Newton's laws of motion and General Relativity

Question

Are Newton's laws of motion valid in general relativity? Particularly, I want to know if his 2nd law of motion is valid i.e. whether acceleration is produced by a force or is it entirely something else in the context of GR.

Answer 1

In General Relativity there is an important concept known as local inertial frame, or freely falling frame. Such a frame can be realised in the ideal limit of a collection of non-interacting particles falling freely and near to one another. One can set up measures of distance and time in a well-defined way, such that the distances between the particles don't change for small times (to be precise they only change to third order in time).

With such a frame of reference to hand, we can define velocity and acceleration, and also (eventually) momentum and force. The relationship between force and momentum is then (in such a local inertial frame) $$ {\bf F} = \frac{d {\rm p}}{dt} $$ It will perhaps surprise readers to learn that this is precisely Newton's second law, in the form he described it in the Principia. In other words the answer is that this law does hold in General Relativity, as long as it is applied correctly, which is to say it describes how non-gravitational forces cause changes of momentum away from the value the momentum would have if the particle were in free-fall. Here the momentum is related to rest-mass $m$ and velocity $\bf v$ through the equation $$ {\bf p} = \gamma m {\bf v} $$ where $$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}. $$

Another way to see the above is to regard it as an example of the Strong Principle of Equivalence (which states, briefly, that ordinary non-gravitational special relativistic physics holds in local inertial frames, for all effects that do not directly relate to spacetime curvature).

Answer 2

In the regime of General Relativity the only thing that matters are Einstein's field equations:$$G_{\mu\nu}=kT_{\mu\nu}.$$ However, in the classical limit, one can retrieve Newton's second law from Einstein's equations. For example, Einstein's theory predicts a curved space-time with geodesics that aren't straight lines as they are in Euclidean space, the equation for the geodesics is $${d^2x^\mu\over ds^2}+\Gamma^\mu_{\alpha\beta}{dx^\alpha\over ds}{dx^\beta\over ds}=0.$$ However, if the gravitational effects are small, i.e. $$g_{\mu\nu}=-\delta_{\mu\nu}+\gamma_{\mu\nu},$$ where $g_{\mu\nu}$, is the metric tensor found from solving Einstein's equations, then we are justified in writing: $$ds^2=-dx^2_{\nu}=dt^2(1-\beta^2).$$ Now if the velocity of the particles be such that they are well below the speed of light, i.e. non-relativistic and the gravitational field very nearly constant in time, we get: $${d^2x^\mu\over dt^2}={\partial\over \partial x_\mu}{1\over 2}\gamma_{00}.$$ Here the $\gamma_{00}$ arise from the assumption of slowly varying gravity on the Christoffel symbols found in the geodesic equation: $$\Gamma^\mu_{\alpha\beta}={1\over 2}\big({\partial \gamma_{\alpha\beta}\over\partial x_\mu}-{\partial\gamma_{\alpha\mu}\over\partial x_\beta}-{\partial\gamma_{\beta\mu}\over\partial x_\alpha}\big).$$ If we interpret the $\gamma_{00}/2$ with gravitational potential $-\phi$, then since $$\mathbf F=-\nabla \phi$$ we have arrived at Newton's second law:$${d^2x^\mu\over dt^2}={\partial\over \partial x_\mu}{1\over 2}\gamma_{00}=-\nabla\phi=F.$$

Thus, so long as the above noted assumptions hold true one can see that General Relativity and Newtonian physics are entirely equivalent to one another.

Answer 3

Answer : No it is not produced by a force. Explanation : Firstly it would be of great benefit to distinguish inertial mass from gravitational mass; despite there being an empirical equivalence.

(If you are familiar with the distinction you may skip the following)

Essentially, inertial mass $m_I$ represents the resistance to applied force $F_I$; Hence Newton's second law. Gravitational mass $m_G$ represents the quantity that strengthens the gravitational force $F_G$.

A particle with inertial mass $m_I$ and gravitational mass $m_G$ accelerating as result of a gravitational force only (that is a particle in free fall), such that: $\mathbf F_I=\mathbf F_G$

$\mathbf{a}=\frac{m_G}{m_I}\mathbf g$

It is an established experimental result that $m_G =m_I=m$ , and this equivalence of inertial mass to gravitational mass leads to the equivalence principle ($\mathbf a=\mathbf g$).

Continued explanation

The Gravitational force on test particle with mass $m$ is given as: $\mathbf F_G=G\frac{m M}{\mathbf r^2}\frac{\mathbf r}{|\mathbf r|} = m\mathbf g$

Let$ |\mathbf r|=r$

where $\mathbf g=G\frac{M}{r^2} \left(\frac{\mathbf r}{|\mathbf r|}\right)=-\frac{d}{dr}\left( \frac{GM}{r}\right)\frac{\mathbf r}{|\mathbf r|} $

This implies that the gravitational acceleration $\mathbf g =-\nabla\phi$

where $\phi=G\frac{M}{r}$ is the gravitational potential; note that it depends only on the mass of the gravitational source, and where we choose to evaluate it in space.

Thus $\mathbf F_G=-m\nabla\phi$

Location is a geometric property,and the distance between two points is subject to the intrinsic geometrical nature of the curve,surface or manifold on which the points are defined.

From differential geometry we learnt that to measure lengths we do not always use straight edges, we rather use a measuring tool tailored for our curve, surface or manifold;we do not use a straight edge to measure the distance between two points on a circle, because the latter has curvature and the former does not.

So the gravitational potential depends on the geometry of the spacetime. In the case of considering dust particles as our gravitational source;these particles do not exert any force on the spacetime rather their collective motion changes their collective mass density, through relativistic effects.

Why is this important?

Well the gravitational potential may generally be found by using Poisson's equation:

$\nabla^2\phi=4\pi G\rho$

Where $\rho$ is the mass density. The mass density changes as result of relativistic effects on both it's defining quantities...these changes result to a change in value of the gravitational potential,and ultimately the gravitational force.

Our test particle would move according to the global geometrical nature of the spacetime defined by the dynamical properties of our gravitational source...that is to say it will globally follow a geodesic path on the spacetime.

To really appreciate this, note the following:

Supoose $n_0$ dust particles confined in a unit volume($V=1$),were to move in the positive $x-direction$ with speed $v$,and each of our dust particles has mass $m_0$.

We have the mass density at rest is: $\rho_0=\frac{n_0 m_0}{V_0}$ And while in motion and preserving particle number:

$\rho=\frac{n_0 m}{V} =\gamma_v^2 \frac{n_0 m_0}{V_0}=\gamma_v^2 \rho_0$ Where,

$m=\gamma_v m_0$ and $V=\frac{V_0}{\gamma_v}=\frac{1}{\gamma_v}$

So if the speed at which the dust particles move changes their collective mass density, and the mass density determines the gravitational potential, which furthermore determines how the gravitational acceleration,which finally ,by virtue of the equivalence principle determines the geodesic path our test particle must follow in the vicinity or neighborhood of our gravitational source...then gravity is caused by special relativistic effects.

Hopefully this makes sense.