# Why do gravitational waves travel "only" at the speed of light but the gravitational scalar potential is instantaneous?

To my old Space SE question Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch? @DavidHammen's excellent answer replies that one should not add a time delay to gravitational effects based on distance when calculating orbits.

This lead to my and others' answers to How to calculate the planets and moons beyond Newtons's gravitational force? which explain that we can approximate GR effects with the following:

$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$

Although I'm not familliar with GR, I'm going to recommend an equation that seems to work well and seems to be supported by several links. It is an approximate relativistic correction to Newtonian gravity that is used in orbital mechanics simulations. You will see various forms in the following links, most with additional terms not shown here:

The following approximation should be added to the Newtonian term:

$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right)$$

Below this answer to Is the zero gravity experienced in ISS the “artificial” kind? I was scolded for saying "gravity moves at the speed of light" and told that in GR the scalar gravitational potential is instantaneous. It seems that I'd embraced that in my answer quoted above, but lost track of that after thinking about the finite speed of gravitational waves.

Question: Why do gravitational waves travel "only" at the speed of light while at the same time the gravitational scalar potential is instantaneous?

My thoughts on this so far:

If one stands near a pair of bodies orbiting around their center of mass with angular frequency $$\omega$$ and holds a sensitive accelerometer and plots the Fourier transform of the signal, one will see peaks at (at least) $$\omega$$ and $$2 \omega$$ with the relative intensities depending on distance and mass ratio.

If I put a bead on a lossy stick, I could extract energy from that.

But to my understanding, this is not Feynman's famous "bead on a stick" argument that gravitational waves have energy. So I'm thinking that there would not be a distance $$d$$-dependent $$d/c$$ delay for this signal.

But Feynman's bead on a stick gravitational wave energy extractor would show a $$d/c$$ delay.

Have I got this right so far?

• Also see physics.stackexchange.com/q/27845/123208 & physics.stackexchange.com/q/535344/123208 and links therein. Commented Jul 28, 2021 at 3:08
• Also related: physics.stackexchange.com/q/156746/21441 Commented Jul 28, 2021 at 10:53
• What is a lossy stick? Commented Jul 30, 2021 at 15:54
• @user36138 start with checking the "gravitational wave energy extractor" link to the Physics SE post which quotes Feynman's argument directly.
– uhoh
Commented Jul 30, 2021 at 22:20
• No help whatsoever! Commented Dec 20, 2021 at 12:10

It is not true that GR describes changes in a scalar gravitational potential as propagating with no delay. It's actually not even true that GR describes gravity as arising from a scalar gravitational potential. Most spacetimes in GR cannot be described by a scalar gravitational potential. The thing that Newtonian gravity describes as a scalar gravitational potential is actually analogous to the metric in GR, and the metric is a tensor, not a scalar.

What is probably causing people to tell you these incorrect things is a misunderstanding of a somewhat more subtle fact about fields, which is that when a source moves inertially, its field points toward it, not toward where it used to be. This applies to other fields, such as the electric field. This does not mean that a disturbance propagates instantaneously when the source does not move inertially.

It's also true that if you try to take relativity into account simply by using a retarded version of the field you get totally wrong results. This is because when you transform a field, you get velocity-dependent forces. This is true both in electromagnetism and in GR.

If you want to get a post-newtonian approximation for purposes such as finding the trajectory of a space probe very accurately, there are methods for doing that, but they're not as simple as just taking a delayed potential or not taking a delayed potential.

What I've written above is just a bare statement of the facts, without much in the way of explanation or proof. The question, linked question, and comment threads on them all seem to show that there are a lot of people trying to come to grips with this at a deeper level, and that this topic can also be extremely confusing to people.

I think the easiest part to explain based on more fundamental principles is the fact that the electric field of an inertially moving point charge points toward the charge, not toward its retarded position. Let's start in frame of reference K, which is the frame in which the charge is at rest. In this frame, the field $$E$$ is purely electric, and it has the familiar Coulomb form. Now we transform to another inertial frame, K', that is in motion relative to K. In this frame, we have both an electric field and a magnetic field, $$(E',B')$$. The directions and magnitudes of $$E'$$ and $$B'$$ can be determined by doing a transformation $$(E,0)\rightarrow(E',B')$$ that is local and linear. Now consider the value of E at two different points along the same electric field line. Since the field lines in K are straight, $$E_1$$ and $$E_2$$ are parallel and can differ only in magnitude. Therefore in K', we have $$E'_1$$ and $$E'_2$$ that are also parallel and have the same ratio of strengths. This shows that the electric field lines in K', like the ones in K, must be straight, not curved as they would be if they were determined by the retarded position of the charge. However, the form of the field is not the same as in K. It is nonuniform and has a nonvanishing curl. This tells us that E' can't be derived from a potential.

The above argument doesn't quite translate straightforward into a similarly rigorous argument in general relativity. For example, we don't have global frames of reference in GR, and there is no such thing as a global Lorentz transformation. However, I think the E&M argument shows that there is nothing mysterious about this fact, which also applies to GR if reinterpreted carefully, e.g., as a statement about a post-newtonian approximation.

Another thing that is pretty straightforward to prove is that this property cannot extend to the case where the source of the field is undergoing an arbitrary acceleration. If so, then we could transmit information instantaneously by wiggling a point charge, but special relativity tells us that the instantaneous propagation of information is impossible. (As a side note, the above logic does not imply anything about constant acceleration, since a constant acceleration is unmodulated and can't transmit information. In fact, there is an argument to be made that a mass undergoing constant acceleration does not radiate. However, this sort of thing is rather subtle and depends on precise definitions of what we mean by a radiation field.)

• Might be worth adding that the post-Newtonian approx. does surprisingly well even in regimes where one would expect it not to! arxiv.org/abs/1102.5192 Commented Jul 28, 2021 at 13:24
• @JDługosz that sounds really interesting and likely to provide further helpful insight, and it seems to shed further direct light on my question as written, since I'd originally started down this path from an orbital mechanics problem. Please consider adding an answer here expanding on that. Thanks!
– uhoh
Commented Jul 28, 2021 at 23:50
• @JDługosz: It sounds to me like you've misunderstood something. Electric fields seem instantaneous for linear motion, and this can be understood via magnetic components. Gravity seems instantaneous for orbital motions. No, this is wrong.
– user15381
Commented Jul 29, 2021 at 1:41
• Here's the link: math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html Commented Jul 29, 2021 at 13:48
• @uhoh I added an answer, more precisely/correctly stated after I re-read the article. Commented Jul 29, 2021 at 14:15

The old Physics FAQ has an article on this.

The interesting parts (emphasis mine):

This cancellation may seem less strange if one notes that a similar effect occurs in electromagnetism. If a charged particle is moving at a constant velocity, it exerts a force that points toward its present position, not its retarded position, even though electromagnetic interactions certainly move at the speed of light. Here, as in general relativity, subtleties in the nature of the interaction "conspire" to disguise the effect of propagation delay. It should be emphasized that in both electromagnetism and general relativity, this effect is not put in ad hoc but comes out of the equations. Also, the cancellation is nearly exact only for constant velocities. If a charged particle or a gravitating mass suddenly accelerates, the change in the electric or gravitational field propagates outward at the speed of light.

Now, in electrodynamics, a charge moving at a constant velocity does not radiate. Technically, the lowest-order radiation is dipole radiation, and the radiated power depends on the second time derivative of the electric dipole moment; two time derivatives give acceleration. So, to the extent that A's motion can be approximated as motion at a constant velocity, A cannot lose angular momentum. For the theory to be consistent, there must therefore be compensating terms that partially cancel the instability of the orbit caused by retardation. This is exactly what happens; a calculation shows that the force on A points not towards B's retarded position, but towards B's "linearly extrapolated" retarded position.

In general relativity, roughly speaking, a mass moving at a constant acceleration does not radiate. Here, the lowest order radiation is quadrupole radiation, and the radiated power depends on the third time derivative of the mass quadrupole moment. (The full picture is slightly more complex, since one cannot have a single, isolated accelerating mass; whatever it is that causes the acceleration also has a gravitational field, and its field must be taken into account.) For consistency, just as in the case of electromagnetism, a cancellation of the effect of retardation must occur, but it must now be even more complete—that is, it must hold to a higher power of v/c. This is exactly what one finds when one solves the equations of motion in general relativity.

In a nutshell, changes to the gravitational field propagate at the speed of light. Yet, the force between the orbiting objects is directed at the current position, not the position you see it at due to propagation delay (the light from it travels at the same speed as gravity).

You can understand this with electric fields: a moving electric charge in the field will experience a magnetic force sideways, and the result is attraction to the linear extrapolated position of the other charged particle.

But you cannot make a stable orbit using electric charge as the attracting force. Not only does the cancellation not extrapolate the accelerated motion of the other particle, but the accelerated particle will radiate. These two effects are linked! In order to lose angular momentum, it has to go someplace; radiation allows it to be transferred to the field (the photons will cary angular momentum).

With gravity, accelerated motion is extrapolated to very high precision, due to effects that can be thought of like gravito-magnetism. And ideal point bodies in orbit (or other gravitational interaction) would radiate only a miniscule amount of gravitational waves due to being accelerated.

• Yet, the force between the orbiting objects is directed at the current position, not the position you see it at due to propagation delay (the light from it travels at the same speed as gravity). This is only true to a certain level of post-Newtonian approximation.
– user15381
Commented Jul 29, 2021 at 14:40
• But you cannot make a stable orbit using electric charge as the attracting force. Here it sounds like you're implying a contrast with gravity. Actually the same is true for gravity. For both gravity and E&M, there are no stable (classical) orbits, and the reason is the same: energy is radiated away. It's just that the quadrupole nature of gravitational radiation makes the effect much weaker than in the case of EM radiation, which can be dipole.
– user15381
Commented Jul 29, 2021 at 14:41
• With gravity, accelerated motion is extrapolated, due to effects that can be thought of like gravito-magnetism. This is sort of true and sort of not true. It depends on subtleties about how one defines a radiation field.
– user15381
Commented Jul 29, 2021 at 14:42
• "In a nutshell..." implies that this is gross, not subtle. It's a conceptual summary, with no math at all being shown. Commented Jul 29, 2021 at 14:44
• And ideal point bodies in orbit (or other gravitational interaction) do not radiate gravitational waves due to being accelerated. (Pulsars radiate energy because they are not points and thus exhibit tidal effects.) No, this is wrong. The rate of gravitational radiation depends on the time variation of the mass quadrupole moment of the system as a whole. The mass quadrupole moment doesn't vanish for a point particle, because it's reckoned from the center of mass. This is pretty easy to see if you just look at closed-form approximations for the rate of radiation from a binary system.
– user15381
Commented Jul 29, 2021 at 14:44