A question about $E=pc$ for massless particles

Question

Since photon has no (rest)mass and

$$E^2=(pc)^2+(mc^2)^2$$

we derive that $E=pc$ for particle with no (rest)mass.

However, if we transform the non-relativistic formula for kinetic energy

$$E_k=\frac{mv^2}{2}$$ $$E_k=\frac{p^2}{2m}$$ $$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}=\lim_{m\rightarrow0}\frac{m^2v^2}{2m}=\lim_{m\rightarrow0}\frac{mv^2}{2}=\frac{pv}{2}$$

we derive that $E_k=\frac{pc}{2}$.

And this is bizarre because I'm expecting $E_k=E$.

I've just started learning physics recently so I'm pretty sure I've got certain part wrong. And here are my hypotheses:

1.$ $ The net energy of a photon $E_{net}=pc$ is equivalent $E_{net}=E_k+E_0=\frac{pc}{2}+E_0$ and thus $E_0=pc-E_k=\frac{pc}{2}$. (If that's true, what exactly is $E_0$? The rest energy of a photon?)

2.$ $ It is incorrect to derive the kinetic energy formula from integrating the equation

$$ \frac{\mathrm dE}{\mathrm dv}=mv $$

I should integrate another equation instead (perhaps $\frac{\mathrm dE}{\mathrm dv}=p$? If that's true, why do we break $p$ into $mv$ when deriving the formula $E_k=\frac{pc}{2}$?)

3.$ $ I got the limit wrong:

$$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}\ne \frac{pc}{2}$$

$$E_k=\lim_{m\rightarrow0}\frac{p^2}{2m}=pc$$

So is any of these hypothesis correct? And why?

Answer 1

Your kinetic energy formula is incorrect. The correct one for a massive particle is $$ KE = m c^2\left[ \frac{1}{\sqrt{ 1 - v^2/c^2}} - 1 \right] $$ with $$ p = \frac{ m v}{\sqrt{1 - v^2/c^2}} $$ Now, to describe a photon, we wish to take two limits $m \to 0$ and $v \to c$. These limits must be taken properly. To be precise, we wish to take this limit in such a way that the momentum remains well-defined and is equal to $p$. This implies $$ \frac{p}{c} = \lim_{v\to c}\lim_{m\to0} \frac{ m}{\sqrt{1 - v^2/c^2}} $$ Applying this same limit to the kinetic energy, we get $$ KE = \lim_{v\to c}\lim_{m\to0} \left( \frac{ m c^2}{\sqrt{ 1 - v^2/c^2}} - m c^2 \right) = \frac{p}{c} c^2 - 0 = p c $$ which is consistent with the equation you first wrote down!

Answer 2

Using the relation between mass, energy and momentum $$ E^2 = p^2c^2 + m^2c^4 $$ $$ E^2 - p^2c^2 = m^2c^4 $$

$$ \begin{align*} E_{kin} &= E - E_0 \\&= \sqrt{m^2c^4 + p^2c^2} - mc^2 \\&= mc^2\left[\sqrt{1 + \frac{p^2}{m^2c^2}} - 1\right] = m c^2\left[ \frac{1}{\sqrt{ 1 - v^2/c^2}} - 1 \right] \\ \\&\approx mc^2\left[\left(1 + \frac12 \frac{p^2}{m^2c^2}\right) - 1\right] \\&= \frac{p^2}{2m} \end{align*} $$

where the approximation is just the first-order Taylor expansion of $\sqrt{1+x}$ at $0$ as $p^2 \ll m^2c^2$ in the non-relativistic case.

Answer 3

There are actually a couple of different ways to look at this. The simple way is to say that you took the wrong limit - but not because of any mathematical error. The problem is that the formula $E_k = \frac{1}{2}mv^2$ or $E_k = \frac{p^2}{2m}$ is itself a low-velocity limit of the true formula,

$$E = \sqrt{p^2 c^2 + m^2 c^4}$$

That calculation is demonstrated in ProgrammingEnthusiast's answer. Hopefully it makes sense that if you take the low-velocity limit of one formula to get another formula, then you plug in a high-velocity to the second formula, it's not going to get you the first formula back. (By analogy: if you take the low-$x$ limit of $\sin x$, getting $x$, and then plug in $x = 2\pi$, it's not going to be the same as $\sin(2\pi)$!)

Another way to look at it is that your second hypothesis is right: the formula $\frac{\mathrm{d}E_k}{\mathrm{d}v} = mv$ isn't valid for photons, and you need to integrate something else. That something else is actually one of Hamilton's equations, which in this case can be written

$$\frac{\partial E}{\partial p} = v$$

Of course, to integrate this you still need $v$ as a function of $p$, so in a sense this is just delaying the important physics by a step. The "important physics" is this: for a low-velocity (non-relativistic) particle, you can use $v = \frac{p}{m}$, but for a high-velocity (relativistic) particle, that equation doesn't hold. You need to use $p = \frac{mv}{\sqrt{1 - v^2/c^2}}$, or solving for $v$,

$$v = \frac{pc}{\sqrt{m^2c^2 + p^2}}$$

You can see that $v = \frac{p}{m}$ is the low-momentum (or high-mass) limit of this equation. Anyway, when you do the integral, you will get out the proper relativistic formula for energy.

Answer 4

The second hypothesis is correct. For relativistic particles (photons being a good example), $E=\frac{1}{2}mv^2$ does not hold up. Instead, you should start from $E=\sqrt{m^2c^4+p^2c^2}$ as you did for the photon. By the way, the relativistic momentum is also given by $p=\gamma mv$ where $\gamma$ is the Lorentz factor: $\gamma=(\sqrt{1-v^2/c^2})^{-1}$.

Finally, if you want to see how the 'Newtonian kinetic energy' $E_k=\frac{1}{2}mv^2$ arises, look at Ron Maimon's awesome answer to this question.