For a static electric field $E$ the conservation of energy gives rise to $$\oint E\cdot ds =0$$ Is there an analogous mathematical expression the conservation of momentum gives rise to?

  • $\begingroup$ Maybe I'm being stupid but surely if you could consider an object that has no net force acting on it, you would get 0 if you just took any loop integral around it: ∮pho⋅ds=0 What would one call the 'force field' be called though? $\endgroup$ – qftme Jun 30 '11 at 22:32
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    $\begingroup$ It might interest you to look at the following question and the answers provided therein. $\endgroup$ – yayu Jun 30 '11 at 23:02
  • $\begingroup$ @yayu: indeed, particulry @Andrew's answer and the associated paper. Thanks $\endgroup$ – qftme Jun 30 '11 at 23:33

We have a thermodynamic understanding of energy violation, in terms of perpetual motion, but momentum violation is somewhat less immediately paradoxical in a changing background, maybe because we have intuition that the field has momentum in this case. But the field momentum is irrelevant for this question. Mechanical momentum should be conserved in a z-independent background ignoring any field momentum considerations.

I'll state the principle of conservation of momentum is as follows: if the z direction is made periodic with period L (for mathematical convenience), and all the physical conditions (the time dependent E and B fields) do not depend on z, it is impossible to get a rod parallel to the z-direction going all the way around the box to move along the z direction only by moving it in the x and y directions, keeping it parallel to the z axis.

In terms of differential forms, the E equation you give is part of

$dF = 0$

When F is independent of time, this reduces to $\nabla \times E = 0$, and to the loop integral condition, but I will try to state it more covariantly in spacetime. Consider a spacetime 2-d surface which is some parametrized loop x(s) y(s) z(s) in the t=0 plane, extended in times by translation. The integral of the two form over the surface is equal to zero, by stokes theorem. But this integral is equal to (the infinite length in time of the surface times) $\int E_x dx + E_y dy + E_z dz$, and you conclude that in time-independent fields, the integral must be zero.

If you believe the Lorentz equation of motion, this is a statement of conservation of mechanical energy, as you noticed.

But since it is stated in differential form language, all you have to do is tip the picture over into the z-axis to get a momentum version. If the F tensor does not depend on z, and z is periodic with period L (just in case), then integrating the F form on a two-surface made by taking a parametrized loop t(s),x(s),y(s), extended by translation along the z axis gives zero. This reduces to the following one dimensional integral:

$\int (E_z dt - B_y dx + B_x dy) = 0$

This integral is equal to zero by the Maxwell equation dF=0.

If you believe the Lorentz equation of motion, this integral must also vanish for reasons of conservation of momentum. If you take a uniformly charged z-extended wire around this loop, it would acquire momentum of this magnitude.

But you can't get a charged wire to move backward in time, so you have to use two wires, and at the proper instant charge them up to equal and opposite charge densities, then move them along the x and y directions as required by the loop, then bring them back and let them neutralize each other.

If the integral above is nonzero, the wires will get a nonzero relative z-momentum, which is impossible in a z-invariant system. The above is just the integrated Lorentz force in the z-direction.

So from conservation of momentum, you conclude that the above is zero. This allows you to see that in the static case, $B_y$ and $B_x$ are derived from a vector potential $A_z$ just as E is derived from a potential $\phi$. The differential form formulation gives you this automatically of course.

Just as the conserved mechanical energy quantity in a static background E field is the kinetic energy plus $q\phi$, the conserved mechanical momentum in a z-independent background is the mechanical z-momentum plus $qA_z$.

  • $\begingroup$ Thanks for your interesting answer. What do you mean by making the z direction periodic? $\endgroup$ – John McVirgooo Aug 23 '11 at 20:08
  • $\begingroup$ I just mean consider a situation where the electric and magnetic field are z-independent, and space is imagined to be a box so that positions z and z+L are identified. I do this so that the box one integrates the 2-form over in the z-direction is finite. $\endgroup$ – Ron Maimon Aug 25 '11 at 3:21
  • $\begingroup$ Your answer considered a particular case for $z$ being periodic and the physical conditions independent of $z$, but didn't make it easy for people less gifted like me to see how your expression could be generalised to any static electric field. If you can edit your answer to show this as a conclusion, I'll make yours the accepted answer. $\endgroup$ – John McVirgooo Aug 25 '11 at 20:58
  • $\begingroup$ But your original question considered a "t independent" conditions. Momentum is to energy as space is to time, so "t independent" turns into z-independent. The answer is giving the analog of Kinetic+potential energy, it is "mechanical momentum plus vector potential". There is no generalization for static electric fields. The generalization involves magnetic fields. $\endgroup$ – Ron Maimon Aug 26 '11 at 0:29
  • $\begingroup$ Jackson and Griffiths don't say anything about this stuff so can you recommend a good book that goes into this? $\endgroup$ – Physiks lover Jun 22 '12 at 18:01

for this particular case, there doesn't appear to be an obvious expression that comes from the conservation of momentum, analagous to that for the conservation of energy. However, Noether's theorem offers a route in determing one of the conserved quantities as being the canonical momentum, which on rearranging gives the Lorentz force law for a static electric field: $$\frac d {dt} \gamma mv = qE$$

  • $\begingroup$ I gave the obvious expression right here.. What's going on? $\endgroup$ – Ron Maimon Aug 25 '11 at 17:08

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