Rescaling Variables HW: Get \frac{\partial \hat{f}(z,x)}{\partial z}

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Homework Statement



Suppose I have the following expression:
(v \frac {\partial}{\partial r} ) f(r,v)

I want to obtain:

\frac {\partial \hat{f}(z,x)}{\partial z}

Homework Equations



x \rightarrow v/v0
z \rightarrow (r-r0)/H
H \rightarrow \frac{k_{b} T} {m g }
\hat{f}(z,x) = x^2 f(r,v)

The Attempt at a Solution


[/B]
x v0 \frac {\partial}{\partial (zH+r0)} \frac{1}{x^2} \hat{f}(z,x)

is this possible?
 
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Where does the x2 come from?

Assuming H and r0 are constant, you can simplify the derivative.
 
mfb said:
Where does the x2 come from?Assuming H and r0 are constant, you can simplify the derivative.
It was just a given a change of variable. I am trying to verify it.
f(r,v) = f( zH+r0, xv0)

Can you show me explicitly?
 
The equations look good apart from that x2. Where does it come from?
 
What do you mean?

I am trying to show

(v \frac {\partial}{\partial r} ) f(r,v) \rightarrow<br /> <br /> \frac {\partial \hat{f}(z,x)}{\partial z}
 
There is an x2 in your equations. Why? What did you calculate that let this factor appear in the equation?

Edit: This one: ##\hat{f}(z,x) = x^2 f(r,v)##
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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