# "And God said ... and the universe was ..." What does this equation mean?

$$\gamma:\mathbb R\rightarrow M$$ is a curve whose image lies in the spacetime $$M$$, so $$\gamma(t)$$ is the event at parameter value $$t$$ along the curve. $$\gamma'(t) \in T_{\gamma(t)}M$$ is the tangent vector to the curve at parameter value $$t$$, and $$\gamma''(t)\in T_{\gamma(t)} M$$ is the rate of change of this tangent vector with respect to the parameter $$t$$. If $$M$$ is a Lorentzian manifold and $$\gamma$$ a timelike curve parameterized by its arc length $$t$$ (which we would call the proper time), then $$\gamma'(t)$$ would be the 4-velocity and $$\gamma''(t)$$ the 4-acceleration.

The right-hand side is simply the expansion of $$\gamma''(t)$$ in coordinates $$x^R:M\rightarrow \mathbb R$$. What we would usually write as $$x^R(t)$$ is, strictly speaking, $$(x^R \circ \gamma)(t)$$. Presumably, $$\delta_R$$ is the unit vector in the $$R$$ direction, which we would typically write as $$\frac{\partial}{\partial x^R}$$.

Given the reference to a higher power, I would guess that the intended market for this shirt is people who have studied GR and want strangers on the street to be aware of that fact. It is perhaps ironic, therefore, that there's no actual physics on the shirt - it is simply the coordinate expression for the second derivative of a curve on a manifold-with-connection. My (cynical) best guess is that some marketing person googled "General Relativity formulas" and picked one they found impressive, but who can say for sure?

Consider the local positional coordinate vector $$\mathbf{r}$$ which can be used to define the location of any arbitrary point in space, but here in our case, since it only depends on a single time parameter, it therefore only describes a curve within a submanifold of said space. Futhermore it acts as a parameterization that is infinitely many times continuously differentiable and maps a non-empty interval of elements of the real numbers onto the prior mentioned compact submanifold. It's acceleration vector is given by the second order derivative with respect to the already mentioned time parameter $$\lambda$$. \begin{align*} \frac{{d}^{2}\mathbf{r}}{d\lambda^{2}}=\frac{d}{d\lambda}\bigg(\frac{dq^{\mu}}{d\lambda}\partial_{\mu}\mathbf{r}\bigg)=\frac{dq^{\mu}}{d\lambda}\frac{dq^{\nu}}{d\lambda}\partial_{\mu}\partial_{\nu}\mathbf{r}+\frac{d^{2}q^{\mu}}{d\lambda^{2}}\partial_{\mu}\mathbf{r} \end{align*} The partial derivative of the coordinate vector $$\partial_{\mu}\mathbf{r}$$ is just defined as the covariant basis vector $$\boldsymbol{\varphi}_{\mu}$$ and it's partial derivative can now be rewritten as the following expression: \begin{align*} \partial_{\mu}\partial_{\nu}\mathbf{r}=\Gamma^{\omega}_{{\mu}{\nu}}\partial_{\omega}\mathbf{r}+L_{{\mu}{\nu}}\mathbf{n} \end{align*} Where $$\Gamma^{\omega}_{{\mu}{\nu}}$$ is the Christoffel symbol and $$\mathbf{n}$$ is the, perpendicular to the surface area, normal vector field. It may seem slightly odd at first since the additional second term $$L_{{\mu}{\nu}}\mathbf{n}$$ doesn't normally appear, nor is written. But we need it to describe the direction of the acceleration vector separately in two linear independent; tangential- and surface normal vector parts. Without this extra term, the acceleration vector would only point towards to a tangential direction. Now our equation becomes: \begin{align*} \frac{{d^2}\mathbf{r}}{d\lambda^{2}}=\bigg(\frac{d^{2}q^{\omega}}{d\lambda^{2}}+\Gamma^{\omega}_{{\mu}{\nu}}\frac{dq^{\mu}}{d\lambda}\frac{dq^{\nu}}{d\lambda}\bigg)\partial_{\omega}\mathbf{r}+L_{{\mu}{\nu}}\frac{dq^{\mu}}{d\lambda}\frac{dq^{\nu}}{d\lambda}\mathbf{n} \end{align*} $$\mathfrak{Q.E.D.}$$