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Nother's theorem for a static E field

  1. Jul 1, 2011 #1
    I can use Noether's theorem and the homogeneity of time to derive the conservation of energy for a static E field, but can I also use the homogeneity of space to derive the conservation of momentum?
     
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  3. Jul 1, 2011 #2

    jambaugh

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    Yes, except it will be the conservation of the canonical momentum:
    P = p + qA
    (-qA?)

    You then see why one needs a vector potential for fully relativistic E-M, it (times charge) is the "potential momentum" extension of potential energy in relativistic mechanics.

    Note, though we can eliminate A for a static E field via choice of gauge, the symmetries you rely upon to apply Noether's theorem are not satisfied by the gauge constraints (which by definition break symmetry). Apply Noether's theorem prior to imposing gauge conditions.
     
  4. Jul 2, 2011 #3
    This looks exactly what I'm looking for, especially the "potential momentum" part which I've never come across before. I feel a little out of my depth :smile: The relativistic EM Lagrangian is given by [itex] - \frac{m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot A} - e\Phi[/itex]

    Using Noether's theorem to create the conserved canonical momentum requires [itex]\frac{e}{c}\mathbf{u \cdot A} - e\Phi[/itex] to be independent of [itex]x[/itex]. This looks possible for a static [itex]E[/itex] dependent upon [itex]x[/itex] by transforming [itex]\mathbf A\rightarrow \mathbf A'= \mathbf A + \nabla\Lambda[/itex]. This gives the conserved canonical momentum as [itex]\frac\partial{\partial \mathbf u} ( \frac{-m_0c^2}{\gamma} + \frac{e}{c}\mathbf{u \cdot (A+\nabla\Lambda}) - e\Phi)[/itex] and with B = 0, becomes [itex] -m_0\mathbf u\gamma +\frac e c \mathbf u \nabla\Lambda[/itex]

    Is this correct?
     
    Last edited: Jul 2, 2011
  5. Jul 4, 2011 #4

    jambaugh

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    I prefer to work in four-vector format: (and use [itex]m=m_0[/itex] avoiding reference to "relativistic mass" all together.)

    You can find a nice brief clarification of the relativistic Lagrangian for choices of metric at:
    http://arxiv.org/pdf/0912.0655.pdf".
    I choose the convention [itex]g_{\mu\nu} \sim diag(1,-1,-1,-1)[/itex] indicating the metric measures proper time (rather than proper distance).

    The action then is the path integral:
    [tex]S = \int_{x(\tau)}[(mc)ds + e/c A_\mu dx^\mu] = \int L d\tau [/tex]
    with [itex]L=mc\sqrt{\dot{x}^\mu \dot{x}_\mu} + e/c A_\mu \dot{x}^\mu[/itex].
    Note that [itex]\tau[/itex] is a path parameter not necessarily proper time.

    The canonical momentum-energy four vector components are:
    [tex] P_\mu =\partial_{\dot{x}^\mu}L= \frac{mc}{\sqrt{{ \dot{x}}_\mu \dot{x}^\mu}} \dot{x}^\mu+ e/c A_\mu= p_\mu + (e/c) A_\mu[/tex]
    The Euler-Lagrange equations are then:[itex]\dot{P}_\mu = (e/c)\dot{x}^\nu A_{\nu,\mu}[/itex] or...
    [tex]\dot{p}_\mu = (e/c) \dot{x}^\nu [A_{\nu,\mu}-A_{\mu,\nu}] = (e/c)\dot{x}^\nu F_{\mu\nu}[/tex]
    OK Now I will attempt toapply Noether's theorem. We need to write the variation of the action in the form:[itex] \delta S = \int Q_\mu \delta\dot{x}^\mu d\tau[/itex] and then [itex]\dot{Q}_\mu = 0[/itex] along paths satisfying the dynamics.

    [itex]\delta S = \int d\tau [P_\mu \delta\dot{x}^\mu + e/c\dot{x}^\nu A_{\nu,\mu} \delta x^\mu][/itex]
    By expanding [itex]e/c \frac{d}{d\tau}(x^\nu A_{\nu,\mu} dx^\mu)[/itex] and substitution with integration by parts (zeroing variation on the boundary) this yields:
    [tex]\delta S = \int d\tau [P_\mu - e/cx^\nu A_{\nu,\mu}]\delta \dot{x}^\mu - x^\nu\dot{x}^\lambda A_{\nu,\mu\lambda}\delta x^\mu[/tex]
    We can take different tracks from here but let us assume that the four-potential is linear in the coordinates eliminating the second order derivatives and thus the quantities:
    [itex]Q_\mu = P_\mu -( e/c )x^\nu A_{\nu,\mu}[/itex] are conserved.

    Now the fact that the four potential is linear gives us:[itex] A_\nu = W_{\nu\mu}x^\mu[/itex] with the [itex]W[/itex]'s constant and that the e-m field tensor components are:
    [tex]F_{\nu\mu} = A_{\mu,\nu}-A_{\nu,\mu} = W_{\mu\nu}-W_{\nu\mu}\equiv 2W_{[\mu\nu]}[/tex]
    The bracketed index indicates the anti-symmetric component of W: [itex]W_{\mu\nu} = W_{[\mu\nu]}+W_{(\mu\nu)}[/itex] the sum of anti-symmetric and symmetric components.

    Now since the symmetric components contribute nothing to the physics we can eliminate them with a gauge transformation. [itex]\Lambda = (1/2)W_{(\mu\nu)}x^\mu x^\nu[/itex], [itex]A_\mu \to A_\mu - \partial_\mu \Lambda[/itex].
    This means that [itex] A_{\nu,\mu} = -A_{\mu,\nu}[/itex] and thus that [itex]x^\nu A_{\nu,\mu} = - x^\nu A_{\mu,\nu} = -A_\mu[/itex].
    So the conserved quantities are:
    [tex]Q_\mu = P_\mu + (e/c)A_\mu = p_\mu + 2(e/c)A_\mu=p_\mu + 2(e/c)W_{[\mu\nu]}x^\nu = p_\mu + (e/c)x^\nu F_{\nu\mu}[/tex]

    So here is a general result for general constant electro-magnetic fields. Note that if A is constant you can just gauge it to 0 as it means zero electromagnetic fields. My earlier post was incorrect in that it is not the canonical momentum which is conserved and qA is not the "potential momentum".

    I think you can generalize this to arbitrary cases and I'm sure it's written up in the texts and available online somewhere.

    (Note: I'm going crosseyed trying to keep everything straight so check my work.)
     
    Last edited by a moderator: Apr 26, 2017
  6. Jul 5, 2011 #5

    jambaugh

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    I should add one detail (actually I'm sure many but one that I just recalled)...

    The space-time translation symmetry manifesting a conservation law in this case (constant F = constant E and B fields) is augmented with a gauge transformation. The e-m four potential needn't be constant, it need only (be linear in the coordinates to) yield a constant e-m field tensor which indicates a group of translations + gauge transformations leaves the action invariant (or rather a representation of the group of translations which includes gauge transformations leaves S invariant).

    While scribbling on paper I also found that you can consider an alternative relativistic Lagrangian yielding the same E-L equations:

    [tex] L = \frac{1}{2} \dot{x}^\mu Q_\mu(x,\dot{x})[/tex]
    where, as earlier, [itex]Q_\mu = p_\mu + 2(e/c) A_\mu[/itex]. Then [itex] S = \frac{1}{2}\int Q_\mu dx^\mu[/itex] as a path integral.
     
  7. Jul 9, 2011 #6
    Since it's a conserved quantity

    [tex] d/d\tau\ (p_\mu + (e/c)x^\nu F_{\nu\mu}) = 0[/tex]
    [tex] d/d\tau\ p_\mu = - (e/c)u^\nu F_{\nu\mu} [/tex]

    Which is just the Lorentz force equation.

    I suspect you can carry out your procedure for time varying fields to give another conserved quanity which also gives rise to the Lorentz force equation.

    Hmm... I think the canonical momentum that includes the gauge change, is invariant; and likewise for qA as the potential momentum. On the other hand, I don't get the expected Lorentz force equation as you did, so perhaps I'm using the wrong conserved quantitiy: Isn't it just [itex] \frac{\partial L}{\partial\mathbf {\dot q }}
    [/itex]?
     
    Last edited: Jul 9, 2011
  8. Jul 10, 2011 #7

    jambaugh

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    Right, to treat it properly you have to go to the field theory with a Lagrangian density and look at the local conservation of stress energy (which is the 4-momentum's current density) of the particle + e-m field. Of course the particle as a point must be expressed using delta function fields. I don't recall if this works out OK in the classical setting as I've only seen it in detail done for quantum particles.

    [itex] \frac{\partial L}{\partial\mathbf {\dot q }}[/itex]
    is indeed the canonical momentum. But my error was in assuming that it is the canonical momentum which is conserved. That was some past sloppiness on my part having not actually worked through the details. I just assumed (and you know what happens when you do that!;)).

    Although the sign's don't quite work out right, you can think of it in terms similar to the four quantities of: Kinetic Energy, Potential Energy, The Hamiltonian, and the Lagrangian.
    Note that H = T+V= L + 2V to the "total momentum" Q= p+2eA = P+eA.

    While short on sleep I tried to define "kinetic momentum", "potential momentum", "Lagrangian momentum" and a "4-Hamiltonian". It is the Hamiltonian which is conserved.

    I'll give it another go today looking closely at the Legendre transformation to see what "holds up". It would seem that since we can parametrize the particle motion w.r.t. an arbitrary observer's time and define a Hamiltonian for each, and conserved energy for each things should match up nicely.

    I'll get back to the thread if I find anything interesting.
     
  9. Jul 10, 2011 #8

    jambaugh

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    Maybe, but maybe not. Remember energy fails to be conserved when you have explicit time dependence, and that Noether's theorem is an "if an only if" statement. Breaking the symmetry should break the conservation of the quantity. But if one drops the qualifier and/or considers an arbitrary enough symmetry (incorporating gauge) for a certain class of time and space varying fields then... maybe.

    I'll get back to you on that as well.
     
  10. Jul 10, 2011 #9

    jambaugh

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    Here's one way to go about it. Work in Kaluza-Klein space-time-phase.

    Define a fifth coordinate [itex](t,x,y,z,\theta) = (x^0,x^1,x^2,x^3,x^4)[/itex].

    Absorb the electro-magnetic four potential into the metric:
    [tex]\eta_{ab} \sim \left( \begin{array}{c|c} g_{\mu\nu} & A_\mu \\ \hline \\ A_\nu & \psi\end{array}\right)[/tex]

    Use the Lagrangian: [itex] L = \frac{m}{2}\eta_{ab}(x)\dot{x}^a\dot{x}^b[/itex]
    And you should get an extra-dimensional analogue of a general relativistic particle.

    Note that charge here is the fifth component of momentum:
    [tex] P_\mu = \frac{\partial}{\partial x^\mu}L = m\eta_{\mu\nu}\dot{x}^\nu + m\eta_{\mu 4}\dot{x}^4 = m\dot{x}_\mu + (m\dot{x}^4)A_\mu[/tex]
    [itex]m\dot{x}^4 = p^4 = q[/itex]

    There is a fifth momentum component which leads to conservation of charge, and one can by choice of coordinates eliminate the [itex]\psi[/itex] term in the metric (choosing [itex]\theta[/itex] to be in a light-like direction.)

    EM Gauge transformations become coordinate transformations, specifically translating the points of zero phase coordinate [itex]\theta=x^4[/itex] depending on space-time coordinates [itex]x^0,\cdots,x^3[/itex].

    Now so far as conservation laws go, you can look at the facts in GR and apply them to this K-K GR + EM scenario. Conservation laws get a little messy, because one must work in the generally covariant treatment and be careful of one's frame dependent definitions.

    One note here, if you want to play with this GR+EM KK unification...be careful about lowering and raising indices since one has a choice of using the extended metric which includes the E-M potential, or the embedded space-time metric.
     
  11. Aug 20, 2011 #10
    Can anyone criticize this solution of mine?

    The Lagrangian for a relativistic particle moving in an electromagnetic field is given by [tex]L = - \frac {mc^2} \gamma + q~( \vec u(t)\cdot \vec A(\vec x,t) -\Phi(\vec x,t))[/tex]Conservation of the canonical momentum requires the Lagrangian to be independent of coordinates, whereas in general it isn't. We therefore need to show it's possible to create this independence with some suitable gauge transformation of the potentials [itex]\Phi[/itex] and [itex]A[/itex]. For the case of a static electric field with [itex]\vec A=0~[/itex], the EM gauge transformations [tex]\vec A= \nabla\lambda(\vec x,t),\quad\phi(\vec x)\to\phi(\vec x) - \frac d {dt} \lambda(\vec x,t) [/tex] gives the Lagrangian [tex]L = -\frac{m_0c^2}\gamma + q~( \vec u\cdot\nabla\lambda(\vec x,t) - \phi(\vec x) + \frac d {dt}\vec\lambda(\vec x, t))[/tex] It isn't generally possible to make this expression independent of coordinates for every [itex]\Phi(\vec x)[/itex] including that for a static Coulomb electric field. However, lets take the simplest case of an electric field directed along the x axis giving a potential and defining a gauge as[tex]\Phi(\vec x) = -Ex\qquad \lambda(\vec x, t) = -Ext[/tex]to give a coordinate independent Lagrangian[tex]L = -\frac{m_0c^2}\gamma - q\vec u\cdot Et[/tex] and the conserved canonical momentum [tex]\frac{\partial L}{\partial\vec u} = (\gamma mu_x - qEt, 0, 0)[/tex] Since this is constant over time, we're finally left with just the Lorentz force on a relativistic charge in a static electric field along the x direction [tex]\frac d {dt}\gamma m\vec u = q(E,0,0)[/tex]

    This can be generalised to any time and coordinate dependent electric field from the principle of locality, causality and memorylessness:
    • The dynamics of a charge depends upon the field at its coordinates only
    • The dynamics of a charge depends upon the field at time t, and not the past or the future
    Therefore, for any electric field [itex]\vec E(\vec x,t)[/itex], the Lorentz force and associated conserved expression is[tex]\frac d {dt}\gamma m\vec u = q\vec E(\vec x,t),\qquad\gamma m\vec u - q\vec E(\vec x,t)t[/tex] and can finally conclude using the gauge [itex]\lambda(\vec x, t) = -\Phi(\vec x, t)t[/itex], the conserved canonical momentum for a relativistic charge in a electric field [itex]\vec E(\vec x, t)~[/itex] is given by [tex]\frac{\partial L}{\partial\vec u} = \gamma m\vec u - q\vec E(\vec x,t)t[/tex]
     
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