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Suppose the functions $e(x)$ and $h(x)$ with units $\left[\frac{V}{m}\right]$ and $\left[\frac{A}{m}\right]$ and their Fourier transforms $E(k)$ and $H(k)$ with units $\left[V\right]$ and $\left[A\right]$. The time avaverage A of $e(x)h(x)$ in function of their Fourier transforms $E(k)$ and $H(k)$ can be written as \begin{align} A &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} e(x)h(x) dx \\ &= \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X} \int\limits_{-\infty}^{\infty} E(k)e^{ikx} dk \int\limits_{-\infty}^{\infty} H(k')e^{ik'x} dk' dx \\ &= \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\left[ \lim\limits_{X\to \infty} \frac{1}{X} \int\limits_{-X}^{X}e^{i(k+k')x} dx \right] dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}E(k) H(k')\delta(k'+k) dk dk' \\ &= 2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk \\ \end{align}

The question: the unit of the time average A is $\left[\frac{VA}{m^2}\right]$ whereas the unit of the final result is $\left[\frac{VA}{m}\right]$. How can this be and how to interpret it?

Example: if you know the Fourier transforms $E(k)$ and $H(k)$, and you want to calculate the average A, then you will use $A=2 \int\limits_{-\infty}^{\infty} E(k) H(-k)dk$ but this will give another unit as expected? So, then you have the wrong result?

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  • $\begingroup$ What is m ? Are you talking about electricity? $\endgroup$ – LL 3.14 Oct 29 '20 at 14:04
  • $\begingroup$ The unit $m$ corresponds to the unit of meter and I am indeed talking about electric and magnetic fields. But the question is quite general, the specific units are not that important... My problem is that the units of the original and final results differ by $s$, and I do not understand why. $\endgroup$ – Fre Oct 29 '20 at 14:08
  • $\begingroup$ It just means you did an error in your dimensional analysis. One of the common reason is that there is a unit hidden in a constant equal to $1$ in some theorem you are using. The best way to see the units is to do a rescaling: change $x$ by $\ell\tilde x$ with no units. And do the same for e and h $\endgroup$ – LL 3.14 Oct 29 '20 at 14:18
  • $\begingroup$ @LL3.14, yes, that is a good suggestion to rescale it to variables without unit, I should have written the question in this way... However, even by doing this, I still end up with an additional "m" in the units after the integrations... So, it is a purely math-related problem according to me... $\endgroup$ – Fre Oct 29 '20 at 15:12
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Your error is in the integral converging to the Dirac delta. Normally it should be $$\lim_{X\to\infty} \int_{-X}^X e^{i(k+k')x}\,\mathrm d x = \delta_{k+k'}$$ so without dividing by $X$. This is at this step that you are loosing a $m$.

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