cfgauss:
I have been studying tensor calculus in relativity a little bit before
school starts, and I have a question that I haven't been able to answer
by myself. I know that we can rewrite Maxwell's equations using the
electromagnetic field tensor,
F^(mu nu) =3D
(0 E^1 E^2 E^3)
(-E^1 0 B^3 -B^2)
(-E^2 -B^3 0 B^1)
(-E^3 B^2 -B^1 0)
And we can re-write Maxwell's equations as, with d meaning curly-d, and
J is the 4-current
d_nu F^(mu nu) =3D 4 pi J^mu
d_lambda F_(mu nu) + d_mu F_(nu lambda) + d_nu F_(lambda mu) =3D 0

With which we can do all kinds of nice special relativity-stuff, and, I
presume, general relativity stuff, too (although I haven't gotten that
far in what I've been studying).

OK, then the quantum mechanical transcription is the same that you
used for non-relativistic quantum mechanics, i.e., start with the
substitutions, E = i\hbar d/dt and p = -i\hbar\grad and put those
into the mass energy relation, E^2 = p^2 + m^2.

That is the relativistic schroedinger equation (also called the
klein-gordon equation for general spin). if you factor the right-hand
side as p^2 + m^2 = (a.p + bm), and solve for `a' and `b', you get the
dirac equation. The solutions to these are not simple scalar wavefunctions.
Since the wave equations carry lorentz indicies, you can apply the
same machinery from relativity. The dirac equation, however has additional
features, since the solutions are spinors, not vectors.

[...]
one in terms of B. Now, it seems like we could form another "quantum
mechanical field tensor" in terms of the components of A and B, just
like we did for E and B, and re-write our differential equations as
tensor equations like we did before, and do more relativity-stuff.
Now, it seems to me that since we've got something in the tensor
language of general relativity, we should be able to do general
relativity with this. But, obviously, we can't, or people would be
doing this. Why don't we do this (or do we, and no one has told me
about it)?

Because you don't need to do that. For example, the wave equation
you get from maxwell's equations is the correct quantum mechanical
wave equation for a massless vector field. Quantum mechanics just
takes over where the classical theory leaves off. In particular, if
you want to talk about photons and the like, you quantize the vector
potential. To work with feynman diagrams, you don't even need to do
that.

At what point does this break and not make any sense
anymore? Can we at least do relativistic quantum mechanics like this
if we wanted to?

Possibly, but even if you could, I'm not sure what it would buy you.
Basically, quantum mechanics is not all _that_ different from the classical
theory, as conceived, so there's no point in trying to make any more
different. If you aren't quantizing the fields and don't need to
obtain creation and anihilation operators, then for many purposes,
you can work with things like, A^u, as if it were just the classical
four-potential.

Bill Hobba:

Another way of looking at is explained in Zee's excellent book on Quantum
Field Theory - Quantum Field Theory in a Nutshell. The modern way of
looking at it to impose a cutoff (technically called a regularization - we
can cutoff energy and even - wow - dimensions). The bottom line is once we
impose this cutoff in both gravity and EM the infinites disappear. For
gravity you may find the following interesting
http://arxiv.org/abs/gr-qc/9512024.

Your book omits an important point, which is that regardless of
what the theory is, renormalization is involved. If that weren't
the case, it would be impossible to do _any_ physics without knowing
the most fundamental theory of everything. Renormalization allows
you to ignore the underlying details in favor of macroscopic
paramaters. The same infinities would occur in something like
a theory of sound waves, if you tried to describe the physics at
intermolecular distances in terms of the bulk parameters.

Maybe because quantum mechanics is non-local. Quantum Mechanics uses
spinors, mathematical objects similar to vectors, but which change sign
under a rotation of 2p radians.

"Bilge" wrote in message
...
Bill Hobba:

Another way of looking at is explained in Zee's excellent book on Quantum
Field Theory - Quantum Field Theory in a Nutshell. The modern way of
looking at it to impose a cutoff (technically called a regularization -
we
can cutoff energy and even - wow - dimensions). The bottom line is once
we
impose this cutoff in both gravity and EM the infinites disappear. For
gravity you may find the following interesting
http://arxiv.org/abs/gr-qc/9512024.

Your book omits an important point, which is that regardless of
what the theory is, renormalization is involved.

It is. However I suspect the problem is not Zees fault - it is my
explanation and comprehension.

If that weren't
the case, it would be impossible to do _any_ physics without knowing
the most fundamental theory of everything. Renormalization allows
you to ignore the underlying details in favor of macroscopic
paramaters. The same infinities would occur in something like
a theory of sound waves, if you tried to describe the physics at
intermolecular distances in terms of the bulk parameters.

Yes indeed. And Zee examines such issues in Chapter 6.8 - under the heading
of Renormalization Group Flow as a Natural Concept in High Energy Physics
and Condensed Matter Physics. I rather like Zees closing remarks from that
chapter:

'In a sense the renormalization group goes back to a basis notion of
physics, that the effective description can and should change as we move
from one length scale to another. For example in hydrodynamics we do not
have to keep track of the detailed interaction among water molecules.
Similarly when we apply the renormalization group flow to the strong
interaction starting at high energies and moving toward low energies the
effective theory goes from a theory of quarks and gluons to a theory of
nucleons and mesons. In this more general picture then we no longer think
of flowing in a space of coupling constants but in the 'space of
Hamiltonians' that some condensed matter physics like to talk about.'

Thanks
Bill

"Orion" wrote in message
oups.com...
Maybe because quantum mechanics is non-local.

Where do you get the idea QM is non local? Bells inequality uses the
concept of local realism - it is that concept that has been disproved - not
locality.

Thanks
Bill

Quantum Mechanics uses
spinors, mathematical objects similar to vectors, but which change sign
under a rotation of 2p radians.

Bill Hobba wrote:
"Orion" wrote in message
oups.com...
Maybe because quantum mechanics is non-local.

Where do you get the idea QM is non local? Bells inequality uses the
concept of local realism - it is that concept that has been disproved - not
locality.

That's a matter of INTERPRETATION and PHILOSOPHY. Other interpretations
suggest nonlocality.

Thanks
Bill

Quantum Mechanics uses
spinors, mathematical objects similar to vectors, but which change sign
under a rotation of 2p radians.

"Schoenfeld" wrote in message
ups.com...

Bill Hobba wrote:
"Orion" wrote in message
oups.com...
Maybe because quantum mechanics is non-local.

Where do you get the idea QM is non local? Bells inequality uses the
concept of local realism - it is that concept that has been disproved -
not
locality.

That's a matter of INTERPRETATION and PHILOSOPHY. Other interpretations
suggest nonlocality.

I certainly agree with that. And the converse is also true - other
interpretations do not as well - see
http://quantum.phys.cmu.edu/quest.html

'Is quantum mechanics nonlocal? This depends on what one means by
"nonlocal." Two separated quantum systems A and B can be in an entangled
state that lacks any classical analog. However, it is better to think of
this as a nonclassical rather than as a nonlocal state, since doing
something to system A cannot have any influence on system B as long as the
two are sufficiently far apart. In particular, quantum theory gives no
support to the notion that the world is infested by mysterious long-range
influences that propagate faster than the speed of light. Claims to the
contrary are based upon an inconsistent or inadequate formulations of
quantum principles, typically with reference to measurements.'

Thanks
Bill


Thanks
Bill

Quantum Mechanics uses
spinors, mathematical objects similar to vectors, but which change sign
under a rotation of 2p radians.

Schoenfeld:

Bill Hobba wrote:
"Orion" wrote in message
oups.com...
Maybe because quantum mechanics is non-local.

Where do you get the idea QM is non local? Bells inequality uses the
concept of local realism - it is that concept that has been disproved - not
locality.

That's a matter of INTERPRETATION and PHILOSOPHY. Other interpretations
suggest nonlocality.

Except that by assuming non-locality, you still give up local
realism. The only difference is that you've chosen a weird way to
do it, by giving some existence to a process that isn't ``real''
using that same definition of realism. The choice of using the
word ``realism'' as an attribute was a poor one, since it suggests
the conflict is nature's problem rather than a problem with the
particular concept of reality chosen by humans.

In article .com,
Orion wrote:
Maybe because quantum mechanics is non-local. Quantum Mechanics uses
spinors, mathematical objects similar to vectors, but which change sign
under a rotation of 2p radians.



It's true that spinor particles need to be rotated by 4*pi to return to
their original state. I know the guy, Sam Werner, who demonstrated that
in a neutron interferometer. But that doesn't make it non-local. That
just makes them spinor particles.

--
"A good plan executed right now is far better than a perfect plan
executed next week."
-Gen. George S. Patton

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.

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.)

---
David Cross
dcross1 AT shaw DOT ca