I have some trouble accepting the Conservation of Energy in electromagnetism. Assume two EM pulses in opposite directions move toward each other and are entirely in reflection symmetry (their field vectors are like $\vec{E}$ and $-\vec{E}$ in all time). when they interfere, there is a moment they should completely cancel each other and there should be $\vec{E}=\vec{0}$ in all space, and according to $$\text{Energy}=\int{\epsilon _0 E^2}\mathrm{d}V$$ the energy in the entire space is zero but before interference, we had energy (sum of each pulse energy). so how conservation of energy is valid here? (we can also express everything I've said about magnetic field too)

  • $\begingroup$ If you have incident and "reflected" wave, you will have a standing wave. The energy is not zero $\endgroup$ Commented Mar 16, 2023 at 14:47
  • $\begingroup$ The general argument follows from the Maxwell equations (Poynting theorem), but one can check for any specific forms of pulses satisfying ME - have you dried it? $\endgroup$
    – Roger V.
    Commented Mar 16, 2023 at 15:10
  • $\begingroup$ You forgot to take into account energy due to magnetic field. $\endgroup$ Commented Mar 16, 2023 at 16:04

1 Answer 1


we can also express everything I've said about magnetic field too

Here is your mistake. In fact, in the situation you have described the magnetic fields must add constructively.

Recall that for an EM wave we have $\vec S = \vec E \times \vec H$ where $\vec S$ defines the direction of propagation. Thus, in your case, since the pulses are moving in opposite directions, we must have $\vec S_1 = -\vec S_2$. So $$\vec E_1 \times \vec H_1 = - \vec E_2 \times \vec H_2$$ And since you have further specified that $\vec E_1=-\vec E_2 = \vec E$ we have $$\vec E \times \vec H_1 = \vec E \times \vec H_2$$$$\vec H_1=\vec H_2$$

So in the very instant when the E fields cancel out the H fields align. The complete destructive interference of the E fields is balanced by the complete constructive interference of the H fields.

Furthermore, in an EM wave half of the energy is in the E field and half of the energy is in the H field, so the total energy of the E and H fields of both waves is 4 times the energy in the E field of one wave. Also, the energy is proportional to the square of the field, so when the E fields cancel out, the H field is doubled, so the energy in the combined H field is 4 times the energy in the E field of one wave, thus conserving energy.

EDIT: in response to @hyportnex comments below I work this out for a specific example using units where $\mu_0=\epsilon_0=c=1$ for simplicity.

Let $$u(x)=\begin{cases} 1 & -\frac{1}{2}<x<\frac{1}{2} \\ 0 & \text{otherwise} \end{cases}$$

Now, suppose $$\vec E_1 = u(x-t) \hat y$$$$\vec H_1 = u(x-t) \hat z$$and$$\vec E_2 = - u(x+t) \hat y$$$$\vec H_2 = u(x+t) \hat z$$You can confirm that these fields satisfy the OP's requirements for two identical waves propagating in opposite directions. You can also confirm that they satisfy Maxwell's equations.

Now, let$$\vec E=\vec E_1+\vec E_2$$$$\vec H = \vec H_1 + \vec H_2$$You can also confirm that this satisfies Maxwell's equations, although it should be obvious by the linearity of Maxwell's equations without doing a lot of math. Nevertheless, you can also confirm that $\left. \vec E\right|_{t=0}=0$ everywhere, which satisfies the OP's requirements that the E field cancel out completely. You can further confirm that $\left. \vec H \right|_{t=0}=2 \left. \vec H_1\right|_{t=0} = 2\left. \vec H_2\right|_{t=0}$ as I stated above.

So in this example, as described in general above, as claimed when the E fields add destructively the H fields add constructively and the energy is conserved.

  • $\begingroup$ How can you have an oscillating $H$ field without inducing an electric field somewhere? $\endgroup$
    – hyportnex
    Commented Mar 16, 2023 at 15:46
  • $\begingroup$ Also, $\vec A cos(\omega t -k z)$ propagates in the $+\hat z$ direction , while $\vec B cos(\omega t +k z +\phi)$ propagates in the $-\hat z$ direction. Irrespective of what $\vec A$ or $\vec B$ are the sum of the two waves is never identically zero, it can form a standing wave though. $\endgroup$
    – hyportnex
    Commented Mar 16, 2023 at 16:14
  • $\begingroup$ @hyportnex I have added a big edit for you. If you need more than that please ask a new question $\endgroup$
    – Dale
    Commented Mar 16, 2023 at 19:10
  • 2
    $\begingroup$ oh, I completely misread the question; it is asking about for a "moment of time". $\endgroup$
    – hyportnex
    Commented Mar 16, 2023 at 19:43
  • 2
    $\begingroup$ @RogerVadim My initial comment was because of a misunderstanding of the question and of Dale's original answer. I missed that the question/answer was about the complete cancellation of the $E$ wave field for an infinitesimal long instant be it a pulse or a sinusoid, and as this answer correctly shows that can indeed happen. What it cannot do is to cancel for a finite length of time unless $H$ is static, because in an oscillating field $\partial H/\partial t \ne 0$ and then $curl E \ne 0$. $\endgroup$
    – hyportnex
    Commented Mar 17, 2023 at 15:35

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