Solve Physics Problems: Splitting Algebraic Equations | Step-by-Step Guide

[Physics Dad]

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)k_bT)^3/2v²e^(-mv2/2k_bT)

formula into the following:

f(v) = aT^-3/2v²e^(-mv2/2k_bT)

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

 

[mastrofoffi]

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

 

[Physics Dad]

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

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

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

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

I just feel like I am getting algebra blindness!

Thanks in advance for your patience!

 

[mastrofoffi]

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

 

[Physics Dad]

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 won't be the last time I am asking but it is great to know that a facility like this exists.