Mixing
Infinitely many conserved currents in any QFT?
Clash Royale CLAN TAG#URR8PPP
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So I have the following curiosity: Consider for example, in QED, the quantity
$$
j^muequivpartial_nu (lambda(x) F^mu nu)
$$
where $lambda(x)$ is an arbitrary scalar function of spacetime, constructed from elements of the theory, e.g.
$$
lambda(x)=A_mu A^mu F_rho sigma F^rho sigma
$$
or anything else you can think of. Then, since $F^mu nu$ is antisymmetric, $partial_mu j^mu=0$ identically.
In fact this can be generalized to any theory; construct from various elements an antisymmetric two-rank tensor, and a scalar, multiply them together, and there is a corresponding conserved current for any choice you make. If these currents are not trivial (e.g. giving only vanishing charges) then it appears that all theories give an infinite landscape of conserved currents. Is this so? Am I missing something? How is this justified logically?
electromagnetism conservation-laws field-theory gauge-theory
edited 4 hours ago
Qmechanic♦
98k121681061
asked 4 hours ago
Valentina
858
add a comment |Â
up vote
2
down vote
favorite
So I have the following curiosity: Consider for example, in QED, the quantity
$$
j^muequivpartial_nu (lambda(x) F^mu nu)
$$
where $lambda(x)$ is an arbitrary scalar function of spacetime, constructed from elements of the theory, e.g.
$$
lambda(x)=A_mu A^mu F_rho sigma F^rho sigma
$$
or anything else you can think of. Then, since $F^mu nu$ is antisymmetric, $partial_mu j^mu=0$ identically.
In fact this can be generalized to any theory; construct from various elements an antisymmetric two-rank tensor, and a scalar, multiply them together, and there is a corresponding conserved current for any choice you make. If these currents are not trivial (e.g. giving only vanishing charges) then it appears that all theories give an infinite landscape of conserved currents. Is this so? Am I missing something? How is this justified logically?
electromagnetism conservation-laws field-theory gauge-theory
edited 4 hours ago
Qmechanic♦
98k121681061
asked 4 hours ago
Valentina
858
Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I have the following curiosity: Consider for example, in QED, the quantity
$$
j^muequivpartial_nu (lambda(x) F^mu nu)
$$
where $lambda(x)$ is an arbitrary scalar function of spacetime, constructed from elements of the theory, e.g.
$$
lambda(x)=A_mu A^mu F_rho sigma F^rho sigma
$$
or anything else you can think of. Then, since $F^mu nu$ is antisymmetric, $partial_mu j^mu=0$ identically.
In fact this can be generalized to any theory; construct from various elements an antisymmetric two-rank tensor, and a scalar, multiply them together, and there is a corresponding conserved current for any choice you make. If these currents are not trivial (e.g. giving only vanishing charges) then it appears that all theories give an infinite landscape of conserved currents. Is this so? Am I missing something? How is this justified logically?
electromagnetism conservation-laws field-theory gauge-theory
edited 4 hours ago
Qmechanic♦
98k121681061
asked 4 hours ago
Valentina
858
So I have the following curiosity: Consider for example, in QED, the quantity
$$
j^muequivpartial_nu (lambda(x) F^mu nu)
$$
where $lambda(x)$ is an arbitrary scalar function of spacetime, constructed from elements of the theory, e.g.
$$
lambda(x)=A_mu A^mu F_rho sigma F^rho sigma
$$
or anything else you can think of. Then, since $F^mu nu$ is antisymmetric, $partial_mu j^mu=0$ identically.
In fact this can be generalized to any theory; construct from various elements an antisymmetric two-rank tensor, and a scalar, multiply them together, and there is a corresponding conserved current for any choice you make. If these currents are not trivial (e.g. giving only vanishing charges) then it appears that all theories give an infinite landscape of conserved currents. Is this so? Am I missing something? How is this justified logically?
electromagnetism conservation-laws field-theory gauge-theory
electromagnetism conservation-laws field-theory gauge-theory
edited 4 hours ago
Qmechanic♦
98k121681061
asked 4 hours ago
Valentina
858
edited 4 hours ago
Qmechanic♦
98k121681061
asked 4 hours ago
Valentina
858
edited 4 hours ago
Qmechanic♦
98k121681061
edited 4 hours ago
Qmechanic♦
98k121681061
edited 4 hours ago
Qmechanic♦
98k121681061
98k121681061
asked 4 hours ago
Valentina
858
asked 4 hours ago
Valentina
858
asked 4 hours ago
Valentina
858
858
Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago
add a comment |Â
Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago
Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
OP wrote (v2):
If these currents are not trivial [...]
In fact most$^dagger$ of them are trivial. The corresponding charges $$Q ~:=~ int_V d^3x~ j^0~=~int_V d^3x~sum_i=1^3d_i(lambda F^i0)~=~int_partial V d^2x~(ldots) ~=~0$$
vanish if the components $lambda F^i0$ fall off fast enough at spatial infinity $partial V$.
--
$^dagger$ The case $lambda=1$ corresponds to ordinary electric current, cf. Maxwell's eqs.
answered 3 hours ago
Qmechanic♦
98k121681061
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
add a comment |Â
up vote
1
down vote
If we have $partial_nu F^munu=0$ on-shell, as happens e.g. in EM without a source current, $F^munuX_nu$ is conserved provided $partial_mu X_nu$ is symmetric, as happens e.g. with $X_nu=partial_nulambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.
answered 2 hours ago
J.G.
8,41421125
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
OP wrote (v2):
If these currents are not trivial [...]
In fact most$^dagger$ of them are trivial. The corresponding charges $$Q ~:=~ int_V d^3x~ j^0~=~int_V d^3x~sum_i=1^3d_i(lambda F^i0)~=~int_partial V d^2x~(ldots) ~=~0$$
vanish if the components $lambda F^i0$ fall off fast enough at spatial infinity $partial V$.
--
$^dagger$ The case $lambda=1$ corresponds to ordinary electric current, cf. Maxwell's eqs.
answered 3 hours ago
Qmechanic♦
98k121681061
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
OP wrote (v2):
If these currents are not trivial [...]
In fact most$^dagger$ of them are trivial. The corresponding charges $$Q ~:=~ int_V d^3x~ j^0~=~int_V d^3x~sum_i=1^3d_i(lambda F^i0)~=~int_partial V d^2x~(ldots) ~=~0$$
vanish if the components $lambda F^i0$ fall off fast enough at spatial infinity $partial V$.
--
$^dagger$ The case $lambda=1$ corresponds to ordinary electric current, cf. Maxwell's eqs.
answered 3 hours ago
Qmechanic♦
98k121681061
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
OP wrote (v2):
If these currents are not trivial [...]
In fact most$^dagger$ of them are trivial. The corresponding charges $$Q ~:=~ int_V d^3x~ j^0~=~int_V d^3x~sum_i=1^3d_i(lambda F^i0)~=~int_partial V d^2x~(ldots) ~=~0$$
vanish if the components $lambda F^i0$ fall off fast enough at spatial infinity $partial V$.
--
$^dagger$ The case $lambda=1$ corresponds to ordinary electric current, cf. Maxwell's eqs.
answered 3 hours ago
Qmechanic♦
98k121681061
OP wrote (v2):
If these currents are not trivial [...]
In fact most$^dagger$ of them are trivial. The corresponding charges $$Q ~:=~ int_V d^3x~ j^0~=~int_V d^3x~sum_i=1^3d_i(lambda F^i0)~=~int_partial V d^2x~(ldots) ~=~0$$
vanish if the components $lambda F^i0$ fall off fast enough at spatial infinity $partial V$.
--
$^dagger$ The case $lambda=1$ corresponds to ordinary electric current, cf. Maxwell's eqs.
answered 3 hours ago
Qmechanic♦
98k121681061
edited 1 hour ago
answered 3 hours ago
Qmechanic♦
98k121681061
answered 3 hours ago
Qmechanic♦
98k121681061
answered 3 hours ago
Qmechanic♦
98k121681061
98k121681061
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
add a comment |Â
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks.
– Valentina
3 hours ago
add a comment |Â
up vote
1
down vote
If we have $partial_nu F^munu=0$ on-shell, as happens e.g. in EM without a source current, $F^munuX_nu$ is conserved provided $partial_mu X_nu$ is symmetric, as happens e.g. with $X_nu=partial_nulambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.
answered 2 hours ago
J.G.
8,41421125
add a comment |Â
up vote
1
down vote
If we have $partial_nu F^munu=0$ on-shell, as happens e.g. in EM without a source current, $F^munuX_nu$ is conserved provided $partial_mu X_nu$ is symmetric, as happens e.g. with $X_nu=partial_nulambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.
answered 2 hours ago
J.G.
8,41421125
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If we have $partial_nu F^munu=0$ on-shell, as happens e.g. in EM without a source current, $F^munuX_nu$ is conserved provided $partial_mu X_nu$ is symmetric, as happens e.g. with $X_nu=partial_nulambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.
answered 2 hours ago
J.G.
8,41421125
If we have $partial_nu F^munu=0$ on-shell, as happens e.g. in EM without a source current, $F^munuX_nu$ is conserved provided $partial_mu X_nu$ is symmetric, as happens e.g. with $X_nu=partial_nulambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.
answered 2 hours ago
J.G.
8,41421125
edited 2 hours ago
answered 2 hours ago
J.G.
8,41421125
answered 2 hours ago
J.G.
8,41421125
answered 2 hours ago
J.G.
8,41421125
8,41421125
add a comment |Â
add a comment |Â
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Maybe this obvious, but why is $partial_mu j^mu=(partial_mulambda)(partial_nu F^munu) = 0$ necessarily?
– WAH
4 hours ago
It is $partial_mu partial_nu (lambda F^mu nu)=-partial_nu partial_mu (lambda F^nu mu)=-partial_mu partial_nu (lambda F^mu nu)$ where in the last equality I just renamed the dummy indices.
– Valentina
4 hours ago
I agree that $(partial_mupartial_nu lambda)F^munu=0$ and $lambda partial_mupartial_nu F^munu=0$. So then $partial_mupartial_nu(lambda F^munu)=(partial_mulambda)(partial_nu F^munu)$. It's not clear to me that the last quantity is always identically 0.
– WAH
4 hours ago