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

How to spot conserved quantities from Lagrangians

  1. May 19, 2013 #1
    Hi guys,

    The title pretty much says it. Say you have a very simple 3D Lagrangian:

    [itex]L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) - V[/itex]

    So How do you tell what is conserved from a generic potential?

    I know for example that if V = V(x,y,z) then the total linear momentum is not conserved, if V = V(x,y) then the z component of linear momentum is conserved...etc.

    I get stuck when angular momentum gets involved. For example, what is conserved if [itex]V = V(x^2+y^2, z) [/itex]...I need to know how to spot the conserved quantities generally. I know that the total energy is conserved unless there is explicit dependence on time, so dont worry about that one...im looking for conservation with respect to angular / linear momentum components.

    Can you guys help me out please? thanks!
  2. jcsd
  3. May 19, 2013 #2
    Well, it all boils down to symmetries (Noether's theorem), but there doesn't seem to be a good systematic way to find all symmetries of a given Lagrangian. In fact, some symmetries are really hard to find in a given coordinate system. Translational symmetries (associated with momentum conservation, as you point out) is often easy to spot in Cartesian coordinates (x,y,z), whereas rotational symmetries (associated with conservation of angular momentum) lends themselves to spherical coordinates.

    As an example, if you can write V=V(r)=V(x^2+y^2+z^2), there is full rotational symmetry and all components of angular momentum is conserved. If there is one axis, about which you have rotational symmetry, then there is an associated conserved component of the angular momentum (the component along that axis). That can be quite difficult to spot though, in the general case. I find that physical intuition is often a better method than just staring at the Lagrangian itself.

    Hope that was of some help.
  4. May 19, 2013 #3
    Howdy! :)

    The answer is The momenta conjugate to a cyclic coordinate is conserved.

    I'll explain what this means below.
    A coordinate is called cyclic or ignorable if it does not appear in the Lagrangian. The momenta conjugate to a cyclic coordinate is conserved. One has to be cautious about the meanings of these terms though. E.g. don’t confuse mechanical momentum with canonical, aka generalized, momentum. For example: Suppose you have a system described by a Lagrangian in which the x-coordinate is cyclic. Then the canonical momentum P which is conjugate to x is not necessarily p = mv. If there is a magnetic field present then P = mv + qA or P = p + qA where p = mv is the mechanical (spatial) momentum. In general P will be conserved while p is not.

    Of course what is conserved in one coordinate system may not be conserved in another coordinate system.

    What do you mean when you say "unless there is explicit dependence on time"? I’d like to make an observation before I move on regarding pet peeve of mine. A vector field F is said to be conservative if curl F = 0. If the field is conservative then it can be expressed as the gradient of a scalar function. However the scalar function might be an explicit function of time. So a vector field might be conservative in the sense that it’s the negative gradient of a scalar function V but it might not be conserved in the sense that the energy is a function of time.

    Absolutely. Since you know what a Lagrangian is then you must have some knowledge of analytical mechanics. The nice thing about analytical mechanics (i.e. Lagrangian and Hamiltonian mechanics) is that there is a formal procedure to determine what momenta are conserved.

    A few very nice texts on analytical mechanics that I’ve read, at least in part, are

    Classical Mechanics - Third Edition by Goldstein, Safko and Poole
    Analytical Mechanics - Fifth Edition by Fowles and Cassiday
    Analytical Mechanics with an Introduction to Dynamical Systems by Josef S. Torek
    The Variational Principles of Mechanics by Cornilius Lanczos
    Classical Dynamics of Particles and Systems by Marion and Thorton

    There’s also a Schaum’s outline on Lagrangian mechanics that you should get at the Library of a book store. I think there are some online too. I’ll see if I can find one later tonight if I remember.

    Basically if you have a potential function and you want to know if the momentum about a certain axis, let's call it the z-axis. is conserved you first apply a change in coordinates to a system in which the z-axis of your cylindrical coordinate system is the z-axis about which you wish to determine whether the angular momentum about that axis is conserved. If the polar angle is absent from the Lagrangian then the angular momentum corresponding to that angle is conserved.
  5. May 19, 2013 #4


    User Avatar
    Science Advisor

    If you have a set of generalized coordinates ##\{q^{1},...,q^{n}\}## on some open subset ##U\subseteq M## where ##M## is the configuration space, and there exists an ##i\in \{1,..,n\}## such that ##\frac{\partial L}{\partial q^{i}} = 0## then the associated generalized momentum ##p_{i} = \frac{\partial L}{\partial \dot{q}^{i}}## is a conserved quantity.

    The problem is that just because ##\frac{\partial L}{\partial q^{j}}\neq 0## for all ##j## doesn't mean that the system has no conserved quantities. It may just mean that you are using a coordinate system in which the symmetry is not apparent from the Lagrangian, in which case you would have to make an intuitive choice of a better coordinate system in which the symmetries have the potential to become more apparent. The problem is in actually finding such a coordinate system. All in all, I agree with Hypersphere that physical intuition works better in most cases.

    EDIT: By the way, there is a way of systematically finding symmetries for special lagrangians (kinetic energy lagrangians basically) which involve killing vector fields but this requires much more mathematics to develop and understand.
    Last edited: May 19, 2013
  6. May 20, 2013 #5
    I make the example for a simple one particle system.

    If V = V(x2+y2, z) you would choose r and z as generalized coordinates. To write the kinetic energy (in polar coordinates) we first write the cartesian coordinates as functions of r, a (= angle of rotation in the x y plane) and z and then derive them with respect to time.

    x(r,a,z) = r cos(a)
    y(r,a,z) = r sin(a)
    z(r,a,z) = z

    x' = dx/dt = r' cos(a) - r a' sin(a)
    y' = r' sin(a) + r a' cos(a)
    z' = z'

    L = (1/2) m (x'2 + y'2 + z'2 ) - V(r,z) = (computations) =

    = (1/2) m(r'2 + r2 a'2 + z'2) - V(r,z)

    Since L doesn't depend on the angle a, @L/@a = 0 so @L/@a' is a constant of motion:

    @L/@a' = m r2 a'2 = constant.
  7. May 21, 2013 #6
    @L/@a' = m r2 a' = constant.

    → angular momentum = constant.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook