2
$\begingroup$

Maybe my question is too specific but I could not find the answer.

Abstract from:

Yield Optimization and Time Structure of Femtosecond Laser Plasma $\kappa \alpha$ Sources

The generation of femtosecond $\kappa \alpha$ x-rays from laser-irradiated plasmas is studied with a view to optimizing photon number and pulse duration. Using analytical and numerical models of hot electron generation and subsequent transport in a range of materials, it is shown that an optimum laser intensity $I_{opt}=7×10^9Z^{4.4}$ exists for maximum Kα yield. Furthermore, it is demonstrated that bulk targets are unsuitable for generating sub-ps x-ray pulses: instead, design criteria are proposed for achieving $\kappa \alpha$ pulse durations ≤ 100 fs using foils of $≈2μm$ thickness.

I have found in Yield Optimization and Time Structure of Femtosecond Laser Plasma $\kappa \alpha$ Sources: PRL one dimensional Maxwell energy distribution of electrons as

$f(E) dE =\frac{1}{\sqrt{EkT}}\exp\left(-\frac{E}{kT}\right) $

This expression diverges for $E\rightarrow 0$.

How one reaches to this expression I could not understand. Usually in 3D case the exponent is multiplied by $\sqrt{E}$.

I don't think that it is a typographical error

They have also written the total energy content as

$En=\int n f\left(E\right)E dE$

From dimensions of this expression it looks like the particle distribution per unit energy.

Is there a specific physical significance of $f\left(E\right)$

Please help.

$\endgroup$
1

1 Answer 1

3
$\begingroup$

Yes, there is. You can look here. It is the Maxwell-Boltzmann distribution of the energy per degree of freedom. This is a chi-square distribution and note that $$\int_0^\infty \frac{1}{\sqrt{EkT}}e^{-\frac{E}{kT}}dE= \frac{2}{\sqrt{kT}}\int_0^\infty e^{-\frac{x^2}{kT}}dx$$ that is finite as it should. The authors of the paper are somewhat cavaliers about normalization factors. Note also, that any power of $E$ grants a finite integral reducing to known Gaussian integrals.

$\endgroup$
6
  • $\begingroup$ Thanks, the integration is indeed finite but what is the physical significance of a distribution having dimension of inverse of energy. $\endgroup$
    – hsinghal
    Feb 2, 2017 at 15:28
  • 1
    $\begingroup$ That is the one-dimensional Maxwell-Boltzmann distribution that when expressed in terms of energy, being $v=\sqrt{2E/m}$, gains the singular contribution $1/\sqrt{E}$. So, it happens also to the distribution of energy for a single degree of freedom. In the paper, the authors claim to work with the one-dimensional Maxwell-Boltzmann distribution and so it goes for the energy, normalization factors apart. $\endgroup$
    – Jon
    Feb 2, 2017 at 16:46
  • $\begingroup$ Usually when we talk about f(E) we say that is probability of finding a particle in energy range E , E+dE. Is this distribution can be used like this or we need to multiply this with E to make it dimension less. $\endgroup$
    – hsinghal
    Feb 2, 2017 at 16:57
  • 1
    $\begingroup$ I give you the full derivation. You should start from Maxwell-Boltzmann distribution in one dimension $f(v)dv=\sqrt{m/2\pi kT}e^{-\frac{mv^2}{2kT}}dv$. Then you note that $v=\sqrt{2E/m}$ implies $dv=dE/\sqrt{2mE}$. Expressing your distribution in terms of $E$ yields $f(E)dE=\sqrt{\frac{m}{2\pi kT}}e^{-\frac{E}{kT}}\frac{dE}{\sqrt{2mE}}$. No further normalization is needed. $\endgroup$
    – Jon
    Feb 2, 2017 at 17:02
  • $\begingroup$ Thanks, although after your earlier explanation the derivation was not required. $\endgroup$
    – hsinghal
    Feb 2, 2017 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.