Why can’t we travel faster than the speed of light? Einstein explained this pheonmenon through his famous equation

\begin{equation}\label{eqn:mass} m=\frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}, \end{equation}

which is the last of his four celebrated works in 1905 (Albert Einstein’s Year of Miracles) titled Does the inertia of a body depend on its energy content?. This paper entails arguably the most famous formula so far: Mass-Energy Equivalence

\[E=mc^2.\]

From \eqref{eqn:mass}, when the velocity of a particle approaches \(c\), the relative mass \(m\) will goes to \(\infty\). Therefore any particle with \(m_0>0\) can never exceed the speed of light.

For the rest of the post, we use high school physics to proof these famous results!

1. Review from high school physics materials

  • Momentum Conservation Principle (动量守恒定律): For a system (with two objectives) has 0 momentum, it always holds: \begin{equation}\label{eqn:mon} 0=P_1-P_2=m_1v_1-m_2v_2, \end{equation} multiply both sides by time t, we further have \begin{equation}\label{eqn:mon_s} 0=m_1S_1-m_2S_2. \end{equation}
  • Electromagnetic waves: \begin{equation}\label{eqn:egw} E=Pc \end{equation}
  • Theorem of Momentum: \begin{equation}\label{eqn:tom} dW=Fdx=dE,\quad F=\frac{dP}{dt}. \end{equation}

2. Proof of \(E=mc^2\)

Einstein’s Thought Experiment. Imagine there is a box with length \(L\), a photon with energy \(E\) goes from one side to the other side. Consider the system of photon and the box.

By \eqref{eqn:egw},

\begin{equation} P_1=\frac{E}{c} \end{equation}

then by \eqref{eqn:mon},

\begin{equation} P_2=P_1=\frac{E}{c}. \end{equation}

Then by the definition of momentum \eqref{eqn:mon}

\begin{equation} v_{\text{box}}=\frac{P_2}{m_2}=\frac{E}{m_2c}. \end{equation}

On the other hand, for photon \(s_1\approx L\), then for box we have

\begin{equation} s_2=v_{\text{box}}\cdot t=\frac{E}{m_2c}\cdot t=\frac{E}{m_2c}\cdot \frac{L}{c} \end{equation}

Let the mass of photon be \(m\), then by Momentum Conservation Principle,

\begin{equation} mL-m_2\cdot \frac{EL}{m_2c^2}=0 \end{equation}

Which gives

\[E=mc^2.\]

This can be generalize to not only photon but all particles!

Quite easy! Right?

Proof of \(m={m_0}/{\sqrt{1-(\frac{v}{c})^2}}\)

Recall \(E=mc^2\), \(P=mv\), hence

\begin{equation}\label{eqn:inter} \frac{E}{P}=\frac{c^2}{v}\Rightarrow E=\frac{Pc^2}{v}. \end{equation}

Now by \eqref{eqn:tom},

\begin{equation} dE=Fdx=\frac{dP}{dt}dx=v\cdot dP, \end{equation} multiply above by \eqref{eqn:inter}, then

\[EdE=\frac{Pc^2}{v}\cdot v\cdot dP=Pc^2\cdot dP.\]

Take integral, we obtian

\[E^2=c^2P^2+const.\]

When \(P=0\), \(E=E_0=m_0c^2\). Therefore we already derived

\begin{equation}\label{eqn:final} E^2=c^2P^2+m_0^2c^4 \end{equation}

In particular, for photon, by \eqref{eqn:egw} \(E=Pc\), which implies the invariant mass (光子的静止质量) \(m_0=0\)! This simly says: For any pariticle that has positive invariant mass, it cannot achieve the speed of light!

Finally, use \eqref{eqn:inter}, we have \(P=\frac{Ev}{c^2}\), plug in \eqref{eqn:final} to obtain

\[E^2=E^2\frac{v^2}{c^2}+m_0^2c^4\Rightarrow E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}.\]

Again, use \(E=mc^2\), we obtain

\[m=\frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}.\]

That’s it! Now you know why \(c\) is the fastest speed we can get!