Light is a Transverse Wave

In the previous module we studied different types of waves. This page will detail the fact that light is a transverse wave.

Electric Field and K-Vector

We first investigate this fact using one of Maxwell's equations e.g., Gauss's equation for electric field in vacuum

$LaTeX: \overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=0$ $\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=0$

and then assume a particularly useful type of wave that we have already discussed i.e., a plane wave. A plane wave propagating in an arbitrary direction can be written as

$LaTeX: \overset{\rightharpoonup }{E}\left(\overset{\rightharpoonup }{r},t\right)=\overset{\rightharpoonup }{E}_0 e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$ $\overset{\rightharpoonup }{E}\left(\overset{\rightharpoonup }{r},t\right)=\overset{\rightharpoonup }{E}_0 e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$

where

$LaTeX: \overset{\rightharpoonup }{E}_0=E_{0,x}\hat{i} + E_{0,y}\hat{j}+ E_{0,z}\hat{k}$ $\overset{\rightharpoonup }{E}_0=E_{0,x}\hat{i} + E_{0,y}\hat{j}+ E_{0,z}\hat{k}$

and

$LaTeX: E_{0,x} \text{, } E_{0,y} \text{, and } E_{0,z}$ $E_{0,x} \text{, } E_{0,y} \text{, and } E_{0,z}$

are complex constants. Using Gauss's equation for electric fields in vacuum, we see that

$LaTeX: \overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}_0 e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}=\overset{\rightharpoonup }{\nabla }\cdot (E_{0,x}\hat{i} + E_{0,y}\hat{j}+ E_{0,z}\hat{k})e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$ $\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}_0 e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}=\overset{\rightharpoonup }{\nabla }\cdot (E_{0,x}\hat{i} + E_{0,y}\hat{j}+ E_{0,z}\hat{k})e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$

distributing the divergence

$LaTeX: \overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=E_{0,x} \frac{\partial}{\partial x} e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}+E_{0,y} \frac{\partial }{\partial y}e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}+E_{0,z} \frac{\partial }{\partial z}e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$ $\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=E_{0,x} \frac{\partial}{\partial x} e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}+E_{0,y} \frac{\partial }{\partial y}e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}+E_{0,z} \frac{\partial }{\partial z}e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}-\omega t\right)}$

and noting

$LaTeX: \overset{\rightharpoonup }{k}=k_x\hat{i} +k_y\hat{j} +k_z\hat{k}$ $\overset{\rightharpoonup }{k}=k_x\hat{i} +k_y\hat{j} +k_z\hat{k}$

and

$LaTeX: \overset{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$ $\overset{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$

we can write $LaTeX: \overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}$ $\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}$ as

$LaTeX: e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}=e^{i \left(k_x x+k_y y+ k_z z- \omega t\right)}$ $e^{i \left(\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{r}- \omega t\right)}=e^{i \left(k_x x+k_y y+ k_z z- \omega t\right)}$

so that

$LaTeX: \overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=E_{0,x} \frac{\partial}{\partial x}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+E_{0,y} \frac{\partial}{\partial y}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+E_{0,z} \frac{\partial}{\partial z}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}$ $\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{E}=E_{0,x} \frac{\partial}{\partial x}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+E_{0,y} \frac{\partial}{\partial y}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+E_{0,z} \frac{\partial}{\partial z}e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}$

carrying out the derivatives

$LaTeX: i E_{0,x} k_x e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+i E_{0,y} k_y e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+i E_{0,z} k_z e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}=0$ $i E_{0,x} k_x e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+i E_{0,y} k_y e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}+i E_{0,z} k_z e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}=0$

noting that the complex exponentials cancel we see that

$LaTeX: i k_x E_{0,x}+i k_y E_{0,y}+i k_z E_{0,z}=0$ $i k_x E_{0,x}+i k_y E_{0,y}+i k_z E_{0,z}=0$

or written again in vector notation,

$LaTeX: \boxed{\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{E}_0=0}$ $\boxed{\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{E}_0=0}$

this clearly indicates that the k-vector and the electric field vector are orthogonal.

Magnetic Field and K-Vector

Our next step is to use Faraday's Law in differential form

$LaTeX: \overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{E}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$ $\overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{E}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$

and again choose a plane wave written as

the LHS of Faraday's law says

$LaTeX: \overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{E}=[(E_{0,z}\partial_y-E_{0,y}\partial_z)\hat{i}-(E_{0,z}\partial_x-E_{0,x}\partial_z)\hat{j}+(E_{0,y}\partial_x-E_{0,x}\partial_y)\hat{k}]e^{i(k_xx+k_yy+k_zz-\omega t)}$ $\overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{E}=[(E_{0,z}\partial_y-E_{0,y}\partial_z)\hat{i}-(E_{0,z}\partial_x-E_{0,x}\partial_z)\hat{j}+(E_{0,y}\partial_x-E_{0,x}\partial_y)\hat{k}]e^{i(k_xx+k_yy+k_zz-\omega t)}$

carrying out these derivatives we see that

$LaTeX: i[(E_{0,z}k_y-E_{0,y}k_z)\hat{i}-(E_{0,z}k_x-E_{0,x}k_z)\hat{j}+(E_{0,y}k_x-E_{0,x}k_y)\hat{k}]e^{i(k_xx+k_yy+k_zz-\omega t)}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$ $i[(E_{0,z}k_y-E_{0,y}k_z)\hat{i}-(E_{0,z}k_x-E_{0,x}k_z)\hat{j}+(E_{0,y}k_x-E_{0,x}k_y)\hat{k}]e^{i(k_xx+k_yy+k_zz-\omega t)}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$

the term in square brackets can be written as a cross product itself, which means

$LaTeX: i \left(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{E}_0 \right) e^{i \left(k_xx+ k_yy+k_zz-\omega t\right)}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$ $i \left(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{E}_0 \right) e^{i \left(k_xx+ k_yy+k_zz-\omega t\right)}=-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}$

the cross product in front is a vector but has no functional dependence (i.e., a constant vector) which means that the functional form of the magnetic field must also be a plane wave inasmuch as

$LaTeX: -\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}=i \overset{\rightharpoonup }{\mathcal{B}}_0 e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}$ $-\frac{\partial \overset{\rightharpoonup }{B}}{\partial t}=i \overset{\rightharpoonup }{\mathcal{B}}_0 e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}$

the only way for this to be true is if

$LaTeX: \overset{\rightharpoonup }{B}=\frac{-i \overset{\rightharpoonup }{\mathcal{B}}_0}{-i \omega }e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}=\frac{\overset{\rightharpoonup }{\mathcal{B}}_0 }{\omega }e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}$ $\overset{\rightharpoonup }{B}=\frac{-i \overset{\rightharpoonup }{\mathcal{B}}_0}{-i \omega }e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}=\frac{\overset{\rightharpoonup }{\mathcal{B}}_0 }{\omega }e^{i \left(k_xx+k_yy+k_zz- \omega t\right)}$

for now, we absorb the 1/ω and define $LaTeX: \overset{\rightharpoonup }{B_0}\equiv \frac{\overset{\rightharpoonup }{\mathcal{B}}_0}{\omega }$ $\overset{\rightharpoonup }{B_0}\equiv \frac{\overset{\rightharpoonup }{\mathcal{B}}_0}{\omega }$ , so that our magnetic field is

$LaTeX: \overset{\rightharpoonup }{B}=\overset{\rightharpoonup }{B}_0 e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}$ $\overset{\rightharpoonup }{B}=\overset{\rightharpoonup }{B}_0 e^{i \left(k_xx+k_yy+k_zz-\omega t\right)}$

Now that it is clear that the magnetic and electric fields are BOTH plane waves if either are. It is also imperative that we understand that because both electric and magnetic fields share the same time dependence they are IN PHASE. We can now immediately use Gauss's law for magnetic fields to write

$LaTeX: \overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{B}=0 ,$ $\overset{\rightharpoonup }{\nabla }\cdot \overset{\rightharpoonup }{B}=0 ,$

which, as in the case of Gauss's law for electric fields gives

$LaTeX: \boxed{\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{B_0}=0}$ $\boxed{\overset{\rightharpoonup }{k}\cdot \overset{\rightharpoonup }{B_0}=0}$

We understand this to mean that the k-vector is also perpendicular to the magnetic field.

Electric and Magnetic Fields

The last relation we investigate here is given by Ampere's law in vacuum, which states

$LaTeX: \overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{B}=\mu _0\epsilon _0 \frac{\partial \overset{\rightharpoonup }{E}}{\partial t} .$ $\overset{\rightharpoonup }{\nabla }\times \overset{\rightharpoonup }{B}=\mu _0\epsilon _0 \frac{\partial \overset{\rightharpoonup }{E}}{\partial t} .$

Now that we have a form for both the electric and magnetic fields in the case of plane waves, we can write

$LaTeX: i \left(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0 \right)=i \mu _0 \epsilon _0\omega \overset{\rightharpoonup }{E}_0$ $i \left(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0 \right)=i \mu _0 \epsilon _0\omega \overset{\rightharpoonup }{E}_0$

this can be written as

$LaTeX: \overset{\rightharpoonup }{E}_0=\frac{1}{\mu _0 \omega \epsilon _0}(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0) =\frac{k_0}{\mu _0 \omega \epsilon _0}(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0 )$ $\overset{\rightharpoonup }{E}_0=\frac{1}{\mu _0 \omega \epsilon _0}(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0) =\frac{k_0}{\mu _0 \omega \epsilon _0}(\overset{\rightharpoonup }{k}\times\overset{\rightharpoonup }{B}_0 )$

where I have defined the k-vector

$LaTeX: \overset{\rightharpoonup }{k}\equiv k_0 \hat{k}_0$ $\overset{\rightharpoonup }{k}\equiv k_0 \hat{k}_0$

as a unit vector in the LaTeX: k_0 $k_0$ -hat direction with magnitude $k_0$ . We also not that the speed of light is given, as before

$LaTeX: c=\frac{1}{\sqrt{\mu _0 \epsilon _0}}$ $c=\frac{1}{\sqrt{\mu _0 \epsilon _0}}$

and

$LaTeX: k_0=\frac{\omega }{c}$ $k_0=\frac{\omega }{c}$

inserting these relations we have

$LaTeX: \boxed{\overset{\rightharpoonup }{E}_0=c \left(\hat{k}_0 \times \overset{\rightharpoonup }{B}_0\right)}$ $\boxed{\overset{\rightharpoonup }{E}_0=c \left(\hat{k}_0 \times \overset{\rightharpoonup }{B}_0\right)}$

the magnitude of the vector on the RHS is given by

$LaTeX: \left| \hat{k}_0 \times\overset{\rightharpoonup }{B}_0\right| =\left| \hat{k}_0\right| \left| \overset{\rightharpoonup }{B}_0\right| \sin (\theta )$ $\left| \hat{k}_0 \times\overset{\rightharpoonup }{B}_0\right| =\left| \hat{k}_0\right| \left| \overset{\rightharpoonup }{B}_0\right| \sin (\theta )$

where θ is the angle between the two vectors. As we have previously shown, $LaTeX: \hat{k}_0$ $\hat{k}_0$ is perpendicular to $LaTeX: \overset{\rightharpoonup }{B}_0$ $\overset{\rightharpoonup }{B}_0$ so sin(θ)=1, which means

$LaTeX: \left| \hat{k}_0 \times\overset{\rightharpoonup }{B}_0\right| = \left| \overset{\rightharpoonup }{B}_0\right|$ $\left| \hat{k}_0 \times\overset{\rightharpoonup }{B}_0\right| = \left| \overset{\rightharpoonup }{B}_0\right|$

the magnitude of the LHS is trivial so we have the relation

$LaTeX: \boxed{\left| \overset{\rightharpoonup }{E}_0\right| =c \left| \overset{\rightharpoonup }{B}_0\right|}$ $\boxed{\left| \overset{\rightharpoonup }{E}_0\right| =c \left| \overset{\rightharpoonup }{B}_0\right|}$

Although this result was derived in the case of plane waves, it is quite general.