Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

My book says : The fact that electromagnetic radiation of energy carried momentum was known from classical theory and from the experiments of Nichols and Hull in 1903. This relation is also consistent with the relativistic expression for a particle with zero rest energy.) Does anyone know what those experiments were, and why classical theory suggests that momentum of light is E/c?

share|cite|improve this question
Also, classical theory didn't imply $p=\frac{E}{c}$, it probably had an extra $\frac{1}{2}$ factor. – Ali Jul 29 '13 at 5:04
@Ali If I calculate the time-averaged radiation pressure of a normally incident plane wave on a perfectly conducting plane mirror, I get $p=\frac{1}{2} \epsilon_0 E_0^2$, where $E_0$ is the electric field amplitude. This is twice the momentum incident per unit area per unit time (the light doubles back). The rate flow of energy per unit area into the mirror is $\frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_0^2$, so this scenario therefore gives $p=\frac{E}{2 c}$, which is your factor of 2. Wow! I hadn't heard of this discrepancy. – WetSavannaAnimal aka Rod Vance Jul 29 '13 at 13:46
What's the name of the book and page number for the reference? – Larry Harson Jul 29 '13 at 15:30
@Ali: No, there is no anomalous factor of 1/2. You're confused. – Ben Crowell Sep 28 '13 at 13:23

2 Answers 2

I don't know the experiments or history, but here are three classical calculations that gives you what you want:

Method 1: Test Charge and Plane Electromagnetic Wave

As in Section 34-10 of the first Feynman Lectures on Physics Volume, think of a plane wave and a test charge at the origin. At this point, let $\mathbf{E} = E \,\hat{\mathbf{x}}$, $\mathbf{B} = \frac{E}{c} \,\hat{\mathbf{y}}$: this is a plane wave solution of Maxwell's equations. Then the charge will be oscillating in the $\hat{\mathbf{x}}$-direction, so suppose at the instant in time under consideration it is moving with velocity $\mathbf{v} = v\,\hat{\mathbf{x}}$. The rate of the electromagnetic field's working on the particle through the electic field is $P = q \,E \,v$, so the charge is absorbing energy from the field at this rate. Note that the electric force on the particle is oscillatory and has a time-average of nought. But, at the same time, the magnetic force on the particle is $\mathbf{F} = q\,\mathbf{v}\wedge \mathbf{B}$ and $F=q\,v\,\frac{E}{c}$. $\mathbf{E}$ and $\mathbf{B}$ are in-phase, and the particle's velocity at steady state bears a constant phase relationship to $\mathbf{B}$, so both $P$ and $\mathbf{F}$ oscillate in-phase at twice the light's frequency and with nonzero time average, and the ratio of this average, from the above, is $\frac{P}{F} = \frac{q \,E \,v}{q\,v\,\frac{E}{c}} = c$. The force is simply the time rate of transfer of momentum. Thus, whenever the electromagnetic field transfers energy $W$ to a particle, it also transfers momentum $\frac{W}{c}$.

Method 2: Inertia Content of Energy

We imagine a pulse of energy $W$ emitted from one end of a spaceship and absorbed at the other. But the inertia content of this pulse is $\frac{W}{c^2}$. So, superficially it seems that the centre of mass of the system is shifting (by dent of the moving energy) whilst the light pulse is in flight. Conservation of momentum would be violated if this supposition is true. However, one can resolve the contradiction if one assumes that the spaceship feels a recoil at the launch of the light pulse. So, if the spaceship's mass is $M$, it has to move off in the opposite direction to the light with speed $c \frac{W}{c^2} \frac{1}{M}$, because this speed will keep the system's centre of mass still. Thus, we get $\frac{W}{c}$ for the momentum of the pulse.

Notice how if we used the converse of method 2 together with method 1 (i.e. knowing the radiation pressure and thus the thrust on the spaceship), we could deduce from first principles that the effective mass of the light needed to uphold momentum conservation is $\frac{E}{c^2}$.

Method 3: Plane Wave Incident on a Metal

This is actually a special case of method 1 (which is more general), but it has an adjustable parameter (the conductivity of the mirror) that can be used to illustrate two separate behaviours. I am using the convention $\partial_t \mapsto -i\,\omega$ where $\omega$ is the field's angular frequency.

Let the $x-y$ plane (i.e. $z=0$) be the face of a mirror: for $z<0$ we have freespace characterized by the freespace electric and magnetic constants $\epsilon_0$ and $\mu_0$, for $z>0$ we have a metal of electric constant $\epsilon$, magnetic constant $\mu$ and conductivity $\sigma$. In freespace there is an incident wave plane wave such that in the plane $z=0$:

Electric field: $\mathbf{E} = E_i \hat{\mathbf{x}}$

Magnetic field: $\mathbf{H} = \sqrt{\frac{\epsilon_0}{\mu_0}}\, E_i \hat{\mathbf{y}}$

and there is also a reflected plane wave ($E_r$ is to be deduced) such that:

Electric field: $\mathbf{E} = E_r \hat{\mathbf{x}}$

Magnetic field: $\mathbf{H} = -\sqrt{\frac{\epsilon_0}{\mu_0}}\, E_r \hat{\mathbf{y}}$

The opposite sign assumed on the magnetic field stands for a wave with wavevector pointing in the $-\hat{\mathbf{z}}$ direction. The incident wave runs in the $+\hat{\mathbf{z}}$. In the metal, the field has the following dependence, as can be proven from finding a plane wave solution to Maxwell's equations (again, E_m is to be deduced):

Electric field: $\mathbf{E} = E_m \,e^{\gamma\,z}\, \hat{\mathbf{x}}$

Magnetic field: $\mathbf{H} = -i\,\frac{\gamma}{\omega\,\mu}\, E_m \,e^{\gamma\,z}\, \hat{\mathbf{y}}$

Current Density: $\mathbf{J} = \sigma\,E_m \,e^{\gamma\,z}\, \hat{\mathbf{x}}$

where the complex wavenumber is:

$\gamma = \frac{-1+i}{\sqrt{2}} \sqrt{\omega\mu\left(\sigma-i\, \omega\,\epsilon\right)}$

is chosen (there are two possible $\pm$ values from Maxwell's equations for $\gamma$) so that the field decays exponentially as $z\rightarrow\infty$.

From the standard electromagnetic boundary conditions at the interface (continuity of tangential components of $\mathbf{E}$ and $\mathbf{H}$ across the interface):

$E_i + E_r = E_m$

$E_i - E_r = \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{\gamma}{i\,\omega\,\mu}\,E_m$


$E_m = \frac{2\,E_i}{1 + \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{\gamma}{i\,\omega\,\mu}}$

Now, the time averaged force per unit area on the metal is:

$\frac{1}{2}\mathrm{Re}\left(\int\limits_0^\infty \mathbf{J} \wedge \mathbf{B}^* . \hat{\mathbf{z}}\right) = \frac{\mathrm{Im}(\gamma)\,\sigma\,|E_m|^2}{4\,\omega\,\mathrm{Re}(\gamma)} = \frac{\mathrm{Im}(\gamma)\,\sigma\,|E_i|^2}{\omega\,\mathrm{Re}(\gamma) \left|1 + \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{\gamma}{i\,\omega\,\mu}\right|^2} $

Now the time averaged power per unit area incident on the interface (by calculating the Poynting vector) is:

$\frac{1}{2}\sqrt{\frac{\epsilon_0}{\mu_0}} |E_i|^2$

so that the momentum transferred to the metal for each unit of incident energy is:

$\frac{2\,\mathrm{Im}(\gamma)\,\sigma}{\omega\,\mathrm{Re}(\gamma) \left|1 + \sqrt{\frac{\mu_0}{\epsilon_0}} \frac{\gamma}{i\,\omega\,\mu}\right|^2} \sqrt{\frac{\mu_0}{\epsilon_0}} $

This is the ratio we seek. For the case $\sigma\rightarrow \infty$ this ratio approaches $\frac{2}{c}\frac{\mu}{\mu_0}$. For $\mu = \mu_0$, this is twice the value calculated in methods 1 and 2 because the infinite conductivity excludes the field from the metal, there is no loss and light is reflected without loss. So, just as there is an impulse $2\,p_z$ transferred to a wall by an elastic bounce of a ball initially with momentum with component $p_z$ into the wall, so too the impulse transferred to the mirror from energy $W$ is twice the momentum of the incoming light, i.e. $\frac{2\,W}{c}$.

Now we let $\sigma\rightarrow 0$ so that $\gamma\approx -i \,\omega\,\sqrt{\mu\,\epsilon} + \frac{1}{2}\sqrt{\frac{\mu}{\epsilon}} \sigma$. I'll finish this case later, but it yields the same answer as methods 1 and 2 when $\mu = \mu_0$ and $\epsilon = \epsilon_0$, i.e. the light passes without reflexion into a weakly conducting medium and is absorbed.

share|cite|improve this answer

Maxwell's equations are not consistent with Galilean relativity. They are implicitly relativistic. In relativity, the definition of mass is $m^2=E^2-p^2$. For a ray of light, $m=0$, so $E=|p|$.

share|cite|improve this answer
So, just for the OP's benefit (although this OP seems to lose interest in his questions and not give too much feedback), there never was a version of electromagnetism that had to be upgraded to account for relativity. Indeed, historically, EM's Lorentz covariance that jarred with everything else and forced special relativity to be born. – WetSavannaAnimal aka Rod Vance Sep 29 '13 at 10:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.