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I would like to compute what the constant acceleration trajectories are in the Minkowski spacetime $(t, x)$ with $d\tau^2 = dt^2 - dx^2$. So given some trajectory $x(t)$ I know the velocity vector is given by $$U = \left( \frac{dt}{d\tau}, \frac{dx}{d\tau} \right)$$ In the previous version of this post I made some great confusion by differentiating $\tau^2 = t^2 - x^2$ to obtain $d\tau$ and thus $dt/d\tau$ etc. I realize now that there's perhaps some poor notation chosen in the circumstances. Here $d\tau$ really should mean the line element, defined as acting on some vector by $d\tau(v) = \sqrt{|\eta(v, v)|}$. This has nothing to do with the differential of the function $\tau = \sqrt{t^2 - x^2}$. In retrospect that is clear -- the differential (thought of as a 1-form) is linear, while the line element clearly is not. Unfortunately the standard notation for both these things is $d\tau$, which is why I got so terribly confused.

Now I realize that to find the velocity vector of some curve $\alpha$ we must first parametrize $\alpha$ by arc-length and then differentiate as usual. By doing that I managed to convince myself that, for the trajectory $x(t)$, its velocity (covariant) vector is $$ U = \gamma(1, \dot{x}) $$ where $\gamma = 1/\sqrt{1 - \dot{x}^2}$ (here I am writing $\dot{x}$ instead of $v$ to emphasize that this trajectory is accelerated). Differentiating again, however, I seem to get $$ dU/d\tau = \left( \frac{1}{2} \gamma^4 \ddot{x} , \frac{1}{2} \gamma^4 \dot{x} \ddot{x} + \gamma \ddot{x} \right)$$

I believe each of these components must be constant functions, but I cannot seem to solve the resulting equations. Have I done something wrong again? If not, how to proceed?

Edit: actually, after Rounak's answer I realized I made some stupid mistakes and I should actually get $$ dU/d\tau = (\gamma^4 \dot{x} \ddot{x}, \gamma^4 \dot{x}^2 \ddot{x} + \gamma^2 \ddot{x}) $$

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  • $\begingroup$ may be this could help en.wikipedia.org/wiki/… $\endgroup$ Jul 17 '20 at 21:46
  • $\begingroup$ Is it not simpler to use that $dt/d\tau = \gamma$? $\endgroup$ Jul 17 '20 at 22:56
  • $\begingroup$ I am realizing that my computations were all completely wrong. Once I finish doing them correctly I will write about why. $\endgroup$
    – Pedro
    Jul 18 '20 at 3:48
  • $\begingroup$ I have edited the post $\endgroup$
    – Pedro
    Jul 18 '20 at 5:48
  • $\begingroup$ Because $u^{\mu }u_{\mu }=c^{2}$ so $ \dfrac{d}{d\tau } \left( u^{\mu }u_{\mu }\right) =0$ $\endgroup$
    – Eli
    Jul 18 '20 at 8:19
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let $\dot{x}=u$ and 'a' is spatial acceleration.

$U= \gamma(1,u)$

Acceleration, $A = dU/d\tau = \gamma dU/dt = \gamma(\dot{\gamma},\dot{\gamma}u+a\gamma)$ --(1)

as $dt/d\tau = \gamma$

$\dot{\gamma} = \sqrt{1-u^2}$ $\implies$ $\dot{\gamma} = \gamma^3ua$

Plug this in (1) and you will get 4-acceleration. If you say that for constant 4-acceleration, each component must be constant, then one can always find some coordinate in which that is not the case. What you should really be trying is acceleration scalar: $A^2 = g_{\alpha\beta} a^\alpha a^\beta$, where $a^\mu$ are the components of $A$ and $g_{\alpha\beta}$ is the metric.

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  • $\begingroup$ You are right with the formulas, I made some simple mistakes while computing. $\endgroup$
    – Pedro
    Jul 18 '20 at 8:57
  • $\begingroup$ With regards to the coordinates, the point is I am working in usual $(t, x)$ coordinates. The Christoffel symbols are all zero, so I believe parallel transport is the usual (trivial) one. So that in these coordinates, saying the covariant derivative of the acceleration is zero would be just saying that each component is a constant function. Am I wrong? $\endgroup$
    – Pedro
    Jul 18 '20 at 9:11
  • $\begingroup$ @Pedro No, in usual (t,x) coordinates (by that I mean, the metric is independent of coordinates and is diagonal), the condition is equivalent to each component being zero. $\endgroup$
    – Rounak
    Jul 18 '20 at 9:26
  • $\begingroup$ Each component of $dA/d\tau$ should be $0$, yes. Which is the same as saying each component of $A$ is a constant function. $\endgroup$
    – Pedro
    Jul 18 '20 at 9:45
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Rounak's answer is correct, so I accepted; I just wish to add some details here. The formula I wrote for $dU/d\tau$ simplifies, with a little bit of algebra, to $$dU/d\tau = \gamma^4 \ddot{x}(\dot{x}, 1)$$ Then indeed if you compute $d\tau^2$ of this vector you end up with $\gamma^8 \ddot{x}^2 (\dot{x}^2 - 1)$. You set that equal to some constant and it's not hard to solve the resulting differential equation to indeed get the hyperbolic motion. This is done e.g. in Landau's book on classical field theory, page 24.

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