16
$\begingroup$

Our high-school teacher told us that the Earth pulls us with some force $F$ and we pull the Earth with the same force $F$. Within Newtonian physics this is true because of Newton's 3rd law, but let’s consider Einsteinian gravity. My mass is small; so I don’t warp space-time much. But Earth’s large mass warps space-time to a far greater extent.

So do I pull Earth with the same force it pulls me? If yes, how?

$\endgroup$
16
$\begingroup$

You both fall toward the common center of mass. Because the mass of the Earth is quite a bit larger than yours, the center of mass is very close to the center of the Earth, but rather far away from you. Thus, as you both fall to the common center, the Earth hardly moves while you fly until you hit the ground.

More specifically, you are talking about the two-body solution. Both bodies curve spacetime and move in this curved spacetime. As you justly stated, your contribution is small and for this reason the Earth movement toward you is very small as well.

However, when you interact with the Earth, the momentum you get equals the momentum the Earth gets. And yes, in the classical view, you attract the Earth with the same force as the Earth attracts you. While your gravity is very weak, the mass of the Earth attracted by it is enormous. Therefore the forces work out to be the same, as expected.

$\endgroup$
  • 12
    $\begingroup$ This answer doesn't seem to include any GR at all. $\endgroup$ – knzhou Oct 4 '17 at 14:48
18
$\begingroup$

For a Newtonian $n$-body system, the weak Newton's 3rd law implies total momentum conservation, but not vice-versa, cf. e.g. this Phys.SE post. However for a 2-body system, which OP asks about, they are equivalent, so OP's question is essentially equivalent to:

How do we see total momentum conservation in full-fledged GR without going to the Newtonian limit?

Answer: That's a great question! Notions, such as, e.g., force, mass, momentum, energy, etc., are notoriously subtle in GR. For a generic spacetime, there is no satisfactory definition of a gravitational stress-energy-momentum tensor, only a pseudotensor.

For an asymptotically flat spacetime, one may define an ADM energy-momentum 4-vector, which plays the role of conserved total energy-momentum associated with the full spacetime (incl. probes).

This answers OP's question in principle, but may not be completely satisfactory: We can easily associate localized energy-momentum to each point-probe$^1$ in the spacetime, but it is less clear how to give an independent definition for the energy-momentum of spacetime minus the probes (other than to declare it to be the difference). I.e. translated back to OP's problem: We don't seem to have an independent definition of Earth's energy-momentum by itself, even if we for simplicity assume that Earth is a black hole with a point-like singularity/matter-distribution.

--

$^1$ Our notion here of a point-probe is a point particle that can be assigned a localized energy-momentum, but unlike a test particle, it can backreact on the spacetime. The notion of probes is not really essential for the discussion. To emulate OP's 2-body system without using probes, consider instead 2 black holes with point-like singularities/matter-distributions. There doesn't seem to be a well-defined notion of energy-momentum associated to each individual black hole. Their individual energy-momenta are fuzzy/delocalized/ill-defined.

$\endgroup$
  • 1
    $\begingroup$ While a general solution is apparently subtle, the situation described may not need one: If earth and person do not move relative to each other they share a common frame of reference which should make things considerably more plain. If we additionally assume that they are the only bodies in the universe (yeah, doesn't make much sense when discussing GR, but still), that frame of reference is even an inertial system. Can we just look at a snapshot of the resulting space time curvature and make a statement which amounts to what the OP is assuming? $\endgroup$ – Peter - Reinstate Monica Oct 4 '17 at 13:09
  • $\begingroup$ Ricci curvature is via EFE directly linked to matter energy-momentum, not gravitational energy-momentum per se. As already mentioned, the definition of the latter is a delicate issue. $\endgroup$ – Qmechanic Oct 4 '17 at 19:19
  • 3
    $\begingroup$ Do you really expect a high-school student to understand this answer? $\endgroup$ – Massimo Ortolano Oct 4 '17 at 21:01
  • 1
    $\begingroup$ @MassimoOrtolano: Nothing can be generalized when it comes to understanding. I've had several events in my childhood where adult people wanted to take away (or refuse to give me) stuff I wanted to learn; what a pile of bullshit. If I am really not able to understand the stuff, I will return it soon enough or ask for supportive material. But if I am able to understand, I fuel my drive to learn and expand my knowledge. Not giving stuff is really a lose-lose where nothing is to be gained from; too much patronizing. $\endgroup$ – phresnel Oct 5 '17 at 10:18
  • 2
    $\begingroup$ Well, OP's GR question is not simple. $\endgroup$ – Qmechanic Oct 5 '17 at 11:38
-1
$\begingroup$

If you are speaking in the context of General Relativity, then there is no role of mass in gravitational dynamics. Rather, every possible interaction is only due to the curvature of the spacetime.

$\endgroup$
  • $\begingroup$ In GR, mass (via the stress-energy tensor) is the source of curvature. $\endgroup$ – user4552 Jan 19 at 14:02
-3
$\begingroup$

We attract earth ,earth also attracts us we both attract each other with same force but we know that attraction between two bodies depends upon their mass,greater the mass of two bodies greater is the force of attraction between them (according to Newton).So as we know that we have very small mass with respect to earth so due to greater mass of earth the impact of its gravity is very strong with respect to our gravity. Due to this its attraction is strong than our one's.

$\endgroup$
  • 3
    $\begingroup$ The OP already understands the Newtonian explanation. They want to know what General Relativity has to say about this scenario. $\endgroup$ – PM 2Ring Jul 13 '19 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.