I often hear the story of how Einstein came up to the conclusion that time would slow down the faster you move, because the speed of light has to remain the same.
My question is, how did Einstein know that measuring the speed of light wouldn't be affected by the speed at which you are moving. Was this common knowledge already before Einstein published his paper on special relativity? If not, what led him to that conclusion?
Best Answer
Besides Michelson and Morley experimental results, Einstein also considered the theoretical aspects. It can be derived from Maxwell's equations that the speed at which electromagnetic waves travel is: $c=\left(\epsilon_{0}\mu_{0}\right)^{-1/2}$. Since light is an electromagnetic wave, that means that the speed of light is equal to the speed of the electromagnetic waves. $\epsilon_{0}$ and $\mu_{0}$ are properties of the vacuum and are constants, so $c$ will also be a constant. Thus from Maxwell's theory of electromagnetism alone we can already see that the speed of light in vacuum should be constant.
On the other hand, Galilean invariance tells us that the laws of motion have the same form in all inertial frames. There is no special inertial frame (as far as Newton's laws are concerned).
Another key element here is Galilean transformation, which was the tool used for transforming from one inertial frame to another. It can be easily seen that considering the first two elements to be valid:
means that we can no longer apply the Galilean transformation, because otherwise we will get a contradiction. Thus at least one of these three "key elements" must be wrong.
It turned out that the last one (Galilean transformation) was wrong. Einstein considered the first two correct and built the special theory of relativity. The correct transformation from one inertial frame to another, in the assumption of the validity of the Maxwell's theory and Galilean invariance, turns out to be Lorentz transformation . It is nice to check that the Lorentz transformation does indeed reduce to the Galilean transformation in the $v\ll c$ limit. That's why, in a sense, Galilean transformation is not wrong, but rather incomplete or a particular case. We can say that Galilean transformation needed to be generalized, and this was acomplished by introducing the invariance of the speed of light and maintaining the Galilean invariance.
How do Maxwell's equations predict that the speed of light is constant
Maxwell's equations in differential form: $$\tag{1}\nabla\cdot \mathbf{E}=\frac{\rho}{\epsilon_{0}}\label{1}$$ $$\tag{2}\nabla\cdot \mathbf{B}=0\label{2}$$ $$\tag{3}\nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}\label{3}$$ $$\tag{4}\nabla\times \mathbf{B}=\mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t}\label{4}$$
We can try to derive a wave equation in vacuum. Since we are considering the vacuum, we do not have charge densities, so equation ($\ref{1}$) becomes: $$\tag{5}\nabla\cdot \mathbf{E}=0\label{5}$$ In vacuum we do not have current densities either, so equation ($\ref{4}$) becomes: $$\tag{6}\nabla\times \mathbf{B}=\mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t}\label{6}$$
Now if we apply the curl to equation ($\ref{3}$), we get: $$\tag{7}\nabla\times\left(\nabla\times\mathbf{E}\right)=-\frac{\partial}{\partial t}\left(\nabla\times\mathbf{B}\right)\label{7}$$
We can use vector identity to evaluate the LHS of equation ($\ref{7}$): $$\tag{8}\nabla\times\left(\nabla\times\mathbf{E}\right)=\nabla\left(\underbrace{\nabla\cdot\mathbf{E}}_{=0}\right)-\nabla^2\mathbf{E}\label{8}$$ $$\tag{9}\nabla\times\left(\nabla\times\mathbf{E}\right)=-\nabla^2\mathbf{E}\label{9}$$
For the RHS of equation ($\ref{7}$), we can replace $\nabla\times\mathbf{B}$ with the expression we have from equation ($\ref{6}$): $$\tag{10}-\frac{\partial}{\partial t}\left(\nabla\times\mathbf{B}\right)=-\frac{\partial}{\partial t}\left(\mu_{0}\epsilon_{0}\frac{\partial \mathbf{E}}{\partial t}\right)=-\mu_{0}\epsilon_{0}\frac{\partial^2 \mathbf{E}}{\partial t^2}\label{10}$$ Putting all together: $$\tag{11}-\nabla^2\mathbf{E}=-\mu_{0}\epsilon_{0}\frac{\partial^2 \mathbf{E}}{\partial t^2}\label{11}$$ $$\tag{12}\nabla^2\mathbf{E}-\mu_{0}\epsilon_{0}\frac{\partial^2 \mathbf{E}}{\partial t^2}\label{12}=0$$ The general form of a wave equation is: $$\tag{13}\nabla^2\mathbf{\Psi}-\frac{1}{v^2}\frac{\partial^2 \mathbf{\Psi}}{\partial t^2}\label{13}=0$$ where $v$ is the velocity of the wave. Equation ($\ref{12}$) decribes an electromagnetic wave moving with velocity $v=\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$. Since light is an electromagnetic wave, that means that light is also propagating at this speed in vacuum. And since both $\epsilon_{0}$ and $\mu_{0}$ are constant, that means that $\frac{1}{\sqrt{\epsilon_{0}\mu_{0}}}$ is also a constant. Hence light moves at a constant speed in vacuum.