Mixing
Are Maxwell's equations “physical”?
$begingroup$
The canonical Maxwell's equations are derivable from the Lagrangian
$${cal L} = -frac{1}{4}F_{munu}F^{munu} $$
by solving the Euler-Lagrange equations.
However: The Lagrangian above is invariant under the gauge transformation
$$A_mu to A_mu - partial_mu Lambda(x) $$
for some scalar fiend $Lambda(x)$ that vanishes at infinity. This implies that there will be redundant degrees of freedom in our equations not motion (i.e. Maxwell's equations).
Therefore, as I understand gauge fixing, this implies that Maxwell's equations (without gauge fixing) can lead to unphysical predictions.
Question: Hence my question is simply are Maxwell's equations (the ones derived from $cal{L}$ above) actually physical, in the sense they do not make unphysical predictions?
Example: The general solution to the equations of motion derived from $cal{L}$ is given by
$$A_mu(x) = sum_{r=0}^3 int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2E_{mathbf{p}}}}left(epsilon^r_mu(mathbf{p}) a^r_mathbf{p}e^{-ipx} + epsilon^{*r}_mu(mathbf{p}) (a^r_mathbf{p})^dagger e^{ipx} right)$$
where we have, at first, 4 polarization states for external photons.
My understanding: is that we can remove one of these degrees of freedom by realizing that $A_0$ is not dynamical, but to remove the other one we have to impose gauge invariance (cf. (2)). This seems to imply that unless we fix a gauge Maxwell's equations will predict a longitudinal polarization for the photon.
lagrangian-formalism gauge-theory maxwell-equations gauge
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
$endgroup$
|
show 2 more comments
$begingroup$
The canonical Maxwell's equations are derivable from the Lagrangian
$${cal L} = -frac{1}{4}F_{munu}F^{munu} $$
by solving the Euler-Lagrange equations.
However: The Lagrangian above is invariant under the gauge transformation
$$A_mu to A_mu - partial_mu Lambda(x) $$
for some scalar fiend $Lambda(x)$ that vanishes at infinity. This implies that there will be redundant degrees of freedom in our equations not motion (i.e. Maxwell's equations).
Therefore, as I understand gauge fixing, this implies that Maxwell's equations (without gauge fixing) can lead to unphysical predictions.
Question: Hence my question is simply are Maxwell's equations (the ones derived from $cal{L}$ above) actually physical, in the sense they do not make unphysical predictions?
Example: The general solution to the equations of motion derived from $cal{L}$ is given by
$$A_mu(x) = sum_{r=0}^3 int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2E_{mathbf{p}}}}left(epsilon^r_mu(mathbf{p}) a^r_mathbf{p}e^{-ipx} + epsilon^{*r}_mu(mathbf{p}) (a^r_mathbf{p})^dagger e^{ipx} right)$$
where we have, at first, 4 polarization states for external photons.
My understanding: is that we can remove one of these degrees of freedom by realizing that $A_0$ is not dynamical, but to remove the other one we have to impose gauge invariance (cf. (2)). This seems to imply that unless we fix a gauge Maxwell's equations will predict a longitudinal polarization for the photon.
lagrangian-formalism gauge-theory maxwell-equations gauge
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
$endgroup$
1
$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
5
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
2
$begingroup$
I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
$endgroup$
– chrylis
Jan 6 at 5:39
|
show 2 more comments
$begingroup$
The canonical Maxwell's equations are derivable from the Lagrangian
$${cal L} = -frac{1}{4}F_{munu}F^{munu} $$
by solving the Euler-Lagrange equations.
However: The Lagrangian above is invariant under the gauge transformation
$$A_mu to A_mu - partial_mu Lambda(x) $$
for some scalar fiend $Lambda(x)$ that vanishes at infinity. This implies that there will be redundant degrees of freedom in our equations not motion (i.e. Maxwell's equations).
Therefore, as I understand gauge fixing, this implies that Maxwell's equations (without gauge fixing) can lead to unphysical predictions.
Question: Hence my question is simply are Maxwell's equations (the ones derived from $cal{L}$ above) actually physical, in the sense they do not make unphysical predictions?
Example: The general solution to the equations of motion derived from $cal{L}$ is given by
$$A_mu(x) = sum_{r=0}^3 int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2E_{mathbf{p}}}}left(epsilon^r_mu(mathbf{p}) a^r_mathbf{p}e^{-ipx} + epsilon^{*r}_mu(mathbf{p}) (a^r_mathbf{p})^dagger e^{ipx} right)$$
where we have, at first, 4 polarization states for external photons.
My understanding: is that we can remove one of these degrees of freedom by realizing that $A_0$ is not dynamical, but to remove the other one we have to impose gauge invariance (cf. (2)). This seems to imply that unless we fix a gauge Maxwell's equations will predict a longitudinal polarization for the photon.
lagrangian-formalism gauge-theory maxwell-equations gauge
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
$endgroup$
The canonical Maxwell's equations are derivable from the Lagrangian
$${cal L} = -frac{1}{4}F_{munu}F^{munu} $$
by solving the Euler-Lagrange equations.
However: The Lagrangian above is invariant under the gauge transformation
$$A_mu to A_mu - partial_mu Lambda(x) $$
for some scalar fiend $Lambda(x)$ that vanishes at infinity. This implies that there will be redundant degrees of freedom in our equations not motion (i.e. Maxwell's equations).
Therefore, as I understand gauge fixing, this implies that Maxwell's equations (without gauge fixing) can lead to unphysical predictions.
Question: Hence my question is simply are Maxwell's equations (the ones derived from $cal{L}$ above) actually physical, in the sense they do not make unphysical predictions?
Example: The general solution to the equations of motion derived from $cal{L}$ is given by
$$A_mu(x) = sum_{r=0}^3 int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2E_{mathbf{p}}}}left(epsilon^r_mu(mathbf{p}) a^r_mathbf{p}e^{-ipx} + epsilon^{*r}_mu(mathbf{p}) (a^r_mathbf{p})^dagger e^{ipx} right)$$
where we have, at first, 4 polarization states for external photons.
My understanding: is that we can remove one of these degrees of freedom by realizing that $A_0$ is not dynamical, but to remove the other one we have to impose gauge invariance (cf. (2)). This seems to imply that unless we fix a gauge Maxwell's equations will predict a longitudinal polarization for the photon.
lagrangian-formalism gauge-theory maxwell-equations gauge
lagrangian-formalism gauge-theory maxwell-equations gauge
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
edited Jan 5 at 21:05
Qmechanic♦
103k121851175
103k121851175
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
asked Jan 5 at 19:30
InertialObserverInertialObserver
2,318624
2,318624
1
$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
5
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
2
$begingroup$
I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
$endgroup$
– chrylis
Jan 6 at 5:39
|
show 2 more comments
1
$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
5
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
2
$begingroup$
I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
$endgroup$
– chrylis
Jan 6 at 5:39
1
1
$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
5
5
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
2
2
$begingroup$
I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
$endgroup$
– chrylis
Jan 6 at 5:39
$begingroup$
I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
$endgroup$
– chrylis
Jan 6 at 5:39
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.
At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.
However, such a theory has a deep problem: the equation of motion $partial_mu F^{munu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.
Note that this isn't true if you add a mass term $m^2 A_mu A^mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.
edited Jan 6 at 15:59
J.G.
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
$endgroup$
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
add a comment |
$begingroup$
You are correct that there are gauge degrees of freedom in the solution for $A_mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical.
The polarization of the classical $A_mu$ has nothing to do with any photons. There are no photons in a classical theory. Maxwell's equations alone classically fully suffice to allow only transverse EM waves, see e.g. this question and its answers.
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
$endgroup$
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
add a comment |
$begingroup$
The two answers above are respectable answers and reflect the present theory. I have a different answer, namely that th epotential is physical, and I have proven that it is the only correct one. 19 years ago I published a paper with a different formulation. It starts out with the Fermi Lagrangian ${cal L}= frac{1}{2} epsilon_0 partial_mu A_nu partial^mu A^nu$, to arrive at the wave equation $partial_mu partial^mu A^nu = mu_0 j^nu$. The potential $A^nu$ is uniquely determined by the requirement of causality. Since the wave equation is bijective, current conservation, $partial_mu j^mu = 0$ implies $partial_mu A^mu = 0$ and vice versa. That is, only the transversal part of the solution space describes the field of conserved current. I show that the Lorentz force follows. The solution of the wave equation can be written in terms of a propagator, which is simply the Green's function of the d'Alembertian operator $partial_mu partial^mu$.
Let me explain why I say that this formulation is the only correct one. In principle my formulation predict exactly the same measurement outcome as long as only electromagnetism is considered. The definitive argument however comes from gravitation. An electromagnetic energy-momentum distribution implies a gravitational field. In my theory the EM distribution is the Nöther distribution of the (Fermi) Lagrangian. This is not the case for the gauge invariant theory. The standard EM distribution of the Belinfante-Rosenfeld Lagrangian is an ad hoc modification of the Nöther form, and by this an ad hoc modification of the gravitational field. This ad hoc modification is necessary because the BR Lagrangian implies a non-gauge invariant, asymmetric (GRT requires symmetry) Nöther form.
There are many advantages to this formulation, as you can see for yourself in the paper.
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
$endgroup$
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
|
show 7 more comments
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3 Answers
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3 Answers
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$begingroup$
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.
At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.
However, such a theory has a deep problem: the equation of motion $partial_mu F^{munu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.
Note that this isn't true if you add a mass term $m^2 A_mu A^mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.
edited Jan 6 at 15:59
J.G.
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
$endgroup$
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
add a comment |
$begingroup$
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.
At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.
However, such a theory has a deep problem: the equation of motion $partial_mu F^{munu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.
Note that this isn't true if you add a mass term $m^2 A_mu A^mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.
edited Jan 6 at 15:59
J.G.
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
$endgroup$
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
add a comment |
$begingroup$
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.
At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.
However, such a theory has a deep problem: the equation of motion $partial_mu F^{munu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.
Note that this isn't true if you add a mass term $m^2 A_mu A^mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.
edited Jan 6 at 15:59
J.G.
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
$endgroup$
Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian.
At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to have a gauge symmetry to avoid redundancy. If you don't have a gauge symmetry, the theory simply won't be classical electromagnetism. Instead, waves with a fixed wavevector will have three independent polarizations.
However, such a theory has a deep problem: the equation of motion $partial_mu F^{munu} = 0$ only suffices to determine the evolution of the fields, not the potentials. That is, the classical theory is not deterministic! This directly corresponds to a loss of unitarity at the quantum level. Both these features are sufficiently bad that some would say gauge symmetry is a requirement for the theory to make sense.
Note that this isn't true if you add a mass term $m^2 A_mu A^mu$ to the Lagrangian; in this case the equation of motion is simply the Klein-Gordan equation for each component. This is the Proca theory, which makes perfect sense without gauge symmetry. However, the problems are reintroduced when the mass goes to zero. As such, gauge symmetry for the massless theory isn't really a choice, and it's not specific to quantization either; it's instead required at the classical level and transferred to the quantum.
edited Jan 6 at 15:59
J.G.
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
edited Jan 6 at 15:59
J.G.
9,17921528
edited Jan 6 at 15:59
J.G.
9,17921528
edited Jan 6 at 15:59
J.G.
9,17921528
9,17921528
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
answered Jan 5 at 20:35
knzhouknzhou
43.3k11118206
43.3k11118206
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
add a comment |
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
$begingroup$
When you say we need to have a gauge symmetry to avoid redundancy you you mean we have to fix a gauge? And when you say add the gauge symmetry to the lagrangian, do you mean the gauge fixing bit?
$endgroup$
– InertialObserver
Jan 5 at 20:37
4
4
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
$begingroup$
@InertialObserver Having a gauge symmetry means saying "these two configurations of different $A_mu$'s are really the same physical thing", hence introducing a redundancy. Gauge fixing is a separate thing, where you take the different configurations that are physically equivalent, and just forbid all but one of them, to remove the redundancy. You can do that at the classical or quantum level, and the stage at which you do it is really up to convenience.
$endgroup$
– knzhou
Jan 5 at 20:45
add a comment |
$begingroup$
You are correct that there are gauge degrees of freedom in the solution for $A_mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical.
The polarization of the classical $A_mu$ has nothing to do with any photons. There are no photons in a classical theory. Maxwell's equations alone classically fully suffice to allow only transverse EM waves, see e.g. this question and its answers.
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
$endgroup$
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
add a comment |
$begingroup$
You are correct that there are gauge degrees of freedom in the solution for $A_mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical.
The polarization of the classical $A_mu$ has nothing to do with any photons. There are no photons in a classical theory. Maxwell's equations alone classically fully suffice to allow only transverse EM waves, see e.g. this question and its answers.
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
$endgroup$
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
add a comment |
$begingroup$
You are correct that there are gauge degrees of freedom in the solution for $A_mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical.
The polarization of the classical $A_mu$ has nothing to do with any photons. There are no photons in a classical theory. Maxwell's equations alone classically fully suffice to allow only transverse EM waves, see e.g. this question and its answers.
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
$endgroup$
You are correct that there are gauge degrees of freedom in the solution for $A_mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical.
The polarization of the classical $A_mu$ has nothing to do with any photons. There are no photons in a classical theory. Maxwell's equations alone classically fully suffice to allow only transverse EM waves, see e.g. this question and its answers.
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
answered Jan 5 at 20:11
ACuriousMind♦ACuriousMind
70.7k17124306
70.7k17124306
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
add a comment |
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
So is imposing gauge invariance to eliminate a degree of freedom a choice rather than necessary?
$endgroup$
– InertialObserver
Jan 5 at 20:13
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
@InertialObserver All physically relevant quantities are gauge invariant to begin with and hence unaffected by whether or not we fix a gauge. Gauge fixing is always a matter of formalistic convenience and not physical "necessity", since the gauge symmetry does not express any "real" symmetry to begin with, but just a redundancy in our formalism.
$endgroup$
– ACuriousMind♦
Jan 6 at 13:25
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
$begingroup$
That’s nice point.. idk how I overlooked that.. is that what we mean when we say that a gauge theory is a theory is really an equivalence class?
$endgroup$
– InertialObserver
Jan 6 at 20:41
add a comment |
$begingroup$
The two answers above are respectable answers and reflect the present theory. I have a different answer, namely that th epotential is physical, and I have proven that it is the only correct one. 19 years ago I published a paper with a different formulation. It starts out with the Fermi Lagrangian ${cal L}= frac{1}{2} epsilon_0 partial_mu A_nu partial^mu A^nu$, to arrive at the wave equation $partial_mu partial^mu A^nu = mu_0 j^nu$. The potential $A^nu$ is uniquely determined by the requirement of causality. Since the wave equation is bijective, current conservation, $partial_mu j^mu = 0$ implies $partial_mu A^mu = 0$ and vice versa. That is, only the transversal part of the solution space describes the field of conserved current. I show that the Lorentz force follows. The solution of the wave equation can be written in terms of a propagator, which is simply the Green's function of the d'Alembertian operator $partial_mu partial^mu$.
Let me explain why I say that this formulation is the only correct one. In principle my formulation predict exactly the same measurement outcome as long as only electromagnetism is considered. The definitive argument however comes from gravitation. An electromagnetic energy-momentum distribution implies a gravitational field. In my theory the EM distribution is the Nöther distribution of the (Fermi) Lagrangian. This is not the case for the gauge invariant theory. The standard EM distribution of the Belinfante-Rosenfeld Lagrangian is an ad hoc modification of the Nöther form, and by this an ad hoc modification of the gravitational field. This ad hoc modification is necessary because the BR Lagrangian implies a non-gauge invariant, asymmetric (GRT requires symmetry) Nöther form.
There are many advantages to this formulation, as you can see for yourself in the paper.
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
$endgroup$
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
|
show 7 more comments
$begingroup$
The two answers above are respectable answers and reflect the present theory. I have a different answer, namely that th epotential is physical, and I have proven that it is the only correct one. 19 years ago I published a paper with a different formulation. It starts out with the Fermi Lagrangian ${cal L}= frac{1}{2} epsilon_0 partial_mu A_nu partial^mu A^nu$, to arrive at the wave equation $partial_mu partial^mu A^nu = mu_0 j^nu$. The potential $A^nu$ is uniquely determined by the requirement of causality. Since the wave equation is bijective, current conservation, $partial_mu j^mu = 0$ implies $partial_mu A^mu = 0$ and vice versa. That is, only the transversal part of the solution space describes the field of conserved current. I show that the Lorentz force follows. The solution of the wave equation can be written in terms of a propagator, which is simply the Green's function of the d'Alembertian operator $partial_mu partial^mu$.
Let me explain why I say that this formulation is the only correct one. In principle my formulation predict exactly the same measurement outcome as long as only electromagnetism is considered. The definitive argument however comes from gravitation. An electromagnetic energy-momentum distribution implies a gravitational field. In my theory the EM distribution is the Nöther distribution of the (Fermi) Lagrangian. This is not the case for the gauge invariant theory. The standard EM distribution of the Belinfante-Rosenfeld Lagrangian is an ad hoc modification of the Nöther form, and by this an ad hoc modification of the gravitational field. This ad hoc modification is necessary because the BR Lagrangian implies a non-gauge invariant, asymmetric (GRT requires symmetry) Nöther form.
There are many advantages to this formulation, as you can see for yourself in the paper.
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
$endgroup$
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
|
show 7 more comments
$begingroup$
The two answers above are respectable answers and reflect the present theory. I have a different answer, namely that th epotential is physical, and I have proven that it is the only correct one. 19 years ago I published a paper with a different formulation. It starts out with the Fermi Lagrangian ${cal L}= frac{1}{2} epsilon_0 partial_mu A_nu partial^mu A^nu$, to arrive at the wave equation $partial_mu partial^mu A^nu = mu_0 j^nu$. The potential $A^nu$ is uniquely determined by the requirement of causality. Since the wave equation is bijective, current conservation, $partial_mu j^mu = 0$ implies $partial_mu A^mu = 0$ and vice versa. That is, only the transversal part of the solution space describes the field of conserved current. I show that the Lorentz force follows. The solution of the wave equation can be written in terms of a propagator, which is simply the Green's function of the d'Alembertian operator $partial_mu partial^mu$.
Let me explain why I say that this formulation is the only correct one. In principle my formulation predict exactly the same measurement outcome as long as only electromagnetism is considered. The definitive argument however comes from gravitation. An electromagnetic energy-momentum distribution implies a gravitational field. In my theory the EM distribution is the Nöther distribution of the (Fermi) Lagrangian. This is not the case for the gauge invariant theory. The standard EM distribution of the Belinfante-Rosenfeld Lagrangian is an ad hoc modification of the Nöther form, and by this an ad hoc modification of the gravitational field. This ad hoc modification is necessary because the BR Lagrangian implies a non-gauge invariant, asymmetric (GRT requires symmetry) Nöther form.
There are many advantages to this formulation, as you can see for yourself in the paper.
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
$endgroup$
The two answers above are respectable answers and reflect the present theory. I have a different answer, namely that th epotential is physical, and I have proven that it is the only correct one. 19 years ago I published a paper with a different formulation. It starts out with the Fermi Lagrangian ${cal L}= frac{1}{2} epsilon_0 partial_mu A_nu partial^mu A^nu$, to arrive at the wave equation $partial_mu partial^mu A^nu = mu_0 j^nu$. The potential $A^nu$ is uniquely determined by the requirement of causality. Since the wave equation is bijective, current conservation, $partial_mu j^mu = 0$ implies $partial_mu A^mu = 0$ and vice versa. That is, only the transversal part of the solution space describes the field of conserved current. I show that the Lorentz force follows. The solution of the wave equation can be written in terms of a propagator, which is simply the Green's function of the d'Alembertian operator $partial_mu partial^mu$.
Let me explain why I say that this formulation is the only correct one. In principle my formulation predict exactly the same measurement outcome as long as only electromagnetism is considered. The definitive argument however comes from gravitation. An electromagnetic energy-momentum distribution implies a gravitational field. In my theory the EM distribution is the Nöther distribution of the (Fermi) Lagrangian. This is not the case for the gauge invariant theory. The standard EM distribution of the Belinfante-Rosenfeld Lagrangian is an ad hoc modification of the Nöther form, and by this an ad hoc modification of the gravitational field. This ad hoc modification is necessary because the BR Lagrangian implies a non-gauge invariant, asymmetric (GRT requires symmetry) Nöther form.
There are many advantages to this formulation, as you can see for yourself in the paper.
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
edited Jan 5 at 21:34
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
answered Jan 5 at 21:26
my2ctsmy2cts
4,8872618
4,8872618
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
|
show 7 more comments
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
I'm not the downvoter, but statements that your answer is "the only correct one" generally attract them. There are advantages and disadvantages to every theory and interpretation. What you interpret as a knockdown blow might be, to somebody else, just a minor annoyance or even a positive feature.
$endgroup$
– knzhou
Jan 5 at 21:35
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
It's an interesting answer, and one that will take me some time to understand. Thanks for your input
$endgroup$
– InertialObserver
Jan 5 at 21:46
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@knzhou I don't follow this. To make statement might can appear a negative sign but I back it up with published proof. As for minor annoyances, any inconsistency, no matter how small, is an inconsistency. The scientific method requires to find such inconsistencies. The 43" perihelion advance per century of some distant planet that took a few centuries and a lot of accurate calculations to extract certainly also qualifies as a minor annoyance. Yet physicists and astronomers took it very seriously with spectacular results.
$endgroup$
– my2cts
Jan 5 at 21:48
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
$begingroup$
@InitialObserver Thank you for your interest !
$endgroup$
– my2cts
Jan 5 at 21:50
1
1
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
$begingroup$
@ G. Smith Thank you for pointing this out. I was unaware of any citations so this is good news to me. The referring papers are in excellent journals. I will look into these. Note that it is irrelevant to the correctness of the paper that I have been cited only twice. What is relevant is the content of my paper.
$endgroup$
– my2cts
Jan 5 at 23:39
|
show 7 more comments
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$begingroup$
Yes, the Maxwell eqs. (in terms of $F$) are gauge-invariant and physical. But you presumably already know that. So what are you asking?
$endgroup$
– Qmechanic♦
Jan 5 at 19:53
$begingroup$
Let me clarify in an edit with an example
$endgroup$
– InertialObserver
Jan 5 at 19:53
5
$begingroup$
I wouldn't call the mode expansion for $A_mu$ a "solution to Maxwell's equations". Maxwell's equations contain the electric and magnetic fields. They don't even mention $A_mu$.
$endgroup$
– knzhou
Jan 5 at 20:06
$begingroup$
@knzhou they do solve maxwell's equations though. The way that it's always presented is as a general solution to the EOM
$endgroup$
– InertialObserver
Jan 5 at 20:07
2
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I'm familiar with Maxwell's demon, but I'm not sure that he's a scalar fiend.
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– chrylis
Jan 6 at 5:39