There is a T-Shirt with this equation on it:

enter image description here

It says: And God said ... and the universe was ...

It looks like something related to General Relativity but I don't know what is it? Could you help me? Why the universe was ...?


$\gamma:\mathbb R\rightarrow M$ 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?

  • $\begingroup$ yeah, some time ago, I think I asked the question in another place and somebedy said something simmilar about the marketing person. But I have seen a lot of people using it with a lot of pride, and that's the weird thing to me, because they don't even know what does the equation mean but they are using it. I actually have the T-shirt, it was a gift, but I don't use it because I don't know what does it mean $\endgroup$ – DieDauphin Sep 3 '20 at 2:15
  • $\begingroup$ and because as you said, there is no actual physics on there. It would be like lying .. $\endgroup$ – DieDauphin Sep 3 '20 at 2:22
  • $\begingroup$ @DieDauphin Well, it's a perfectly nice bit of differential geometry, it just seems to have very little to do with the commands of a deity. If the curve in question is a straight line (or rather, the nearest you can get to a straight line on a curved surface), then the quantity on the t-shirt will be equal to zero. Now that you know what it means, you can wear the shirt with pride! $\endgroup$ – J. Murray Sep 3 '20 at 2:33
  • $\begingroup$ I still need to study the math, I am not very good in differential geometry, but once i've become more confident on it, I may consider using it, thanks $\endgroup$ – DieDauphin Sep 3 '20 at 2:37
  • 2
    $\begingroup$ +1 for the use of "...simply," $\endgroup$ – Oscar Bravo Sep 3 '20 at 7:32

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*} $\mathfrak{Q.E.D.}$


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