# Generalizing a relativistic kinematics formula for spatial-acceleration dependence

I'm starting from this expression

$$\alpha dt = \gamma^3 dv$$

where $\alpha$ is proper acceleration of a point particle, $dv$ and $dt$ are coordinate differentials of velocity and time, and $\gamma$ is the relativistic factor of the particle being subject to the acceleration

If $\alpha$ is constant, one arrives at the usual expression:

$$t_f - t_0 = \frac{1}{\alpha} (\frac{v_f}{\sqrt{1-\frac{v_f^2}{c^2}}} - \frac{v_0}{\sqrt{1-\frac{v_0^2}{c^2}}})$$

Now, I have some dependence of acceleration to position; $\alpha(x) = f(x)$ and I'm not sure how to integrate it in order to obtain a similar expression relating time, velocity and position. For instance, I tried the following for the left-side differential:

$$\alpha dt = \alpha(x) \frac{dt}{dx} {dx} = \int{ \frac{ \alpha(x) }{v(x)} dx }$$

and leaving the right-hand side untouched. When I do this I get a weird integral with velocity in both sides, and I'm not sure how to continue.

Update

so, after a while of staring at this, i noticed an error i was doing, and actually the 2nd derivative of acceleration should look like

$$\frac{d^2 x}{d \tau^2} = \gamma^4 \frac{d^2 x}{dt^2}$$

So far so good, but the approach on the comments doesn't seems to work. I try with a simple force potential: $\alpha(x) = -k x^3$, but is not clear how to work from it

$$-k x^3 = \gamma^4 \frac{d^2 x}{dt^2}$$

$$-k x^3 (1 - 2 \frac{1}{c^2} (\frac{dx}{dt})^2 + \frac{1}{c^4} (\frac{dx}{dt})^4 ) = \frac{d^2 x}{dt^2}$$

It looks like a (nonlinear) differential equation. I just want to be sure i'm on the right track. This is what it takes to solve this kind of problem? Or should some numerical integration/summation be enough? What throws me off in particular, is that if i would replace $-k x^3$ with just $\alpha_0$ (constant acceleration case), the $\gamma^4$ would still look pretty ugly and i wouldn't know how to solve the problem even in that case using the above expression, when we already know that a closed formula exists

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Hi diffeomorphism, I added the homework tag to your question because it seems to apply. See the homework policy and homework tag for details. –  Brandon Enright Jun 1 '13 at 17:44
It is not homework, but i have no problem whatsoever with the tag –  diffeomorphism Jun 1 '13 at 17:54
You have : $\alpha(x) = \frac{d^2x}{d \tau^2}$, so try to solve this equation, and express $x$ as a function of $\tau$. –  Trimok Jun 1 '13 at 18:09
thanks, let me try –  diffeomorphism Jun 1 '13 at 18:21
This is one of the more interesting and challenging homework like question. The OP clearly says what he tried and what he does not understand, so this should not be closed. –  Dilaton Jun 1 '13 at 21:45

## 2 Answers

Let's start with $$\frac{\text{d}v}{\left(1-v^2/c^2\right)^{3/2}} = \alpha\text{d}t = \frac{\alpha(x)}{v}\text{d}x.$$ As Clem suggested, multiply both sides by $v$, so that $$\int_{v_0}^{v}\frac{v'\text{d}v'}{\left(1-v'^2/c^2\right)^{3/2}} = \int_{x_0}^x\alpha(x')\text{d}x' = A(x),$$ which yields $$\frac{1}{\sqrt{1-v^2/c^2}} - \frac{1}{\sqrt{1-v_0^2/c^2}} = \frac{A(x)}{c^2},$$ or (assuming $v_0\geqslant 0$) $$v = c\frac{\sqrt{\left(A(x)/c^2+\gamma_0\right)^2-1}}{A(x)/c^2+\gamma_0} = \frac{\text{d}x}{\text{d}t},$$ with $$\gamma_0 = \frac{1}{\sqrt{1-v_0^2/c^2}}.$$ Thus $$\int_{x_0}^x\frac{A(x')/c^2+\gamma_0}{\sqrt{\left(A(x')/c^2+\gamma_0\right)^2-1}}\text{d}x' = c(t-t_0).$$ For a given $\alpha(x)$, you can solve this to get $x(t)$, and thus $A(t)=A(x(t))$, which will give you $v(t)$. If the acceleration is constant, then $A(x)=\alpha(x-x_0)$, and the expressions should reduce to the familiar formulae.

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doh! i feel pretty stupid now. How is possible that my brain never thought about moving the dividing v(x) to the other side, in fact it made the integral much easier! –  diffeomorphism Jun 2 '13 at 17:10

In your equation $\alpha dt = \alpha(x) \frac{dt}{dx} {dx}$, just multiply each side by v before you integrate.

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doh! i feel pretty stupid now. How is possible that my brain never thought about moving the dividing v(x) to the other side, in fact it made the integral much easier! –  diffeomorphism Jun 2 '13 at 17:10