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Stupid differential algebra problem

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data

    For the life of me, I can't decide whether the method below is correct and it's making me feel really stupid.

    2. Relevant equations

    Simple matter of an algebraic identity. Is:

    [tex]\frac{\partial L}{\partial(\frac{d\overline{A}}{dx_{\mu}})} = \frac{\partial L}{\partial (\frac{dA_{x}}{dx}})} + \frac{\partial L}{\partial (\frac{dA_{y}}{dx}})} + \frac{\partial L}{\partial (\frac{dA_{z}}{dx}})} + \frac{\partial L}{\partial (\frac{dA_{x}}{dy}})} + ...[/tex]

    ... as a decomposition of a vector derivative, correct?

    3. The attempt at a solution

    Not really a problem as such, just confirming whether my algebra is correct. It's the lagrangian formulation of maxwell's equations, if anyone's interested.

    edit: sodding TeX. I can't get those right brackets to format correctly. You get the idea.
  2. jcsd
  3. Mar 3, 2008 #2
    Nobody? I can't see this being a difficult question for those in the know; it's just that I can't for the life of me figure out whether or not this is valid.
  4. Mar 4, 2008 #3
    If you mean: [tex]\frac{\partial L}{\partial(\partial_\mu A_\nu)}[/tex]?

    Then there is no summation to be done -- you end up with a 2nd rank tensor.

    If you mean: [tex]\frac{\partial L}{\partial(\partial_\mu A^\mu)}[/tex]?

    Then you would sum over the mu's, and get 4 terms (for 1 time and 3 space).
  5. Mar 4, 2008 #4
    I don't understand the difference. The problem I'm solving uses precisely the notation I gave, apart from the brackets being in the wrong place. I don't know where this stuff about tensors comes in; it's not in the question nor in the method I'm trying to use from class. This lecturer in particular is notoriously sloppy with his notation and terminology so it's probable that I haven't seen the Euler-Lagrange equation expressed properly.
  6. Mar 5, 2008 #5
    The rules of index manipulations is that you only have a sum when you have repeated indices. In the expression you wrote down, there is only one [tex]\mu[/tex] -- so there is no summation to be done.
  7. Mar 6, 2008 #6
    Sorry for being so dense - it's partly me not knowing my mathematics rigorously enough and partly my lecturers not being bothered.

    As far as I understand the problem, what I'm trying to do is differentiate a lagrangian [tex]L[/tex] with respect to four spacetime derivatives - d/dxt, d/dx, d/dy, d/dz - of the three-vector [tex]\overline{A}[/tex].

    As I understand your response - what you're saying is that the way I've expressed the left-hand side, it comes out as a tensor, is that right? If so, I'm stuffed, since I'm not entirely sure what a tensor is. In any case - this wasn't the intention. If it's coming out that way, I haven't expressed the question properly since I know there should be multiple terms. Given that I don't really know what I'm talking about, I'm not sure where to go from here since I don't know how to ask the question correctly.
    Last edited: Mar 6, 2008
  8. Mar 7, 2008 #7
    Mandl & Shaw's QFT expresses the relationship I'm after more correctly as:

    [tex]\frac{\partial L}{\partial \phi_{r}} - \frac{\partial}{\partial x^{\alpha}}(\frac{\partial L}{\partial \phi _{r, \alpha}})= 0[/tex]

    In this case, I believe I'm beginning to understand. The r is the field index; since I'm only dealing with one field this is inconsequential. The alpha is presuambly then the 4 coordinates xt, x, y, z and (r,a) implies a sum of:

    [tex]\frac{\partial}{\partial x}(\frac{\partial L}{\partial \frac{\partial \phi}{\partial x}}) + \frac{\partial}{\partial y}(\frac{\partial L}{\partial \frac{\partial \phi}{\partial y}}) + ...[/tex]

    Is that correct? I do hope so, because i) it greatly simplifies my working and ii) means I understand it.

    I'm blaming this problem in advance on bad notation.
  9. Mar 9, 2008 #8
    I'm going to repeat the question since I still haven't been able to find a satisfactory answer:

    The best expression of the Euler-Lagrange equation I've been able to find so far is [tex]\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - \frac{\partial \mathcal{L}}{\partial \psi} = 0[/tex] where [tex]\partial_\mu = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)[/tex].

    Now: When you take [tex]\psi[/tex] to be the 3-vector [tex]\overline{A}=A_{x}\overline{e}_{x}+A_{y}\overline{e}_{y}+A_{z}\overline{e}_{z}[/tex], the term [tex]\partial_\mu \overline{A}[/tex] is at the very least equal to (for example) [tex]\frac{\partial \overline{A}}{\partial x} = \hat{i}\frac{\partial A_{x}}{\partial x} + \hat{j}\frac{\partial A_{y}}{\partial x} + \hat{k}\frac{\partial A_{z}}{\partial x}[/tex]

    What I am asking is:

    Regardless of what one assumes or does not assume about the form of [tex]A_{y}[/tex] and [tex]A_{z}[/tex], you now have a sum of component derivatives on the bottom of the first equation, inside the partial d. Does the whole term seperate out into partial derivatives of L with respect to each component derivative of A, like so:

    [tex]\frac{\partial}{\partial x}(\frac{\partial L}{\partial(\frac{\partial \overline{A}}{\partial x})}) = \frac{\partial}{\partial x}(\frac{\partial L}{\partial(\frac{\partial A_{x}}{\partial x})}) + \frac{\partial}{\partial x}(\frac{\partial L}{\partial(\frac{\partial A_{y}}{\partial x})}) + \frac{\partial}{\partial x}(\frac{\partial L}{\partial(\frac{\partial A_{z}}{\partial x})})[/tex]

    ..or not? If not, why not?
  10. Mar 9, 2008 #9


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    In [tex]\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - \frac{\partial \mathcal{L}}{\partial \psi} = 0[/tex]

    You can't treat independently the two indices! If [itex]\mu [/itex] is 1, it must be one at both places so you can never have
    [tex] \partial_x \frac{\partial L}{\partial \vec{A}} [/tex]
  11. Mar 9, 2008 #10


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    I am confused. If you are working with spacetime derivatives you should not have a three-vector, you should have a four-vector [itex] A_\mu [/itex]. So, are you working with a relativistic notation or are you working with a nonrelativistic formulation?
  12. Mar 9, 2008 #11
    It's a QFT problem so it's four-dimensional, but the four-vector [tex]\tilde{A}[/tex] is decomposed into [tex]\overline{A} + A_{t}\overline{e}_{0}[/tex]. The Lagrangian formulation of electromagnetism in the method I've been given is to apply the euler-lagrange equation to each of these components seperately.

    Edit to answer another issue:

    Your last statement in post #9 refers to a term I've never written. I'm not sure what you're saying there, since it's not part of the question.
    Last edited: Mar 9, 2008
  13. Mar 9, 2008 #12


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    Then what do you mean by
    [tex]\frac{\partial}{\partial x}(\frac{\partial L}{\partial(\frac{\partial \overline{A}}{\partial x})}) [/tex] ???????????

    What does this overline symbol means????
  14. Mar 9, 2008 #13


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    Yes, this is correct.
  15. Mar 9, 2008 #14
    The bar is 3-vector notation. Are you familiar with this method? The field A and the derivatives of the field are treated as being independent of one another, so the term you quoted from me and the term you equated it to do not infer one another.

    RE Post #13: The piece you quoted from me is the exact same equation as the one you're telling me is incorrect, simply substitute [tex]\overline{A}[/tex] for [tex]\phi[/tex] and you have precisely the decomposition I'm referring to in post #8.
  16. Mar 9, 2008 #15


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    No, I am not familiar with that notation. It seems very confusing to me. The A appearing in the E-L should have an index [itex] \nu [/itex] or an index "i"! What does "a bar indicates three-vector notation" mean? What is the quantity [itex] \overline{A}[/itex]?

    I will simply say this and hope that this will be helpful..

    The E-L applied to electromagnetism is

    [tex] \partial_\mu \bigl( \frac{\partial L}{\partial(\partial_\mu A_\nu)} \bigr) - \frac{\partial L}{\partial A_\nu} = 0 [/tex]

    This is a set of four equations, one for each value of [itex] \nu [/itex].
    For each value of nu, the index mu is summed over (Einstein's summation convention).
    So for nu=0, you get

    [tex] \partial_0 \bigl( \frac{\partial L}{\partial(\partial_0 A_0 )} \bigr) + \partial_1 \bigl( \frac{\partial L}{\partial(\partial_1 A_0 )} \bigr) + \partial_2 \bigl( \frac{\partial L}{\partial(\partial_2 A_0 )} \bigr) + \partial_3 \bigl( \frac{\partial L}{\partial(\partial_3 A_0 )} \bigr) - \frac{\partial L}{\partial A_0} = 0 [/tex]

    And then you repeat with nu=1,2,3. That's all there is to it.
  17. Mar 9, 2008 #16


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    For phi you used an index "r" [itex] \phi_r [/itex] which makes sense to me. For A you used an overline which is not clearly defined. Why not use a label too?
  18. Mar 9, 2008 #17
    The quoted identity is for a general method comprising r fields - the case I'm applying it to is the one where the fields are A (r=1) and At (r=2). For r=1 the field is a three-component vector, and there is no escaping that.
  19. Mar 9, 2008 #18


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    Ok but then the "r=1" field contains really three components, there is no way escaping that either. So It seems to me that it would make more sense to folow the usual notation of having an equation for A_0 and an equation for A_i where it would be understood that i=1,2,3. It would seem to me much clearer to write the equation for the three-vector part as

    [tex] \partial_\mu \bigl(\frac{\partial L}{\partial (\partial_\mu A_i)}\bigr) - \frac{\partial L}{\partial A_i} = 0 [/tex]
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