from the wordline without introducing anisotropies of higher order than monopolar and dipolar (something that doesn't look impossible to me, unless this automatically introduces higher order anisotropies in the flux somehow), there should be no gravitational waves emitted at any point during the procedure, thanks to the mechanism Damour describes above.

If you change the flux, you also change the rate of change of the mass, which will change the radiative term in the equation. And that change will propagate outward at the speed of light from the point on the rocket's worldline where the change was made. Whether you want to call that a "gravitational wave" or not is a matter of terminology, not physics.

Clearly, though, the gravitational field far away would not be able to change instantly, to account for the different acceleration of the rocket, wouldn't it?

Of course not. Any change in the rocket's behavior will cause a change in the field that will propagate at the speed of light. See above.

 

[PeterDonis]

the direction of the "radiative term" in the last equation I wrote above is toward the retarded position of the rocket ${n}^\textit{i}$ (correct?), while the other non-radiative terms are pointing, effectively, toward the "instantaneous position". How is then correct to say that there is no aberration?

First, Carlip never says there is no aberration. He is quite clear that most of the aberration one would naively expect is cancelled, for the reasons he gives, but not all; there is a tiny bit left over which is not cancelled.

Second, Carlip makes clear that by "aberration" he means aberration in the "Newtonian force", i.e., the direction in which that force points (or, if you like, the direction in which the "field" that produces the force points). In other words, he is talking about aberration of the non-radiative terms. Aberration of radiation is something different; for example, the electromagnetic aberration he describes is not the same thing as the aberration of starlight due to the Earth's orbital motion about the Sun.

Possibly it would have been better to use a different term than "aberration" to describe what Carlip is talking about. He chose the term "aberration" because it was the one used by Van Flandern to argue that the propagation speed of gravity must be much greater than $c$ (because of the observed absence of aberration of the gravitational force).

 

[Vanadium 50]

I guess that naively one would expect the quadrupole moment to change. Afterall the total system of photon flux + mass accelerating due to recoil is changing shape.

Rocket goes one way, light the other, right? Quadrupole moment stays the same.

 

[pervect]

Rocket goes one way, light the other, right? Quadrupole moment stays the same.

Are you sure? The quardupole moment is $\Sigma \rho x^i x^j$ according to my reference (MTW). Let $x^1$ be the direction the rocket accelerates, then $I^{11}$ = 0 at t=0, and $I^{11} > 0$ for t>0.

I've left out the bit where we reduce the quadrupole moment to make it trace-free, but we still have a time varying quadrupole moment, I believe.

 

[PeterDonis]

I've left out the bit where we reduce the quadrupole moment to make it trace-free

Unfortunately, doing that in this case is not a good idea. Since the position vector $x$ only has one nonzero component, there are no off diagonal terms in the bivector $x^i x^j$, so the bivector is equal to its trace. That means the trace-free part is zero.

 

[Vanadium 50]

Are you sure?

I thought so.

Everything is in a line. At most I have a dipole.

If the light was shot out off-axis, e.g. in a nozzle or "skirt" configuration, then I would agree you have a quadrupole.

 

[pervect]

Unfortunately, doing that in this case is not a good idea. Since the position vector $x$ only has one nonzero component, there are no off diagonal terms in the bivector $x^i x^j$, so the bivector is equal to its trace. That means the trace-free part is zero.

If we start with, say, diag(1,0,0), the trace free version is diag(2/3, -1/3, -1/3).

 

[PAllen]

I recall that Bill Kinnersley who used to post here stated that if Kirk threw boots out of the Enterprise, causing it to accelerate, that there would be GW. Further, that his eponymous photon rocket was a special case, and that a generic photon thrust pattern would radiate.

 

[Povel]

I thought so.

Everything is in a line. At most I have a dipole.

If the light was shot out off-axis, e.g. in a nozzle or "skirt" configuration, then I would agree you have a quadrupole.

Sorry for the ignorance, but a pole is also "on a line", but it has a non-null quadrupole moment that changes if the length of the pole changes, no?
What is the difference between it and the mass-energy distribution of the (perfectly collimated) photon rocket + emitted photons?