I believe we can use Newtonian Physics to make incredibly good predictions about the movement of celestial bodies as long as they are not too fast/massive and there are only two of them (well, we can make predictions for more bodies, but not the general case, right?)

And I think the only number we need in order to make predictions, besides the positions/speeds/masses of the bodies, is $G$ (i.e. Newton's gravitational constant), and all the math we need is what we call calculus.

But we can make even better predictions if we use General Relativity, and some mathematical objects called tensors, and the speed of light $c$, right? Or do we still need $G$ in GR?

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    $\begingroup$ The answer is yes; we still need $G$. More information can be found here: en.wikipedia.org/wiki/Einstein_field_equations $\endgroup$ – Hunter Jan 13 '14 at 13:40
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    $\begingroup$ G is a constant in Einstein's field equations, as @Hunter linked to. $\endgroup$ – lionelbrits Jan 13 '14 at 13:41
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    $\begingroup$ Because it pops up regularly in GR, it's often taken to be 1 to make writing equations more simple (see this wiki article for more info). $\endgroup$ – Kyle Kanos Jan 13 '14 at 14:09
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    $\begingroup$ > well, we can make predictions for more bodies, but not the general case, right? Classical N-body problem has been solved analytically by Q. Wang, who generalized solution of 3-body problem by Sundman. Although these solutions are useless practically because of very slow convergence of the series, N-body problems still can be (and are) treated numerically. $\endgroup$ – Ruslan Jan 13 '14 at 14:45

G is required in the equations, though usually we choose a unit system where it is set to one for convenience.

In the GR field equations it usually appears on the side with the stress-energy tensor, along with some dimensionless constant.

One way to look at it is that we still need a fundamental constant to relate the three units of mass, length, and time. We cannot write a meaningful expression with mass on one side (unless it is in a ratio, which it isn't) without a fundamental constant capable of getting rid of the mass unit. The metric generated (an expression without mass units) is dependant on the energy distribution (an expression with mass units), so G must pop in somewhere. Of course, it could be a different constant like $\hbar$ , but G makes the most sense intutively and is indeed the actual constant found in the equations.

Besides, GR should limit to newtonian gravity in the classical limit, so some form of G ought to be present.

  • $\begingroup$ > some form of G ought to be present I think OP means that it might have reduced into a combination of something more fundamental. $\endgroup$ – Ruslan Jan 13 '14 at 14:47
  • $\begingroup$ I had the illusion, based on my ignorance, that c would have been enough. So sad. $\endgroup$ – Pablo Grisafi Jan 13 '14 at 18:38
  • $\begingroup$ @PabloGrisafi: there's no way of getting around needing some sort of conversion factor for "how much gravity is generated by X amount of mass", in the roughest sense. $\endgroup$ – Jerry Schirmer Jan 13 '14 at 19:15

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