I was doing a search on Google to verify something, and I found a bunch of Web sites describing Einstein's equation.

Except that they weren't.

If you really want to know, Einstein's equation is $R_{ab}-Rg_{ab}/2=8\pi T_{ab}$, where $R_{ab}$ is the Ricci-tensor, $R$ is the scalar curvature (both quantities are derived from the Riemann curvature tensor, which in turn is a function of the metric tensor, $g_{ab}$, describing the "intrinsic curvature" of the spacetime manifold), and $T_{ab}$ is the stress-energy-momentum tensor characterizing any matter and energy fields that fill the spacetime. What this equation basically tells you is that the presence of matter and energy (the right side of the equation) determine the curvature of spacetime (the left side). |

For the record, $E=mc^2$ is not, repeat, NOT Einstein's equation. Oh, it was derived by Einstein alright^{*}, it's just not what physicists normally call Einstein's equation. is simply the residual "rest energy" of an object in special relativity. In Newtonian mechanics, this quantity is undetermined; what matters is the relative energy levels between two states, not the absolute energy of any given state. In contrast, in special relativity this rest energy is well defined.

The fundamental assumption of special relativity is that a transformation from one coordinate system to another leaves the relativistic interval, $ds^2=c^2dt^2-dx^2-dy^2-dz^2$, unchanged. Dividing both sides with $dt^2$, this equation can be rewritten as $ds^2/dt^2=c^2-v^2$, where $v$ is the ordinary velocity.

The motion of a particle is governed by the principle of least action. The "action" is the time integral of a function called the *Lagrangian*, between the start and end position of the particle's path.

In special relativity, the action of a free particle of mass $m$ is simply $S=\int-mc~ds$. This is really just the relativistic version of the statement that a freely moving particle will always choose the "shortest" path between two points.

Taking the previous expression into account, the action can be rewritten in the form of a time integral as $\int-mc\sqrt{c^2-v^2}~dt$. From this, the Lagrangian of a free particle is $L=-mc\sqrt{c^2-v^2}$.

The momentum of a particle is defined as $\vec{p}=\partial L/\partial\vec{v}=m\vec{v}/\sqrt{1-v^2/c^2}$. The energy of a particle is defined as $E=\vec{p}\vec{v}-L$. These two quantities are derived from the Lagrangian using symmetry considerations: assuming that the Lagrangian remains invariant under a spatial or time translation, it can be observed that these two quantities remain conserved. (Briefly: The principle of least action, combined with the assumption that space is homogeneous, demands that under small variations of the spatial coordinates, $\vec{x}$, the variation of $L$ will be zero, or $\delta L=\delta\vec{x}\cdot\partial L/\partial\vec{x}=0$. Since $\delta\vec{x}$ can be arbitrary, $\partial L/\partial\vec{x}$ must be zero. But $\partial L/\partial\vec{x}=d(\partial L/\partial\vec{v})/dt$, so $\vec{p}=\partial L/\partial\vec{v}$ is a conserved quantity. Conversely, time homogeneity means that $L$ does not explicitly depend on $t$, so it will be a function of coordinates and velocities. Then, $dL/dt=(\partial L/\partial\vec{x})(d\vec{x}/dt)+(\partial L/\partial\vec{v})(d\vec{v}/dt)$. But $\partial L/\partial\vec{x}=d(\partial L/\partial\vec{v})/dt$, so $dL/dt=\vec{v}\cdot d(\partial L/\partial\vec{v})/dt+(\partial L/\partial\vec{v})(d\vec{v}/dt)$. Or, $d(\vec{p}\vec{v}-L)/dt=0$, so $E=\vec{p}\vec{v}-L$ is conserved under time translations.)

The rest is a straightforward calculation:

\[E=\frac{mv^2}{\sqrt{1-\frac{v^2}{c^2}}}+\frac{mc^2\left(1-\frac{v^2}{c^2}\right)}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}.\]

When the velocity is small, higher-order terms can be ignored and we're left with the expression $E\simeq mc^2+mv^2/2$. This is consistent with the Newtonian expression of energy, $E=mv^2/2+C$, where $C$ was an arbitrary integration constant; in the relativistic case, we're no longer free to choose any $C$, instead the "rest energy" of the particle is well defined: when $v=0$, $E=mc^2$.

To summarize, $E=mc^2$ is the consequence of four assumptions: that the relativistic interval, $ds$, is invariant under a change of coordinate systems, that space is homogeneous, that time is homogeneous, and that the motion of a free particle is governed by the requirement that the action, $S=\int-m~ds$, remains minimal between two points.

That the rest energy is well defined does suggest that an actual, physical relationship exists between matter and energy. Interesting, to be sure, and may be of some significance if you were to build an antimatter bomb (not an atomic bomb, as some writers suggest) but it's really just a minor consequence of a simple equation, nothing more. And it is not, I repeat, NOT the one physicists call Einstein's equation.

I suspect one reason $E=mc^2$ became "Einstein's equation" is because it's easy to remember, and even a mediocre high school education is enough to make one understand what it means. Nothing wrong with that, just make sure you also know that this is *not* a fundamental equation, this is *not* why physicists revere Einstein and his work, this is *not* what makes the theory of relativity perhaps the most powerful physical theory known to man. It is just a simple result of a simple calculation.

Keep this in mind the next time you come across a writing that describes how Einstein set out to "discover" $E=mc^2$, or worse yet, writings like the one I recently saw that "expose" Einstein as a "fraud" or a "plagiarist" because purportedly, his "derivation" of $E=mc^2$ was "flawed".

^{*}I was horrified to learn recently that this article was seen by some as an attempt to "prove" that $E=mc^2$ was not derived by Einstein. That is not what I am saying here! Indeed, I even decided to change the title (originally, it said "*E* = *mc*² is not Einstein's Equation") because frankly, the last thing I had in mind when I wrote this was to provide fuel for the ramblings of anti-Semitic crackpots.

The point I am making here is not that $E=mc^2$ is not Einstein's work (it most certainly is) but that $E=mc^2$ is NOT the equation that pops into most physicists' minds when you mention "Einstein's equation". See the sidebar above. Of course the pedantic might point out that what I call Einstein's equation is really Einstein's field equation(s), and thus it's okay to call $E=mc^2$ Einstein's equation, but I never much subscribed to pedantry, and in any case, both my Wald and my Landau & Lifshitz call the field equation Einstein's equation, and you cannot get much more pedantic than Landau & Lifshitz!