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Step from Mass Point Mechanics to Field Theory

  1. Aug 29, 2014 #1
    At the moment I am trying to understand classical field theory and there's a conceptual problem I encountered, which bothers me a lot and I don't seem to be able to resolve the issue. When making the step to classical field theory, many texts start as follows:
    First they recall the/a action in mass point mechanics and write something like this
    $$S = \int L \, d t \, ,$$
    then they say that we don't want to look at particles anymore but at (say scalar) fields ##\phi## and introduce the Lagrangian density ##\mathcal L## as a function of the "field variables" ##\phi## and its derivatives, say ##\partial_t \phi, \partial_x \phi## as follows
    $$L = \int_{- \infty}^{\infty} \mathcal L \, d x \, . $$
    Now here's my problem: This equation is rubbish, plain mathematical nonsense.
    As Ben Niehoff already clarified in a similiar question I asked some time ago ( https://www.physicsforums.com/showthread.php?t=751858 ),
    there's an action for curves ##\gamma## and an action for fields ##\phi## and the two are, at least on a mathematical level, two seperate concepts. If we want to be a bit more rigorous, we thus write the mass point action like
    $$S(\gamma, \tau_1, \tau_2) = \int_{\tau_1}^{\tau_2} L ( \gamma, \dot \gamma) \, d \tau$$
    and the field action of a (real) scalar field ##\phi## in analogy
    $$\mathcal S (\phi, \phi_\text{boundary}) = \int_{\text{interior spacetime}} \mathcal L(\phi, d \phi)$$
    where ##\mathcal L(\phi, d \phi)## is now an actual density on the spacetime ( http://en.wikipedia.org/wiki/Density_on_a_manifold ).
    This clarifies why the equation I was talking about is nonsense: ##L## and ##\mathcal L## have two distinct inputs, i.e. live on different spaces and the equation is not invariant under coordinate change.
    So my questions are:
    1) What is the correct step/transition from Mass Point Mechanics to Field Theory?
    2) Why do we only consider first order derivatives of the field in the Lagrangian density?
    3) How do I get to the Hamiltonian formalism in the field theory in a coordinate independent manner?
     
  2. jcsd
  3. Sep 4, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Sep 4, 2014 #3
    I think my problem stems from the fact that I did not distinguish between the Newtonian and relativistic case carefully enough as I thought they could be cast in a unified coordinate-independent language, but this only works for point particles.

    Let us consider the example in pp. 12 of this document, where they explain the step to field theory by considering a homogenous string in one Euclidean dimension, modelling it as a collection of equal point masses connected by equal massless strings and then letting the distance between the masses go to zero while keeping the mass density and force strengths fixed.
    In the discrete case they get
    $$\ddot \phi_i + \frac{k}{m} (2 \phi_i - \phi_{i+1} - \phi_{i-1})= 0 \, ,$$
    where the ##\phi_i##s give the value of the displacement of the ##i##th point mass of mass ##m## from the equilibrium position and ##k## is the spring constant. In the continuum case, they get
    $$\ddot \phi_i - \frac{Y}{\mu} \frac{\partial^2 \phi}{\partial x ^2}= 0$$
    where ##\mu## and ##Y## are the continuum analogues of ##k## and ##m##. How they are obtained is explained in the document.

    Now, in the Newtonian case it makes sense to consider an "external time parameter" ##\tau## and then the last equation can be rewritten into a coordinate independent form, also suggesting an obvious generalization to ##3## or ##n## dimensions:
    $$\ddot \phi - \frac{Y}{\mu} \Delta \phi = 0 \, .$$
    The dot denotes, as usual, time derivatives and the ##\Delta## is the Laplace-Beltrami operator for the Riemannian manifold ##(\mathbb R ^n, \delta = \delta_{ij} \, d x^i \otimes d x^j)##, i.e. the ordinary Laplacian. The bottom line is that this is a perfectly legitimate equation and that we can get it out of an action of the form
    $$\mathcal S ( \phi, \phi_{\text{boundary}}(\tau_0), \phi_{\text{boundary}}(\tau_1), \tau_0, \tau_1 ) =\int_{\tau_0}^{\tau_1} \int_{\text{interior space}} \mathcal L (\phi, d \phi, \dot \phi,\tau) \, d \tau \, , $$
    where ##\mathcal L## is a density on ##(\mathbb R ^n, \delta)##.

    In the relativistic case, it is not meaningful to introduce ##\tau##, that is the proper time, for fields, making the above example useless in this case. So how does one get around that?
     
    Last edited: Sep 4, 2014
  5. Sep 5, 2014 #4
    Is this the source of the problem of time?
     
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