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Differential Operator

  1. Dec 13, 2005 #1
    This may seem like an odd question, but why are there two different ops for the normal and partial derivatives? i.e., [tex]\frac{d}{dx}[/tex] and [tex]\frac{\partial}{\partial x}[/tex]? I don't see a difference if only one is used, since we are always differentiating wrt a single variable anyway.
     
  2. jcsd
  3. Dec 13, 2005 #2
    think of it....

    in classical mechanics, you can have functions like the action [itex]L[/itex] wich is defined in the form

    [tex]L=L(x,y,z,t)[/tex],

    but [itex]x=x(t)[/itex], [itex]y=y(t)[/itex], [itex]z=z(t)[/itex], so

    [tex]\frac{d L}{d t}=\frac {\partial L}{\partial x} \dot{x}(t)+\frac {\partial L}{\partial y} \dot{y}(t)+\frac {\partial L}{\partial z} \dot{z}(t)[/tex]

    wich clearly is different from [itex]\partial L/\partial t[/itex].


    EDIT:

    Sorry, my mistake... the derivative is missing one term. It should be read

    [tex]\frac{d L}{d t}=\frac {\partial L}{\partial x} \dot{x}(t)+\frac {\partial L}{\partial y} \dot{y}(t)+\frac {\partial L}{\partial z} \dot{z}(t)+\frac{\partial L}{\partial t}[/tex]
     
    Last edited: Dec 13, 2005
  4. Dec 13, 2005 #3
    Interesting. I never encountered those. Then again, I'm not in physics.
     
  5. Dec 13, 2005 #4
    The above are functions present in Hamiltonian systems, wich are a big subject of study for mathematitians too... Specially in P.D.E.

    EDIT:

    Not to mention Calculus of Variations.
     
  6. Dec 13, 2005 #5

    matt grime

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    one came first, d/dx, and the other is a generalization of it, but asking what d/dx of some object is is strictly different from asking what partial d by dx of it is since the former assumes that the other variables (if there are any) are a function of x too. That is to say that if f(x)=x+y then

    [tex]\frac{\partial f}{\partial x}[/tex]

    makes sense but

    [tex] \frac{df}{dx}[/tex]

    doesn't
     
  7. Dec 13, 2005 #6
    Wait, if [tex]f(x)=x+y[/tex], wouldn't [tex]\frac{df}{dx}[/tex] make sense since y is a constant? That is, [tex]\frac{df}{dx}[/tex] does not make sense if it was [tex]f(x,y)=x+y[/tex]?
     
  8. Dec 13, 2005 #7

    matt grime

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    And what if y weren't a constant? come on, put the pieces together, you should be able to correct the obvious mistakes that people make! Dear God.
     
  9. Dec 13, 2005 #8
    Fascinating.
     
  10. Dec 14, 2005 #9

    matt grime

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    Here's another reason for the disticntion.

    I give you y, jsut y, now differentiate it with respect to x. What's the answer? dy/dx or 0?

    I suppose it is unfair of me to expect you to recognize silly errors from catastrpohically bad ones, not to mention hypocritical perhaps.
     
    Last edited: Dec 14, 2005
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