3-force in General Relativity

Question

Let's consider the definition of 3-force in different theories:

Newtonian mechanics

In Newtonian mechanics, we can define the force as a cause of changing momentum $p=m\vec{v}$ with time. According to the Second Newton law: $$\vec{F} = \frac{d\vec{p}}{dt},$$ or in the coordinate form: $$F^i = \frac{dp^i}{dt},$$ where $i = 1\ldots 3$, and $t$ -- is the absolute time here.

Special Relativity (SR) (let $c = 1$)

In Special Relativity the time is not absolute and it is better to formulate SR in Minkowsky space-time with metric: $$ds = dt^2 - \delta_{ik}dx^idx^{k},$$ $i,k = 1 \ldots 3$.

Firstly, we can introduce 4-displacement vector $dx^{\mu}$: $$dx^{\mu} = (dt,-dx,-dy,-dz).$$

Secondly, we can introduce a 4-momentum vector: $$p^{\mu} = m\frac{dx^{\mu}}{ds}$$

And we can introduce 4-force as $$\mathcal{F}^{\mu} = \frac{dp^\mu}{ds}.$$

Now we want to introduce 3-force.

Firstly, the spatial components of 4-momentum are: $$p^i = m\frac{dx^i}{ds} = m\gamma \frac{dx^i}{dt}.$$

Secondly, the spatial components of 4-force are not 3-force: $$\mathcal{F}^i = \frac{dp^i}{ds} = \gamma \frac{dp^i}{dt},$$ where $\gamma^2 = \frac{1}{1 - v^2}$ and $t$ -- is coordinate time (not proper time).

Thirdly, to determine the 3-force, we need to remember the definition of force in Newtonian mechanics, there the force is equal to the change 3-momentum with time $F^i = \frac{dp^i}{dt}$, and from the last equation we conclude the relation between spatial components of 4-force and 3-force: $$F^i = \frac{1}{\gamma}\mathcal{F}^i = \frac{1}{\gamma}\frac{dp^i}{ds}.$$

General Relativity (GR)

In General Relativity (GR) the almost general form of metrics has a form: $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ and 4-force is the covatiant derivative of 4-momentum (covariant derivative necessary for 4-force to be a vector), and its spatial components are: $$\mathcal{F^i} = m\frac{Dp^i}{ds}.$$

How can I define 3-force in curved space?

What does that have to do with anything? It's true for each component, spatial or temporal. — J. Murray, Oct 14 '17 at 19:56
@J. Murray You want to say Landau was wrong? Because in his book he defines 3-force as $Dp^{\alpha}/ds$. — Sergio, Oct 18 '17 at 06:12
I suggest you read what I wrote. $\frac{D p^\alpha}{ds}$ is equal to zero along a geodesic in spacetime. When do particles follow spacetime geodesics? — J. Murray, Oct 18 '17 at 06:15
@J.Murray But we still define an pseudo force of gravity, as described in book of Landau. — Sergio, Oct 18 '17 at 06:24
I still don't understand your question. The force $f^\alpha = c\sqrt{1-\frac{v^2}{c^2}}\frac{Dp^\alpha}{ds}$ is equal to zero along a spacetime geodesic. If you want to break up the covariant derivative $\frac{Dp^\alpha}{ds}$ into $\frac{dp^\alpha}{ds} + m\Gamma^\alpha_{\mu \nu} \frac{du^\mu}{ds}\frac{du^\nu}{ds}$ and then treat the latter term as a pseudoforce, you're more than welcome to. — J. Murray, Oct 18 '17 at 06:34
@J.Murray I have added the image from Landau book with problem. In the problem he define 3-force of gravity — Sergio, Oct 18 '17 at 06:37

3-force in General Relativity

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