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Working with infintesimals

  1. Jul 30, 2008 #1
    Hello, I did notice another thread that has a similar question but my question does not seem to fit perfectly into the OP's question so rather than hijack that thread I'll start a new one.

    I have found myself manipulating infintesimals algebraically. Recently I have read things that give me reason to believe this may be an incorrect thing to do that may or may not lead to correct results.

    For instance, say

    W = [tex]\int[/tex]F*ds = [tex]\int[/tex]mdv/dt*vdt

    I might simply say that dv/dt by dt is dv giving me

    = [tex]\int[/tex]mvdv

    But apparently this may not be a valid operation? I don't understand how it wouldn't be since, as I understand it, dv is simply a change in v (a number) that corresponds to a change in t of dt (dt being another number).

    From what I've found it seems "non-standard analysis" may give me insight as to what is going on here, but I do not want to spend time being side-tracked by this if it isn't worthwhile enough - perhaps my time would be better spent studying PDEs or fourier analysis. Especially if all non-standard analysis is going to give me is reassurance that what I'm doing is, in fact, a valid operation (I can take their word for it).
  2. jcsd
  3. Jul 30, 2008 #2
    i'm sorry but i don't know why ds here is? i'm suspicious of your transformation in the first place so please expound on it a little more.
  4. Jul 30, 2008 #3


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    Well he meant to say [tex]d\textbf{s}[/tex] is the infinitesimal displacement he's using when evaluating the work done along a path. So a = dv/dt, and ds=vdt. Then he substituted everything back into the integral and it now becomes a line integral in "velocity space".
  5. Jul 30, 2008 #4
    The statement W = [tex]\int[/tex]F*ds

    is simply work done on a particle along the path s where F is the force applied to the particle (it is a vector) and ds is the displacement of the particle (also a vector) over the period of time dt.

    I should have defined it - it was intended to be a familiar statement that I could use to illustrate the operations that I am concerned about.
    Last edited: Jul 30, 2008
  6. Jul 31, 2008 #5
    but there in lies the problem because [itex] \frac{ds}{dt} \neq v [/itex] but [itex] \frac{ds}{dt} = |v|[/itex] and so I don't think that transformation follows. I mean for regular differentials dx = x'(t) dt then you can divide the dt but since but that's not how ds is defined you run in to problems. Hopefully someone can flesh that out a little more.
  7. Jul 31, 2008 #6
    Hi, Nick R

    I'm not 100% sure, it's this way, but I whink this operation is legitimate :)

    You got after all the transformations W = (mv^2)/2+C. Why this should not be correct? The unit is the same - Joule. This is also the formula for the kinetic energy. And since mechanical work and energy is the same you just transformed them.

    What might be tricky here is the whole work consists of kinetic and potential energy.

    Now, consider following: Int[F]ds = Int[ma]ds = Int[mg]ds = mgs+K = mgh+K (where we set s=h)

    Look what these 2 expressions give us: (mv^2)/2+C = mgh+K or (mv^2)/2 - mgh = const But we said that work and energy is the same, so it's just

    [tex]\Delta E=const[/tex] - the conservation of energy
  8. Jul 31, 2008 #7
    actually now that i think about it in the case of dx = v dt the transformation is fine.
  9. Jul 31, 2008 #8
    I guess what I'm asking is - is it always OK to use infinitesimals as found in Leibniz's notation as if they were just very small real numbers? Or will this ever get me in trouble/yield incorrect results?
  10. Aug 3, 2008 #9
    What you did up there it is legitimate,you inherently used the theorem in differentials,that the ratio of two differentials (dy/dx) equals y'(x),i.e the derivative of the function y(x),even if x is a function of another variable i,e x=x(t)
    Note also now that dy/dx =dy/dx where the notation on the L,H.S is
    another way for expressing y'(x) and the line ( / ) between dy and dx does not mean division
    Note also that infinitesimals are variables which have 0as their limit,while differentials are mere quantities like fractions. differentials can be infinitesimals.
    Finally Osgood's theorem can account for the substitution under the integral sign
  11. Aug 3, 2008 #10
    The integration will give you the kinetic energy i.e. 1/2mv^2
  12. Aug 3, 2008 #11
    It is almost always okay, but there are some mentors on these forums who will contrive examples involving multivariable partial derivatives where it fails.

    In the case you are presenting, what is really happening is that you are going backwards on the chain-rule, transforming the integral by what is called 'u-substitution' in the US. It is just that you are doing it in general, without an explicit function formula, so in a sense it requires more mathematical maturity.
  13. Aug 3, 2008 #12
    can you be more precise about this?
  14. Aug 3, 2008 #13
    Sure, just take the following derivative:

    [tex]\frac{d}{dt} v(t)^2 [/tex]

    According to the chain rule, we first differentiate the outside function and then multiply by the derivative of the inside function:

    [tex]2 v(t) \frac{d}{dt} v(t) [/tex]

    So when integrating with respect to time an expression that is similar to the second one, you obviously recover the first one by the fundamental theorem of calculus. The steps look like this:

    [tex] \int m v \frac{dv}{dt} dt = \int \frac{d}{dt}(\frac{1}{2} m v^2) dt = \frac{1}{2} m v^2 [/tex]
  15. Aug 3, 2008 #14


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    Hi Nick! :smile:
    Straight answer:


    The chain rule wouldn't work if you couldn't. :wink:
  16. Aug 3, 2008 #15
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