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Symmetry and conservation law

  1. Nov 23, 2012 #1
    This is my first post, so hello everybody. I don't have university background and english is not my native language, so please forgive me if what I'm writing is hard to understand sometimes. I'll do my best to be clear.
    I've always loved physics in general, but recently came to conclusion that in order to really understand it I have to get into details and learn the formalism. So, I started with classical mechanics (where else could I start?). I'm making my way through calculus of variations and Lagrangian formulation, which has been great experience for me so far. However, when trying to get my head around the conservation laws and how they arise from symmetries I got stucked at one point.
    Let me focus on conservation of (linear) momentum, which is the consequence of the homogeneity of space. This is how I understand it. Our real-life experience is that no part of the physical space is special, meaning that the laws of physics (for isolated system) are everywhere the same. We can move the system one meter to the right, start it with the same boundary conditions (positions and velocities) and it will perform the same motion. This makes sense.
    What follows is the tricky part. All papers deriving the conservation law from homogeneity of space say more or less the following:
    "Because space is homogenous the (infinitesimal) translation of coordinates will not change the Lagrangian."
    The problem is that I can't convince myself this is necessarilly the case! To me the only conclusion from homogeneity of space is that:
    "The new Lagrangian (being the result of translation) is such, that the equations of motion (expressed in new coordinates) are the same as the original ones"
    In other words, the translated system trajectory is the same as that of the system in its original location.
    These two statements don't seem to be equivalent to me. Of course, if Lagrangian is unchanged the trajectory won't change either. However, the opposite is not true. It seems to me that I can easily come up with different Lagrangian which results in the same equations of motion. For example, one can add a term which is time derivative of a function (of positions and velocities) to the Lagrangian:

    [itex]\mathcal{L}^{'} = \mathcal{L} + \frac{d}{dt} F(q,\dot{q})[/itex]

    The change in the action over the trajectory from such Lagrangian change is:

    [itex]\delta{S} = \delta \int_{t1}^{t2} \mathcal{L}dt = \int_{t1}^{t2} \frac{d}{dt} F(q,\dot{q}) dt = F(q,\dot{q}) \bigg|_{t1}^{t2}[/itex]​

    Which is constant and only depends on the endpoints (and not the trajectory itself). So, the new trajectory with such new Lagrangian will be the same as the original (because it minimizes the action - adding a constant does not change the condition for stationarity).
    So, the question is, isn't the conclusion (arising from homogeneity of space) that the Lagrangian does not change under translation too strong? All we observe in nature is that the motion of the translated system is the same. However, we can get the same motion with different Lagrangian! Where do I make mistake?
  2. jcsd
  3. Nov 23, 2012 #2


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    You lost the [itex]\delta[/itex].

    [tex]\delta S' = \delta S + \delta F(q,\dot{q})\bigg |_{t_1}^{t_2}[/tex]

    And [itex]\delta F[/itex] isn't necessarily trivial under variations of trajectory. At very least, [itex]\dot{q}(t_2)[/itex] is not fixed by boundary conditions.

    But anyways, the statement "Lagrangian is invariant under translation," means that if you shift the Lagrangian itself and the solution, the shifted solution is a solution to shifted Lagrangian. In other words, yes, equations of motion remain the same.

    This is all made far more explicit if you look at where it comes from, namely, Noether's Theorem.

    Edit: Scratch that. I should be more careful when I answer questions this late. You don't shift the Lagrangian. Just the variables. For example, if you have two masses on a spring, the potential energy depends only on difference of positions. So if you shift both coordinate variables by same amount, the total Lagrangian remains exactly the same. In contrast, if you have an mgy term for gravitational potential, shifting y results in change in Lagrangian. But that makes sense. Vertical momentum is not conserved, because we are ignoring momentum transfer to Earth when we write potential this way.
    Last edited: Nov 23, 2012
  4. Nov 23, 2012 #3
    Thanks for your answer, K^2!
    Now that I'm looking at my post again I realize I didn't formulate it properly. Speaking about time derivative of function F, which, when added to Langrangian, does not change the equations of motion, I should have written
    [itex]F(q,t)[/itex] ​
    instead of
    [itex]F(q,\dot{q})[/itex] ​
    In such case
    [itex]\frac{d}{dt}F(q,t)[/itex] ​
    will vanish when variating the Lagrangian (because time instants and position of the both ends of trajectory are fixed).
    In such case the endpoint velocity does not matter. Each Lagrangian constructed this way is equivalent (results in the same equations of motion and trajectory). When deriving the conservation laws, why do we assume that translation does not change the Lagrangian at all, then? I intuitively feel that this might have something to do with the freedom of choosing the inertial frame, but my intuition is not trained yet, so just a guess.
  5. Nov 23, 2012 #4


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    Like I said in the edit, it's the translation of coordinates that doesn't change the Lagrangian. I suppose, a concrete example would be better. Say, I have two masses, m1 and m2 at locations x1 and x2 connected by harmonic potential.

    [tex]L(x_1,x_2,\dot{x}_1,\dot{x_2}) = \frac{1}{2}(m_1 \dot{x}_1^2+m_2 \dot{x}_2^2) - \frac{k}{2}(x_2 - x_1)^2[/tex]

    Now, let's say I introduced new coordinates, x'1=x1+c, x'2=x2+c.

    [tex]L(x'_1,x'_2,\dot{x}'_1,\dot{x}'_2) = \frac{1}{2}(m_1 \dot{x}\prime_1^2+m_2 \dot{x}\prime_2^2) - \frac{k}{2}(x'_2 - x'_1)^2 = \frac{1}{2}(m_1 \dot{x}_1^2+m_2 \dot{x}_2^2) - \frac{k}{2}(x_2 + c - x_1 - c)^2 = L(x_1,x_2,\dot{x}_1,\dot{x_2})[/tex]

    So the Lagrangian is exactly the same after coordinate got shifted. Not just the total action, but the actual value of Lagrangian at every point.
  6. Nov 23, 2012 #5
    I fully agree with your example. The Lagrangian is valid, it describes isolated system (no explicit time dependence - potential only depends on distance between system components and not the absolute positions). In such case it's obvious that translation will not change the Lagrangian.
    Your example made me aware of one important thing, though (which, in fact, addresses my doubts)! Even though we have certain freedom in constructing the Lagrangian (we can always add total time derivative of function of positions and time to the Lagrangian and end up with the same trajectory), we DO WANT to construct the Lagrangian for isolated system in its simplest form, e.g. independent (explicitly) from time and absolute positions, which is probably equivalent to saying we construct it in an inertial frame of reference.

    To show an example of the systems I was thinking about consider:
    [itex]\mathcal{L} = \frac{1}{2}m \dot{x}^2 + \dot{x}x[/itex]​
    It can be shown that even though the translation changes the Lagrangian, it does not alter the equations of motion - original trajectory after translation is still valid (minimizes the action), because the change in the action due to translation is (by [itex]\delta S[/itex] I mean how much the action differs between translated and original system for the same trajectory, assuming the translation is small):
    [itex]\delta \mathcal{S} = \int_{t1}^{t2} \delta \mathcal{L} dt = \int_{t1}^{t2} \bigg( \frac{d \mathcal{L}}{dx} \delta x + \frac{d \mathcal{L}}{d \dot{x}} \delta \dot{x} \bigg) dt[/itex]​
    Now, since [itex]\delta \dot{x}[/itex] is 0 the second term in integral is zero, too, so we can write:
    [itex]\delta \mathcal{S} = \delta x \int_{t1}^{t2} \frac{d \mathcal{L}}{dx} dt = \delta x \int_{t1}^{t2} \dot{x} dt = \delta x \int_{t1}^{t2} \frac{dx}{dt} dt = \delta x \bigg(x \bigg|_{t1}^{t2} \bigg)[/itex]​
    This is constant and will vanish when variating the trajectory.
    So, the Lagrangian has different form after translation, yet it yields the same equations of motion. However, as I just realized, it does not describe isolated system (at least not in an inertial frame), due to explicit dependence on position in the second term. It might be equivalent to some isolated system expressed in non-inertial frame (I'm not sure). Anyway - it doesn't matter. You helped me understand this. Now I think I know why we say that translation does not change the Lagrangian when deriving conservation laws. It's because the original Lagrangian does not depend explicitly on positions and time. Thanks!
  7. Nov 23, 2012 #6
    If lagrangian does not depend on time explicitly,then energy is conserved.If it does not on position,momentum is conserved.These things rather arises directly from euler-lagrange eqn.
  8. Nov 23, 2012 #7
    Yes, exactly. I was just surprised to see that there can be Lagrangians which depend explicitly on position or time, which still yield the same equations of motions after translation - like the one in my example. Now I wonder if it's because they describe the isolated systems (and thus 'conservative' in an inertial frame) in non-inertial frame of reference? This is all new to me, so sometimes I might be doing things backwards... Thanks for your patience in pointing me into right direction!
  9. Nov 23, 2012 #8


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    Actually, it's slightly more complicated. You didn't just add potential. You've modified the kinetic term.

    [tex]p = \frac{\partial L}{\partial \dot{x}} = m\dot{x} + x[/tex]

    So momentum of your particle is position-dependent. But lets take a look at total energy.

    [tex]H = \dot{x}p - L = \frac{1}{2}m\dot{x}^2[/tex]

    It looks like a Hamiltonian of a free particle. Of course, we need it in terms of momentum, so that's going to get messy.

    [tex]H = \frac{p^2}{2m} - \frac{xp}{m} +\frac{2x-x^2}{2m}[/tex]

    So it's a particle whose kinetic energy is function of p and x in a potential.

    In other words, you did not simply create a different Lagrangian with same trajectory. You came up with a completely new set of physics where under a particular potential, you end up with the same equations of motion as free particle under normal physics. There is no surprise, then, that in general, the conservation laws are going to be very different with these new physics.
  10. Nov 23, 2012 #9
    This really starts to make some sense to me. Thank you! I greatly appreciate your help and time you spent on it.
    Indeed, I not only added potential (which could possibly be identified with some fictitious force acting on the system due to non-inertial frame), but also modified the kinetic term! No chance this can describe isolated system in non-inertial frame.
  11. Nov 24, 2012 #10
    Looks like I need to resurrect the thread that seemed to be closed!
    In my first post I asked whether demanding that Lagrangian does not change under transformation that is expected to leave laws of physics unchanged is correct way of thinking. The reason for my question was that there is certain 'redundancy' in Lagrangian - in other words the Lagrangian describing the physical system undergoing certain motion is not unique.
    My original argument was: all we know empirically is that such transformation leaves the motion of the system unchanged. So, we should rather demand that it's the equations of motion that don't change rather than the Lagrangian itself (because there are more than one Lagrangians resulting in the same motion). I got some great answers that really helped me understand the topic. However, now I got confused again!
    There is a transform that leaves the motion unchanged that seems to motivate my doubts. It's the well known Galilean transform. It leaves the time coordinate unchanged, but changes the spacial coordinate (which depends on relative velocity between frames). Applying such transform clearly changes the Lagrangian (and the action), even if it was originally properly expressed in inertial frame for isolated system (e.g. not explicitly dependent on time and absolute positions). What we get is different Lagrangian here, which nevertheless results in the same equations of motion (because the added term is total time derivative of a function, which vanishes when variating the action). Now, how would you comment on this? It looks to me like my initial argument was justified and valid! We got the transform that does not change the laws of physics, yet it changes the Lagrangian. It's only the equations of motion that don't change here.

    EDIT: Of course, the opposite way still works - if certain transformation does not change the Lagrangian, the motion won't change either. Also, we can expect the conservation of certain quantity to be associated with such transformation. But it seems that there are some transformations that change the Lagrangian, yet still result in the same equations of motion. How can we know (or demand) that certain transformation that leaves the laws of physics unchanged does not change the Lagrangian? The only evidence we get from real world observations and measurements is the motion. I still can't get my head around it.
    Or is it the following:
    - if a transform does not change the Lagrangian (or should I say "action") then we can associate the conserved quantity with it.
    - if a transform changes the Lagrangian but results in the same motion, we cannot derive the new conservation law from it
    Last edited: Nov 24, 2012
  12. Nov 24, 2012 #11


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    Classical Lagrangian is not invariant under Galilean Transformation. It is not an underlying symmetry of classical mechanics. There are no associated conservation laws. As a result, the solution is not actually identical. Solution of Galilean-transformed Lagrangian simply happens to be similarly transformed. Another example would be transformation to a rotating coordinate system. Again, I can transform the solution to match, but the Lagrangian itself will be wildly different and include centrifugal and Coriolis terms. And again, there is no associated conserved quantity because there is no symmetry.

    In contrast, consider Lagrangian from relativistic field theory. (Classical or Quantum) Such a Lagrangian is invariant under Lorentz Transformation of coordinates. If you consider group of all available linear coordinate transformations, including shifts, rotations, and Lorentz boosts, you do get an associated conserved quantity. That quantity is the Stress-Energy Tensor, which implies energy and momentum conservation in a more frame-invariant manner.

    Think of it this way. If you transformed coordinates in Lagrangian and similarly in the solution, how can it possibly not match? The point of invariance is that you do not have to adjust both. Transforming coordinates in Lagrangian gives you back exactly the same Lagrangian, and you do not have to transform the solution.
  13. Nov 26, 2012 #12
    That's what I thought, more or less. So it's only the real symmetries with respect to Lagrangian that can be associated with conserved quantities.

    Unfortunately I don't know much about Lagrangian formulation of relativistic field theory. Hope to know more before I die :)
    Anyway, I know what the Lagrangian for relativistic mechanics looks like. As it's expressed in terms of proper time, it is automatically invariant under Lorentz transformation. So, for every kind of Lorentz transformation (translation, rotation, boost) there must be associated conserved quantity. True?

    [EDIT] I meant the Lagrangian describing free relativistic particle here...

    I think I know where you're coming from. We're not interested in describing the same system in different coordinates - this is just different description. What really yields new laws is the symmetry - the coordinates transforms that leave our original solutions unchanged.

    I really need to think it over well to gain some confidence. And the best way to think about things is to take a bath - that's exactly what I'm gonna do now!
  14. Nov 26, 2012 #13


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    Sort of. Because of how these conserved quantities transform under coordinate transformations (energy and momentum are both frame-dependent), it ends up being much more straight forward to simply lump these all together into a single conserved quantity. Kind of like if you take translation -> momentum, time -> energy, and then you can look at general invariance under translation in space-time and get conservation of four-momentum. Well, if you look at translation in space-time, rotation in space, and boost (boost is a hyper rotation in time, basically) you get conservation of Stress-Energy Tensor. That is effectively a relativistic generalization of conservation of energy, momentum, and angular momentum all lumped into a single quantity.

    You could probably define a conserved quantity just for Lorentz boost, for example, but I doubt it would be something particularly meaningful. It really makes more sense to treat all these coordinate transformations together.

    The rest of it, that's pretty much how I understand it. So if there is more to it, I can't really help you.