## PHY450H1S. Relativistic Electrodynamics Tutorial 3 (TA: Simon Freedman). Relativistic motion in constant uniform electric or magnetic fields.

Posted by peeterjoot on February 4, 2011

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# Motion in an constant uniform Electric field.

Given

We want to solve the problem

Unlike second year classical physics, we will use relativistic momentum, so for only a constant electric field, our Lorentz force equation to solve becomes

In components this is

Integrating the component we have

If we let , square and rearrange a bit we have

For

Now for the components, with , our equation to solve is

Squaring this one we have

and

Observe that our energy is

and for

We can then write

Some messy substitution, using 1.8, yields

Solving for we have

Can solve with hyperbolic substitution or

Now we have something of the form

so our final solution for is

or

Now for we have

A final bit of substitution, including a sort of odd seeming parametrization of in terms of in terms of , we have

## Checks

FIXME: check the checks.

(a parabola)

## An alternate way.

There’s also a tricky way (as in the text), with

We can solve this for

With the cross product zero, has only a component in the direction of , and we can invert to yield

This implies

and one can work from there as well.

# Motion in an constant uniform Magnetic field.

## Work by the magnetic field

Note that the magnetic field does no work

Because and are necessarily perpendicular we are reminded that the magnetic field does no work (even in this relativistic sense).

## Initial energy of the particle

Because no work is done, the particle’s energy is only the initial time value

Simon asked if we’d calculated this (i.e. the Hamiltonian in class). We’d calculated the conservation for time invariance, the Hamiltonian (and called it ). We’d also calculated the Hamiltonian for the free particle

We had not done this calculation for the Lorentz force Lagrangian, so lets do it now. Recall that this Lagrangian was

with generalized momentum of

Our Hamiltonian is thus

which gives us

So we see that our “energy”, defined as a quantity that is conserved, as a result of the symmetry of time translation invariance, has a component due to the electric field (but not the vector potential field ), plus the free particle “energy”.

Is this right? With and being functions of space and time, perhaps we need to be more careful with this argument. Perhaps this actually only applies to a statics case where and are constant.

Since it was hinted to us that the energy component of the Lorentz force equation was proportional to , and we can peek ahead to find that , let’s compare to that

which is

So if we have

I’d guess that we have

which is, using 2.40

Can the left hand side be integrated to yield ? Yes, but only in the statics case when , and for which we have

FIXME: My suspicion is that the result 2.43, is generally true, but that we have dropped terms from the Hamiltonian calculation that need to be retained when and are functions of time.

## Expressing the field and the force equation.

We will align our field with the axis, and write

or, in components

Because the energy is only due to the initial value, we write

implies

write

Evaluating the delta

Looks like circular motion, so it’s natural to use complex variables. With

Using this we have

which comes out nicely

for

Real and imaginary parts

Integrating

Which is a helix.

FIXME: PICTURE.

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