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Lagrangian density

  1. Oct 5, 2005 #1
    Here's the problem. For a neutral vector field [tex]V_{\mu}[/tex] we have the Lagrangian density

    [tex]\mathcal{L} = -\frac{1}{2}(\partial_{\mu}V_{\nu})(\partial^{\mu}V^{\nu})+\frac{1}{2}(\partial_{\mu}V^{\mu})(\partial_{\nu}V^{\nu})+\frac{1}{2}m^2V_{\mu}V^{\mu}[/tex]

    We are then going to use the Euler-Lagrange equations to show that (for [tex]m\neq 0[/tex])

    [tex]\partial_{\mu}\partial^{\mu}V^{\nu}+m^2V^{\nu} = 0 \quad;\quad \partial^{\mu}V_{\mu} = 0[/tex]

    Now, the Euler-Lagrange equation (as I found in the textbook) is

    [tex]\frac{\partial \mathcal{L}}{\partial \varphi}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\varphi)} = 0[/tex]

    My problem here, with what at first glance would appear to be a rather simple problem, is that I'm confused by all the indices! I haven't a lot of experience working like this, in fact this is all new to me, so I don't quite know how or where to start.

    I can, however, give a specific example of what I don't understand, just to get things started.

    The first part seemed at first easy enough, as the Lagrangian only contains one part with [tex]V_{\mu}[/tex] which is


    and from what I've understood [tex]V_{mu}V^{\mu}[/tex] is just the square of each of the components of the vector, so that when you derivate it with respect to [tex]V_{\mu}[/tex], I thought you'd get something like


    but the text says [tex]m^2 V^{\nu}[/tex] which brings up two question, first of all, why is it [tex]\nu[/tex] and not [tex]\mu[/tex], and why is it an upper index, rather than a lower one?

    Furthermore I was wondering if anyone could recommend a good book introductory book about tensor algebra and such, preferably one intended for physicists rather than mathematicians. (It doesn't have to be a book on just about tensors, as long as it contains a good introduction to tensors.)
  2. jcsd
  3. Oct 5, 2005 #2


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    It matters whether you differentiate the scalar [itex] V_{\mu}V^{\mu} [/itex] wrt the covector [itex] V_{\nu} [/itex] or wrt the vector [itex] V^{\nu} [/itex]. That's why the indices must be treated with great care.

  4. May 16, 2010 #3


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    Hi Spinny,

    I worked my way through Schaum's Tensor Calculus which I found very good except for quite a few typos.


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