2

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.

Community
  • 1
hsinghal
  • 2,571
  • Related: https://physics.stackexchange.com/q/769018/226902 (Does the 1D hard-rod gas converge to the 1D Maxwell-Boltzmann distribution?) – Quillo Jun 21 '23 at 11:52

1 Answers1

3

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.

Jon
  • 3,818
  • Thanks, the integration is indeed finite but what is the physical significance of a distribution having dimension of inverse of energy. – hsinghal Feb 02 '17 at 15:28
  • 1
    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. – Jon Feb 02 '17 at 16:46
  • 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. – hsinghal Feb 02 '17 at 16:57
  • 1
    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. – Jon Feb 02 '17 at 17:02
  • Thanks, although after your earlier explanation the derivation was not required. – hsinghal Feb 02 '17 at 17:05
  • You are welcome. It will be helpful for somebody else. – Jon Feb 02 '17 at 17:10