rpenner
5th November 2009 - 01:37 AM
QUOTE (Rebe+Nov 5 2009, 01:13 AM)
I'm trying to verify that the operator
R^μν = g^μν □ - ∂^μ ∂^ν
of the Maxwell equation in an arbitrary gauge R^μν A_ν = j^ν can't be inverted, but I'm a little stuck.
Some help?
Now is the d'Alembertian or the four-gradient? Guessing...
R^μν A_ν = g^μν □ A_ν - ∂^μ ∂^ν A_ν = g^μν ∂_ξ ∂^ξ A_ν - ∂^μ ∂^ν A_ν
Oops out of time.
// continuing.
Maxwell's equations can be written as:
∂_u ∂^u A_ν - ∂_ν ∂^μ A_μ = j_ν
But if A_ξ = ∂_ξ a, then j_ξ = ∂_u ∂^u ∂_ν a - ∂_ν ∂^μ ∂_μ a = 0 for any a.
Thus anything in the form ∂_ξ a is in the null subspace for the "maxwell operator"
OK, my tensor manipulation skills are next to nil, but trying the same thing, don't we have:
For R^μν ∂_ν a we have g^μν ∂_ξ ∂^ξ ∂_ν a - ∂^μ ∂^ν ∂_ν a
= ∂_ν ∂^ν ∂^μ a - ∂^μ ∂^ν ∂_ν a
= 0 ?
rpenner
5th November 2009 - 11:09 PM
Well this thread is going nowhere fast. I've yet to even sit in on a class where they use these notations. And clearly this forum is unfriendly to matrix maths, let alone tensors.
Prometheus, AlphaNumeric, are you going to chime in on how off-base I am?
prometheus
6th November 2009 - 07:48 AM
The best way to do this is to show that the operator in question is a projection operator, ie that R^2 = R, and once you've done that it's easy to show that, if a projection operator has an inverse then it must be the identity. Since your operator is clearly not the identity it cannot have an inverse.
RPenner, your solution doesn't really show that R has no inverse, but it is a consequence of the fact that it doesn't. This crops up in gauge theories because when you try to find the Greens functions of the theory you're trying to solve R G = i delta which is essentially assuming that R does have an inverse, so it's pretty important that it doesn't.
rpenner
6th November 2009 - 08:11 AM
Thanks.
qwtyu
8th November 2009 - 01:05 AM
QUOTE (prometheus+Nov 6 2009, 07:48 AM)
The best way to do this is to show that the operator in question is a projection operator, ie that R^2 = R, and once you've done that it's easy to show that, if a projection operator has an inverse then it must be the identity. Since your operator is clearly not the identity it cannot have an inverse.
RPenner, your solution doesn't really show that R has no inverse, but it is a consequence of the fact that it doesn't. This crops up in gauge theories because when you try to find the Greens functions of the theory you're trying to solve R G = i delta which is essentially assuming that R
does have an inverse, so it's pretty important that it doesn't.
QUOTE
This crops up in gauge theories because when you try to find the Greens functions of the theory you're trying to solve R G = i delta which is essentially assuming that R does have an inverse, so it's pretty important that it doesn't.
Can you please explain this?
rpenner
8th November 2009 - 04:13 AM
has an inverse.
The matrix
CODE
1 0
1 0
clearly does not.
The generalization of this is that some legal (two-slot) tensor operators have inverses and some do not.
We call the second class "projection operators" since at least part of the allowable base vector set gets "projected" to zero. Other terminology for this condition is a non-trival kernel.
qwtyu
8th November 2009 - 01:25 PM
QUOTE (rpenner+Nov 8 2009, 04:13 AM)
has an inverse.
The matrix
CODE
1 0
1 0
clearly does not.
The generalization of this is that some legal (two-slot) tensor operators have inverses and some do not.
We call the second class "projection operators" since at least part of the allowable base vector set gets "projected" to zero. Other terminology for this condition is a non-trival kernel.
Yes, but I was wondering under what condition would a physical theory not be bijective.
I really do not know with this case. So, I was asking.