HW set 1- Problem 5.3 b

[Ashveen Tewarie]

Can someone provide some help with problem 5.3 (b)
I thought that determining the mp KE is the same as determining the mp
Root Mean Squared speed but this doen't seem to work. Thank you!

[Josh Duckett]

Glad somebody else decided to join...I haven't gotten to 5.3 yet but
the mp is determined by differentiating the distribution function like
they did to get equation 5.11 for the velocities - the rms will be
something different.

Have you gotten 5.2 worked out? I'm having trouble getting all three
of the conditions in part b to work out along with something that
checks out in part a. What did you get for your distribution
function? Did you convert to spherical polar coordinates like they
did in the text?

Have you taken a Hassan course before? I wanted to take this class
but was dreading him...I never really understand what he is doing in
class the way he teaches.

[Josh Duckett]

Where did you start to get the KE in 5.3a. I'm not getting anything
that seems to be closing on a solution.

[Ashveen Tewarie]

I started 5.3a with the integrand 1/2 mC^2 * chi(C). This results in
an integral that looks like the equation on page 47 where the RMS is
determined. It's only multiplied by m/2. But this does not give e most
probable of kT/2. But it gives e average = 3/2 *kT (This integral was
also worked during the lectures). So my procedure is almost the same
as determining the mp value for C^2. It doesn't seem to work. Did you
finalize 5.1 (c)?

[Josh Duckett]

Alright - I'll take a look at that and see if I can work out anything for
the mp. I haven't gotten 5.1C either - having trouble with the KE stuff.

Did you get all the solutions worked out for 5.2? I can't get them all to
match what the book solutions are. For Part A I got (m/2PIkT)^(1/2)
*exp((-m/2kT)Ct^2) which works for Part A and gives unity when you do the
-infinity to infinity integral verification. But it doesn't work out for
the Part B values. What did you get for the distribution curve?

[Ashveen Tewarie]

Josh,
I am now at work, I will check later on today what distribution curve
I got for problem 5.2.
But I would like to inform you that it's a 2D problem, which means
that you will have to derive the velocity distr. function again, as we
did during the lectures. This time dC=d(pi*c^2) instead of dC= d
(4/3*pi*c^3). You only have to consider 2 velocity components dc1 and
dc2.Earlier today I e-mailed Dr Hassan regarding 5.3. I am still
waiting for a reply. Hopefully I'll get it today.

[Josh Duckett]

Yea I think I figured it out - started my derivation too late - had to
go back another step. I am getting....

f = (beta^2)/pi * exp(-(beta^2)*(c')^2)

beta^2 = m/(3*k*T)

I am about to go through the rest of the problem but I think thats
right. I went back and started my derivation with what he did on page
4 of Lecture 6 notes but only using C1' and C2' - is that what you did?

[Josh Duckett]

oops...typo...The f I gave above is right - just a change to the beta

Beta^2 = m/(2*k*T)

Now on to try and figure out the energy stuff - having any luck?

[Ashveen Tewarie]

Yes, that's the same velocity distribution function I have.
I am now working on 5.3 again. I'm not sur if what I'm doing is
correct.
The mp KE I'm getting differs from the result given in the text. Have
you tried 5.1 (c)?

[Josh Duckett]

I got the e average worked out too but not the e_mp. Are you getting
sqrt(8kt/m) for e_mp

[Josh Duckett]

I was able to get a bit closer - I got e_mp = 2kT...I took (1/2)mC^2
* CHI(C)...then I subbed in epsilon anywhere there was(1/2)mC^2...I
ended up with

8*pi*m^(1/2) * (1/2*pi*k*T)^(3/2)*epsilon^2*exp((-1/k*t)*Epsilon)

I took the derivative of that with respect to epsilon and got
2kT...its closer - you have any ideas from that?

[Josh Duckett]

Have you figured out anything for 5.3c or d?

[Josh Duckett]

I keep getting 1/3 for 5.1c and that doesn't make any sense or match
the answer - have you gotten anywhere?

[Ashveen Tewarie]

Unfortunately I'm still where I was yesterday. It doesn't seem to work
what I'm doing. Having some difficulties in finding the integral. I
will look at it for the last time. Hopefully I can come up with sth
that works out well.

[Ashveen Tewarie]

Yes, that's the same answer I've got at this moment. e m.p = 2kT. This
is 4x the answer it should be! Not sure what I should do with this....