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A Action for a point particle

  1. Jun 7, 2017 #1
    The action for a relativistic point particle is proportional to the infinitesimal invariant length ds. Is there some more intuitive explanation for this?

    The action is defined by taking the time intergal over the difference between the kinetic energy and the potential energy. So how does this relate to the integral over the invariant length?
  2. jcsd
  3. Jun 7, 2017 #2


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    You mean: the Lagrangian in classical mechanics is KE - PE and the Lagrangian in Relativity (if there is no potential) is proper time or invariant length. Why?

    I'm not sure there is an intuitive answer. But, if you take away the potential in classical physics, you get motion at constant velocity in a straight line. And, in SR, constant velocity in a straight line maximises the invariant length. And, if you extend this principle to GR and curved spacetime it seems to generalise nicely.
  4. Jun 7, 2017 #3


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    I'd suggest thinking of the infinitesimal invariant length ds as infinitesimal proper time, so that the principle of extremal action becomes equivalent to the principle of extremal aging, i.e. that a particle moves in such a way as to extremize proper time.

    That's true in non-relativistic mechanics, it's not true in relativistic mechanics. My paraphrase of what Golstein has to say on the topic (which I've reviewed recently for a different thread) is that a relativistic Lagrangian is one that gives the proper relativistic equations of motion.

    Basically, L=T-V only works for non-relativistic mechanics, so it's just not adequate as a definition of the Lagrangian.
  5. Jun 8, 2017 #4


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    If you analyse the problem to find a single-particle Poincare invariant action, using the coordinate time as the parameter applying Noether's theorem you get (up to a multiplicative constant, ##A##)
    $$L=-A \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
    where the dot indicates the derivative wrt. coordinate time. Now
    $$L=-A \sqrt{c^2-\dot{\vec{x}}^2}=-A c \sqrt{1-\dot{\vec{x}}^2/c^2}.$$
    For ##|\dot{\vec{x}}| \ll c## you get
    $$L \simeq -A c \left (1-\frac{1}{2c^2} \dot{\vec{x}}^2 \right).$$
    Comparing to the non-relativistic case you have to set
    $$A=m c,$$
    because then the action gets
    $$L \simeq =-m c^2 + \frac{m}{2} \dot{\vec{x}}^2.$$
    The relativistic free-particle Lagrangian thus finally is
    $$L=-m c \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
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