David Cross wrote:
This is, again, a GR-related question as I saw this show up in the GR class.
So, the basic deal is this: if you can add the gradient of a scalar potential
to a vector field, nothing happens when you run it through, if I remember
right, the curl of the whole thing.

Yes. In 3d: curl grad f = 0 is an identity for any real function f(.) on
the manifold (basically reflecting the fact that partial derivatives
commute).

While normally considered in flat space, this actually holds in any
(semi-)Riemannian 3-d space, regardless of curvature.


The questions are, how does gauge invariance conflict with the Galilean
Transformations, and how exactly does it result in the Lorentz
Transformations? (The class got as far as the proof that Maxwell's equations
lack Galilean invariance unless you insert a gauge transformation - which I
freely admit I didn't entirely follow.)

I don't think they actually conflict, but there most definitely is an
enormous difference in how "natural" things are.


In 4-d spacetime, start with a 1-form potential A, so the Maxwell field
tensor F is:
F = dA
where d is the exterior derivative; clearly F is a 2-form. F is, of
course, the field tensor that appears in Maxwell's equations.

A gauge transform is simply adding the exterior derivative of any scalar
function f(.) on the manifold:
A' = A + df
F' = dA' = d(A+df) = dA = F

Some comments on this quite natural sequence:
A) nowhere does this distinguish between space and time.
B) this all holds in any (semi-)Riemannian space(time) of any
dimension = 2, flat or curved. In fact, this doesn't depend
on either the metric or the orientability of the manifold
(the 3-d identity: curl grad f = 0 depends on both).
C) this all holds completely independent of any coordinates. IOW:
every equation above is a tensor equation (all forms are tensors).
D) this all holds independent of Maxwell's equations.
E) this is basically a re-statement in more general terms of the 3-d
identity: curl grad f = 0.
F) the basis of a gauge transform is the exterior calculus identity:
ddf = 0 -- that is what makes this so "natural".



By contrast, in a Galilean world with 3-d space and a separate time, the
basis of gauge transforms is Maxwell's equations and a surprising and
apparently ad-hoc additional requirement. That is, in order to apply
curl grad f = 0 to the vector potential, you must rather surprisingly
also add df/dt to the scalar potential -- that just falls out of
nowhere, but is required in order to keep Maxwell's equations satisfied.
This is a rather "unnatural" coupling of vector and scalar potentials,
and also an "unnatural" coupling of space and time -- ostensibly they
are separate in this Galilean world, but here spatial derivatives of f
are intimately linked with time derivatives.


Because of the generality of the above sequence in 4d, I doubt this can
be used to determine the form of the Lorentz transform. After all, the
identity ddf = 0 is independent of coordinates, so _any_ coordinate
transform can be used, not just Lorentz transforms. The unique thing
about Lorentz transforms is that they relate INERTIAL frames, and that
alone determines their form up to a discrete choice (among the Euclid,
Galileo, and Lorentz groups -- experimental observations of course
select the latter).


Tom Roberts

On Wed, 14 Sep 2005 13:06:52 -0500, Tom Roberts wrote:

David Cross wrote:
This is, again, a GR-related question as I saw this show up in the GR class.
So, the basic deal is this: if you can add the gradient of a scalar potential
to a vector field, nothing happens when you run it through, if I remember
right, the curl of the whole thing.

Yes. In 3d: curl grad f = 0 is an identity for any real function f(.) on
the manifold (basically reflecting the fact that partial derivatives
commute).

Ah, good. I was going on memory there, not having had my notes in front of me.


By contrast, in a Galilean world with 3-d space and a separate time, the
basis of gauge transforms is Maxwell's equations and a surprising and
apparently ad-hoc additional requirement. That is, in order to apply
curl grad f = 0 to the vector potential, you must rather surprisingly
also add df/dt to the scalar potential -- that just falls out of
nowhere, but is required in order to keep Maxwell's equations satisfied.
This is a rather "unnatural" coupling of vector and scalar potentials,
and also an "unnatural" coupling of space and time -- ostensibly they
are separate in this Galilean world, but here spatial derivatives of f
are intimately linked with time derivatives.

Now that you mention it, when the derivation was done in class the scalar
potentials were rather arbitrarily introduced wth some hand-waving as to why
you needed a gauge transformation to preserve invariance of the fields.

Because of the generality of the above sequence in 4d, I doubt this can
be used to determine the form of the Lorentz transform. After all, the
identity ddf = 0 is independent of coordinates, so _any_ coordinate
transform can be used, not just Lorentz transforms. The unique thing
about Lorentz transforms is that they relate INERTIAL frames, and that
alone determines their form up to a discrete choice (among the Euclid,
Galileo, and Lorentz groups -- experimental observations of course
select the latter).

Unfortunately I had to snip quite a bit in the interest of brevity.

I think I may have been using sloppy wording; my impression was that the
Lorentz transformations could be derived from gauge transformations, and
because electromagnetism does not obey the Galilean transformations, a gauge
transformation would not be possible in a Galilean world.

I'll be happy to be corrected on that if my impression is incorrect. I haven't
seen a lot of gauge stuff yet, so I don't know when you should use one or not.
---
David Cross
dcross1 AT shaw DOT ca

David Cross wrote:
I haven't
seen a lot of gauge stuff yet, so I don't know when you should use one or not.

As the underlying theory is gauge invariant, the choice is always yours.
Most people use that freedom to make a given computation easier, so for
radiation problems they use the Lorentz gauge, and for electrostatics
they use the Coulomb gauge, etc.


Tom Roberts

David Cross:

Now that you mention it, when the derivation was done in class the
scalar potentials were rather arbitrarily introduced wth some hand-
waving as to why you needed a gauge transformation to preserve
invariance of the fields.

Here's why. The fields satisfy the relation,

d\phi/dt + div A = 0

We can write that in terms of a component transverse to the propagation
direction, a longitudinal component and a scalar component.

div A_trans = 0 by definition, leaving

div A_long + d\phi/dt = 0

Since light has only transverse polarizations, it has only two degrees
of freedom, yet is described by 4 numbers. You can choose A_long and
\phi to be anything that satisfies the above, because the first equation
is _always_ satisfied. Light is always transverse. The other two
components are artifacts (at least classically).

Tom Roberts wrote:
[snip]
The basic difference between QED and this approach to quantum gravity is
that in QED the field quanta (photons) are neutral, so there are only
loops on the charged particle lines.

Photons get self correction terms. It's just that the first
one involves pair creation/destruction, so it's down by
two factors of alpha. The first self correction term for
an electron involves the emission and reabsorption of a
photon, and so only one factor of alpha.

In GR, however, the "charge" is
energy, and that is also carried by the field quanta. So instead of just
loops on the particle lines, you get loops on the quanta lines as well,
and loops on the loops, and loops on the loops on the loops.... The
divergence is of a whole different order than in QED.

You get loops on loops on loops in QED as well.

The big deal in QED is that you can show that all of those loops
can be swept under one rug. Or rather, under a finite number of
rugs. What's that buzz phrase? Dyson-Schwinger equations? Not sure.
Been a looooong time. But the idea is, at each order of alpha,
you can show that by modifying the "bare" values of just the
mass and charge of the electron, plus the mass of the photon,
you can "hide" the infinities. So, a finite set of measurements
fixes the entire theory.

A similar thing happens in, for example, QCD. There the gluon is
the exchange particle, and it carries colour charge. So there is
a colour interaction betwen gluons. But, again, you can show
(with quite a bit more work) that each order of the loop expansion
can have its infinities hidden by adjusting the "bare" values of
the colour charge for each particle, plus the mass for each particle.
A finite set of measurements fixes the entire theory.

In the canonical method of quantizing general relativity, you find
that this fails. You wind up having a new kind of infinity at each
step in the expansion. It's not because gravitons carry "charge"
as such. It's because adding a new order of the loop expansion will
produce new divergences that are independent of the bare values of
the mass and charge on each object. So you need a new adjustable
bare parameter at each order of the expansion. Generally that means
you would need to measure a new interaction for each order of the
expansion you wanted to make work.
Socks

Puppet_Sock:

The big deal in QED is that you can show that all of those loops
can be swept under one rug. Or rather, under a finite number of
rugs. What's that buzz phrase? Dyson-Schwinger equations? Not sure.
Been a looooong time. But the idea is, at each order of alpha,
you can show that by modifying the "bare" values of just the
mass and charge of the electron, plus the mass of the photon,
you can "hide" the infinities. So, a finite set of measurements
fixes the entire theory.

What you get is that the divergences cancel order by order,
so that the number you get at each order is finite (ward identities).
That doesnt guarantee that the series converges in the limit of an
infinite number of terms, and indeed it most likely does not.

The question is then, whther qed is a theory which is independent of
anything else. It's quite likely that it isn't, since the only interacting
theories which are permissible in a relativistic theory are those which
are asymptotically free. But, since qed is just the low energy limit of
a theory which is believed to be asymptotically free, qed need only be
finite up to an energy known as the landau pole with the landau pole
being large (relative to the point that qed is no longer the correct
description of the phenomena). I recall a crude estimate of something
like 10^250 GeV, which exceeds the energy of the universe and is well
above the point at which qcd is already needed to explain the muon magnetic
moment).