# GR: What is the curved spacetime analogue of Newton 2nd law?

I am following Carroll, he states that the geodesic equation is the the generalization to curved spacetime $\vec f = m \vec a$, for $\vec f = \vec 0$. This leads me to wonder what is the correct to generalize this further to:

1. Paths that are not geodesics
2. Cases where $\vec f \neq 0$

Suppose for example that an observer is sitting at on the surface of a neutron star in Schwarzschild geometry. I imagine that (1) this observer does not follow a geodesic since he is not falling toward smaller values of $r$ like test-particles would and (2) there is something present which the observer would interpret as a force, $\vec f = \vec f_g \neq 0$. What would be the acceleration such an observer feels. In particular, this means that I do not want to arrive at newtons law of gravitation.

What generalization of the geodesic equation is appropriate in situations like this, and how would it be applied? In any possible answer, I would prefer if point (1) and (2) would be addressed separately if possible.

Newton's second law is (with a minor rearrangement):

$$\frac{d^2x^i}{dt^2} = \frac{F^i}{m} = A^i \tag{1}$$

where I've used index notation because of course we'll be doing that when we move to GR, and $A^i$ is the force per unit mass.

The equivalent to this in GR is the equation for the four-acceleration:

$$A^\alpha = \frac{d^2x^\alpha}{d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}\dfrac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} \tag{2}$$

where $\mathbf A$ is now the four-force per unit mass.

In your question (1) you ask what is the force when the observer isn't on a geodesic. In this case you feed in the observed (non-geodesic) trajectory, i.e. the observed values for $d^2x^\alpha/d\tau^2$ and $dx^\alpha/d\tau$, and equation (2) gives you the four-force per unit mass $A^\alpha$. The force the observer feels is just the norm of $\mathbf A$. For example you ask about doing this for an observer on the surface of a neutron star, and this calculation is nicely described in: What is the weight equation through general relativity?

In question (2) you ask what is the motion given some non-zero four-force, and again this is given by equation (2). To calculate the motion just figure out what the four-force (per unit mass) is in your coordinates, feed it into the equation and solve the resulting (probably horrible) equations.

While we're here, note that the geodesic equation that you describe in your question is just the special case of the four-force being zero i.e.

$$0 = \frac{d^2x^\alpha}{d\tau^2} + \Gamma^\alpha_{\,\,\mu\nu}\dfrac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau}$$

Finally note that in flat spacetime, where the Christoffel symbols are zero, equation (2) simplifies to:

$$A^\alpha = \frac{d^2x^\alpha}{d\tau^2}$$

which is just the special relativistic form of Newton's second law, and in the low speed limit where $t \approx \tau$ this simplifies further to:

$$A^i = \frac{d^2x^i}{dt^2}$$

which is where we came in.

• Thanks for your reply. (i) Are you able to address my points (1) and (2) separately? And (ii) you assumed that the observer is on a geodesic. The observer is at rest on the surface on the neutron star, such that (1), the path is not a geodesic. – Mikkel Rev Jun 13 '17 at 15:17
• Derive the geodesic equation and find out. – Mikkel Rev Jun 13 '17 at 15:19
• I take back what I said. I didn't read it carefully enough. Please let me read it a few more times. – Mikkel Rev Jun 13 '17 at 15:25
• @MariusJonsson: OK, how does that look now? – John Rennie Jun 13 '17 at 15:28
• @MariusJonsson: no, I mean the usual norm $A^2=g_{\alpha\beta} A^\alpha A^\beta$. However in the rest frame of the observer $A^0$ is zero, so in fact it reduces to $A_iA^i$. For details see chapter 6 of Gravitation by Morris, Thorne and Wheeler. Also do read the question I linked as twistor59 does a good job of explaining the calculation. – John Rennie Jun 13 '17 at 15:59

Newton's law comes from the geodesic equation $$\frac{d^2x^\mu}{ds^2}~+~\Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}~=~0$$ We look for the component with $x^\mu~=~r$ and we use the diagonal Schwarzschild metric $$\frac{d^2r}{ds^2}~+~\Gamma^r_{00}\frac{dx^0}{ds}\frac{dx^0}{ds}~+~\Gamma^r_{ii}\frac{dx^i}{ds}\frac{dx^i}{ds}~=~0,$$ The last term is $O(1/c^2)$ smaller and so is dropped. For low velocities we make the further approximation that $ds~\simeq~dt$. The Christoffel symbol $\Gamma^r_{00}$ is evaulated so the geodesic equation gives $$\frac{d^2r}{dt^2}~+~\frac{GM}{r^2}~=~0,$$ which is Newwton's second law of motion in a gravity field.

If one want to go further, curvature calculations give the tidal acceleration. This is the physical content of Weyl curvature. By these calculations it can be shown that Newton's laws emerge from general relativity in the limit $c~\rightarrow~\infty$ of where the speed of light is much larger than any physics of interest.

• Thanks for your reply. (i) Are you able to address my points (1) and (2) separately? And (ii) you assumed that the observer is on a geodesic. The observer is at rest on the surface on the neutron star, such that (1), the path is not a geodesic. – Mikkel Rev Jun 13 '17 at 15:08
• The geodesic equation can be written as $\frac{D^2x^\mu}{ds^2}~=~0$ in a covariant form. A more compact notation is $\nabla_uU~=~0$. If there is some other force $F^\mu~=~ma^\mu$ then $\frac{D^2x^\mu}{ds^2}~=~a^\mu$. – Lawrence B. Crowell Jun 13 '17 at 16:26