# Electricity and magnetism/E = m c²

Light exerts pressure on the surfaces that reflect or absorb it. We can calculate this radiation pressure with Maxwell's equations (1865) because it is a magnetic force exerted by the incident wave on the electric charges that it sets in motion. Lebedev observed and measured this light pressure, in 1900.
The pressure exerted by the light shows that it has mass. We can calculate the mass of light using Maxwell's equations. We find $m={\frac {E}{c^{2}}}$ where $E$ is its energy.
It is often believed that $E=mc^{2}$ is a law of the theory of relativity, but this is false. This chapter gives a proof of $E=mc^{2}$ for light from Maxwell's and Newton's equations.
The theory of relativity (1905) shows that all energy has mass, even the kinetic energy of moving masses, not just light. The mass of kinetic energy is not taken into account in Newton's equations.

## The mass of light

Light has mass.
Proof: let us consider two photons of the same energy which bounce horizontally between two mirrors in a stationary box. With each bounce, a photon exerts pressure on the mirror. This is the radiation pressure, which can be calculated from Maxwell's equations. If we set the box in motion by exerting a force $F$, the acceleration of the box is $a={\frac {F}{m}}$, according to Newtonian physics. The speed of the box modifies the pressure that the photons exert on its walls. If the speed of the box is from left to right, a photon which goes from right to left exerts a greater pressure towards the left than the pressure exerted towards the right by a photon of the same energy which goes towards the right. When the box is accelerated, it loses part of its momentum which it gives to the photons which bounce off its walls. It is thus slowed down, transiently, by the photons it contains. A box full of photons is therefore less accelerated than the same empty box subjected to the same force. So a box full of photons has a greater mass than the same empty box. So photons have mass.
If we filled a box with 1 kg of photons, we would give it an energy approximately equal to that released by a nuclear bomb. When a nuclear bomb explodes a significant part of its initial mass is transformed into light, particularly X-rays.
The speed $c$ of light is almost equal to $300$ $000$ $km/s=3.10^{8}$ $m/s$. So $c^{2}=(3.10^{8})^{2}m^{2}/s^{2}$ is almost equal to $10^{17}m^{2}/s^{2}$. $10$ $MJ=10^{7}$ $J$ is approximately the energy provided by daily bread. This is a physicist's order of magnitude, not the recommendation of a nutritionist doctor. For a physicist, one is almost equal to two, not for a doctor. The energy released by a nuclear bomb, approximately $1$ $kg.10^{17}m^{2}/s^{2}=10^{17}J$ is therefore roughly equal to the energy consumed in their food by all human beings during a day. This calculation is made for uranium bombs. Hydrogen bombs are much more powerful.
Photons are sometimes said to have zero mass, but this is false, because their energy is mass. We also say that their rest mass is zero, but this is also false, because there is no frame of reference where they are at rest, so their rest mass is not defined. In the equation $E^{2}=m_{0}^{2}c^{4}+p^{2}c^{2}$, $m_{0}=0$ for photons because $E=pc</math>but<math>m_{0}$ has no physical meaning.
There is no particle of zero mass, neither at rest nor in motion. Everything that exists in the Universe always has mass. So-called zero-mass particles like photons are not massless particles, but restless particles. There is never an observer for whom they are still.
The Higgs boson is sometimes said to give their mass to particles which without it would be massless. But it's wrong. It gives rest to particles which without it would never find rest.

## Rest mass

The masses of all particles, whether without rest or with rest, always depend on the frame of reference where they are measured. But particles with rest have privileged frames of reference, those where they are at rest. In these frames of reference, they always have the same rest mass $m_{0}$. It is a constant specific to the particle, which is conserved throughout its existence. It is the same for all observers, provided that they measure it with the formula
$m=\gamma m_{0}$
where $m$ is the measured mass, $\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$, $v$ is the speed of the particle and $c$ that of light. ${\frac {1}{\gamma }}$ is the coefficient of contraction of lengths in the direction of their movement.
Restless particles have no rest mass because they have no frame of reference where they are at rest.

## Everything that exists has momentum

Everything that exists physically always has momentum.
Proof: what exists physically always acts on other beings which exist physically. A being that never acted on other physical beings would never have any effect and could never be observed. It would have no physical existence. When a being acts on another, it modifies its movement, therefore its momentum $\mathbf {p} =m\mathbf {v}$ for a body of mass $m$ and speed $\mathbf {v}$. But the total momentum is always conserved. If one body increases the momentum of another, it loses momentum. If one body decreases the momentum of another, it gains momentum. Therefore a body without momentum cannot act on another and cannot exist physically.
The momentum of a photon and any restless particle can be calculated with the formula $\mathbf {p} =m\mathbf {v}$ as for all other particles:
$\mathbf {p} =m\mathbf {c}$
where $m$ is the mass of the particle and $\mathbf {c}$ its velocity vector. The magnitude $c$ of the velocity vector $\mathbf {c}$ is the same for all restless particles, in all frames of reference. It's the speed of light.
Since everything that exists has momentum, everything that exists has mass.

## E = pc

If we accept the fundamental equations of the photon, among the first in quantum physics, given by Einstein (1905):
$E=h\nu$
$p={\frac {h}{\lambda }}$
where $h$ is Planck's constant,
we immediately obtain
$E=pc$
because
$c=\lambda \nu$
where $\lambda$ is the wavelength of the photon and $\nu$ its frequency.
We can also prove $E=pc$ directly from Maxwell's equations, without going through quantum physics, by calculating the energy and momentum of the electromagnetic field. This calculation is given at the end of this chapter using the Poynting vector of an electromagnetic wave.
Einstein would not have stated his fundamental equations if he had not first understood Maxwell.

## E = mc²

If we admit that the momentum of a photon is $p=mc$ where $m$ is its mass, we immediately obtain from $E=pc$ that
$E=mc^{2}$
is the energy of a photon of mass $m$.
Calculating the mass of a photon trapped in a box shows that indeed
$m={\frac {p}{c}}$
is the mass of a photon of momentum $p$.

## The Doppler effect

To understand the calculation of the mass of light, one needs to know the Doppler effect.
We can tell if the police are approaching or moving away by listening to their siren. It is more acute when they approach than when they move away. This is the Doppler effect for sound waves.
Consider a machine gun that fires $f$ bullets per second projected at speed $V$. $f$ is the firing frequency. Consider a paper target that moves at speed $v$ relative to the machine gun. The frequency $f'$ of bullet impacts on the target is not the same as the firing frequency $f$, because the bullets take less and less time to reach their target, if $v$ is aimed towards the machine gun, and increasingly longer, if $v$ is aimed in the opposite direction. This is the Doppler effect for a machine gun.
$f'=f(1-{\frac {v}{V}})$
Proof: during a duration $T$, the target travels a distance $|v|T$, where $|v|$ is the absolute value of $v$. The balls are spaced a distance $L={\frac {|V|}{f}}$. So there is ${\frac {|v|T}{L}}=|{\frac {v}{V}}|fT$ balls on the distance $|v|T$. For the $T$ duration, the machine gun fired $fT$ bullets. If the target is aimed at the balls, ${\frac {v}{V}}<0$, and it encounters $fT+|{\frac {v}{V}}|fT=f(1-{\frac {v}{V}})T=f'T$ balls. If the target moves away from the bullets, ${\frac {v}{V}}>0$, and it encounters $fT-|{\frac {v}{V}}|fT=f(1-{\frac {v}{V}})T=f'T$ balls.
$f'=f(1-{\frac {v}{V}})$ is the formula for the non-relativistic Doppler effect. Its proof presupposes the velocity addition theorem of Newtonian physics. A relativistic calculation is more exact, but gives almost the same result, even if $V=c$, provided that $v$ is much smaller than $c$.
If the frequency of light decreases, its color shifts toward red. If the frequency increases, the color is shifted towards blue. The Doppler effect for light changes its color:
The farther away the galaxies are from us, the more their light is redshifted, so the faster they move away from us. Hubble's observation of this red shift is therefore proof of the expansion of the Universe.

## Radiation pressure

Light exerts pressure on the surfaces which absorb or reflect it. It's easy to understand if we take its mass into account, because it behaves like a ball that we catch, if it is absorbed, or which bounces, if it is reflected.
Consider a ball of mass $m$ and speed $v$ which bounces on a wall which remains stationary. We assume that the speed is perpendicular to the wall. If the ball is perfectly bouncy, its speed after the bounce is exactly equal and opposite to its speed before the bounce. The variation of its momentum is
$p_{1}-p_{2}=mv_{1}-mv_{2}=mv+mv=2mv$
The variation of momentum $dp$ for a time $dt$ is $Fdt$ where $F$ is the force exerted on the ball:
$F={\frac {dp}{dt}}$
So
$p_{2}-p_{1}=\int dp=\int Fdt$
The force $F$ exerted by the wall on the particle is the opposite of the force $-F$ exerted by the particle on the wall. So
$p_{1}-p_{2}=2mv=\int -Fdt$
is the integral of the force exerted by the ball on the wall. This integral is proportional to the integral of the pressure exerted by the ball on the wall. The pressure exerted by the ball on the wall is therefore proportional to the product of its mass and its speed.
The effect of the pressure exerted by the ball depends only on its momentum. If we vary its mass and its speed without varying their product, the integral of the pressure is the same. A photon of momentum $p$ has the same effect as a ball of speed $v$ and mass $m={\frac {p}{v}}$. This suggests giving photons a mass $m={\frac {p}{c}}$ such that
$p=mc$
If we move towards a ball, we experience greater pressure from it when we receive it, or when it bounces, than if we are standing still. Likewise, the pressure exerted by light on a surface depends on the speed of the surface receiving this pressure.
The radiation pressure on a mirror depends on the speed of the mirror in the same way as the pressure exerted by a bouncing mass on a moving wall.
Proof :
Let R be a reference frame where the wall is stationary and R' a reference frame which goes at speed $-V$ with respect to R. In R', the speed of the wall is $V$ and that of the ball is $u=v+V$ before the bounce and at the speed $-v+V$ after the bounce.
Its momentum variation is therefore always
$p1-p2=m(v-V-(-v-V)=2mv=2m(u-V)$
For the same speed $u$ of the bouncing ball, the pressure exerted on the wall increases as $u-V=u(1-{\frac {V}{u}})$.
The pressure exerted by a photon depends on its momentum $p$ in the same way as a bouncing ball. According to the fundamental equations of the photon
$p={\frac {h}{\lambda }}={\frac {h\nu }{c}}$
According to the Doppler effect formula, if $\nu _{0}$ is the frequency of a photon, $\nu _{0}(1-{\frac {V}{c}})$ is its frequency in a frame of reference of a mirror which encounters it at speed $V$
$p={\frac {h\nu _{0}(1-{\frac {V}{c}})}{c}}$
If the momentum of the photon is fixed at $p={\frac {h\nu _{0}}{c}}$ the pressure it exerts on a wall which meets it at the speed $V$ varies as $1-{\frac {V}{c}}$ in the same way that the pressure exerted by a bouncing ball of speed $u$ varies as $1-\ frac{V}{u}$.
This proof is given with the equations of Newtonian physics and the formula for the non-relativistic Doppler effect. The relativistic calculation is more exact and leads to the same theorem.

## Calculation of the mass of a photon trapped in a box

An horizontal force $F$ is exerted from left to right, for a short duration $\Delta$, on a box initially at rest, which contains two photons, initially of the same energy $E$, which bounce horizontally on its walls.
We assume that the force is exerted at the instant $t=0$ where the two photons cross in the middle of the box, far from the walls, and that the duration $\Delta t$ is short enough that the photons do not have time to reach the mirrors during $\Delta t$. While it is exerted, the force $F$ therefore sets the box in motion as if it were empty, since the photons are far from the walls.
Let $m$ be the mass of the empty box. Its speed just after the application of the force $F$ is
$v(\Delta t)=\int _{0}^{\Delta t}dv=\int _{0}^{\Delta t}adt=\int _{0}^{\Delta t}{\frac {F}{m}}dt$
where $a={\frac {dv}{dt}}$ is the acceleration of the box.
When the photons bounce off the mirrors, they slow down the box, because the Doppler effect works in opposite directions for each of them. The photon going from right to left exerts greater pressure than the photon going from left to right.
The laws of conservation of energy and momentum make it possible to calculate the speed of the box $v(\infty )$ when it is no longer slowed down by the photons it contains.
Let $E_{b}(t)$ be the kinetic energy of the empty box, and $p_{b}(t)$ its momentum. Let $E_{l}(t)$ be the total energy of the two photons, and $p_{l}(t)$ their total momentum.
$E_{b}(0)=0$, $p_{b}(0)=0$
$E_{l}(0)=2E$, $p_{l}(0)=0$
$E_{b}(\Delta t)={\frac {1}{2}}mv^{2}(\Delta t)$, $p_{b}(\Delta t)=mv(\Delta t)$
$E_{l}(\Delta t)=2E$, $p_{l}(\Delta t)=0$
Energy conservation requires
$E_{b}(\infty )+E_{l}(\infty )=E_{b}(\Delta t)+E_{l}(\Delta t)={\frac {1}{2}}mv^{2}(\Delta t)+2E$
The conservation of momentum imposes
$p_{b}(\infty )+p_{l}(\infty )=p_{b}(\Delta t)+p_{l}(\Delta t)=mv(\Delta t)$
Let $p'_{0}$ be the momentum of a photon in the box in a frame of reference R' where it is at rest when braking is completed. The momentum of this photon in the reference frame R where the box is moving is
$p_{0}=p'_{0}(1+{\frac {v(\infty )}{c}})$ or $p_{0}=-p'_{0}(1-{\frac {v(\infty )}{c}})$
depending on whether the photon goes from right to left or from left to right.
So
$p_{l}(\infty )=2p'_{0}{\frac {v(\infty )}{c}}$
$p_{b}(\infty )=mv(\infty )$
So
$mv(\infty )+2p'_{0}{\frac {v(\infty )}{c}}=mv(\Delta t)$
Let $m'$ be the mass of the box with its two photons.
$v(\infty )=\int _{0}^{\Delta t}{\frac {F}{m'}}dt={\frac {m}{m'}}\int _{0}^{\Delta t}{\frac {F}{m}}dt={\frac {m}{m'}}v(\Delta t)$
$m'-m$ is the mass of the two photons in the box.
$m'-m=m({\frac {v(\Delta t)}{v(\infty )}}-1)={\frac {mv(\Delta t)-mv(\infty )}{v(\infty )}}=2{\frac {p'_{0}}{c}}$
$E=pc$ therefore
$m'-m={\frac {2E'}{c^{2}}}$
where $E'$ is the energy of a photon in R', after braking the box.
If $v(\infty )$ tends to zero, $E'$ tends to $E$. The mass $m_{p}$ of a photon is therefore
$m_{p}={\frac {E}{c^{2}}}$
where $E$ is its energy.
When $v(\infty )$ is not infinitesimal, this non-relativistic calculation gives a false and absurd result: the initial mass of the photons depends on the force applied to the box. But this calculation ignores the mass of the kinetic energy of the box. This is why we must take the limit where this kinetic energy tends towards zero, to obtain the correct result.

## The momentum of the electromagnetic field

When an electromagnetic field exerts the Lorentz force on a charge, it varies its momentum. The conservation of the total momentum requires that the momentum of the field varies at the same time and in the opposite direction as the momentum of the particle. We can thus calculate, from Maxwell's equations, that the momentum density of the field is obtained from the Poynting vector:
$\mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B}$
$\mathbf {E}$ is the electric field, $\mathbf {B}$ the magnetic field and $\mu _{0}$ a constant which depends on the choice of units. $\mu _{0}={\frac {1}{\epsilon _{0}c^{2}}}$
The momentum density ${\frac {\mathbf {p} }{v}}$ of an electromagnetic field $\mathbf {E}$, $\mathbf {B}$ is
${\frac {\mathbf {p} }{v}}={\frac {1}{c^{2}}}\mathbf {S} =\epsilon _{0}\mathbf {E} \times \mathbf {B}$
For an electromagnetic wave, $B={\frac {E}{c}}$ and $\mathbf {B}$ is perpendicular to $\mathbf {E}$ , so
${\frac {\mathbf {p} }{v}}={\frac {1}{c}}\epsilon _{0}E^{2}$
From Maxwell's equations, Lorentz equation and the conservation of energy, one can calculate that the energy density of the electromagnetic field is
$u={\frac {1}{2}}(\epsilon _{0}\mathbf {E} ^{2}+{\frac {1}{\mu _{0}}}\mathbf {B} ^{2})$
For an electromagnetic wave:
$u={\frac {1}{2}}(\epsilon _{0}+{\frac {1}{\mu _{0}c^{2}}})E^{2}=\epsilon _{0}E^{2}$
so
$u={\frac {pc}{v}}$
where $p$ is the magnitude of the momentum $\mathbf {p}$ of the field in the volume $v$.
If now $E$ is the field energy in the same volume:
$E=vu=pc$
The intensity of the electric field radiated by a dipole is represented by color. The arrows represent the Poynting vector.

## See also

Relativity
Gravitation