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I found many questions regarding relativistic velocity addition and the relative formula

$$V_{x} = \cfrac {V_{x}^{'} + v}{1+\cfrac{vV_{x}^{'}}{c^{2}}}$$

but I found none concerning velocity addition in the presence of gravity. Can you tell me what is the formula to find the final velocity of a body freefalling on a massive planet. A link would be enough and a example would be much appreciated. ( like a rock travelling at $0.866c$ in a gravitational field of the strength of the sun or any other value.)

EDIT:

Can someone explain why (like JohnRennie says) we need special relativity or this question is a duplicate of the one about a black hole?

I am referring to an ordinary case of a comet approaching the earth and heading to the Sun or anoter massive planet or neutron star. Why can't we use the usual method of adding the PE to the KE to find the final velocity through the total KE?

user157860
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2 Answers2

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In General Relativity ("with gravity") you can always transform to a frame in which the metric is of the form $dt^2 - dx^2 - dy^2 - dz^2$. In that frame, the usual special r elativity formulas hold. (The standard caveat is that the experiments measuring velocities should not extend "too far" in space and time.)

For example, if you started with Schwarzschild coordinates, then used a locally-SR frame, added your velocities, you can then transform back to Schwarzschild coordinates.

johndecker
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For a relativistic particle under constant acceleration, $a$, the Lagrangian is formulated as:

$$\mathcal{L} = T - V = mc^2\sqrt{1 - \dot{x}^2 / c^2} - max$$

Using the Euler-Lagrange Equations:

$$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{d}{dt} \left[ -\frac{m\dot{x}}{\sqrt{1-\dot{x}^2/c^2}}\right]$$

$$\frac{\partial \mathcal{L}}{\partial x} = - ma$$

$$\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} = \frac{\partial \mathcal{L}}{\partial x}$$

$$\frac{d}{dt} \left[ \frac{m\dot{x}}{\sqrt{1-\dot{x}^2/c^2}}\right] = ma$$

$$\frac{\dot{x}}{\sqrt{1-\dot{x}^2/c^2}} = at + K$$

$$\frac{v}{\sqrt{1-v^2/c^2}} = at + k$$

If $u$ is the initial velocity when $t=0$, then:

$$k = \frac{u}{\sqrt{1-u^2/c^2}}$$

So, we can write the equation as:

$$\frac{v}{\sqrt{1-v^2/c^2}} = \frac{u}{\sqrt{1-u^2/c^2}} + at$$

Or, rearranged for $v$:

$$v = \pm c \sqrt{\frac{a^2t^2 + 2akt + k^2}{c^2 + a^2 t^2 + 2akt + k^2}}$$

Note: when $v \ll c$ and $u\ll c$, $c$ can be considered to be $\infty$, so the equation reduces to the standard, non-relativistic equation, $v = u+at$.

Beta Decay
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