# Where in his equations does Newton make the assumption that gravity travels instantaneously?

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1. Oct 16, 2014

### Amin2014

In other words, where does this assumption become mathematically explicit? Is it because there is no parameter representing time in Newton's universal law of gravity?

If so, what about other force laws like Coulomb's law for charges? I don't see any time embedded in that.

Last edited: Oct 16, 2014
2. Oct 16, 2014

### A.T.

Yes, the law of gravity itself says nothing about a delay. And that's how it must be interpreted to get elliptical orbits.

3. Oct 16, 2014

### Staff: Mentor

Yes, the assumption is in the way that $r$ appears in Newton's law with no time dependence, and yes, the same is true of Coulomb's law. Imagine that a charged particle or a gravitating mass were to suddenly appear anywhere in the universe at time $T$ - if we take those two laws at face value, there would be a non-zero force everywhere in the universe at any time after $T$. That doesn't make a lot of sense, so instead of taking them at face value we accept that they don't apply in time-varying situations.

Maxwell's laws, discovered in 1861, supplied the necessary understanding of time dependence for electricity and magnetism. General relativity, a half-century later, did the same for gravitation.

4. Oct 16, 2014

### Amin2014

THANK YOU! So the more general equations reduce EXACTLY to the two mentioned equations for time-invariant situations?

5. Oct 16, 2014

### Staff: Mentor

They do not. For example, the Schwarzschild solution to the Einstein field equations of general relativity describes the static time-independent gravitational field of a spherical mass, just as does Newton's $F=Gm_1m_2/r^2$ - but the solution is slightly different and as a result the orbits predicted by Newton's theory do not quite match those predicted by GR and actually observed (google for "Mercury orbit precession").