# Computing Riemann tensor components for Alcubierre metric

I am currently trying to compute the Riemann tensor components for the Alcubierre metric, and already, on the computation of the first component, I'm running into some issues.

The trouble component in question is $$R^x_{\;txt}$$ , so I've started with the formula: $$R^x_{\;txt} = \partial_x\Gamma^x_{\;tt} -\partial_t\Gamma^x_{\;xt}+\Gamma^x_{\;x\mu}\Gamma^\mu_{\;tt}-\Gamma^x_{\;t\mu}\Gamma^\mu_{\;xt}$$ Using the Christoffel symbols provided by Mueller and Grave's Catalogue of Spacetimes, I started computing the first term, $$\partial_x\Gamma^x_{\;tt}$$: \begin{align} \partial_x\Gamma^x_{\;tt} &= \partial_x\frac{f^3f_xv_s^4-c^2ff_xv_s^2-c^2f_tv_s}{c^2} \\ &=\frac1{c^2}\Big(\partial_xf^3f_xv_s^4-\partial_xc^2ff_xv_s^2-\partial_xc^2f_tv_s\Big) \end{align} And, again isolating the first term and computing further: $$\partial_xf^3f_xv_s^4 = f_xv_s^4\partial_xf^3 + f^3v_s^4\partial_xf_x + f^3f_x\partial_xv_s^4$$ If my math is correct.

The area I'm having trouble is in the partial derivative $$\partial_xv_s$$. I can't quite understand how to compute it. For reference, $$v_s$$ is defined by Alcubierre as a function of time: $$v_s(t) = \frac{dx_s(t)}{dt}$$ and $$x_s(t)$$ is simply described as an "arbitrary function of time", describing the trajectory of the hypothetical spacecraft in the scenario of the metric.

I don't quite see how the actual $$x$$ axis relates to this function, and so I'm at a loss as to how I should treat a derivative of the function with respect to $$x$$. What am I missing?

• For simplicity, to start, you could just treat the case of a warpdrive spacecraft moving uniformly and take it to be a constant. Jan 24, 2020 at 21:11
• @G.Smith I do want my end result to work with changing speed, wouldn't that simplification prohibit that? Jan 25, 2020 at 2:11
• I’m tempted to say that it is simply a function of $t$ and not of the spatial coordinates, so $\partial_x v_s$ is zero but $\partial_t v_s$ is nonzero. But I’m not sure if I’m thinking about it the right way, having never thought about this metric. But it seems similar to the Friedmann metric, where $a(t)$ is just a function of $t$. Jan 25, 2020 at 2:32
• Sorry, I don’t understand what “applying $x_s$ to $x$” is supposed to mean. Jan 26, 2020 at 5:20
• OK. I’m convinced that $\partial_ x v_s=0$. Jan 26, 2020 at 18:19

## 1 Answer

As G. Smith helped answer, $$v_s(t)$$ is a function that depends solely on time, and as such does not change across the x direction, so $$\partial_xv_s(t) = 0$$. The function is defined as the derivative of a function named $$x_s(t)$$, which also depends only on time, and it's the name of this function which mislead me.