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Noethers theorem, physics and mathematics

  1. Mar 28, 2015 #1


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    First of all, I'm sorry for the long post. I've tried to make it self consistent, but some prior knowledge will be assumed in both physics and differential geometry.
    Also maybe some math will be displayed wrongly, it's my first time here...

    In physics people usually describe Noether's theorem by calculating the variation of the Lagrangian, which for a symmetry of the theory will need to be equal to a total derivative:

    \delta \mathcal{L} = \epsilon^A \partial_{\mu} (\frac{\delta \mathcal{L}}{\delta \partial_{\mu}\phi^i} \Delta_A \phi^i) = \epsilon^A \partial_{\mu} K_A^{\mu}.

    Then a Noether Current can be defined as
    J^{\mu}_A = -\frac{\delta \mathcal{L}}{\delta \partial_{\mu}\phi^i} \Delta_A \phi^i + K_A^{\mu}.

    For fields which abide the equations of motion this current is then conserved: $\partial_{\mu} J^{\mu}_A = 0$. This is just because the variation of the action will be set to zero using Stokes and assuming the fluctuations on the boundaries will go to zero fast enough.

    In mathematics it is done a little different. So far I've tried to describe everything as good as possible such that a reader with some physics background is familiar with everything. Now comes a little differential geometry and I will assume that some basic stuff is known, so if you're not too familiar with it I guess you shouldn't let this question bore you.

    A Liouville vectorfield
    D \in \mathcal{X}(TQ),
    where Q is some smooth manifold, is defined such that her flow has the following form:

    \Phi_t : TQ &\rightarrow TQ \\
    (q^i,v^i) &\mapsto (\bar{q}^i = q^i, \bar{v}^i = e^t v^i)

    The energy function associated to a function $L$ is then defined as

    E_L = D(L) -L.

    The Lagrangian vector field associated to a regular function L on TQ is the the vector field $\Gamma$ such that

    i_{\Gamma}\omega_L = -dE_L
    \omega_L = d\theta_L
    is the Poincare-Cartan 2-form associated to L, and d is the exterior derivative. Finally we can now introduce Noether's theorem;

    Let Y be a vector field on TQ, such that:
    \mathcal{L}_Y \theta_L = df
    for some function f, and:
    Y(E_L) = 0.
    Then Y is a symmetry of $\Gamma$ and associated to Y there is a first integral F, determined by
    F = f - <Y,\theta_L>.
    And with every first integral F of $\Gamma$ there is a symmetry Y which abides the above mentioned properties.

    Now comes the question...

    Although in physics we are usually interested in invariance of the action, especially in field theory (for example for space-time symmetries), the Noether theorem I just described gives a one to one correspondence between the symmetry of the Lagrangian and the conserved quantities. Now I know that what I just described is valid for mechanics and not field theory, and when you want to describe field theory in this geometric formalism you need to look at jet bundles. Is it then so that when you use jet bundles you will eventually get a correspondence between the action and conserved quantities?
    And am I wrong to think that actually in mechanics the symmetries will be described with the action as well?
  2. jcsd
  3. Mar 28, 2015 #2
    While I have not understood this stuff completely myself (gotta learn this one day), I think this might help:


    EDIT: In the general theory, when everything is "nice" (smooth, you have actual lie groups etc..), you typically have a base manifold ##\mathcal Q## and a (or more) smooth section ##\phi \in \Gamma^\infty \left( \mathcal Q, E \right)## of a vector bundle ##E## that is equipped with a connection ##\nabla## (out of which one can construct a differential ##d##). The action ##\mathcal S## is then a complex/real functional on that space of sections defined as follows:
    $$\mathcal S \left( \phi \right) = \int_{\Omega \subseteq \mathcal Q} \mathcal L \left( \phi, \nabla \phi \right) \quad . $$
    Here ##\mathcal L## is a map that takes the section and ''derivatives'' (beware of the relativity principle!) as input and spits out a complex/real density.

    To get any conserved quantities you need a Lie group action (horizontal, vertical or both) on ##E##, that leaves this integral invariant. Note that a similiar construction can be put up for principal bundles instead of vector bundles.

    Hope this helps.
    Last edited: Mar 28, 2015
  4. Mar 29, 2015 #3


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    Thank you for your reply.

    And thank you for the link to the discussion on stackexchange.

    I think that what you said, and what is said in the stackexchange discussion, that Lie group actions are needed, is indeed the formulation that is usually used in physics. But the theorem I described I think is on a higher than that, in the sense that it does not need to make use of any action of a Lie group.

    I'm not sure, but maybe the following application of the theorem helps to find the link with the formulation in physics:

    X \in \mathcal{X}(Q)
    such that X^c(L) = 0 (the complete lift), then Y := X^c is a Noether symmetry (as described in my first post) and the first integral is of the form
    F = X^v (L)
    where X^v is the vertical lift of X.
  5. Mar 29, 2015 #4
    Well, as far as I understand it - and I am not very deep in the subject, in the ''standard mass point mechanics without crazy constraints'' case you have a Lagrangian ##L \in C^\infty \left( T \mathcal Q , \mathbb R \right)## and some kind of natural symplectic structure ##\omega## on the tangent bundle. Then this gives you a Hamiltonian system on the tangent bundle and for every smooth function ##f## on the tangent bundle you can take the corresponding Hamiltonian vector field ##X_f##, which should be a constant of motion if and only if its flow preserves the Lagrangian ##L##:

    $$\mathcal L _ {X_f} L = 0$$

    I don't know how that relates to Noether though and so I can't construct the generalization for ''field theory''.
  6. Apr 19, 2015 #5
    I have continued thinking about this, because I find it quite interesting myself. It's really hard to find references that explain this stuff in a rigorous, coordinate free and understandable way.

    As far as I have figured it out, the variation of the action only gives you the Euler-Lagrange equations if you have some way to transform the integral of the stuff you don't like into an integral over the boundary of your integration domain, which you require to vanish.

    For example, if the action is a functional on the space of differentiable curves (then the Lagrangian takes the corresponding curve on the tangent bundle and spits out a function on the domain of the curve), i.e.

    $$\mathcal S \left( \gamma \right) := \int\limits_{(\tau_1, \tau_2) \subseteq R} L \left( \gamma \left( \tau \right), \frac{d}{d \tau} \right) \, d \tau $$

    , you just use that ##df/d\tau \, d\tau = d f## and you get out "the bad part", because ##f## is constant on ##\partial[\tau_1, \tau_2] = \lbrace \tau_1, \tau_2 \rbrace##.

    Something similiar can be done in the case of tensor fields. If ##\phi, \chi## are tensor fields of same valence on an orientable pseudo-Riemannian manifold ##\left( \mathcal Q, g \right)## with canonical volume form ##\mu## and Levi-Civita connection ##\nabla##, then we can define the action as
    $$\mathcal S \left( \phi \right) = \int\limits_{\Omega} L \left( \phi, \nabla \phi \right) \, \mu$$
    where, for convenience, ##\Omega## is compact.

    The variation now works in complete analogy to http://en.wikipedia.org/wiki/Functional_derivative#Formula assuming that the variation tensor field ##\chi## vanishes on the boundary ##\partial \Omega##. Using that for vector fields ##X## we have ##\mathcal L _X \mu = \text{div} X \, \mu## with ##\text{div} X = \text{tr} (\nabla X)## and Stoke's theorem, we arrive at

    $$\frac{\partial L}{\partial \phi} = \text{div} \frac{\partial L}{\partial \left( \nabla \phi \right)}$$

    Having this, one can in principle set up a Noether's theorem, whose proof is analogous to the standard one. I haven't gotten to this yet though. :)


    I have also thought about the other approach. To get a kind of Lagrangian mechanics on the tangent bundle, you need a 1-form ##\theta## ("momentum") and a function ##L## ("Lagrangian") on the tangent bundle, then you can consider vector fields ##X## that satisfy
    $$\mathcal L _X \theta = d L$$
    whose integral curves represent the solution of this equation of motion. My statement from above that the Lagrangian is preserved under the flow of ##X## was wrong, because it does not follow from the above equation. Indeed the Lagrangian is usually not a constant of motion.
    Last edited: Apr 19, 2015
  7. Apr 19, 2015 #6
    I'm not sure if you are aware of it or not but in my opinion one of the best sources that deals with variational problems and proves, among many other things, Noether theorems in the very general setting of local functionals defined for the infinite jet bundle of any fibered manifold (so in the simplest case of a particle moving on the line say, you think of the motions of the particle as sections of the trivial bundle [itex] \mathbb{R}\times \mathbb{R}\to \mathbb{R}[/itex] but the theorems automatically hold for vector/tensor fields or even something like Chern Simons or Yang Mills theories where the fields are connections on a principal bundle since these are all examples of sections of fiber bundles) is the following book by Ian Anderson: http://math.uni.lu/~jubin/seminar/bicomplex.pdf.

    In it is constructed the variational bicomplex which I think is a very interesting way of encoding these variational type problems into a bicomplex (it is essentially the differential forms on the infinite jet bundle but bigraded and with two differentials that kind of split the exterior derivative into horizontal and vertical components) and then you can use the tools of homological algebra like cohomology and spectral sequence arguments to try and solve the problems which started out as more analysis/differential equations type questions.
  8. Apr 20, 2015 #7


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    Thanks for the input.

    I have seen how to find the Euler-Lagrange equations for fields (there might be several different approaches), by using second order jet bundles, and indeed sections. It is like follows.

    Suppose that the Lagrangian L

    L \in C^{\inf}(J^1 \pi)

    and C a compact m-dimensional submanifold of your underlying manifold M. Then if \phi is a local section of \pi, and \phi is an extremal of L, then \phi satisfies

    (j^2\phi)*(\frac{\partial L}{\partial u^a} - \frac{d}{dxî}\frac{\partial L}{\partial u_i^a})

    where \phi is an extremal of L when

    \int (j^1\phi)*d_{X^1}L \Omega

    With X a vertical vector field on the bundle, which vanishes on the border.

    (This is from 'The geometry of jet bundles' by David Saunders; http://bookshub.org/book/9780521369...es-English-Subsequent-Edition-by-D-J-Saunders , I'm sorry for the link to some site which sells the book, it was just the first link I found on the book.)

    (And I'm sorry for these \phi in the sentences)
  9. Apr 22, 2015 #8


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    So I'm sorry if I was unclear before, but my main question was about the link between the different formulations of Noethers theorem in physics and mathematics, or if they are actually not equivalent and it is simply the case of using the same terms in different contexts.
    Especially because in physics the action is the main object and in differential geometry it is the Lagrangian in concern for the symmetries, as far as I've read/learned.
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