Are Electromagnetic Waves Always Transverse? Full Explanation

In this insight, we will explore classical electrodynamics and examine whether electromagnetic (EM) waves are always transverse. We use Jefimenko’s equations and Poynting’s Theorem and conclude:

1. EM waves are always transverse by the weak definition, where the fields are perpendicular to the propagation vector.
2. EM waves are not always transverse by the strong definition, where we also require the E and B fields to be perpendicular to each other.

For what follows, we’ll assume the region of interest is vacuum. Similar conclusions generally hold for linear, isotropic, nondispersive materials. This assumption simplifies the relation between fields and the Poynting vector.

We also assume there are no boundaries (for example, waveguide walls) that impose special conditions on the fields and can cause the propagation direction to differ from the Poynting-vector direction.

1. Fields perpendicular to the direction of propagation

Looking at Jefimenko’s equations — the general solution to Maxwell’s equations for arbitrary charge and current densities — we have:

$$\mathbf{E}(\mathbf{r},t)=\frac{1}{4\pi\epsilon_0}\int \left [\frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^3}\rho(\mathbf{r’},t_r)+\frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^2}\frac{1}{c}\frac{\partial\rho(\mathbf{r’},t_r)}{\partial t}-\frac{1}{c^2}\frac{1}{|\mathbf{r-r’}|}\frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}\right ]d^3\mathbf{r’}$$

$$\mathbf{B}(\mathbf{r},t)=-\frac{\mu_0}{4\pi}\int \left [ \frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^3}\times\mathbf{J}(\mathbf{r’},t_r)+\frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^2}\frac{1}{c}\times\frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}\right ]d^3\mathbf{r’}$$

Poynting vector and transversality

At first glance, those integrals do not immediately reveal the propagation direction of the $\mathbf{E}$ and $\mathbf{B}$ fields. However, Poynting’s theorem helps: the usual physical definition of a wave’s propagation direction is the direction of energy flow. For an electromagnetic field the energy flux is the Poynting vector $\mathbf{S}=\mathbf{E}\times\mathbf{B}$.

Using the vector triple-product and properties of the cross product, one obtains

$$\mathbf{E}\cdot \mathbf{S}=(\mathbf{E}\times\mathbf{E})\cdot \mathbf{B}=0$$

and similarly for $\mathbf{B}$. Therefore $\mathbf{E}$ and $\mathbf{B}$ are both perpendicular to $\mathbf{S}$. In this weak sense — perpendicular to the propagation (energy-flow) vector — an EM wave is always transverse:

$$\mathbf{E}\perp\mathbf{S},\qquad \mathbf{B}\perp\mathbf{S}.$$

2. Fields perpendicular to each other?

Can we prove that $\mathbf{E}$ and $\mathbf{B}$ are always perpendicular to each other (the strong definition)? In the general case, we cannot. Although the integrands in Jefimenko’s equations may look locally orthogonal — for example, the term

$$\frac{1}{c^2}\frac{1}{|\mathbf{r-r’}|}\frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}$$

from the E-field integrand appears perpendicular to

$$\frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^2}\frac{1}{c}\times\frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}$$

from the B-field integrand, orthogonality need not survive the integration. In general, the dot product of two vector integrals is not equal to the integral of the dot product:

$$\text{If }\mathbf{A}\cdot\mathbf{B}=0\text{ pointwise, it does not imply }\int\mathbf{A}\,dV\cdot\int\mathbf{B}\,dV=\int\mathbf{A}\cdot\mathbf{B}\,dV=0.$$

Thus nothing in the general Jefimenko form guarantees $\mathbf{E}\cdot\mathbf{B}=0$ everywhere.

Far-field (radiation-zone) approximation

To make progress, we often use the far-field (radiation-zone) approximation. Place the origin inside the source volume V (the region where $\rho$ and $\mathbf{J}$ are nonzero), assume the source is finite, and evaluate the fields at points $\mathbf{r}$ such that

$$|\mathbf{r}|\gg|\mathbf{r’}|\quad\text{for all }\mathbf{r’}\in V.$$

Then to leading order one may approximate

$$\frac{\mathbf{r-r’}}{|\mathbf{r-r’}|^i}\approx\frac{\mathbf{r}}{|\mathbf{r}|^i}.$$

This moves the common geometric factors outside the integrals (integration is over $\mathbf{r’}$), and Jefimenko’s equations reduce to:

$$\mathbf{E}(\mathbf{r},t)=\frac{1}{4\pi\epsilon_0}\left [\frac{\mathbf{r}}{|\mathbf{r}|^3}\int \rho(\mathbf{r’},t_r) \,d^3\mathbf{r’}+\frac{\mathbf{r}}{c|\mathbf{r}|^2}\int\frac{\partial \rho(\mathbf{r’},t_r)}{\partial t}\, d^3\mathbf{r’}-\frac{1}{c^2|\mathbf{r}|}\int\frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}\,d^3\mathbf{r’}\right ]$$

$$\mathbf{B}(\mathbf{r},t)=-\frac{\mu_0}{4\pi}\left [ \frac{\mathbf{r}}{|\mathbf{r}|^3}\times \int \mathbf{J}(\mathbf{r’},t_r)\,d^3\mathbf{r’}+\frac{\mathbf{r}}{|\mathbf{r}|^2}\times \int \frac{\partial \mathbf{J}(\mathbf{r’},t_r)}{\partial t}\,d^3\mathbf{r’}\right ]$$

These expressions are valid only in the far-field region (not near the sources). Under this approximation, and with the additional assumption that time derivatives of the current do not change its direction (i.e. $\partial\mathbf{J}/\partial t\parallel\mathbf{J}$), one can show that $\mathbf{E}\cdot\mathbf{B}=0$ in the radiation zone. That is, far from the source the radiated fields are orthogonal to each other and to the propagation direction.

3. Conclusions

Using Poynting’s theorem we have shown that the electric and magnetic fields are always perpendicular to the direction of energy flow (the Poynting vector). In this weak sense, EM waves are always transverse.

However, in the general near-source case the electric and magnetic fields need not be perpendicular to each other. Orthogonality of $\mathbf{E}$ and $\mathbf{B}$ arises in the far-field (radiation zone) under the usual approximations.

Thus, far from a finite source, an EM wave is fully transverse regardless of the detailed charge and current distributions that generated it (provided the source is finite and the far-field approximations apply).

Comment thread