Here are some depictions of electromagnetic wave, similar to the depictions in other places:

enter image description here enter image description here enter image description here

Isn't there an error? It is logical to presume that the electric field should have maximum when magnetic field is at zero and vise versa, so that there is no moment when the both vectors are zero at the same time. Otherwise one comes to a conclusion that the total energy of the system becomes zero, then grows to maximum, then becomes zero again which contradicts the conservation law.


The depictions you're seeing are correct, the electric and magnetic fields both reach their amplitudes and zeroes in the same locations. Rafael's answer and certain comments on it are completely correct; energy conservation does not require that the energy density be the same at every point on the electromagnetic wave. The points where there is no field do not carry any energy. But there is never a time when the fields go to zero everywhere. In fact, the wave always maintains the same shape of peaks and valleys (for an ideal single-frequency wave in a perfect classical vacuum), so the same amount of energy is always there. It just moves.

To add to Rafael's excellent answer, here's an explicit example. Consider a sinusoidal electromagnetic wave propagating in the $z$ direction. It will have an electric field given by

$$\mathbf{E}(\mathbf{r},t) = E_0\hat{\mathbf{x}}\sin(kz - \omega t)$$

Take the curl of this and you get

$$\nabla\times\mathbf{E}(\mathbf{r},t) = \left(\hat{\mathbf{z}}\frac{\partial}{\partial y} - \hat{\mathbf{y}}\frac{\partial}{\partial z}\right)E_0\sin(kz - \omega t) = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$

Using one of Maxwell's equations, $\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$, you get

$$-\frac{\partial\mathbf{B}(\mathbf{r},t)}{\partial t} = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$

Integrate this with respect to time to find the magnetic field,

$$\mathbf{B}(\mathbf{r},t) = -\frac{E_0 k}{\omega}\hat{\mathbf{y}}\sin(kz - \omega t)$$

Comparing this with the expression for $\mathbf{E}(\mathbf{r},t)$, you find that $\mathbf{B}$ is directly proportional to $\mathbf{E}$. When and where one is zero, the other will also be zero; when and where one reaches its maximum/minimum, so does the other.

For an electromagnetic wave in free space, conservation of energy is expressed by Poynting's theorem,

$$\frac{\partial u}{\partial t} = -\nabla\cdot\mathbf{S}$$

The left side of this gives you the rate of change of energy density in time, where

$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right)$$

and the right side tells you the electromagnetic energy flux density, in terms of the Poynting vector,

$$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$

Poynting's theorem just says that the rate at which the energy density at a point changes is the opposite of the rate at which energy density flows away from that point.

If you plug in the explicit expressions for the wave in my example, after a bit of algebra you find

$$\frac{\partial u}{\partial t} = -\omega E_0^2\left(\epsilon_0 + \frac{k^2}{\mu_0\omega^2}\right)\sin(kz - \omega t)\cos(kz - \omega t) = -\epsilon_0\omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$

(using $c = \omega/k$) and

$$\nabla\cdot\mathbf{S} = \frac{2}{\mu_0}\frac{k^2}{\omega}E^2 \sin(kz - \omega t)\cos(kz - \omega t) = \epsilon_0 \omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$

thus confirming that the equality in Poynting's theorem holds, and therefore that EM energy is conserved.

Notice that the expressions for both sides of the equation include the factor $\sin\bigl(2(kz - \omega t)\bigr)$ - they're not constant. This mathematically shows you the structure of the energy in an EM wave. It's not just a uniform "column of energy;" the amount of energy contained in the wave varies sinusoidally from point to point ($S$ tells you that), and as the wave passes a particular point in space, the amount of energy it has at that point varies sinusoidally in time ($u$ tells you that). But those changes in energy with respect to space and time don't just come out of nowhere. They're precisely synchronized in the manner specified by Poynting's theorem, so that the changes in energy at a point are accounted for by the flux to and from neighboring points.


There is no contradiction with conservation law, you are just applying it wrongly. To do it right, you have to consider a small closed region and check whether the variation of energy inside this domain is the same as (minus) the energy flux through its boundaries. In the case of a electromagnetic wave, when the energy density decreases inside the region, it just means that the energy has left the region you are considering.

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    $\begingroup$ On the other hand, if the phases of B and E are shifted, the total energy is constant because $(\sin t)^2+(\cos t)^2$=1 which is what is expected. $\endgroup$ – Anixx Jan 28 '11 at 18:08
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    $\begingroup$ Dear Anixx, please listen to Rafael, he is completely right. There is absolutely no reason why the total energy density should be constant. $\endgroup$ – Luboš Motl Jan 28 '11 at 18:11
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    $\begingroup$ Energy in electromagnetic wave moves from magnetic field to magnetic field and vise versa like in a spring it moves from kinetic form to potential. The total energy remains constant. When energy of electric field is at maximum, the energy of magnetic is zero. Energy cannot apper from nothing. $\endgroup$ – Anixx Jan 28 '11 at 18:37
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    $\begingroup$ Energy of magnetic field is $c_1B^2$ and energy of electric field is $c_2E^2$. They cannot both be at maximum and then come to a minimum. $\endgroup$ – Anixx Jan 28 '11 at 18:40
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    $\begingroup$ And also look at the Maxwell equations. $\operatorname{rot} \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ This equation clearly states that electric field E is proportional to minus derivative of B by time and the fourth Maxwell equation says that B is proportional to derivative of E by time. So if E is sine, B is cosine. $\endgroup$ – Anixx Jan 28 '11 at 18:49

Look at the original, Heinrich Hertz, 1889! If E=B=0 on a plane, S=0 => no energy Transport through that plane. I agree with Annix: "The pictures may be depicting a static wave, but certainly, not a propagating wave."


In a (small) vicinity of any given point the instantaneous total energy $u=u(t)$ varies in time. The average total energy $\langle u\rangle=\frac{E_0^2}{2c \mu_0}$ does not:

$$\frac{d \langle u\rangle}{dt}=0.$$

Thus, energy is conserved for any small vicinity for any point.


Without the detailed mathematics you need to use is Maxwell's equations, for example

$$\nabla \times \mathbf E = -\dfrac{\partial \mathbf B}{\partial t}$$

and a sketch of an electromagnetic wave travelling in the positive z-direction.

enter image description here

In simple terms the equation tells you that the rate of change of the electric field with position in the z-direction is equal to the rate of change of flux with time.

Consider position $L$ with the electric field and the magnetic field being at maximum positive values.
The gradient of the electric field in the z-direction is zero.
The sketch shows the magnetic field at a slightly later time with the magnetic field hardly changing hardly at all and in the limit not at all.
So you have $0$ (gradient of the electric field equal to $0$ (minus rate of magnetic field).
At this position the magnetic field is at a maximum but it is the rate of change of the magnetic field which is important.

Consider position $M$ where the gradient of the electric field in the z-direction is at a maximum magnitude but negative.
This is where the magnetic field is changing most rapidly with time.
Even though the magnetic field/flux is zero at this position it is the rate of change of magnetic field which is important and note that it is a positive increase. With the rate of change of magnetic field a maximum multiplying it by minus one equates it to the maximum negative value of the rate of change with position in the z-direction of electric field.

A similar analysis will show that the sketch graph is consistent with another of Maxwell's equations.

$$\nabla \times \mathbf B = \mu_0 \epsilon_0 \dfrac{\partial \mathbf E}{\partial t}$$


protected by ACuriousMind Apr 19 '15 at 17:16

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