Problem in understanding energy-mass equivalence

Question

From whatever approach (either from modern approach by Hamiltonian or from Einstein's original approach by considering momentum), we can derive that $p_{i} = \gamma mv_{i}$ and $$H = \gamma mc^2 = \sqrt{|p|^2c^2+m^2c^4}.$$ One usually set $|p|=0$, from which we get mass-energy equivalence. To relate classical mechanics where $\beta\approx0$, one could use Talyor's expansion to recover the classical mechanics where $H\approx mc^2+\frac{1}{2}mv^2$.

My problem here is that the whole derivation seems to me some sort of a re-definition of what we mean by energy, in terms of mass, especially in the step where $|p|=0$ was set. I know it is a stupid question but why we can always set $|p|=0$ for any particle except those moving with speed of light? Also for photons, why do we set $m=0$ to reconcile with the problem where $c$ is constant in every frame of reference, for which it could also be possible that different frame of reference getting different value of $H$ so light could also have mass, just the mass observed was different (because $H$ could vary from frame to frame)?

Lastly, why do we even bother stating $E=mc^2$? I saw that when recovering classical mechanics, the term $mc^2$ is always there for any particle, so it can be safely ignored when comparing the change of energy (because only change of energy is of interest), then what is the point of $E=mc^2$? Do we simply redefine what energy is in terms of mass, so we can convert the units by multiplying $c^2$, or is there anything empirical in the equation at the time where Einstein had discovered it (I know there were some experiments proving this fact, but at the time of Einstein there was no experiment, so why did he keep this equation on this paper?)

Answer 1

I consider the relation

$$E = \sqrt{ \vec{p}^2 c^2 + (m c^2)^2}$$

by far not "banal" because it reflects the invariance of the 4-vector length in the Minkowski-metric. I will be more explicit: Energy and momentum can be grouped together into a what is called in special relativity 4-vector, here called 4-momentum vector $p^{\mu} = (p^0,p^1,p^2,p^3)= (E/c,\vec{p})$.

As the squares of 3-vectors in euclidian space don't change -- i.e. are invariant under 3-dimensional rotations like in the Newton's mechanics
$\vec{p}^2= p^2_x+ p^2_y + p^2_z$. squares $X^2$ of 4-vectors $X^{\mu}$ in Minkowski space are invariant under "rotations" of the latter -- i.e. invariant under Lorentz-transformations:

$$p^2:=\sum\limits_{\mu=0}^3 p^{\mu}p_{\mu} = (\frac{E}{c})^2 - \vec{p}^2 = m^2c^2$$

the "banal" term $ (mc)^2 $ does not change under Lorentz-transformations which actually shows that the definition $p^{\mu} = (p^0,p^x,p^y,p^z)= (E/c,\vec{p})$ makes sense. In Euclidian space the 4-momentum $p^{\mu}p_{\mu}$ would only contain positive terms, however, in the Minkowski-space, in which Lorentz-transformations operate however the length of 4-vector contains a minus sign due to its special metric and makes the introduction of 2 types of 4-vectors reasonable: $p^{\mu} =(E/c,\vec{p})$ and $p_{\mu} =(E/c,-\vec{p})$ The evaluation of

$\sum\limits_{\mu=0}^3 p^{\mu}p_{\mu}$

according to the given definitions exactly yields the equation above. It could be argued that it is just a stronger formalisation of the same relation. However, the importance of invariant quantities in relativistic physics cannot be stressed enough. I give another example

The same concept applies to $x^{\mu}=(t,\vec{x})$ (and $x_{\mu}=(t,-\vec{x})$ ) which can also be recognized as 4-vector whose 4-length $x^{\mu} x_{\mu}$ respectively $\Delta x^{\mu}\Delta x_{\mu}$ is invariant under Lorentz-transformations. Einstein discovered in particular this invariance and constructed from this recognition the Lorentz-transformations.

$(\Delta s)^2\equiv \Delta x^{\mu}\Delta x_{\mu}= c^2 (t_2 - t_1)^2 - (x_2-x_1)^2$

What photons concerns, so it was found that the well-known dispersion relation of light $\vec{k}^2 =(\frac{\omega}{c})^2$ in vacuum looks like

$$\sum\limits_{\mu=0}^3 k^{\mu}k_{\mu} = (\frac{\omega}{c})^2 - \vec{k}^2 = 0$$

or together with Plancks relationship $E =\hbar \omega$ and $\vec{p} = \hbar \vec{k}$ like

$$\sum\limits_{\mu=0}^3 p^{\mu}p_{\mu} = (\frac{E}{c})^2 - \vec{p}^2 = 0$$

so we recognize that $(\omega/c, \vec{k})$ can also be written as a 4-vector $k^{\mu}=(\omega/c, \vec{k})$ just with the little distinction that the 4-length of this vector $k^2: = \sum\limits_{m=0}^3 k^{\mu}k_{\mu}=0$, which is of course also invariant under Lorentz-transformations. The dispersion relation for light was of course known long before the discovery of special relativity, now that we see how nicely it fits into the theory with light quanta of rest mass zero, it is so natural distinguish between massless particles with $m=0$ and $m\neq 0$. In particular the relation

$$(\frac{E}{c})^2 - \vec{p}^2 = m^2c^2$$

serves as dispersion relation for almost all elementary particles with rest mass in the standard model as does $$(\frac{\omega}{c})^2 - \vec{k}^2 = 0$$ for massless particles like photons.

EDIT: Apart from the importance to construct 4-vectors in a way that their 4-length (squared) is Lorentz invariant, the definition of the energy is chosen to fit in a 4-vector so that it transforms under Lorentz transformation as the $0^{th}$ component of a 4-vector. If the another definition was preferred, for instance $E_{kin}= (\gamma-1)mc^2$ (which has to property to reproduce in the Newton's limit $E_{kin}\approx\frac{1}{2}mv^2$ ) we don't know easily how to transform it under Lorentz transformations. However, this feature is important if in scattering experiments one wants to switch quickly from the CMSystem to the labor -system or vice-versa.

Answer 2

why we can always set $|p|=0$ for any particle except those moving with speed of light?

The idea is, if the particle is slower than speed of light, there is a reference frame where this particle is at rest. If the particle is at rest $|\mathbf p| = 0$.

Also for photons, why do we set $m=0$ to reconcile with the problem where $c$ is constant in every frame of reference, for which it could also be possible that different frame of reference getting different value of $H$ so light could also have mass, just the mass observed was different (because $H$ could vary from frame to frame)?

Poynting energy and momentum of EM wave obey, based on Maxwell's equations, the relation $$ E = |\mathbf p|c. $$ This means that four-vector of energy-momentum $[E/c,\mathbf p]$ of the EM wave has zero Minkowski norm: $$ E^2/c^2 - |\mathbf p|^2 = 0 $$ So to make this consistent with the idea that mass is defined as norm of the energy-momentum four-vector, the mass has to be zero.

when recovering classical mechanics, the term $mc^2$ is always there for any particle, so it can be safely ignored when comparing the change of energy (because only change of energy is of interest), then what is the point of $E=mc^2$?

In mechanics of particles that do not lose mass, the point is that defining energy as $\gamma mc^2$ (and thus having energy $mc^2$ for particle at rest) makes the 4-tuple $[E/c,\mathbf p]$ a four-vector. If we used different definition of energy, such as $E=\gamma mc^2 - mc^2$ (which gives zero energy when the particle is at rest), the 4-tuple would not be a four-vector.

Do we simply redefine what energy is in terms of mass, so we can convert the units by multiplying $c^2$ ...

Special relativistic mechanics did redefine total mechanical energy of moving particle; instead of $\frac{1}{2}mv^2$, it is $\gamma mc^2$. But the reason was not the desire to express energy as a function of mass; after all, there is also speed-dependent factor $\gamma$, which does not depend on mass. The reason is the above advantage of having a four-vector (which means energy and momentum transform as time and position coordinates). Also, when we step outside of pure mechanics into EM theory, Einstein discovered that bodies that radiate away EM energy $L$ should decrease their mass by $L/c^2$. The new definition of body energy $E=\gamma m c^2$ makes that consistent with the law of conservation of energy.

Answer 3

“Why can we always set $p=0$...” you can’t, they just did that to recover the famous $E=mc^2$ relation. For all classical exchanges of energy (potential to kinetic for instance) this $mc^2$ term can indeed be ignored (if the $m$ stands for rest mass), but rest energy can in fact be exchanged for other types of energy. For instance a electron and positron that disappear to create two photons. So it isn’t merely a constant added to the energy.

Of course at the time of Einstein they didn’t know this, but the equation is exactly what predicted it, and that’s why he didn’t just ignore it. He also did not just find this energy-mass relation and then said “this means that mass can be interchanged with other types of energy”, that would indeed be illogical; he used a thought experiment in which a body doesn’t change its velocity, but does change its kinetic energy, which could only mean a change in mass. You can read it in his article “does the inertia of a body depend on its energy content?”. Which is a short article and relatively easy to read.

Edit: forgot your question about light. For light the energy-mass equivalence equation simply does not apply, because its derivation is based entirely on particles having mass. The energy of a photon is simply $hf$ with $f$ the frequency and $h$ Planck’s constant. This relation was postulated by Max Planck, but it can be proven theoretically in a rather difficult way. I explained this once here: Proof of de Broglie wavelength for electron