Question on RMS Mean Free Path

[warhammer]

Homework Statement

The distribution of free path is given as N = N(o) e^ (-x/lambda) where lambda is the mean free path, and N is the number of molecules that survive collision on travelling a distance x, and N(o)is the no. of molecules at distance x = 0. Show that the root mean square (rms) free path is given by √2*lambda

Relevant Equations

lambda (rms)= v(rms) * t(rms) where t(rms) is the relaxation time.

lambda (rms)= v(rms) * t(rms) -- 1

Now I assume here that t(rms)=1/(√2*n*π*d^2*v(rms))

But this cancels the v(rms) term when used in eq (1) so the mean free path and the RMS free path would actually be the same (even later on when used in the aforementioned Survival Equation)

I would like to state that I have not understood what is meant by v(rms) clearly.. I would be really obliged if someone would explain the concept behind the question as well as provide the guidance to complete the question.

(I even tried to use the original derivation that was used for Survival Eqn in context of Maxwellian gas but that also provided no real insight)

 

[mjc123]

You are over-thinking things. x has an exponential distribution, for which it is a known property that the mean value of x is λ and the variance is λ² so the mean value of x² is 2λ².

 

[warhammer]

You are over-thinking things. x has an exponential distribution, for which it is a known property that the mean value of x is λ and the variance is λ² so the mean value of x² is 2λ².

Thank you for your response. However I'm unable to gauge more clearly how it would entail for the succeeding step in this particular case here. Please provide a hint or elaborate on this.

 

[warhammer]

I would like to state that I have not understood what is meant by v(rms) clearly.. I would be really obliged if someone would explain the concept behind the question as well as provide the guidance to complete the question.

A minor typo here. Please read it as Lambda(rms) or alternatively rms value of free path rather than v(rms).

 

[kuruman]

If you have a distribution $N(x)$, the mean value of $x$ over an interval $a\leq x \leq b$ is $$\langle x \rangle = \frac{\int_a^b x N(x)~dx}{\int_a^b N(x)~dx}.$$ The mean-squared is $$\langle x^2 \rangle = \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}.$$The root-mean-squared is $$x_{\text{rms}}=\sqrt{\langle x^2 \rangle} =\sqrt{ \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}}.$$Just use the definition.

 

[warhammer]

If you have a distribution $N(x)$, the mean value of $x$ over an interval $a\leq x \leq b$ is $$\langle x \rangle = \frac{\int_a^b x N(x)~dx}{\int_a^b N(x)~dx}.$$ The mean-squared is $$\langle x^2 \rangle = \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}.$$The root-mean-squared is $$x_{\text{rms}}=\sqrt{\langle x^2 \rangle} =\sqrt{ \frac{\int_a^b x^2 N(x)~dx}{\int_a^b N(x)~dx}}.$$Just use the definition.

Sir what you're expressing quantitatively is not coming out and the remaining resultant text is indecipherable.

Request you to express it again..

 

[kuruman]

I have expressed it as clearly as I can. Just substitute and do the integral. If it doesn't "come out", please post what does come out and we will take it from there.

 

[warhammer]

I have expressed it as clearly as I can. Just substitute and do the integral. If it doesn't "come out", please post what does come out and we will take it from there.

First of all I would like to apologise for the silly question from my side. I was viewing the PF Site on my mobile and it was not rendering the relevant mathematics, just the written words. It was only after I opened the same via another device I was able to finally see the mathematics

Now I inputted everything via the survival equation & integrated the function using the "by parts technique". This resulted in the mean squared value to be 2*(lambda)^2. Using the same in the RMS definition specified by you, I obtained that RMS value=√2*lambda.

The answer has matched now but I would love if you could explain the mathematical definitions qualitatively if possible (the specific integrands in numerator and denominator, and I hope you would excuse this very trivial and silly probing).

 

[kuruman]

Apology accepted. Look up weighted average (or weighted mean). There are plenty of links and videos explaining it and it makes no sense to repeat that stuff when it already exists.

 

[warhammer]

Apology accepted. Look up weighted average (or weighted mean). There are plenty of links and videos explaining it and it makes no sense to repeat that stuff when it already exists.

Thank you tons for your help sir. You've been most kind!