No information can travel faster than light, so does the force between Earth and Sun point in the direction that the Earth was 8 minutes ago or in the direction that the Earth is now?
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1$\begingroup$ Related question: physics.stackexchange.com/q/73061 $\endgroup$– G. SmithCommented Mar 9, 2020 at 5:05
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1$\begingroup$ Also related: physics.stackexchange.com/q/5456/123208 Please don't just look at the top voted answers there, IMHO most of the answers on that page have interesting insights on this question. $\endgroup$– PM 2RingCommented Mar 9, 2020 at 6:04
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$\begingroup$ The answer is "kinda yes", but the notion of where a moving, relativistically distant object is "right now" is not well defined. $\endgroup$– lvellaCommented Mar 9, 2020 at 14:38
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2 Answers
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… so the force beetwen Earth and Sun points in the direction that the earth was 8 or points in the direction that the earth is now?
No. Gravitational acceleration is directed towards the position of the source quadratically extrapolated from its “retarded position” toward its instantaneous position (that is using velocity and acceleration from that moment in a truncated Taylor expansion for the current position), up to small nonlinear terms and corrections of higher order in velocities.
No information can travel faster than light …
This does not mean that the force must be directed toward retarded position of the second body since the “information” transmitted may also include velocity and acceleration.
In nonrelativistic situations, such as motion of Earth around the Sun, this “quadratic extrapolation” matches the instantaneous position with a high degree of precision. For Earth–Sun system the discrepancy (which could be approximated by the next term of Taylor expansion $\approx \left|\frac{1}{6}\frac{d^3 \mathbf{r}}{dt^3}\,\tau^3\right|$) would be about several centimeters. Whereas discrepancy between instantaneous and retarded positions is about $15\,000\,\text{km}$. So the effects of such “gravitational aberration” are nonexistent within the Solar system.
One should keep in mind that all discussions of finite speed for gravity propagation must proceed within the framework of general relativity which has inherent ambiguity in identifying concepts of Newtonian physics such as force direction. While in Newtonian physics gravitational force between isolated bodies is defined unambiguously in GR we instead have the connection ($\nabla$) and curved spacetime. In order to extract from it Newtonian force and position of the bodies we must choose a specific reference frame, and impose a gauge conditions, which provides a certain freedom in the definitions.
Also note, that for the somewhat similar situation of two charges orbiting each other the direction of electrostatic force is given not by quadratic but by linear extrapolation from retarded position (using only velocity and not acceleration). The difference is in dipole character of EM radiation versus quadrupole nature of gravitational radiation.
For more information see the following (rather technical) paper:
Carlip, S. (2000). Aberration and the speed of gravity. Physics Letters A, 267(2-3), 81-87, DOI:10.1016/S0375-9601(00)00101-8, arXiv:gr-qc/9909087
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$\begingroup$ Why focus on quadratic extrapolations rather than using higher degree approximations or the full Taylor series? Is it just that the quadratic is "close enough" for most purposes? $\endgroup$ Commented Mar 10, 2020 at 22:21
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1$\begingroup$ @sfmiller940: this quadratic extrapolation is a property of gravity (and ultimately traceable to the spin $s=2$ of gravitational field and conservation of energy and angular momentum). If we consider bound systems held by some other field the degree of extrapolation could be different. And this extrapolation is “close enough” only for nonrelativistic dynamics (as in Solar system). For Hulse-Taylor binary pulsar or during final stages of black hole inspirals discrepancies would be significant. $\endgroup$– A.V.S.Commented Mar 11, 2020 at 4:08
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This is more complicated than it first looks. Two forces. The force from the sun on the earth, and the force from the earth on the sun.
The force on the sun from the earth -- which has rather small effect -- has three components. There is the force that the earth would have had if it was stationary in our frame. Second there is the force due to the velocity of the earth 8 minutes ago. These two can be added to give a force that comes from the direction the earth would be now, if it had continued to travel at constant velocity for 8 minutes.
The third force, due to the earth's acceleration, appears to come not from where the earth would have been now, but from where the earth actually was 8 minutes ago.
None of these have a direct relation to where the earth is now, since in those 8 minutes many other forces could act on the earth.
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$\begingroup$ This is wrong. For EM field “effective position” is linearly extrapolated. For gravitational field this “effective position” is extrapolated quadratically, taking into account not only velocity but also acceleration. $\endgroup$– A.V.S.Commented Mar 9, 2020 at 14:43
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5$\begingroup$ The force on the sun from the earth -- kind of small Wrong. Have you ever checked Newton gravity law ? It states that gravity force is proportional to the product of masses : $F=G{\frac {m_{1}m_{2}}{r^{2}}}$. So force from sun to earth and from earth to sun are equal in magnitude. $\endgroup$ Commented Mar 9, 2020 at 14:57
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1$\begingroup$ You're right the forces are equal. I should have said that the sun's response to that force is small, compared to the earth's response. $\endgroup$– J ThomasCommented Mar 9, 2020 at 22:26
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$\begingroup$ @A.V.S It sounds like there's something here I ought to learn about. Would you suggest a link? $\endgroup$– J ThomasCommented Mar 9, 2020 at 22:50
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1$\begingroup$ Carlip's paper in my answer would do. But to see that EM and gravity are different in that regard one could also use post-Newtonian expansion: system of charges is conservative up to $1/c^2$ terms while system of gravitating bodies is conservative up to $1/c^4$. Radiative effects appear at $1/c^3$ for EM and at $1/c^5$ for gravity. $\endgroup$– A.V.S.Commented Mar 10, 2020 at 5:05