Ultra-Relativistic and Non-Relativistic cases for energy of a particle

Question

Let us suppose that we have a particle of energy $$E = (m^2c^4 + p^2c^2)^{1/2}$$

Can we say that for ultra-relativistic limit $p \rightarrow \infty$

$E = pc$ ?

Or in the non-relativistic case $p \rightarrow 0$,

$E = mc^2$ ?

Answer 1

For the non-relativistic case ($mc^2 \gg pc$) you have simplified too much.

A better approximation would be $$\begin{align} E &= (m^2c^4 + p^2c^2)^{1/2} \\ &= mc^2\left(1 + \frac{p^2c^2}{m^2c^4}\right)^{1/2} \\ &= mc^2\left(1 + \frac{p^2}{m^2c^2}\right)^{1/2} \\ &\approx mc^2\left(1 + \frac{p^2}{2m^2c^2}\right) \\ &= mc^2 + \frac{p^2}{2m}. \end{align}$$

Here you recognize $\frac{p^2}{2m}$ as the kinetic energy as known from Newtonian mechanics.

$mc^2$ is an additional constant (the rest energy) which causes no deviation from the Newtonian equations of motion.

Answer 2

This might be a useful energy-momentum diagram (below).

A particle's 4-momentum is drawn, as well its $E$ and $p$ components in this frame.

When describing limits, it's good to explicitly specify
(in addition to what is being varied) that which is being held constant.

So, if you keep the [rest] mass $m$ fixed, then increasing $p$ means that $E$ increases in such a way to keep $E^2-(pc)^2=(mc^2)^2$ fixed---that is, up along the "mass shell" (the hyperbola [or generally hyperboloid]).

It might be good to note that the components of the 4-momentum in this frame can be described by
$E=mc^2\cosh\theta$
and
$pc=mc^2\sinh\theta$
where $\theta$ is the Minkowski-angle (called the rapidity)
between the 4-momentum-vector and the vertical Energy axis.
So, $(v/c)=\tanh\theta$, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\cosh\theta$, and $\gamma v=\frac{v}{\sqrt{1-(v/c)^2}}=\sinh\theta$.

While keeping $m$ fixed, large $p$ corresponds to a large $\theta$.
Since $\cosh\theta=\frac{1}{2}(e^\theta+e^{-\theta}$)
and $\sinh\theta=\frac{1}{2}(e^\theta-e^{-\theta}$), as $\theta$ gets large, $e^{-\theta}$ gets small.
So, $\cosh\theta$ and $\sinh\theta$ approach each other---they approach $e^\theta/2$.
(In fact, it turns out that $\cosh\theta-\sinh\theta=e^{-\theta}$, and we said that $e^{-\theta}$ is getting small with larger $\theta$.)

[By the way, $e^\theta=\cosh\theta+\sinh\theta=\cosh\theta(1+\tanh\theta)=\gamma(1+(v/c))=\sqrt{\frac{1+(v/c)}{1-(v/c)}}$, the Doppler factor.]

[This image came from an old post of mine at Relativity and Momentum of photons ]

Answer 3

The answer is yes for both cases. Keep in mind that ultrarelativistic means an energy much larger than $mc^2 $.

Answer 4

In the ultra-relativistic limit,$$E=(p^2c^2+m^2c^4)^{1/2}=pc\left(1+\frac{m^2c^2}{p^2}\right)^{1/2}\approx pc\left(1+\frac{m^2c^2}{2p^2}\right)=pc+\frac{m^2c^3}{2p}.$$If we neglect first-order corrections, we can describe this limit with $E\sim pc$, which means $\lim_{\frac{p}{mc}\to\infty}\frac{E}{pc}=1$. Similarly, the non-relativistic limit can be written as $E\sim mc^2$, meaning $\lim_{\frac{mc}{p}\to\infty}\frac{E}{mc^2}=1$.