How to spot conserved quantities from Lagrangians

In summary: You take the derivative of the potential with respect to x and then use the principle of extremum to find the extremum of the resulting function. This is a very general principle and it can be used to determine a lot of conserved quantities.
  • #1
Dixanadu
254
2
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 don't worry about that one...im looking for conservation with respect to angular / linear momentum components.

Can you guys help me out please? thanks!
 
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  • #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.
 
  • #3
Dixanadu said:
Hi guys,
Howdy! :)

Dixanadu said:
The title pretty much says it.
The answer is The momenta conjugate to a cyclic coordinate is conserved.

I'll explain what this means below.
Dixanadu said:
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?
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.

Dixanadu said:
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 don't worry about that one...
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.

Dixanadu said:
...im looking for conservation with respect to angular / linear momentum components.

Can you guys help me out please? thanks!
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.
 
  • #4
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.
 
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  • #5
Dixanadu said:
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 don't worry about that one...im looking for conservation with respect to angular / linear momentum components.
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.
 
  • #6
lightarrow said:
@L/@a' = m r2 a'2 = constant.

@L/@a' = m r2 a' = constant.

→ angular momentum = constant.
 

1. How can I identify conserved quantities from a Lagrangian?

To identify conserved quantities from a Lagrangian, you can use the Noether's theorem. This theorem states that for every continuous symmetry of a system, there exists a corresponding conserved quantity. By identifying the symmetries of the system and using the corresponding conserved quantities, you can determine the conserved quantities from the Lagrangian.

2. Can conserved quantities be derived from any Lagrangian?

Yes, conserved quantities can be derived from any Lagrangian. As long as the Lagrangian is invariant under a continuous symmetry, there will be a corresponding conserved quantity. However, not all symmetries of a system will result in conserved quantities.

3. How do I know if a conserved quantity is physically relevant?

To determine if a conserved quantity is physically relevant, you should check if it corresponds to a known physical quantity, such as energy, momentum, or angular momentum. If the conserved quantity has a physical interpretation, then it is considered relevant.

4. Can conserved quantities change over time?

No, conserved quantities cannot change over time. This is because they are associated with symmetries of a system, and these symmetries do not change over time. As a result, the conserved quantities remain constant throughout the evolution of the system.

5. How are conserved quantities used in physics?

Conserved quantities play a crucial role in physics. They allow us to make predictions about the behavior of a system, such as the conservation of energy, momentum, and angular momentum. They also help us understand the underlying symmetries of a system and can be used to simplify complex physical problems.

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