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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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  • Possible duplicate of Will an object always fall at an infinite speed in a black hole? – John Rennie Jul 16 '18 at 07:26
  • Though it isn't obvious from the title of the question, my answer to the above gives the equation for an object falling in a strong (relativistic) gravitational field. – John Rennie Jul 16 '18 at 07:27
  • @JohnRennie, thanks, is this ($ v = \sqrt{\frac{r_s}{r}}c $ ) the equation you refer to? Could you take the trouble to write a short answer just showing how you would proceed? can you consider a ship passing the earth's orbit at .866 c and heading toward the sun, is this case enough to produce relativistic effects? and what about if the pull is not less than 1 cm/s^2 (as it's the pull of the sun) but , say, i Km/s^2, what is final speed when it impacts the sun or other body? – user157860 Jul 16 '18 at 08:23
  • Why the downvote? what is wrong with this question? – user157860 Jul 18 '18 at 06:07

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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  • This is not helpful, can you give a formula and say how to apply it? – user157860 Jul 16 '18 at 08:26
  • @user157860 The metric $ds^2 = dt^2 - dx^2 - dy^2 - dz^2$ is a way of measuring miniscule distances on spacetime and can be generalised to $ds^2 = \sum\limits_{\mu,\nu}g_{\mu\nu}dx^{\mu}dx^{\nu}$. Here, $g_{\mu\nu}$ is called the metric tensor and basically describes the spacetime surface. Now, what johndecker is saying is that even when gravity is present, you can always perform a coordinate transform so that the metric tensor looks the same as it would in special relativity with no gravity. As such, the velocities will always add as $w = \frac{u+v}{1+uv/c^2}$. – Beta Decay Jul 16 '18 at 10:42
  • @BetaDecay, can you add a velocity and an acceleration? – user157860 Jul 16 '18 at 13:55
  • @user157860 You can only add things together if their units are the same. The SI units of velocity are $\mathrm{ms}^{-1}$ and the SI units of acceleration are $\mathrm{ms}^{-2}$. So, no, you cannot add a velocity and an acceleration. – Beta Decay Jul 16 '18 at 13:57
  • @BetaDecay, so what is the use of the formula in your previous comment? What about the formula by John Rennie? do you understand his comments? – user157860 Jul 16 '18 at 14:01
  • @user157860 I'm not sure what you mean. $w$, $u$ and $v$ are all velocities. $u$ and $v$ are the velocities which are being summed together and $w$ is the result. $c$ is the speed of light. – Beta Decay Jul 16 '18 at 14:03
  • @BetaDecay, we have a rock with velocity v=.866 c and the pull of a star g= 1000cm/s^2, how can you with all transform change that substance? how do you find v' after a second or other lapse of time? – user157860 Jul 16 '18 at 14:06
  • Let us continue this discussion in chat. – Beta Decay Jul 16 '18 at 14:07
  • @BetaDecay, I found a question regarding this problem but it's incomplete, can you have a look at your convenience and then ping me at our chat? Thanks – user157860 Jul 27 '18 at 15:54
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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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  • Thanks a lot for your answer, how does your formula compare with JohnRennie's? I edited my question, can you give the final velocity for c ? – user157860 Jul 18 '18 at 06:03
  • @user157860 Sorry, I've written the equation in terms of time. I'm not completely sure how to calculate the velocity as it falls through a distance $s$... – Beta Decay Jul 18 '18 at 09:52
  • Also, the choice of sign is opposite of the usual convention (see i.e. a relativistic particle with constant force $F=ma$). The conventional sign choice emphasizes that expressing the Lagrangian as $\mathcal{L} = T-V$ is really only valid for Newtonian mechanics where the kinetic energy is quadratic. More generally, one needs to verify that the Lagrangian gives the correct equations of motion, then use it to simplify coordinate transforms, constraints, etc. – mforbes Apr 08 '21 at 09:57