Is there 'friction' in spacetime?

Question

So, if all the bodies are embedded in space-time and moves through it, is there some kind of 'friction' with space time of the planets? For example, the Earth suffers friction when moving near the sun due the curvature and General Relativity and loses energy?

If a planet loses energy due to friction can this energy loss be measured?

Yes there is a similar to this: https://iopscience.iop.org/article/10.1088/1742-6596/845/1/012003/pdf — cmd, Mar 03 '24 at 20:09

score 86 · Accepted Answer · edited Dec 31 '16 at 14:23

86

I think the question suggests you are thinking of space-time as if it were e.g. a substance, like a fluid, that we move through. That's not how we view space-time, at least in pure general relativity.

But the question you ask is a deceptively simple one and it raises some complex questions. And I don't think we actually can answer them exactly because I'm not sure we have a definitive answer to the most basic question hidden in your answer: What is space-time?

is there some kind of 'friction' with space time of the planets?

There is a "kind" of friction, but perhaps "interaction" would be a better choice of word, as I'd prefer to avoid the notion of classical friction forces.

We say that when an object moves through space time it distorts space time - stretches it, compresses it. Mass creates distortions we describe as gravity.

It's a little deeper than that.

We also know, thanks to the wonderful LIGO experiments, that these gravitational effects do distort space in a wave-like way. And an object can lose energy (has to, in fact) when it creates such waves.

Which leads us to this:

if a planet loses energy due to friction can this energy loss be measured?

No (I suppose I should say, not at our technological level). It's tiny.

The gravitational waves we have measured (which represent the closest thing to your friction loss) are due to the collisions of huge black holes, and the disturbance they make is so small that LIGO scientists are pushing the boundaries of measurement to detect them at all. A planet is a tiny thing compared to those black holes and it barely makes a dent, as it were, in space time by comparison.

But it's worth saying that our current understanding of space-time is a little basic. We don't have a clear idea of how the quantum world fits into the grand scale of relativistic space-time. At present we have two models, one of a small scale space-time filled with a sea of virtual particles and the other of a pure, clean empty space time with the odd idealized gravitational mass in it. We don't have a single theory connecting them, so we don't really have a proper theory of space-time (or perhaps something deeper than that is needed - no one knows).

edited Dec 31 '16 at 14:23

Jens

3,659

answered Dec 29 '16 at 15:10

StephenG - Help Ukraine

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A superb answer. You make me wonder about timescales. LIGO, and before that the Hulse-Taylor binary pulsar, proved that there are systems whose lifetime against gravitational decay is $10^{9\text{--}10}$ years. It would be interesting to compute the comparable lifetime for the Sun-Jupiter system, or Earth-Moon, and see whether their lifetime against collapse due to gravitational-wave emission is more like $10^{100}$ years, or more like $10^{1000}$ years. – rob Dec 29 '16 at 15:34
Maybe hidden variables make up the sea of spacetime, surrounding particles (of course not surrounding ín space). If so, quantum mechanics is a consequence of this structure and gravity can't be quantized in that case. – Deschele Schilder Dec 29 '16 at 19:07
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I think gravitational waves can be seen as the gravitational equivalent to electromagnetic "bremsstrahlung". The name ("breaking radiation") implies an effect resembling friction; after all, the mass is slowed down. But while friction heats a medium, bremsstrahlung or gravitational waves do not; all kinetic energy lost by the moving masses is radiated away. – Peter - Reinstate Monica Dec 30 '16 at 17:59
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Another difference is that linear movement is not slowed down at all, as opposed to movement underlying normal friction. Space time does (to our knowledge) not offer resistance to moving through it as such -- it's acceleration which is lossy because it creates waves. – Peter - Reinstate Monica Dec 30 '16 at 18:03
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@PeterA.Schneider *"braking radiation" – jpmc26 Dec 31 '16 at 08:18
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@jpmc26 AH, yes. It's not breaking anything I guess ;-). – Peter - Reinstate Monica Dec 31 '16 at 14:12
Mass creates gravitational distortion of the fabric of spacetime whether or not the mass is in motion (gravitational effects do not vary with velocity). The motion of mass is what creates gravitational waves - if the mass is sufficiently great - but the consequent energy loss is tiny. If energy was lost to any form of friction, howsoever termed, the orbital velocity of each planet would be continuously reducing, which is not observed to occur, though small effects may be masked by a planet's orbital velocity continuously varying between apogee and perigee due to the physics of eliptical orbits – Ed999 Mar 05 '17 at 17:39
@PeterA.Schneider I'd never read it that way. I always got the impression that the "Bremse" in "Bremsstrahlung" referred to whatever it was you put in the path of a moving charge, rather than implying a friction. I agree, the two are very alike, although, as far as I understand, this is one of the points where GR differs quite markedly from Gravitoelectromagnetism. In GEM, gravitational radiation is exactly the same as bremsstrahlung, so you can use the Larmor or similar formulas. And you get a power level that is way too high for e.g. Hulse Taylor. Also, you don't get dipoles in GR, but... – Selene Routley Aug 07 '17 at 13:35
.... in GEM, which has gravitational Maxwell equations, you most certainly can. This is one of the few points where the two theories have been experimentally told apart. BTW: the radiation in GEM is still very small: IIRC it's about 3GW for Earth, which sounds a great deal but it would be utterly unmeasurable when you work out how long it would take for a significant effect on Earth's orbit. – Selene Routley Aug 07 '17 at 13:37
@WetSavanna Since bremsstrahlung includes cyclotron or synchrotron radiation where nothing is really "in the way" at all I think it is a reasonable assumption that "brems..." refers just to the "deceleration", i.e. more generally acceleration, somewhat paradoxically ;-). The rest of your comment is beyond my level of knowledge. – Peter - Reinstate Monica Aug 08 '17 at 22:00
Re "We cannot measure a planet's energy loss": What exactly would be the semantics of that measurement? Capture all lost energy and "weigh" it? Of course not. So the "measurement" will in fact always be a calculation; and we -- or rather somebody -- can very well (I assume) calculate earth's energy loss over time. The difference with the black holes is that we cannot detect even a trace of it, but we are reasonably sure it's there, like if we had left a flashlight on the moon burning. – Peter - Reinstate Monica Aug 08 '17 at 22:09
@PeterA.Schneider The only way we know how AFAIK is to measure the spin-down or decrease of a planet's orbital speed as we do with the Hulse-Taylor system. So there's a definite in-principle way. The other way would be to use an array LIGO-type sensors at some distance from the planet: again that is what LIGO has done for mutually orbiting (then crashing) black holes. But the amplitude of the gravitational waves would be impossibly small for any Solar planet. – Selene Routley Aug 08 '17 at 22:19
@PeterA.Schneider Sorry about the rest of my comment, then. I'm simply saying that, although gravitational waves and EM bremsstrahlung are alike, experimental facts about grav waves are one of the ways we can tell general relativity and some of the other gravity theories apart. In particular, GEM is exactly like building gravitational Maxwells equations out of the Newton's inverse square law, in exactly the same way as the EM Maxwell's equations generalize Coulomb's law. It's a very cool theory, simple, and, for the most experimental results to within the accuracy we ... – Selene Routley Aug 09 '17 at 00:06
... can measure, the same as GR. But there can be no GR dipole antennas, and the Larmor formula gives measurably different results from GR. I'll also have to look up who came up with the word bremsstrahlung: my impression was it was one of the 19th century guys crashing electrons into X-ray targets - maybe even Röntgen himself - hence the bremsy name. It's always had a car crashy kind of sound to me. – Selene Routley Aug 09 '17 at 00:09

score -2 · Answer 2 · answered Jan 06 '17 at 00:40

Besides the visible matter in space, space is "full of" electromagnetic radiation (EMR). However, since the density of the EMR is very small, the density of a planet very large, and the charge of the planet essentially neutral, there is an extremely small "interaction" between a planet and EMR. It takes a very massive object to generate barely detectable "friction waves" - in space.

score -9 · Answer 3 · answered Dec 29 '16 at 22:38

We cannot gain an understanding of the nature of Einsteinian spacetime at a planetary scale. All matter, including planets, are composed of particles, and to answer this question we should consider the link between spacetime and particles, such as quarks.

Clearly, a particle has mass, and thus inertia. But what is inertia? It is presumably what the questioner means when he talks of 'friction'. A particle has inertia, or friction if you like, because it is bound to the spacetime field postulated by James Clark Maxwell in the 19th century, in Maxwell's equations.

Modern theory looks upon inertia as being due to a coupling or bond, which attaches a quark to the spacetime field. So yes, all particles have inertia, and the inertia is the reason why a particle has mass: mass being merely a measurement of the amount of energy which is required to break the coupling bond.

Is there 'friction' in spacetime?

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