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Differential descriptions

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution


    Suppose you see something like

    [tex]\frac{\partial^2 \phi}{\partial X^{\mu}\partial X^{\nu}}[/tex]

    What I am interested here is the denominator. Is the term [tex]\partial X^{\mu}\partial X^{\nu}[/tex] simply X taken to the second derivative? I know one can write it like

    [tex]\partial X^{\mu} (\frac{\partial \phi}{\partial X^{\nu}})[/tex]

    But what do we calculate when they are lumped together like [tex]\partial X^{\mu}\partial X^{\nu}[/tex]?
     
    Last edited by a moderator: May 9, 2012
  2. jcsd
  3. May 8, 2012 #2
    What I mean is, is [tex]\partial X^{\mu}\partial X^{\nu}[/tex] equal to [tex]\partial^2 X^2[/tex]?
     
  4. May 8, 2012 #3

    Mark44

    Staff: Mentor

    I'll take a shot at it. Your notation suggests that ##\phi## is a function of at least two variables: X##\mu## and X##\nu##

    This notation:
    [tex]\frac{\partial^2 \phi}{\partial X^{\mu}\partial X^{\nu}}[/tex]

    means

    [tex]\frac{\partial}{\partial X^{\mu}}\left(\frac{\partial \phi}{\partial X^{\nu}}\right)[/tex]

    In other words, take the partial of ##\phi## respect to X##\nu## and then take the partial of that function with respect to X##\mu##.
     
    Last edited: May 9, 2012
  5. May 8, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it is not. [itex]X^\mu[/itex] and [itex]X^\nu[/itex] are different variables. What have is similar to
    [tex]\frac{\partial^2 \phi}{\partial x\partial y}[/tex]
    where x and y are the two variables in the xy-plane. The point is that if you are working in a situation where you have three or four or even more dimensions, it is simpler to write "[itex]X^1[/itex]" and "[itex]X^2[/itex]" rather than "x" and "y", for example.

     
    Last edited by a moderator: May 9, 2012
  6. May 9, 2012 #5
    Thanks, was unsure... but I guessed you approach. Thanks again!
     
  7. May 9, 2012 #6
    Ok, I have a new question of the same type. Suppose then you might have

    [tex]\frac{\partial \phi}{\partial X^j \partial X^j}[/tex]

    This seems to be purporting to the same direction [tex]X^{j}[/tex] yes? So how is this interpretated?

    Assuming you can bring it out again like so

    [tex]\frac{\partial}{X^j}(\frac{\partial \phi}{ \partial X^j})[/tex]

    Does anything get squared now?
     
  8. May 9, 2012 #7

    Mark44

    Staff: Mentor

    No, this doesn't make any sense.
    It would have to be
    $$ \frac{\partial^2\phi}{\partial X^j \partial X^j}$$

    This is the second partial of ##\phi## with respect to Xj.
    It would be interpreted as the partial with respect to Xj of the partial of ##\phi## with respect to Xj.
    This isn't quite right.

    $$ \frac{\partial}{\partial X^j}(\frac{\partial \phi}{ \partial X^j})$$

    Compare this to the related Leibniz notation for the second derivative.
    ##\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)##
     
  9. May 9, 2012 #8
    Sorry I missed out the [tex]\partial^2[/tex]. I knew it should have been there...

    anyway, this is what I am asking. Just in short, does this [tex]\partial X^j \partial X^j[/tex] mean [tex]\partial X^{j^2}[/tex]?

    For

    [tex]\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}[/tex] you can make it

    [tex]\frac{\partial}{\partial X^{\mu}}(\frac{\partial \phi}{\partial X^{\nu}}[/tex]

    why can't I do this for something like

    [tex]\frac{\partial^2 \phi}{\partial X^{j} \partial X^{j}}[/tex]

    Is the denominator really [tex]\partial X^{j^2}[/tex] because it seem from HallsOfIvy that the denominator in

    [tex]\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}}[/tex]

    Are seperate variables and so they cannot be squared. With the new case I have given, this is not obvious. You never answered my question very helpfully.
     
  10. May 9, 2012 #9
    Let me numb the question down:

    HallsofIvy said that

    [tex]\frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = \frac{\partial^2}{\partial X^{\mu}} \frac{\partial \phi} {\partial X^{\nu}}[/tex]

    I knew this and it is simple enough. I asked if the product [tex]X^{\mu}X^{\nu}[/tex] was some squared value, in which it was [tex]X^2[/tex] because they were separate variables. Now I am asking the same of

    [tex]\frac{\partial^2 \phi}{\partial X^{j} \partial X^{j}}[/tex]

    Are they separate variables? If not, then can one express it as being squared?
     
    Last edited by a moderator: May 9, 2012
  11. May 9, 2012 #10

    Mark44

    Staff: Mentor

    No, they're not separate, and yes, you can express that derivative as
    $$ \frac{\partial^2 \phi}{\partial X^{j2}}$$

    That looks pretty messy with the j index as a superscript. That's probably a good reason to write the indexes as subscripts, as X1, X2, and so on.

    Doing that, the partial would look like this:
    $$ \frac{\partial^2 \phi}{\partial X_j^2}$$
     
    Last edited: May 9, 2012
  12. May 9, 2012 #11
    Right, that's good then. Thank you, one last question. In respect to these [tex]X_{j}^{2}[/tex] 's, when can it not be applied to the Leibniz rule?

    This has caused a new confusion.

    [tex]\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial y}{\partial x}[/tex]

    especially this part [tex]\frac{\partial^2 y}{\partial x^2}[/tex] seems like an identical form of

    [tex]\frac{\partial^2 \phi}{\partial X^{2}_{j}}[/tex]

    Thank you
     
  13. May 9, 2012 #12

    Mark44

    Staff: Mentor

    He didn't say that. Nor did I. Here's what was said:
    $$ \frac{\partial^2 \phi}{\partial X^{\mu} \partial X^{\nu}} = \frac{\partial}{\partial X^{\mu}} \frac{\partial \phi} {\partial X^{\nu}}$$

    Notice on the right that it is NOT the 2nd partial of the partial - it's the partial of the partial. I removed an exponent of 2 that you had.
     
  14. May 9, 2012 #13
    Sorry, I carried that on by a paste.



    This has caused a new confusion.

    [tex]\frac{\partial^2 y}{\partial x^2} = \frac{\partial}{\partial x} (\frac{\partial y}{\partial x})[/tex]

    especially this part [tex]\frac{\partial^2 y}{\partial x^2}[/tex] seems like an identical form of

    [tex]\frac{\partial^2 \phi}{\partial X^{2}_{j}}[/tex]
     
  15. May 9, 2012 #14
    Now why can't the latter be expressed using Leibniz rule?
     
  16. May 9, 2012 #15

    Mark44

    Staff: Mentor

    As long as each partial is taken with respect to the same independent variable.
    Yes. Let's keep to the ordinary derivative notation, since it's easier to type.

    $$ \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2 y}{dx^2}$$

    Working with partial derivatives, it would look pretty much the same. Again, I'm assuming that you're taking the partial with respect to the same variable.
     
  17. May 9, 2012 #16
    You said it couldn't, even in the correct format. It has me a bit confused see...
     
  18. May 9, 2012 #17

    Mark44

    Staff: Mentor

    I don't understand what you're asking.
     
  19. May 9, 2012 #18
    You use some wicked latexing... I have never seen the latex you use... anyway, you qouted the expression I gave you and you said:

    This isn't quite right.

    [tex]\frac{\partial}{\partial X^j}(\frac{\partial \phi}{ \partial X^j})[/tex]

    Compare this to the related Leibniz notation for the second derivative.
    [tex]\frac{d^2 y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right)[/tex]
     
  20. May 9, 2012 #19

    Mark44

    Staff: Mentor

    Click any of the things I wrote and you can see how I did it.
    This is the same as

    $$\frac{\partial^2 \phi}{\partial X_j^2}$$

    But you can't collapse the expression if you're doing this:
    $$ \frac{\partial}{\partial X_j} \frac{\partial \phi}{\partial X_i}$$

    Here the partials are with respect to different variables.
     
  21. May 9, 2012 #20
    So, what you are saying is I need the paranthesis? That is the only difference I see...
     
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