Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In the beginning, God said that the four dimensional divergence of an antisymmetric second rank tensor equals zero and there was light.

Can someone explain what is the meaning of this quote by Michio Kaku?

share|cite|improve this question

migrated from Jan 21 '13 at 2:16

This question came from our site for people studying math at any level and professionals in related fields.

In short, the quote is a reference to the special relativistic formulation of Maxwell's equations (without source terms), which in turn governs light. – Qmechanic Jan 21 '13 at 8:20

The antisymmetric second-rank tensor being referenced is the electromagnetic field tensor. It is defined as follows. Let $\varphi$ be the electrostatic potential (a scalar field), and let $\underline{A}$ be the magnetic potential (a 3-vector) from classical E&M. Concatenate them into a 4-vector $\vec{A}$. Now define the tensor of interest as the exterior derivative of $\vec{A}$: $$ \mathbf{F} = \mathrm{d}\vec{A}. $$ We can write this component-wise with partial derivatives: $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu. $$ You can see that if thought of as a matrix, the components $F_{\mu\nu}$ of $\mathbf{F}$ are antisymmetic.

Now the use of this is that the four equations that govern classical electromagnetism (and hence light) are equivalent to: $$ \partial_\nu F^{\mu\nu} = J^\mu $$ ($\vec{J}$ is the 4-current composed of electric charge concatenated with 3-current) and $$ \partial_{[\alpha} F_{\mu\nu]} = 0 $$ (the brackets denote the summing all permutations of indices with a sign given by the parity of the permutation).

Note that depending on your unit system there may be constants like $c$ or $\mu_0$ floating around in these equations.

share|cite|improve this answer
What's the advantage of starting with the requirement that $F$ can be written as $\mathbb{d}A$? And if you start that way, why write down $ \partial_{[\alpha} F_{\mu\nu]} = 0 $ as well? – NikolajK Jan 22 '13 at 8:26

It is similar to the t-shirt designs of a couple decades ago with Maxwell' Equations in place of "Let there be light" in the famous biblical quote, except this time some designer decided to use the slick spacetime tensor way of describing electromagnetic fields. In relativity, the electromagnetic field is described by a 4D antisymmetric tensor. Its divergence, meaning the antisymmetrized derivative, is zero in vacuum. (It's nonzero at charges and currents, but I guess God didn't invent charge until the next day.)

share|cite|improve this answer

From this transcript, the full quote is

And so, if you go to Berkley, where I got my PhD, you can buy a t-shirt which says, “In the beginning God said, the four-dimensional divergence of an antisymmetric, second rank tensor equals zero, and there was light, and it was good. And on the seventh day he rested.” Ladies and gentlemen, this is the equation for light.

A rank two tensor is commonly known as a matrix, and a rank one tensor is commonly known as a vector.

An anti-symmetric tensor is a tensor in which exchanging two indices negates the tensor; for example $a_{ji}=-a_{ij}$.

According to Wikipedia, the divergence of a second order (rank) tensor is a first order (rank) tensor (I've extrapolated the result there to four dimensions) $$ \nabla\cdot\epsilon=\begin{bmatrix} \frac{\partial\epsilon_{xx}}{\partial x}+\frac{\partial\epsilon_{yx}}{\partial y}+\frac{\partial\epsilon_{zx}}{\partial z}+\frac{\partial\epsilon_{wx}}{\partial w}\\ \frac{\partial\epsilon_{xy}}{\partial x}+\frac{\partial\epsilon_{yy}}{\partial y}+\frac{\partial\epsilon_{zy}}{\partial z}+\frac{\partial\epsilon_{wy}}{\partial w}\\ \frac{\partial\epsilon_{xz}}{\partial x}+\frac{\partial\epsilon_{yz}}{\partial y}+\frac{\partial\epsilon_{zz}}{\partial z}+\frac{\partial\epsilon_{wz}}{\partial w}\\ \frac{\partial\epsilon_{xw}}{\partial x}+\frac{\partial\epsilon_{yw}}{\partial y}+\frac{\partial\epsilon_{zw}}{\partial z}+\frac{\partial\epsilon_{ww}}{\partial w} \end{bmatrix} $$ As for the physical meaning, that is more a topic for physics.

share|cite|improve this answer

protected by Qmechanic Dec 3 '15 at 7:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?