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I Algebraic Manipulation

  1. Oct 27, 2016 #1

    Physics Dad

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    Gold Member

    Hi,

    I am having a real senior moment and can't quite get my head around a physics problem.

    I need to split an equation into two parts, in turn creating a coefficient and my mind has gone totally blank!

    I need to split the Maxwell-Boltzmann distribution formula from:

    f(v) = 4π(m/2(pi)kbT)3/2v2e(-mv2/2kbT)

    formula into the following:

    f(v) = aT-3/2v2e(-mv2/2kbT)

    where a is a coefficient to be determined.

    I understand that I am isolating variables in order to create a coefficient (a) but every time I get going I lead myself down a merry path and end up with something different!

    Any help would be gratefully received
     
    Last edited: Oct 27, 2016
  2. jcsd
  3. Oct 27, 2016 #2
    well, the second part of the formula(after ##v^2##) is equal in both forms so we only have to consider ##4\pi(\frac{m}{2\pi K_bT})^{\frac{3}{2}} = aT^{\frac{-3}{2}}##
    remember that ##\frac{1}{T^{x}} = T^{-x}## and there you are
     
  4. Oct 27, 2016 #3

    Physics Dad

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    I am getting an answer but I don't have confidence in my reasoning behind it.

    Basically, if 4π(m/(2kbT))3/2=aT-3/2

    am I right in thinking that T-3/2 becomes 1/T3/2?

    if so, after doing a bit of rearranging magic from here, I get the answer that a=4π(m/2kb)3/2

    I just feel like I am getting algebra blindness!!

    Thanks in advance for your patience!
     
    Last edited: Oct 27, 2016
  5. Oct 27, 2016 #4
    yep, i also want you to notice that it is really immediate to see ##4\pi(\frac{m}{2\pi K_bT})^{3/2} = 4\pi(\frac{m}{2\pi K_b})^{3/2}T^{-3/2} = aT^{-3/2}## so the ##T^{-3/2}## just get canceled out and there's no magic at all :P

    so you basically missed a ##\pi##, probably just for distraction, but you're there:
    ##a = 4\pi(\frac{m}{2\pi K_b})^{3/2}##
    now ##4\pi \frac{1}{(2\pi)^{3/2}}## can be further simplified, give it a try
     
  6. Oct 27, 2016 #5

    Physics Dad

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    Thanks a lot, and yes, you're right, I missed the π from the denominator.

    I know it is obvious to see the cancellation, but I wanted to make certain by expanding and then just cancelling down to be sure.

    Your assistance has been really greatly appreciated. I am sure it wont be the last time I am asking but it is great to know that a facility like this exists.
     
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