In general, the intensity of an electric field is given by $$ I = \frac{c\epsilon_0}{2}E_0^2 $$ where $E_0$ is the peak amplitude of the electric field. Let's say we have an electric field $$ E(t) = E_0 \sin(\omega t). $$ Without having any other potential, the electric field is connected to its corresponding vector potential $A(t)$ via $$ E(t) = -\frac{1}{c}\dot{A}(t). $$ If we compute the vector potential by above electric field, we get $$ A(t) = -c\int_0^t\,dt'\,E(t') = E_0\frac{c}{\omega}\cos(\omega t). $$ We could compute the intensity for a given vector potential via $$ I=\frac{c\epsilon_0}{2}E_0^2=\frac{c\epsilon_0}{2}\frac{A_0^2}{c^2}=\frac{\epsilon_0}{2c}A_0^2 $$ $\textbf{but}$ I wonder how to define the peak amplitude $A_0$ since $\cos(\omega t)/\omega > 1$ is possible. A choice would of course be $$ A_0 := E_0\frac{c}{\omega} $$ but that's not very general since the time-dependent function could be anything. What I mean in general is that, if an intensity is given, the time-dependent parts $f(t)$ and $F(t)=\int_0^t\,dt'\,f(t')$ in $$ E(t)=E_0 f(t)~~~, ~~~ A(t) = A_0 F(t) = -c E_0 F(t) $$ should only be in [-1,1] or not? One could also think of using $\text{max}_t A(t)$, resp. $\text{max}_t E(t)$ to compute the intensity but that's not how I have seen it in several computer codes (which is the main reason I ask this).

  • $\begingroup$ "$\cos(\omega t)/\omega > 1$" this inequality makes no sense. $\endgroup$ – hyportnex Jan 24 at 15:59

One could also think of using [...] $\text{max}_t E(t)$ to compute the intensity

That's the definition of peak intensity. Anything else is OK if it can be derived from it, and incorrect if it differs from what can be derived from that.

That includes using $\text{max}_t A(t)$, unless you're in some restricted situation (say, a monochromatic beam, or some similarly minor variation) where it coincides with what you can get from $\text{max}_t E(t)$.

but that's not how I have seen it in several computer codes (which is the main reason I ask this).

If those codes are considering only restricted situations in which their alternative approaches coincide with the correct results, then they're doing it right. If they're using alternative approaches that do not coincide with what can be derived with the correct definition, then they're doing it wrong.

Since you do not give any precise details about what you did see in those codes, and what the underlying assumptions are about the configurations that they're built to handle, there is nothing more we can say at present.

  • $\begingroup$ Dear Emilio, thanks for your answer. That is also what makes most sense to me. I was confused because, in above mentioned software, they used $A(t)=A_0\int\,dt\, \sin(\omega_1 t)\sin(\omega_2 t)^2:=A_0\, F(t)$ where $F(t)$ can exceed 1 and $A_0$ was computed by a given intensity. $\endgroup$ – Lukk Jan 28 at 11:53
  • $\begingroup$ @Lukk It's extremely hard to comment further without more detailed context. If they're using something that doesn't quite correspond to the actual peak intensity then that certainly sounds fishy ─ though, on the other hand, to be frank, the experimental uncertainties in the intensities of pulsed lasers can often be so short that the differences can be completely moot. Still, without more context, there's nothing more I can say. $\endgroup$ – Emilio Pisanty Jan 28 at 12:43

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