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Cancelling differentials

  1. May 22, 2008 #1
    often times you will see naive people cancel differentials to obtain whatever it is they want for example :

    [tex] \frac{dx}{dy} \frac{dy}{dz} = \frac{dx}{dz} [/tex]

    now i know this isn't rigorous and my question is actually about partials. is there ever an occassion/space/set/etc where i can do this:

    [tex] \frac{\partial x}{\partial y} \frac{\partial x}{\partial z} = \frac{ \partial^2 x}{\partial y \partial z}[/tex]
     
  2. jcsd
  3. May 22, 2008 #2
    I would look at this as a partial differential equation for which we know there are particular solutions, e.g. x = y z. If you found the general solution, then you would know all the functions for which it is true.

    However, there won't be general conditions such as "if x(y,z) is continuous/smooth/etc", unless the general condition you are looking for is "let x,y,z be independent variables...":smile:
     
  4. May 22, 2008 #3
    Well, you can use separation of variables to show that any function of the form

    [tex]f(x,y) = -ln(X(x)+Y(y)) + C[/tex]

    is a solution to the differential equation you posted, where Y and X are arbitrary functions and C is an arbitrary constant. There's also the trivial solution (with f=0). I'm not sure if there are others, but if there are, there probably aren't many.
     
  5. May 22, 2008 #4

    nicksauce

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    From the standpoint of dimensional analysis, the equation doesn't really make any sense.

    Suppose x is metres, t and u are seconds.
    Then d^2 x / (dudt) = d/dt (dx/du) has units of m/s^2
    But (dx/du) (dx/dt) has units of m^2 / s^2.

    So I don't think it will ever be true in general.
     
  6. May 22, 2008 #5
    That's true, there are no physical quantities which obey this relationship. However, as I showed above, there are (dimensionless) mathematical functions which do.
     
  7. May 22, 2008 #6

    tiny-tim

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    Hi ice109! :smile:
    So long as they're genuine ds and not ∂s (so x is a function of y only, and y is a function of z only), that's fine. :smile:
    [tex]\frac{ \partial^2 x}{\partial y \partial z} = \frac{\partial }{\partial y}(\frac{\partial x}{\partial z}) = \frac{\partial }{\partial z}(\frac{\partial x}{\partial y})[/tex]
     
  8. May 22, 2008 #7

    matt grime

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    You start off by treating differentials as things you can multiply and divide. But notice that you've then moved on to multiply dx by dx and think it gives d^2x, rather than dx^2. That little thing should warn you off doing that - the choice of where to put the powers is quite deliberate.

    There might well be occasions where the two things you equate are actually equal (though nothing non trivial springs to mind), but it isn't true in general. Surely we all agree x=yz is about as nice a function of y and z as can hope for. But you're asking if yz=1.
     
    Last edited: May 22, 2008
  9. May 22, 2008 #8

    mathwonk

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    In some sense, you can cancel first differentials, as long as you are in one dimension.

    i.e. at each point p of the real line, there is a one dimensional tangent space and a one dimensional cotangent space.

    for each smooth function f, the differential df is a function whose value at each point p, the the linear function on the tangent space at p, taking the tangent vector h = x-p, to f'(p)h.

    i.e. df(p) is an element of the cotangent space at p.

    Thus if dg at p is the linear function taking h to g'(p)h, then df/dg at p takes every h to f'(p)/g'(p) = {df/dx}/{dg/dx}.

    so these are slightly different, but very close, since df/dg is thus a function whose value at p is the constant function f'(p)/g'(p). and it is standard to equate a constant function with its constant value.

    matt has pointed out the differences in trying such manipulations for second differentials.
     
  10. May 22, 2008 #9

    Defennder

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    I really don't see how your second expression is analogous to the first. It also doesn't seem related to the thread title at all. In the first, the differentials appear to "cancel out" to give the resulting expression, but in the second, nothing on the LHS appear to cancel out. Did you mean [tex] \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = \frac{ \partial x}{\partial z}[/tex] instead?
     
  11. May 23, 2008 #10

    lurflurf

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    here is a humorous clasical example
    suppose f(x,y,z)=0
    then
    [tex]\displaystyle{ \left(\frac{\partial z}{\partial x}\right)_y \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x= -1}[/tex]
     
  12. May 23, 2008 #11

    mathwonk

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    yes it used to mystify me as a stuent when i noticed that if f(x,y) = 0,

    then (∂f/∂x) dx + (∂f/∂y)dy = 0,

    hence dy/dx = - (∂f/∂x)/(∂f/∂y).

    but why not? the curly ∂'s are not first differentials, their symbols are just similar.
     
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