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Silly question about units of Momentum Density from NG Action

  1. Feb 11, 2010 #1
    Ok, Im thinking that I should probably move this to the Coursework area. Im not sure how to do that or if I even can....

    My bad....

    Excuse the question but im stuck on this for the past hour.

    I studying Zweibach's "A first Course in String Theory". From the Nambu-Goto Action he derives the wave equation which consists of the term

    [tex] \frac{\partial \mathcal{L}}{\partial \dot{X} ^{\mu}} &= -\frac{T_{0}}{c}\frac{\dot{X} \cdot X')X'_{\mu}-(x')^{2}\dot{X}_{\mu}} {\sqrt{(\dot{X} \cdot X')^{2} - (\dot{X})^{2}(X')^{2}}} \equiv \mathcal{P}^{\tau}_{\mu} [/tex]

    Now this [tex] \mathcal{P}^{\tau}_{\mu} [/tex] is referred to as the momentum density and is used as such later on in the book:

    [tex] p_\mu(\tau) = \int ^{\sigma _{1}}_{0} \mathcal{P}^{\tau}_{\mu}(\tau , \sigma) d\sigma [/tex]

    Im confused because I would expect [tex] \mathcal{P}^{\tau}_{\mu} [/tex] to have units of Mass divided by time, if \sigma has units of length, or units of momentum, if \sigma is dimensionless, so that when you integrate over \sigma you get units of momentum.

    But as far as I can see [tex] \mathcal{P}^{\tau}_{\mu} [/tex] has units of mass times length. I get this becuase the lagrangian density has units of ML^2/T and it is differentiated with respect to velocity, thus it takes units of ML.

    Can someone help with this?


    Last edited: Feb 11, 2010
  2. jcsd
  3. Feb 12, 2010 #2
    Correct me if I'm wrong, but since you're not doing functional differentiation, I expect your L to be the Lagranian density in one dimension. Since a string is one-dimensional, this is

    energy/length = (ML^2/T^2)/L = ML/T^2

    The differentiated L on your left (momentum density) side would then have dimension

    (ML/T^2) / (L/T) = M/T

    Total momentum, integrated along the string length would have dimension

    (M/T)*L = M*L/T

    which seems right, knowing that momentum can be expressed as mass*velocity.

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