Why is a relativistic calculation needed unless $pc$ is much smaller than the rest energy of a particle?

Question

After introducing the de Broglie wavelength equation, my textbook gives a rather simple example where it asks to find the kinetic energy of a proton whose de Broglie wavelength is 1 fm. In the solution to this problem, it states that "A relativistic calculation is needed unless $pc$ for the proton is much smaller than the proton rest energy."

Could someone please explain why this is the necessary condition? I'm not sure what the quantity '$pc$' represents or means here. I know that for massless particles like photons, the total energy $E$ is equal to $pc$. I'm not sure what it means for particles having rest mass like protons.

Answer 1

In relativity, the energy $E$ of a particle is related to its mass $m$ and momentum $p$ by \begin{equation} E = \sqrt{m^2c^4 + p^2 c^2 } \end{equation} Now let's think about the non-relativistic limit, $c\rightarrow \infty$. To do this, we will expand the square root
\begin{equation} E = m c^2 \sqrt{1 + \frac{p^2}{m^2 c^2}} = mc^2 + \frac{p^2}{2m} + \cdots \end{equation} where the $\cdots$ refer to terms that vanish in the limit $c\rightarrow\infty$.

The first term is the famous equation $E=mc^2$. In non-relativistic physics, the mass is a constant, so this is just a constant term in the energy we can ignore.

The second term $E=\frac{p^2}{2m}$ is the non-relativistic expression for kinetic energy.

You can see that non-relativistic physics is a good approximation to relativistic physics when we can ignore the higher order terms. Thinking back to how we expanded the square root, this amounts to the condition $p c \ll m c^2$.

Answer 2

The momentum of a particle of mass $m$ moving at $v$ is $\gamma m v$ where $\gamma$ is given by the following.

$$\gamma = \frac{1}{ \sqrt{ 1 - \frac{v^2}{c^2} } }$$

Let's always take $v$ to be non-negative. If you need a direction then put it in later. Since $v <= c $, $\gamma$ is a real number greater than or equal to 1. For $v$ very small compared to $c$ it is only slightly larger than 1.

The rest energy of a proton is just $m c^2$. So the comparison they are making is the following.

$$\gamma m v c << mc^2 $$

So dividing out the common factors you get the following.

$$\gamma v << c $$

And for $v$ small compared to $c$, this is just $v << c$.