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Variation of a quantity

  1. Jul 4, 2014 #1


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    Suppose I have such an equation:
    [itex]A= f(B)[/itex]
    so [itex]A[/itex] is a function of [itex]B[/itex].

    Can I really use the fact that a variation of [itex]A[/itex] is like taking the differential of it?
    [itex] \delta A= dA [/itex]

    so that:

    [itex] \delta A = \frac{d f(B)}{dB} \delta B[/itex]
  2. jcsd
  3. Jul 4, 2014 #2


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    Superficially it looks OK. You need to look at the definition involved for variation.
  4. Jul 4, 2014 #3


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    the reason I ask is because I have a confusion with one more way to see variation.
    Suppose I vary [itex]B[/itex], so that [itex]A[/itex] is going to be varied too:

    [itex] A+ \delta A = f(B+\delta B) [/itex]
    And suppose [itex] f(x) = \frac{1}{\sqrt{x}} [/itex]


    [itex] A( 1+ \frac{\delta A}{A}) = \frac{1}{\sqrt{B+\delta B}}= \frac{1}{\sqrt{B}} \frac{1}{\sqrt{1+\frac{\delta B}{B}}}≈ A -A \frac{1}{2} \frac{\delta B}{B}[/itex]

    [itex] \frac{\delta A}{A}= - \frac{1}{2} \frac{\delta B}{B} [/itex]

    This result would imply that:
    [itex] \frac{d f(B)}{dB} = -\frac{1}{2}\frac{A}{B} [/itex]

    in my OP...

    with a simple check, we have:

    [itex] \frac{d (1/ \sqrt{B})}{dB} = -\frac{1}{2} \frac{1}{B^{3/2}}[/itex]

    This method works fine for such an easy function as [itex]f(x)=\frac{1}{\sqrt{x}} [/itex]
    Now suppose I have instead:
    [itex] f(x)= \frac{1}{\sqrt{x}} + c [/itex]

    with [itex]c[/itex] a constant. The same method won't work because I will end up with something like:

    [itex] \frac{\delta A}{A} = - \frac{1}{A} \frac{1}{2} \frac{1}{\sqrt{B}} \frac{\delta B}{B} [/itex]

    unfortunately now [itex] \frac{1}{A}[/itex] can't be cancelled with the [itex]\frac{1}{\sqrt{x}}[/itex] ... Only maybe, up to some error, if I consider [itex]c[/itex] very small... The derivative however will kill [itex]c[/itex] and give some other result (the same as for without [itex]c[/itex] which I think is more physical- isn't it? a constant shouldn't destroy the variations)..
    Last edited: Jul 4, 2014
  5. Jul 4, 2014 #4


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    I think I found my misconception... the variation is still independent of [itex]c[/itex]... it's 1/A which gives me that dependence. So there's no way to escape from c rather than neglecting it :) that's fine...i guess
  6. Jul 4, 2014 #5


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    You can basically use Taylor expansion to show your equality.

    If ##A=f(B)## and we add a small quantity ##\delta B## to ##B## then by Taylor's theorem we will have:

    $$A'\equiv A+\delta A=f(B+\delta B)=f(B)+\frac{df}{dB}\delta B+\mathcal{O}(\delta B^2)$$

    This gives as the limit ##\delta B\rightarrow 0##:

    $$\delta A=\frac{df}{dB}\delta B$$

    Of course, this is usually much more interesting when we consider functionals rather than simple functions.
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