Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Connection between cross product and partial derivative

  1. Jan 25, 2010 #1
    Hi. I have been looking at differential forms, and that inspired me to consider a partial derivative as a ratio between cross products. Please tell me if the following makes sense. Say we have cartesian coordinates (x,y) and polar coordinates ([itex]\rho[/itex], [itex]\phi[/itex]). I want to calculate [itex]\left(\frac{\partial x}{\partial y}\right)_{\rho}[/itex], i.e. the partial derivative of x wrt. y with [itex]\rho[/itex] constant. I do it as follows

    \left(\frac{\partial x}{\partial y}\right)_{\rho}=\left|\frac{\hat x\times\hat\rho}{\hat y\times\hat\rho}\right|=-\frac{\sin\phi}{\cos\phi}

    Is this OK ? I've never encountered it before except in differential forms where I have seen partial derivatives written as wedge products.

    Edit: This is the article I have been reading: http://www.av8n.com/physics/partial-derivative.htm#sec-freex
    Last edited: Jan 25, 2010
  2. jcsd
  3. Jan 26, 2010 #2
    Hi there!

    I suppose what you've been thinking about isn't quite correct:

    here are some arguments:

    1. In the precise definition of partial derivative there's no ratio present, it's the differential quotient under a certain limit process. That's why the expression [tex]\frac{\partial x}{\partial y}[/tex] cannot be interpreted (as far as I know in no mathematical formulation) as a fraction. (better write it down as [tex]\frac{\partial }{\partial y} x[/tex] where the operator properties are more evident.)

    What you really can interpret as a ratio is the total derivative: [tex]\frac{d}{dy}x=\frac{dx}{dy}[/tex], since dx or dy can be thought of as differential 1-forms, or total derivatives.

    2. The basic cross product is not defined for 2 dimensions. Any definition of it on R^2 would involve a determinat-like expression, which is directly connected to the wedge product. Therefore, you can't just do polar coordinates (2D) using a cross product on the vectors - you have to expand (embed) them in R^3, which will then give you the familair expression using the wedge.

    Check once again the definition of the wedge product, the mathematical one!, cuz physicists do lots of lousy things sometimes.

    PS: it's not an easy topic, but once you've caught the main idea, it would be extremely helpful for you, since differential forms is the modern language of vector analysis and you'll need it everywhere in physics.
    Last edited: Jan 26, 2010
  4. Jan 26, 2010 #3
    If I have a function f(x,y) why can't I write

    \frac{\partial f}{\partial x}=\frac{df\wedge dy}{dx\wedge dy}


    This makes sense to me, because

    df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy
    and so
    df\wedge dy=\frac{\partial f}{\partial x}dx\wedge dy[/tex]

    and inserting this into the fraction you get the right result.
  5. Jan 26, 2010 #4
    Ok, the last two equations are both correct.

    The point is, the differential form (1- or 2-form) is not defined to be [tex]dx\wedge dy[/tex] but the whole expression [tex]\frac{\partial f}{\partial x}dx\wedge dy[/tex]. There's NO multiplication dot between the partial derivative and dx^dy. You have to look at it as at a different mathematical object.

    e.g: If you read advanced theory of differential forms, you'll encounter Stokes' thm. in the following way:

    "Let [tex]\omega[/tex] be a differential k-form and G be a closed region and let [tex]\partial G[/tex] denote the boundary of G. Then we have:

    [tex]\displaystyle{\int_{\partial G}}\omega=\displaystyle{\int_{G}}d\omega[/tex]"

    I'm giving this example to show you that there's no 'differential' after omega under the integral sign, as normal. This is because the dx or dV=dx^dy^dz is an inseparable part of the differential form omega by definition.

    And this is why I think your expression is not well-defined.
  6. Jan 26, 2010 #5

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    This article is using the wedge product symbol in a very non-standard way. Differential k-forms are not in general a division algebra and it does not make sense to write expressions like

    [tex]\frac{a \wedge b}{c \wedge d}[/tex]

    What the author should have written instead are Jacobian determinants, which look like this:

    [tex]\frac{\partial (x, y)}{\partial (z, w)} = \det \left| \begin{array}{cc} \frac{\partial x}{\partial z} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial z} & \frac{\partial y}{\partial w} \end{array} \right|[/tex]

    And in general, yes, one can write

    [tex]\left( \frac{\partial A}{\partial B} \right)_C = \frac{\partial (A, C)}{\partial (B, C)}[/tex]

    Jacobians also have useful properties, such as

    [tex]\frac{\partial (x, y)}{\partial (z, w)} \frac{\partial (A, B)}{\partial (C, D)} = \frac{\partial (x, y)}{\partial (C, D)} \frac{\partial (A, B)}{\partial (z, w)}[/tex]

    which can be used to derive various thermodynamics identities.

    Yes, there absolutely IS a multiplication. dx^dy is a basis vector on the space [itex]\Lambda^2 (V^*)[/itex]; the expression [itex]\alpha \, dx \wedge dy[/itex] is simply multiplying the basis element dx^dy by the scalar [itex]\alpha[/itex].

    The problem is that you can't divide by a vector; i.e., an expression like

    [tex]\frac{1}{dx \wedge dy}[/tex]

    is meaningless.
  7. Jan 26, 2010 #6
    I must admit, that's absolutely true! (otherwise even the 1-dimensional substitution rule would be ill-defined)

    The expansion in basis with some coefficients is actually a very good explanation.

    Sorry, for misleading you for a moment,

  8. Jan 27, 2010 #7
    Isn't [itex]dx\wedge dy[/itex] a function of two vectors that returns a scalar? So why is it wrong to divide by it, if it's just a function? Also, isn't the wedge product already defined as a determinant? I.e.
    \omega\wedge\nu (v_1,v_2)=\det \left| \begin{array}{cc} \omega(v_1) & \nu(v_1) \\ \omega(v_2) & \nu(v_2) \end{array} \right|

    (in 2 dimensions).
  9. Jan 27, 2010 #8

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    That is only true AFTER you plug vectors into the k-form. Before you do that, the k-form is an object which exists in a vector space, and is not part of a division algebra.

    Also, the determinant you have there is not a Jacobian determinant. Notice that it does not involve any partial derivatives; it is merely a bunch of functions.

    Also note that you don't HAVE to fill all the slots in a k-form. You can, for example, fill one slot of a 3-form; the result will be a 2-form. This is called the "interior product".
  10. Jan 27, 2010 #9
    Isn't there a close connection between differential forms and jacobi determinants? Say you want to find the area of a circle by integration, so you parameterize it using [itex]x=r\cos\phi[/itex] and [itex]y=r\sin\phi[/itex] and your integral is [itex] \int\int_{\mathcal C} dx\wedge dy[/itex]

    Now [itex]dx=\cos(\phi) dr-r\sin(\phi)d\phi[/itex] and [itex]dy=\sin(\phi) dr+r\cos(\phi)d\phi[/itex] so [itex] dx\wedge dy = rdr\wedge d\phi[/itex] and the result of the integral is [itex]\int_0^R\int_0^{2\pi}rdrd\phi=\pi R^2[/itex]. That is, we found the jacobian [itex] r [/itex] of the parametrization by using the forms. So the 2-form must be closely related to a jacobian.

    I don't know too much about this.. Is there some book you can recommend? Something that's not too hard...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook