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I've always seen EM waves been drawn as such:

EM Wave

The magnetic and electric amplitudes are drawn equally, right?

However, their equations tell a different story. $\textbf{E}(z,t) = E_0 \cos(kz - \omega t + \delta) \hat{\textbf{x}}$ for the electric field, but $\textbf{B}(z,t) = \frac{1}{c} E_0 \cos(kz - \omega t + \delta) \hat{\textbf{y}}$ for the magnetic field. All else being equal, the B field should be drawn as much smaller than the E field.

Am I correct in this, or am I missing something?

Also, is this telling us about the strength (or something) of the magnetic field compared to the electric field for a light wave?

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  • $\begingroup$ There are no units on the plot. Poor practice! $\endgroup$
    – ProfRob
    Commented Nov 11, 2016 at 8:36
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    $\begingroup$ Yes: you're missing the fact that you can't compare apples and oranges. $E_0$ and $E_0/c$ cannot possibly have the same units or dimensionality (and this should jump out immediately as obvious), so asking whether $E_0>E_0/c$ is true is exactly equivalent to asking whether a nanosecond is bigger or smaller than thirty centimeters. $\endgroup$
    – Emilio Pisanty
    Commented Nov 15, 2016 at 18:22

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In reality, the electric field and the magnetic field have different units in some unit systems and identical units in others, so it would never be possible to scale them "correctly." No matter what you do, there will be someone who claims that things need to be drawn differently. Instead of trying to be "correct," we choose to draw these graphs in a way which demonstrates the effects of interest.

If we were to draw the B field 300,000 times smaller than the E field (just so the numbers worked out in the units of your choosing), the graphic would not be helpful for assisting people in understanding. However, we can just as easily adjust the units of the B field until, numerically, the results are interesting. All we're doing in that case in handwaving to change the numbers and units... the actual quantities do not change.

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  • $\begingroup$ In reality, the electric and magnetic field have the same units, so it can be possible to scale them correctly. $\endgroup$
    – Kyle Kanos
    Commented Nov 11, 2016 at 14:07
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    $\begingroup$ @KyleKanos That is entirely dependent on the system of units, and it is not the case for the OP's choice of system, so an unequivocal statement like that is misleading here. Nevertheless, when one chooses a system in which $E$ and $B$ are commensurate, the $E$ and $B$ amplitudes for a plane wave are exactly equal, validating the OP's plot. $\endgroup$
    – Emilio Pisanty
    Commented Nov 15, 2016 at 18:23
  • $\begingroup$ @EmilioPisanty: My comment mirrors Cort's first sentence, pointing out that this answer is, at best, misleading because it ignores the fact that there is more than just SI in which B & E can be drawn as OPs plot has it. $\endgroup$
    – Kyle Kanos
    Commented Nov 15, 2016 at 18:25
  • $\begingroup$ @KyleKanos See edited answer. $\endgroup$
    – Emilio Pisanty
    Commented Nov 15, 2016 at 19:09
  • $\begingroup$ Thanks for the back and forth! And here I thought I had a simple answer to offer! $\endgroup$
    – Cort Ammon
    Commented Nov 15, 2016 at 19:11
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Electric and magnetic fields are actually two parts of a matrix called the 'electromagnetic field tensor'. It's what we use when we treat electricity and magnetism relativistically.

One thing that falls out is there are some quantities that are the same in every frame. One is
$B^2-\frac{E^2}{c^2}$

So it makes sense to compare B not with E, but $\frac{E}{c}$.

https://en.wikipedia.org/wiki/Electromagnetic_tensor

The diagrams are actually correct, but to place the electric and magnetic fields on the same graph, with the same units, we need to graph B and E/c.

It's a little like asking which is longer, 1 meter or 1 second. Before relativity, the question was pretty meaningless, but after special relativity we have x and ct as two parts of a 4-vector. In that case 1 second is the same as $3 \times 10^8$ m.

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  • $\begingroup$ How does this answer the question? $\endgroup$
    – Kalpak Gupta
    Commented Nov 11, 2016 at 8:38
  • $\begingroup$ In the case of electricity and magnetism, instead of a 4-vector we have a 4x4 matrix. The elements are the components of B and E/c. So if we want to compare electric and magnetic fields using the same units, we need to use E/c. $\endgroup$
    – David Elm
    Commented Nov 12, 2016 at 7:55
  • $\begingroup$ Thanks for clarifying. This should be explicitly mentioned in the answer I think! $\endgroup$
    – Kalpak Gupta
    Commented Nov 15, 2016 at 7:21
  • $\begingroup$ In that case 1 second is the same as $3\times10^8$ m Not really. While in appropriate units (i.e., natural units) they share the same unit (e.g., 1/eV) and are not "the same." $\endgroup$
    – Kyle Kanos
    Commented Nov 15, 2016 at 18:19
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Good question and good answers. Worth adding that also the electric field energy and the magnetic field energy are going to be exactly equal in such plane EM wave. It is just the traditional definitions of different units for E and B that make B seem thousands of times smaller. Since the energies are the same it is good they both get represented fairly in typical picture.

Note the in-phase nature of E and B means the total energy density oscillates like sin-squared going along the waveform for this linearly polarised case. However for circular polarization the energy density is uniform, seemingly more natural and funnily enough corresponding to photon pure spin state.

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