# Question about general relativity.

I know from my study of Newtonian mechanics that the force acting on any particle in the vicinity of any object is mass of the particle times the gravitational field at that point. My question is that is it possible to calculate the force acting on the particle using Newtonian mechanics by taking the mass of the particle and taking the value of gravity well (in this case the gravitational field) predicted by general relativity? Kindly correct me if I'm misunderstanding something. Please keep the explanation simple. Thank you.

• Probably the best we can do in that situation is use GR to work out the value of g locally, and then go from there using Newtonian Mechanics. – Mozibur Ullah Aug 26 '17 at 9:52
• @Mozibur Ullah that's what I thought at first but would that yield the right result? – Munj Patel Aug 26 '17 at 10:32
• I think so, otherwise GR would have been rejected as a theory of gravity straight-away. Perhaps a more precise question is to ask how, starting from GR, can we deduce that g is roughly 10 $m/s^2$ at the surface of the Earth? – Mozibur Ullah Aug 26 '17 at 11:14
• Roughly, yes. We work out the value of $g$ locally at the surface of the earth by GR, then multiply by the mass $m$ of the object, and this will give as per Newton's second law, the force exerted on the object by the gravitational field. – Mozibur Ullah Aug 26 '17 at 11:24
• A more descriptive title would be helpful. – garyp Aug 26 '17 at 12:34

## 1 Answer

In a sense, yes.

In general relativity, one of the fundamental postulates is that free bodies travel along geodesics (the paths of shortest distance in a curved geometry) of the spacetime. The caveat is that here, "free" means "free from all external influences except gravity", the reason being that no object is free from gravitational effects and so we might as well regard gravity to be inherent from the start.

Particles travelling on geodesics $x^{\mu}(\lambda)$ obey what is called the geodesic equation $$\frac{dx^{\nu}}{d\lambda}\frac{\partial}{\partial x^{\nu}}\bigg(\frac{dx^{\mu}}{d\lambda}\bigg) + \Gamma^{\mu}_{\nu\sigma}\frac{dx^{\nu}}{d\lambda}\frac{dx^{\sigma}}{d\lambda} = 0$$ Here $\lambda$ is simply some parameter, called an affine parameter if we're being technical, and $\mu$ labeling coordinates in spacetime along the geodesic, so this is really the "$\mu$th coordinate of the path in spacetime", but we abuse notation a bit. The symbols $\Gamma$ are called the connection coefficients, or Christoffel symbols, which describe how our basis vectors change in spacetime. Thus the equation tells that the total directional derivative of the tangent vector to the worldline (particle trajectory) along the worldline (remember, the $\Gamma$'s keep track of the fact that the basis vectors change too) should vanish. In other worlds, as we travel along the curve, the tangent vector doesn't change, it is kept "parallel" (we say that the tangent vector is parallel transported). This is a property we would expect from straight lines in flat space, and it ends up being the feature that carries over to the geodesics determined by the metric (another caveat, is that this "straight path = shortest path" phenomenon occurs because we say that the derivative operator on the spacetime, which takes into account how the basis vectors change and gives us the $\Gamma$'s, is the unique one determined by the spacetime metric, which is an object which gives us an inner product on vectors. This spacetime metric, in a way that we will see soon, is what gives us "gravity" in the sense that it is a "potential" for the "gravitational field" determined by the $\Gamma$'s, since the $\Gamma$'s are linear functions of the derivatives of the metric).

The crucial thing now is that time is an observer dependent quantity in GR. There is no preferred or absolute notion of time otherwise. But any observer, who travels along a geodesic, has a notion of time, given by what she reads off her watch as she travels along her worldline. This is called her proper time $\tau$. We may take this to be the parameter $\lambda$ of her worldline, so that $$\frac{d}{d\lambda} = \frac{d}{d\tau} = \frac{dx^{\nu}}{d\tau}\frac{\partial}{\partial x^{\nu}}$$ thus (denoting by a dot derivatives with respect to $\tau$), the geodesic equation becomes $$\ddot{x}^{\mu} = - \Gamma^{\mu}_{\nu\sigma}\dot{x}^{\nu}\dot{x}^{\sigma}$$ multiplying both sides by $m$ gives "Newton's Law" with gravitational force $$-m\,\Gamma^{\mu}_{\nu\sigma}\dot{x}^{\nu}\dot{x}^{\sigma}$$ hence we may interpret the Christoffel symbol twice contracted with the four-velocity $\frac{dx^{\mu}}{d\tau}$, $- \Gamma^{\mu}_{\nu\sigma}\dot{x}^{\nu}\dot{x}^{\sigma}$ as the "gravitational potential determined by general relativity".

• It is not entirely correct though. To fully account for Newtonian dynamics it is best to cast the geodesic equation to a form where $\lambda$ is coordinate time, not proper time (which is also easy to do). This way we could model the motion in terms of Newtonian-like trajectory $\vec{x}(t)$, which is also the most useful form for computer simulations of orbits. But I guess it depends on your taste which picture to call "Newtonian". – Prof. Legolasov Aug 28 '17 at 3:48