<p>The old <a href="https://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html" rel="nofollow noreferrer">Physics FAQ</a> has an article on this.</p>
<p>The interesting parts (emphasis mine):</p>
<blockquote>
<p>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 &quot;conspire&quot; 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.</p>
<p>Now, <strong>in electrodynamics</strong>, 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; <strong>a calculation shows that the force on A points not towards B's retarded position, but towards B's &quot;linearly extrapolated&quot; retarded position.</strong></p>
<p>In <strong>general relativity, roughly speaking, a mass moving at a constant acceleration does not radiate</strong>.  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 <strong>cancellation of the effect of retardation must occur, but it must now be even more complete</strong>—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.</p>
</blockquote>
<p>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 <em>see</em> it at due to propagation delay (the light from it travels at the same speed as gravity).</p>
<p>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.</p>
<p>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).</p>
<p>With gravity, <em>accelerated</em> motion is extrapolated to very high precision, due to effects that can be thought of like <a href="https://en.wikipedia.org/wiki/Gravitoelectromagnetism" rel="nofollow noreferrer">gravito-magnetism</a>.  And ideal point bodies in orbit (or other gravitational interaction) would radiate only a miniscule amount of gravitational waves due to being accelerated.</p>

CC BY-SA 4.0
2021-07-29T14:44:08.430Z
"In a nutshell..." implies that this is gross, not subtle.  It's a conceptual summary, with no math at all being shown.
JDługosz
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2021-07-29T14:13:32.430Z
2021-07-29T14:49:31.907Z
2021-07-29T14:49:31.907Z
JDługosz
<p>To my old Space SE question <em>Besides retarded gravitation, anything else to worry about when calculating MU69's orbit from scratch?</em> <a href="https://space.stackexchange.com/a/21472/12102">@DavidHammen's excellent answer</a> replies that one should not add a time delay to gravitational effects based on distance when calculating orbits.</p>
<p>This lead to <a href="https://space.stackexchange.com/a/23409/12102">my and others' answers to <em>How to calculate the planets and moons beyond Newtons's gravitational force?</em></a> which explain that we can approximate GR effects with the following:</p>
<blockquote>
<p><span class="math-container">$$\mathbf{a_{Newton}} = -GM \frac{\mathbf{r}}{|r|^3},$$</span></p>
</blockquote>
<blockquote>
<p>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:</p>
<ul>
<li><a href="https://physics.stackexchange.com/q/313146/83380">https://physics.stackexchange.com/q/313146/83380</a></li>
<li>Eq.    1 in <a href="https://www.lpi.usra.edu/books/CometsII/7009.pdf" rel="noreferrer">https://www.lpi.usra.edu/books/CometsII/7009.pdf</a></li>
<li>Eq.   27 in <a href="https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf" rel="noreferrer">https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf</a></li>
<li>Eq. 4-26 in <a href="https://descanso.jpl.nasa.gov/monograph/series2/Descanso2_all.pdf" rel="noreferrer">https://descanso.jpl.nasa.gov/monograph/series2/Descanso2_all.pdf</a></li>
<li>also see David Hammen's discussion/advice in <a href="https://space.stackexchange.com/a/21472/12102">this answer</a>.</li>
<li>Eq. 3.11 in <a href="http://adsabs.harvard.edu/full/1994AJ....107.1885S" rel="noreferrer">http://adsabs.harvard.edu/full/1994AJ....107.1885S</a> (click &quot;print&quot;) see <a href="https://astronomy.stackexchange.com/a/25304/7982">this answer</a></li>
<li>Eq. A.8 in G. F. Rubilar1 &amp; A. Eckart 2001 <a href="https://www.aanda.org/articles/aa/abs/2001/28/aah2613/aah2613.html" rel="noreferrer">Periastron shifts of stellar orbits near the Galactic Center</a>, except that the <span class="math-container">$\mathbf{r}/r^3$</span> Newtonian is included rather than treated separately.</li>
<li>Eq. 2 in M. Parsa et al. 2017 <a href="https://www.eso.org/public/archives/announcements/pdf/ann17051a.pdf" rel="noreferrer">Investigating the Relativistic Motion of the Stars Near the Supermassive Black Hole in the Galactic Center</a>, except that the <span class="math-container">$\mathbf{r}/r^3$</span> Newtonian is included rather than treated separately.</li>
</ul>
<p>The following approximation should be <em>added to</em> the Newtonian term:</p>
</blockquote>
<p><span class="math-container">$$\mathbf{a_{GR}} = GM \frac{1}{c^2 |r|^3}\left(4 GM \frac{\mathbf{r}}{|r|} - (\mathbf{v} \cdot \mathbf{v}) \mathbf{r} + 4 (\mathbf{r} \cdot \mathbf{v}) \mathbf{v} \right)$$</span></p>
<p>Below <a href="https://space.stackexchange.com/a/54215/12102">this answer to <em>Is the zero gravity experienced in ISS the “artificial” kind?</em></a> I was scolded for saying &quot;gravity moves at the speed of light&quot; 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.</p>
<p><strong>Question:</strong>  Why do gravitational waves travel &quot;only&quot; at the speed of light while at the same time the gravitational scalar potential is instantaneous?</p>
<p>My thoughts on this so far:</p>
<p>If one stands near a pair of bodies orbiting around their center of mass with angular frequency <span class="math-container">$\omega$</span> and holds a sensitive accelerometer and plots the Fourier transform of the signal, one will see peaks at (at least) <span class="math-container">$\omega$</span> and <span class="math-container">$2 \omega$</span> with the relative intensities depending on distance and mass ratio.</p>
<p>If I put a bead on a lossy stick, I could extract energy from that.</p>
<p>But to my understanding, this is <strong>not</strong> Feynman's famous &quot;bead on a stick&quot; argument that gravitational waves have energy. So I'm thinking that there would <strong>not</strong> be a distance <span class="math-container">$d$</span>-dependent <span class="math-container">$d/c$</span> delay for this signal.</p>
<p>But Feynman's bead on a stick <a href="https://physics.stackexchange.com/a/262962/83380">gravitational wave energy extractor</a> would show a <span class="math-container">$d/c$</span> delay.</p>
<p>Have I got this right so far?</p>

CC BY-SA 4.0
2021-07-28T03:08:22.313Z
Also see https://physics.stackexchange.com/q/27845/123208 & https://physics.stackexchange.com/q/535344/123208 and links therein.
PM 2Ring
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2021-07-28T10:53:46.803Z
Also related: https://physics.stackexchange.com/q/156746/21441
Ruslan
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2021-07-30T15:54:53.647Z
What is a lossy stick?
user36138
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2021-07-30T22:20:15.030Z
@user36138 start with checking the "gravitational wave energy extractor" link to the Physics SE post which quotes Feynman's argument directly.
uhoh
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2021-12-20T12:10:42.190Z
No help whatsoever!
user36138
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2021-12-20T15:47:07.480Z
@user36138 then I think you have an excellent new Stack Exchange question! I'm not sure if it's better asked here or in Physics SE. I think here would be a lot more welcoming for an "I don't understand this at all" type question than Physics, but sometimes they are really nice and helpful too!
uhoh
CC BY-SA 4.0
2021-12-21T16:23:23.087Z
I am under the mis-apprehension that the purpose of this forum is to spread understanding, - particularly not to using esoteric jargon.
user36138
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2021-12-21T23:02:15.170Z
@user36138 Oh, I'm all for spreading understanding; what I meant was that that is such a good question that it deserves to be a new question post, so that full answer posts can be used to answer it and many different users have the opportunity to compete to write the best answer. Stack Exchange comments are limited in size and purpose and can be deleted in the future. I don't think I can explain enough here, but a bead on an ideal stick can slide freely without friction, but if there was some friction, then the forces that moved the bead up and down would do work and the stick would get hot.
uhoh
CC BY-SA 4.0
2021-12-21T23:06:16.083Z
@user36138 that means that whatever was producing the accelerating force would loose energy. It's not a formal physics term, it's just a contraction of several ideas into one sentence. We sometimes say "lossy transmission line" when talking about low quality coax cable, where some of the power in the line is turning to heat for example. "lossy" just means something that does a bad job at allowing things to move without loss, be it a bead on a stick or charge moving through a cable.
uhoh
<gravity><cosmology><gravitational-waves><newtonian-gravity>
Why do gravitational waves travel "only" at the speed of light but the gravitational scalar potential is instantaneous?
uhoh
# Why do gravitational waves travel "only" at the speed of light but the gravitational scalar potential is instantaneous?

(Asked by: username_0 on 2021-07-28)

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](https://space.stackexchange.com/a/21472/12102) 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?*](https://space.stackexchange.com/a/23409/12102) 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:
>
> * <https://physics.stackexchange.com/q/313146/83380>
> * Eq. 1 in <https://www.lpi.usra.edu/books/CometsII/7009.pdf>
> * Eq. 27 in <https://ipnpr.jpl.nasa.gov/progress_report/42-196/196C.pdf>
> * Eq. 4-26 in <https://descanso.jpl.nasa.gov/monograph/series2/Descanso2_all.pdf>
> * also see David Hammen's discussion/advice in [this answer](https://space.stackexchange.com/a/21472/12102).
> * Eq. 3.11 in <http://adsabs.harvard.edu/full/1994AJ....107.1885S> (click "print") see [this answer](https://astronomy.stackexchange.com/a/25304/7982)
> * Eq. A.8 in G. F. Rubilar1 & A. Eckart 2001 [Periastron shifts of stellar orbits near the Galactic Center](https://www.aanda.org/articles/aa/abs/2001/28/aah2613/aah2613.html), except that the $\mathbf{r}/r^3$ Newtonian is included rather than treated separately.
> * Eq. 2 in M. Parsa et al. 2017 [Investigating the Relativistic Motion of the Stars Near the Supermassive Black Hole in the Galactic Center](https://www.eso.org/public/archives/announcements/pdf/ann17051a.pdf), except that the $\mathbf{r}/r^3$ Newtonian is included rather than treated separately.
>
> 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?*](https://space.stackexchange.com/a/54215/12102) 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](https://physics.stackexchange.com/a/262962/83380) would show a $d/c$ delay.

Have I got this right so far?

**Tags:** gravity, cosmology, gravitational-waves, newtonian-gravity

> **Comments:**
> **username_1**: Also see https://physics.stackexchange.com/q/27845/123208 & https://physics.stackexchange.com/q/535344/123208 and links therein.  
> **username_2**: Also related: https://physics.stackexchange.com/q/156746/21441  
> **username_3**: What is a lossy stick?  
> **username_0**: @username_3 start with checking the "gravitational wave energy extractor" link to the Physics SE post which quotes Feynman's argument directly.  
> **username_3**: No help whatsoever!  
> **username_0**: @username_3 then I think you have an excellent new Stack Exchange question! I'm not sure if it's better asked here or in Physics SE. I think here would be a lot more welcoming for an "I don't understand this at all" type question than Physics, but sometimes they are really nice and helpful too!  
> **username_3**: I am under the mis-apprehension that the purpose of this forum is to spread understanding, - particularly not to using esoteric jargon.  
> **username_0**: @username_3 Oh, I'm all for spreading understanding; what I meant was that that is such a good question that it deserves to be a new question post, so that full answer posts can be used to answer it and many different users have the opportunity to compete to write the best answer. Stack Exchange comments are limited in size and purpose and can be deleted in the future. I don't think I can explain enough here, but a bead on an ideal stick can slide freely without friction, but if there was some friction, then the forces that moved the bead up and down would do work and the stick would get hot.  
> **username_0**: @username_3 that means that whatever was producing the accelerating force would loose energy. It's not a formal physics term, it's just a contraction of several ideas into one sentence. We sometimes say "lossy transmission line" when talking about low quality coax cable, where some of the power in the line is turning to heat for example. "lossy" just means something that does a bad job at allowing things to move without loss, be it a bead on a stick or charge moving through a cable.  

---

## 1 answers

### Answer by username_4 on 2021-07-29. Score: 2

The old [Physics FAQ](https://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html) 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](https://en.wikipedia.org/wiki/Gravitoelectromagnetism). And ideal point bodies in orbit (or other gravitational interaction) would radiate only a miniscule amount of gravitational waves due to being accelerated.

> **Comments:**
> **username_4**: "In a nutshell..." implies that this is gross, not subtle.  It's a conceptual summary, with no math at all being shown.  

---

