Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Many particle SR lagrangian

  1. May 6, 2014 #1
    Hello I was reading something the other day and wondered what a two-particle lagrangian would look like in SR. I'm not exactly sure what lorentz scalar we can write down for the two particles.
  2. jcsd
  3. May 6, 2014 #2


    User Avatar
    Science Advisor

    For a single particle, we can write the action integral in either of two ways:

    I = ∫ L dτ where τ is the particle's proper time, and in this case L is a Lorentz invariant.

    Otherwise, in terms of the coordinate time t,

    I = ∫L' dt

    In the latter case L' is not Lorentz invariant. For a point particle with mass m, an expression that gives the right equations of motion is

    L' = mc2√(1 - β2) - V

    where V(x) is an external potential. See for example the chapter in Goldstein.

    For N particles this form can be easily generalized:

    L' = ∑ mic2√(1 - βi2) - V

    where V(x1, x2, ...) is the interaction potential.
  4. May 6, 2014 #3
    Doesn't the potential break the lorentz invariance?
    Also once we plug the free terms back into the integral, we have an integration over two seperate proper times.
    All this I find very disturbing.
  5. May 6, 2014 #4


    User Avatar
    Science Advisor

    Most problems with an interaction potential are not Lorentz invariant.
    For example V = 1/|x1 - x2|
    :confused: There's just the one integral over coordinate time, which is why I didn't use the first form that involves proper time.
  6. May 6, 2014 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    And this is why we turn to field theory in order to describe interacting particles.

    As you have noticed, the simple approach with a potential and only one integral over a global time does not work because it has to be Lorentz invariant. However, by letting the particle interactions be carried by a field such as the electromagnetic one, it is possible to regain a theory which has the same form in all frames.

    If you do not want to do this while keeping your theory invariant, you can only have point interaction terms between the particles (i.e., things like delta(x1(t)-x2(t)) times some prefactor to make the thing transform the correct way).
  7. May 6, 2014 #6
    I'm looking for strictly lorentz invariant lagrangians.
    Well what exactly do we integrate over with multiple particles?
    Do we simply have separate integrals integrating with respect to the different proper times?
    This seems to break the formalism where we have the whole system under one integral sign.
    Is this perhaps a hint to the breakdown of particles in relavistic physics?
  8. May 6, 2014 #7


    User Avatar
    Science Advisor

    There is just one integral, I = ∫L' dt where L' = ∑ mic2√(1 - βi2) - V

    As Orodruin says, the more usual formulation is in terms of field theory.

    Here's a Wikipedia page that might be close to what you want, describing a Lorentz invariant formulation for the interaction of two Dirac particles.
  9. May 6, 2014 #8


    User Avatar
    Gold Member

    I am not able to understand well the question... but the action for N particles interacting through the EM field is written as:
    [itex] S= -\sum_{i=1}^{N} \int dτ_{i} (-m_{i}c \sqrt{n_{ab} \dot{x}_{i}^{a}(τ_{i})\dot{x}_{i}^{b}(τ_{i})} + q_{i} A_{a}(x_{i}(τ_{i}))\dot{x}^{a}_{i}(τ_{i})) -\frac{1}{16 \pi} \int d^{4}x F_{ab}(x)F^{ab}(x) [/itex]
    where [itex]x_{i}[/itex] is the curves of each particle i (with mass [itex]m_{i}[/itex] and charge [itex]q_{i}[/itex], parametrized by arbitrary parameter [itex]τ_{i}[/itex] each.... The [itex]A_{a}[/itex] is the electromagnetic potential and [itex]F_{ab}= \partial_{a} A_{b} - \partial_{b} A_{a}[/itex] the EM field strength tensor (or in SR they call it the antisymmetric EM tensor)...

    This action is Lorentz Invariant....The form of the Lagrangian in fact, isn't in general of physical significance... what's important is the equations of motion...
    In the case of such an action, the equations of motion give you the same result as having particles interacting with the EM field in a lorentz covariant form.... (you get the inhomog. Maxwell equations as well as the Lorentz force law)
    Last edited: May 6, 2014
  10. May 6, 2014 #9
    I understand this, however what disturbs me is that we cannot write down a lagrangian without breaking lorentz symmetry. (The fact that different particles are integrated with respect to different proper times)
    Perhaps that's irrelevant.
    Last edited: May 6, 2014
  11. May 6, 2014 #10


    User Avatar
    Gold Member

    the thing you integrate is the "lagrangian"....
    For a free particle the lagrangian is just the:
    [itex]S=-mc \int ds= -mc \int dt \frac{ds}{dt}= \int dt L[/itex]
    which is the velocity on a curve
    that's why the term in the square root appears...
  12. May 6, 2014 #11


    User Avatar
    Science Advisor
    Gold Member

    What do you mean by this? The action Chrisver posted is a Lorentz invariant... and so is the Lagrangian.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook