<p><strong>No.</strong> The idea of relativistic mass is an outdated concept. There are many questions over on the <a href="https://physics.stackexchange.com/search?q=mass%20in%20special%20relativity">Physics Stack Exchange about this</a>.  Mass is an invariant quantity in GR and SR. It does not change due to the motion of a particle.  The energy of the particle may change, but the mass doesn't.</p>
<p><strong>But</strong> things like the spin angular momentum of a body do effect its gravitational field.  So it can matter for orbits.</p>
<p>Special relativity (SR) is a subset of general relativity (GR).  It describes a special case of the full theory.  If you complete a calculation using full GR, SR is included without you doing anything else.</p>
<p>Any actual calculation will use an approximate model.  If you say the Earth follows a <a href="https://en.wikipedia.org/wiki/Geodesics_in_general_relativity" rel="nofollow noreferrer">geodesic</a> of the Sun's spacetime, that geodesic includes SR, but you are neglecting the effects of the Earth's mass on the common spacetime of the system.  For the Earth--Sun system, this probably doesn't matter.  The sun is rotating, but for most calculations we still model its gravitational field using the <a href="https://en.wikipedia.org/wiki/Schwarzschild_metric" rel="nofollow noreferrer">Schwarzschild metric</a>, for a non-rotating point mass.</p>
<p>If the Earth or Sun were rotating fast enough, then we would need to include their spin angular momenta in calculations.  For instance, we could model the background spacetime the Earth moves through with the <a href="https://en.wikipedia.org/wiki/Kerr_metric" rel="nofollow noreferrer">Kerr metric</a> instead of the Schwarzschild metric.</p>
<p>There are all sorts of effects that technically matter, but we ignore.  When modeling a system, we always have a target precision goal in mind.  Based on that goal, we will ignore effects that don't matter to the precision we care about.</p>
<p>When building a GR orbital model, the tool to use is the perturbative <a href="https://en.wikipedia.org/wiki/Post-Newtonian_expansion" rel="nofollow noreferrer">post-Newtonian expansion</a>.  This approximates the motion of the bodies as a power series in <span class="math-container">$\frac{GM}{c^2 r} \sim \frac{v^2}{c^2}$</span>.  Successively higher powers in the expansion are less important since <span class="math-container">$\frac{GM}{c^2 r}&lt;1$</span>.  You can simply truncate the series when the required precision is reached.</p>
<p>In this formalism it's easy to sort effects by what post-Newtonian order they first appear.  The spin-orbit and spin-spin coupling effects come in at &quot;second post-Newtonian&quot; order, <span class="math-container">$\mathcal{O}\left(\frac{GM}{c^2 r}\right)^2$</span>, or higher.</p>
<p>We can measure the position of the moon incredibly precisely (<span class="math-container">$\sim 1$</span> cm) thanks to <a href="https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment" rel="nofollow noreferrer">laser ranging</a>.  To model the Earth--Moon system to the required precision, several usually ignored effects matter. For instance, the fact that they are both extended bodies of non-uniform density matters.  General relativity also matters.  The state of the art orbital model used in lunar laser ranging only needs to extend to the first post-Newtonian order (see <a href="https://link.springer.com/article/10.1007/s00190-019-01296-0" rel="nofollow noreferrer">Müller et al, 2019</a>).  So even in this highly precise use case, the spins of the Earth and Moon don't matter to the precision needed to compare to measurements.</p>

CC BY-SA 4.0
2022-04-25T16:34:49.977Z
2022-04-25T16:34:49.977Z
Paul T.
<p>Thanks @ConnorGarcia for your Answer. Avoiding <em>'Tidal Locking</em>' for now - the rotational angular momentum <em>Lrot</em> is 10,000,000 times smaller, so negligible, but could it still be a factor in eccentricity? I realise this is a controversial question, but, does the angular momentum between two rotating planets 'literally' take Relativity into account? i.e. if Planet-Earth with 'clockwise' rotation: <strong>Se</strong>, is measured relative to 'clockwise' rotation of Planet Mars: <strong>Sm</strong>, then their nearest equatorial 'purely-relative' rotation speed will be: (<strong>Se + Sm</strong>) (for Mars and Earth, this is approximately <em>twice</em> their individual rotation speed relative to a '<em>Rest</em>' frame). If two fictitious planets had the same rotation speed <strong>S</strong> and anti-parallel rotation (one clockwise and one anticlockwise), they would - by similar logic - have 'relatively' <strong>Zero</strong> rotation speed (along the shortest distance between them). <em>General Relativity</em> influences the precession of the perihelion of Mercury, but I'm wondering about <em>Special Relativity</em>) My question is: if these 'Relative' speeds were used in the Lorentz transformation, then the masses would change - by a tiny amount.  Could this deflect the orbits by a minute amount over a long period?</p>

CC BY-SA 4.0
2022-04-25T21:25:18.137Z
The recent edit makes the question look like nonsense. Editing questions is not intended for thanking for answers (no matter how polite) or asking new questions: it will only make everything unclear.
Anders Sandberg
<rotation><general-relativity><special-relativity>
Do relativistically measured rotation speeds (Special Relativity) change planetary mass slightly through Lorentz transformation?
Rob Clennell
# Do relativistically measured rotation speeds (Special Relativity) change planetary mass slightly through Lorentz transformation?

(Asked by: username_0 on 2021-07-08)

Thanks @ConnorGarcia for your Answer. Avoiding *'Tidal Locking*' for now - the rotational angular momentum *Lrot* is 10,000,000 times smaller, so negligible, but could it still be a factor in eccentricity? I realise this is a controversial question, but, does the angular momentum between two rotating planets 'literally' take Relativity into account? i.e. if Planet-Earth with 'clockwise' rotation: **Se**, is measured relative to 'clockwise' rotation of Planet Mars: **Sm**, then their nearest equatorial 'purely-relative' rotation speed will be: (**Se + Sm**) (for Mars and Earth, this is approximately *twice* their individual rotation speed relative to a '*Rest*' frame). If two fictitious planets had the same rotation speed **S** and anti-parallel rotation (one clockwise and one anticlockwise), they would - by similar logic - have 'relatively' **Zero** rotation speed (along the shortest distance between them). *General Relativity* influences the precession of the perihelion of Mercury, but I'm wondering about *Special Relativity*) My question is: if these 'Relative' speeds were used in the Lorentz transformation, then the masses would change - by a tiny amount. Could this deflect the orbits by a minute amount over a long period?

**Tags:** rotation, general-relativity, special-relativity

> **Comments:**
> **username_1**: The recent edit makes the question look like nonsense. Editing questions is not intended for thanking for answers (no matter how polite) or asking new questions: it will only make everything unclear.  

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## 1 answers

### Answer by username_2 on 2022-04-25. Score: 1

**No.** The idea of relativistic mass is an outdated concept. There are many questions over on the [Physics Stack Exchange about this](https://physics.stackexchange.com/search?q=mass%20in%20special%20relativity). Mass is an invariant quantity in GR and SR. It does not change due to the motion of a particle. The energy of the particle may change, but the mass doesn't.

**But** things like the spin angular momentum of a body do effect its gravitational field. So it can matter for orbits.

Special relativity (SR) is a subset of general relativity (GR). It describes a special case of the full theory. If you complete a calculation using full GR, SR is included without you doing anything else.

Any actual calculation will use an approximate model. If you say the Earth follows a [geodesic](https://en.wikipedia.org/wiki/Geodesics_in_general_relativity) of the Sun's spacetime, that geodesic includes SR, but you are neglecting the effects of the Earth's mass on the common spacetime of the system. For the Earth--Sun system, this probably doesn't matter. The sun is rotating, but for most calculations we still model its gravitational field using the [Schwarzschild metric](https://en.wikipedia.org/wiki/Schwarzschild_metric), for a non-rotating point mass.

If the Earth or Sun were rotating fast enough, then we would need to include their spin angular momenta in calculations. For instance, we could model the background spacetime the Earth moves through with the [Kerr metric](https://en.wikipedia.org/wiki/Kerr_metric) instead of the Schwarzschild metric.

There are all sorts of effects that technically matter, but we ignore. When modeling a system, we always have a target precision goal in mind. Based on that goal, we will ignore effects that don't matter to the precision we care about.

When building a GR orbital model, the tool to use is the perturbative [post-Newtonian expansion](https://en.wikipedia.org/wiki/Post-Newtonian_expansion). This approximates the motion of the bodies as a power series in $\frac{GM}{c^2 r} \sim \frac{v^2}{c^2}$. Successively higher powers in the expansion are less important since $\frac{GM}{c^2 r}<1$. You can simply truncate the series when the required precision is reached.

In this formalism it's easy to sort effects by what post-Newtonian order they first appear. The spin-orbit and spin-spin coupling effects come in at "second post-Newtonian" order, $\mathcal{O}\left(\frac{GM}{c^2 r}\right)^2$, or higher.

We can measure the position of the moon incredibly precisely ($\sim 1$ cm) thanks to [laser ranging](https://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment). To model the Earth--Moon system to the required precision, several usually ignored effects matter. For instance, the fact that they are both extended bodies of non-uniform density matters. General relativity also matters. The state of the art orbital model used in lunar laser ranging only needs to extend to the first post-Newtonian order (see [Müller et al, 2019](https://link.springer.com/article/10.1007/s00190-019-01296-0)). So even in this highly precise use case, the spins of the Earth and Moon don't matter to the precision needed to compare to measurements.


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