# Prove EM Waves Are Transverse In Nature

Why we say that EM waves are transverse in nature? I have seen some proofs regarding my question but they all calculate flux through imaginary cube. Here is My REAL problem that I can't here imagine infinitesimal area for calculating flux because em line of force will intersect (perpendicular or not) surface at only one point so $E.ds$ will be zero so even flux through one surface of cube will always be zero. I am Bit Confused. I DON'T KNOW VECTOR CALCULUS BUT KNOW CALCULUS.

Why we say that em waves are transverse in nature?

In a region empty of electric charge, we have, from Maxwell's equations:

$$\nabla \cdot \vec E = \nabla \cdot \vec B = 0$$

Since you don't yet know vector calculus, let's rewrite these divergence equations as so:

$$\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 0$$

$$\frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z} = 0$$

Now, assume an electromagnetic wave is propagating in the $z$ direction so that the space and time variation of the field components are of the form

$$\cos(kz - \omega t)$$

Since the spatial variation is zero in the $x$ and $y$ directions, our equations become

$$\frac{\partial E_z}{\partial z} = 0$$

$$\frac{\partial B_z}{\partial z} = 0$$

Which means that electric and magnetic field components in the $z$ direction, the direction of propagation, must be constant with respect to $z$.

In other words, only the electric and magnetic field components transverse to the direction of propagation vary with respect to $z$. i.e., the electromagnetic wave is transverse.

Why spatial Variations are zero in both x and y directions.

We stipulated that the field components are of the form $\cos(kz - \omega t)$ which means the wave is propagating in the $z$ direction.

Clearly, the partial derivative of $\cos(kz - \omega t)$ with respect to $x$ and $y$ is zero.

• @VishvajeetPatil, I've updated my answer. – Alfred Centauri Mar 12 '14 at 20:40
• I deleted my above comment because I understand before you said it to me.Sorry For late Response.Thanks. – Vishvajeet Patil Mar 13 '14 at 3:09

You say the

em line of force will intersect (if perpendicular) surface at only one point.

This is not the case for a plane wave, which is the simplest case one usually considers. I think I know what's going on to make you think this. This may or may not be the issue that helps you.

Take this typical picture of an EM wave. It is so very tempting, yet incorrect, to think that the E and B field are only defined (non-zero) on that axis. No. Here's a better picture that only shows one of the two fields. It's called a plane wave because the E and B fields have the same value everywhere on some plane. So if you imagine this wave incident on some cube, the dot product $\vec{E}\cdot d\vec{S}$ has a value at more than just one point.

• Why can't the first picture be right? Why, as you say the plane wave is simplest? – Vishvajeet Patil Mar 12 '14 at 15:13
• The first picture is right. It just hides a lot of the information. You should interpret the first picture as actually representing the second. – BMS Mar 12 '14 at 15:53
• But why plane em wave is simplest?Please clarify in detail.Does what you are saying may somewhere related to particle nature in depth – Vishvajeet Patil Mar 12 '14 at 15:54
• That sounds like a different question. There should be plenty of internet material on it. If you can't find it, try asking as another question. – BMS Mar 12 '14 at 15:56
• Plane waves are the simplest because all other waves can be written as sums of plane waves. BMS already answered your question. Try to be courteous. – ZachMcDargh Mar 12 '14 at 16:50

I am going to try to "unconfuse" you.

Let's start with a picture of ocean waves; as seen from the side, they look like "sine" waves (just like BMS's red EM wave); as seen from the top, the crests and troughs make lines (not points). The same thing happens with the E and B waves (make lines), and since they are perpendicular to each other, they form a plane (two lines make a plane). This is why the flux is evaluated over a surface area (not a point).

• Would you infer that the flux of sound in a fluid must be evaluated along 1-dimensional path of sound propagation (think of velocity)? Or over a point (think of pressure), indeed? – Incnis Mrsi Oct 24 '14 at 18:33