# Are energy and momentum separable for EM waves

Everywhere I get to read statements like EM radiations and EM fields carry "energy as well as momentum " . I wonder if energy and momentum are inalienable . Is it possible for a field to carry only energy without having any momentum term ? I mean isn't it sufficient to say that fields or EM waves carry energy or Doesn't the ability to transfer energy, of the fields, automatically imply that they can change the momentum of another particle while transferring energy to them?

## 1 Answer

The energy density in an electromagnetic field is given by $$u = \frac{1}{2}\left( \epsilon E^2 + \frac{B^2}{\mu} \right)$$

The momentum carried by the electromagnetic fields is derived from the Poynting vector, which is the power per unit area. $$\vec{N} = \frac{1}{\mu} ( \vec{E} \times \vec{B} )$$

Momentum per unit area, per unit time is just the Poynting vector divided by the speed of light.

Poynting's theorem tell us that the divergence of $\vec{N}$ is equal to the work done by the EM fields (on charges) plus the rate of change of energy density.

Now, what do you mean by "carry energy". If you mean take it from one place to another, then this requires a non-zero Poynting vector, with a non-zero divergence and this will have a non-zero momentum flux.

However, if you mean can the EM fields store energy without there being a momentum flux, then the answer is yes. Just arrange for the net Poynting vector to be zero, but with non-zero E- and/or B-fields.

This is even possible for EM waves if you arrange to form a standing wave. The time-averaged Poynting vector would be zero, but there would be a non-zero energy density.

e.g. $$\vec{E} = E_0 \cos(kz)\cos(\omega t)\ \vec{x}$$ $$\vec{B} = \frac{E_0}{c} \sin(kz) \sin(\omega t)\ \vec{y}$$ is a standing wave formed by two equal amplitude plane waves travelling in opposite directions along the z-axis.

The net energy density (let's assume vacuum, so $\epsilon = \epsilon_0$, $\mu=\mu_0$ and $c^2 = (\mu_0 \epsilon_0)^{-1}$) is given by $$u = \frac{1}{2}\epsilon_0 E_0^{2} (\cos^2(kz)\cos^2(\omega t) + \sin^2(kz) \sin^2(\omega t))$$

Taking the time average and noting that $\cos^2(kz)+\sin^2(kz)=1$, we get $$\bar{u} = \frac{\epsilon_0 E_0{^2}}{4},$$ which is non-zero and independent of $z$.

On the other hand, the Poynting vector is $$\vec{N} = \frac{1}{\mu_0}\cos(kz)\sin(kz)\cos(\omega t)\sin(\omega t)\ \vec{z}$$ $$\vec{N} = \frac{E_0^{2}}{4\mu_0 c}\sin(2kz) \sin(2 \omega t)\ \vec{z}$$

But the time-average of this is zero.