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Differentiating a scalar potential

  • Thread starter Safinaz
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  • #1
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Hello,

I have this potential:

## V(\chi) = \frac { F ’’ (\chi) [ 2 F(\chi) - \chi F’ (\chi) ]}{ (F’(\chi))^3} ##

How to get

## \frac{ d V(\chi)}{d \chi} = \frac{ \chi F’’ + F’ - F’ }{ F’^2} - 2 \frac{ \chi F’ -F }{ F’^3} F’’ ~~~~~~(*)##

My trail,

## V( \chi) = 2 F F’’ F’^{-3} - \chi F’’ F’^{-2} ##

## \frac{ d V(\chi)}{d \chi} = 2 F’’ F’^{-2} + 2 F F’’’ F’^{-3} - 6 F F’’^2 F’^{-4} - F’’ F’^{-2} - \chi F’’’ F’^{-2} + 2 \chi F”^2 F’^{-3} ##,

which is not close to (*) !!
 

Answers and Replies

  • #2
RPinPA
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Can you explain where each of your terms came from? I assume you're using the product rule but you have a product of three terms, so the application of the product rule is going to be more complex. Can you show more steps?

Also, your expression has ##F'''## which I'd expect given that ##V(x)## has terms in ##F''##. Yet ##F'''## doesn't appear in what you say is the answer. Also, the first term in (*) has F' - F' in the numerator. I assume that is a typo and one of those has the wrong number of primes on it.

So can you also check what the correct answer is supposed to be?
 
  • #3
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  • #4
RPinPA
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What you wrote as ##V(\chi)## is equation (5) for ##dV/d\chi##, the derivative of (4) with respect to ##\chi##.
What you wrote as (*) is the term in square brackets in (9), which supposedly represents ##dV/d\chi## (ignoring the constant up front as you did). Thus the authors are saying it's the same as (5), not the derivative of (5).

So all they're doing is algebraic rearranging of the expression in (5), no extra differentiating happened.
 
Last edited:
  • #5
RPinPA
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I showed a derivation which does indeed go from (5) to the expression in [*] by algebraic manipulation, but I have deleted it temporarily. I don't think this is homework since you're just trying to read an online paper, but you did post it in a homework forum, and the policy of this site is not to give answers to homework.

You can repeat what I did by taking the expression in (*) and rearranging it to get to equation (5). Then simply do the steps in reverse.
 
  • #6
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What you wrote as ##V(\chi)## is equation (5) for ##dV/d\chi##, the derivative of (4) with respect to ##\chi##.
What you wrote as (*) is the term in square brackets in (9), which supposedly represents ##dV/d\chi## (ignoring the constant up front as you did). Thus the authors are saying it's the same as (5), not the derivative of (5).

So all they're doing is algebraic rearranging of the expression in (5), no extra differentiating happened.
Yep.. this is true .. thanks.
 

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