THE UNIVERSITY OF BRITISH COLUMBIA
Wed. 1 Feb. 2006 - finish by Wed. 8 Feb.
This is a relatively [pun intended] short Assignment, since
the first Midterm Exam is on Monday February 6
(in class, 50 minutes). Nevertheless it will count
the same as other Assignments, if you choose to tackle it.
For the exam, you may bring your own 1-page summary sheet
with any hard-to-remember equations etc.
The exam will cover all of Chapters 7 and 8,
sections 10.1 of Chapter 10, and all of Chapter 12.
There will not be anything on the Midterm explicitly about
(second problem below) but applications of Eqs. (12.108)
are fair game.
- If we stay in the Lorentz gauge (
is a genuine 4-vector and therefore is
a Lorentz scalar. Use this fact to show that, in any frame,
, where is the
charge density in the frame where .
If is really a Lorentz scalar, then it has the same
value in any reference frame as it does in the frame where ,
namely (in Griffiths' notation) . In general,
. Setting these equal
and rearranging terms gives the desired result. This is just
a warm-up exercise in "invariant thinking".
- Use the formal definition of the FIELD TENSOR
in Eq. (12.118)1 and the rule for its Lorentz transformation,2
to derive the equations analogous to Eqs. (12.108)
describing the transformation of and
under a "boost" into a reference frame moving at velocity u
(with the usual corresponding definitions of and )
in the positive direction.3ANSWER:
This one drove me nuts until I broke the "matrix multiplication"
habit I learned in High School. Here we don't find the
th component of the product of two tensors by summing
the products of the elements of the th row of the first
with the elements of the th column of the second;
instead we sum the products of the elements of the th row of the first
with the elements of the th row of the second.
Doing this twice [first with and then with
has been defined for compactness.
- A capacitor made from two square parallel plates a on a side
and d apart is given a charge -Q on the upper plate and +Q
on the lower plate. Let the origin of coordinates be in the centre
of the capacitor, the edges of the plates parallel to and
and the gap in the direction.
Using Lorentz transformations, find and
inside the capacitor
- in a frame moving at a velocity u in the direction;
For this we can just plug
into Eqs. (12.108) to get the only nonzero
where as usual
- in a frame moving at a velocity u in the direction.
Here we can use the analogous equations derived in the previous problem
to get the only nonzero component
. That is,
the fields have not changed. The only difference in
geometry is that the plates are closer together due to Lorentz
contraction; but that does not affect the electric field, as we know.
- For the former case, compare your result with what you would
expect from simply transforming the dimensions of the plates
into the moving frame and treating their motion as sheets of current.
Lorentz contraction shrinks distances parallel to the motion,
but not perpendicular. Thus the plates are no longer square;
the edge in the direction is a factor of shorter
in the primed frame while the other edge still has a length a.
This compresses the net charge Q (an invariant) into an area
smaller by the same factor, increasing '
and consequently E'z by a factor of .
Meanwhile the plates are now sheets of current
for the top and bottom plates, respectively.4 As for the electric field,
. We know that
two opposite sheets of current produce a uniform magnetic field
between them given by
Multiplying by and noting that
In a simple case like this, we get the right answer by simply taking
Lorentz contraction into account. So one might wonder if all the
elaborate formalism is ever any real help in solving practical problems.
The answer is probably, "It depends on how flawlessy you can visualize
all the effects without losing track of any." Equations like (12.108)
allow you to transform into the frame of your choice
"without much thought" even for rather complicated situations
(e.g. when the fields vary with position).
But without relativity we would have no way of making E&M
obey NEWTON'S LAWS, of which we are understandably fond.
It emerges naturally that magnetic fields are an obligatory
extension of COULOMB'S LAW when we transform consistently
into moving frames; and that knowledge is worth having!
Jess H. Brewer