[electromagnetics] Lorentz Gauge

In summary, the potential based Lorentz gauge is necessary for deriving the potential field of a charged particle in a magnetic field. The second partial derivative of the vector potential must be zero in order for the potential to be based in the Lorentz gauge.
  • #1
kidsasd987
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I am trying to derive the potential based Lorentz gauge, but I am not sure if I am on the right track.
Why the second partial derivative of vector potential must be 0?

Please correct me where I got this wrong.
 
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  • #2
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Final version.
 

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  • #3
Let's start from scratch, i.e., Maxwell's equations and use convenient Heaviside Lorentz units. The Maxwell equations divide into two groups, the homogeneous ones
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
and the inhomogeneous ones
$$-\frac{1}{c} \partial_t \vec{E} + \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
The homogeneous Maxwell equations are just constraints on the fields, and it is often more convenient to reduce the degrees of freedom. This is achieved by introducing first a vector potential for ##\vec{B}##. According to Helmholtz's fundamental theorem of vector calculus because the divergence of ##\vec{B}## vanishes, it must be a pure curl of a vector field,
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
this you plug into the first homogeneous equation (the Fraday law of induction) leading to
$$\vec{\nabla} \times \left (\frac{1}{c} \partial_t \vec{A} + \vec{E} \right)=0.$$
Again using Helmholtz's fundamental theorem, you conclude that the expression in the parenthesis must be the gradient of a scalar field,
$$\frac{1}{c} \partial_t \vec{A} + \vec{E}=-\vec{\nabla} \Phi.$$
The choice of the signs is just convention. Now you have solved the homogeneous equations by introducing the potentials via
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{E}=-\frac{1}{c} \partial_t \vec{A} -\vec{\nabla} \Phi.$$
Now for a given field configuration ##\vec{E}## and ##\vec{B}## these potentials are not unique. You can add a gradient to the vector potential without changing the magnetic field, i.e.,
$$\vec{A}'=\vec{A}-\nabla \chi$$
leads to the same magnetic field, because ##\vec{\nabla} \times (\vec{\nabla} \chi)=0## for any scalar field ##\chi##. To also make the electric field right, you have to introduce a new scalar potential too, i.e., you want
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}'-\vec{\nabla} \Phi'=-\frac{1}{c} \partial_t \vec{A} + \vec{\nabla} \left (\frac{1}{c} \partial_t \chi-\Phi' \right),$$
i.e., you can choose ##\Phi'## such that
$$\frac{1}{c} \partial_t \chi - \Phi'=-\Phi \; \Rightarrow \; \Phi'=\Phi+\frac{1}{c} \partial_t \chi.$$
Thus with any scalar field ##\chi## the potentials
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi$$
describe the same physical situation as the original potentials. This is called gauge invariance of electrodynamics.

This means you have some freedom to choose your potentials conveniently. To see what's most convenient we use the inhomogeneous Maxwell equations. Plugging in the potentials gives
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})-\frac{1}{c} \partial_t \left (-\frac{1}{c} \partial_t \vec{A}- \vec{\nabla} \Phi \right )=\frac{1}{c} \vec{j}.$$
The first expression can be simplified (in Cartesian coordinates!) to
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A},$$
i.e., the Ampere-Maxwell Law becomes
$$\Box \vec{A} + \vec{\nabla} \left (\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A} \right)=\frac{1}{c} \vec{j},$$
where
$$\Box=\frac{1}{c^2} \partial_t^2 -\Delta$$
is the D'Alembert operator. Obviously the above equations become most convenient, when we choose the potentials such that the parentheses vanish, because then the equation of motion for the vector potential separates into three wave equations for each component separately. We are free to so because of the gauge invariance, i.e., we can impose the constraint
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,$$
which is a socalled gauge-fixing constraint. It's known as the Lorenz gauge (in the older literature somwhat injustly called the Lorentz gauge, mixing up the historical order, because it was the Danish physicist Ludvig Lorenz who found this particular gauge fixing rather than the Dutch physicist Hendrik A. Lorentz).

Now fortunately also the remaining equation becomes simple with the choice of the Lorenz gauge, because
$$-\vec{\nabla} \cdot \left (\frac{1}{c} \partial_t \vec{A}+\vec{\nabla} \Phi \right )=\rho.$$
Now we have because of the Lorenz-gauge condition
$$\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A}=-\frac{1}{c^2} \partial_t^2 \Phi,$$
and thus
$$\Box \Phi=\rho.$$
In the Lorenz gauge thus also the scalar potential separates from the vector potential and fulfills a simple inhomogeneous wave equation, which in the usual physical situations, where you like to calculate the emission of electromagnetic waves from given charge-current distributions, is solved by the retarded Green's function, leading to the retarded potentials.
 
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  • #5
I think those who are responsible for giving justice to Lorenz are Jackson and Okun with this very interesting article in RMP:

Jackson, J. D., Okun, L. B.: Historical roots of gauge invariance, Rev. Mod. Phys. 73(3), 663–680, 2001
http://link.aps.org/doi/10.1103/RevModPhys.73.663

or as preprint

https://publications.lbl.gov/islandora/object/ir%3A116656/datastream/PDF/download/citation.pdf
 
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  • #6
vanhees71 said:
Let's start from scratch, i.e., Maxwell's equations and use convenient Heaviside Lorentz units. The Maxwell equations divide into two groups, the homogeneous ones
$$\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
and the inhomogeneous ones
$$-\frac{1}{c} \partial_t \vec{E} + \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
The homogeneous Maxwell equations are just constraints on the fields, and it is often more convenient to reduce the degrees of freedom. This is achieved by introducing first a vector potential for ##\vec{B}##. According to Helmholtz's fundamental theorem of vector calculus because the divergence of ##\vec{B}## vanishes, it must be a pure curl of a vector field,
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
this you plug into the first homogeneous equation (the Fraday law of induction) leading to
$$\vec{\nabla} \times \left (\frac{1}{c} \partial_t \vec{A} + \vec{E} \right)=0.$$
Again using Helmholtz's fundamental theorem, you conclude that the expression in the parenthesis must be the gradient of a scalar field,
$$\frac{1}{c} \partial_t \vec{A} + \vec{E}=-\vec{\nabla} \Phi.$$
The choice of the signs is just convention. Now you have solved the homogeneous equations by introducing the potentials via
$$\vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{E}=-\frac{1}{c} \partial_t \vec{A} -\vec{\nabla} \Phi.$$
Now for a given field configuration ##\vec{E}## and ##\vec{B}## these potentials are not unique. You can add a gradient to the vector potential without changing the magnetic field, i.e.,
$$\vec{A}'=\vec{A}-\nabla \chi$$
leads to the same magnetic field, because ##\vec{\nabla} \times (\vec{\nabla} \chi)=0## for any scalar field ##\chi##. To also make the electric field right, you have to introduce a new scalar potential too, i.e., you want
$$\vec{E}=-\frac{1}{c} \partial_t \vec{A}'-\vec{\nabla} \Phi'=-\frac{1}{c} \partial_t \vec{A} + \vec{\nabla} \left (\frac{1}{c} \partial_t \chi-\Phi' \right),$$
i.e., you can choose ##\Phi'## such that
$$\frac{1}{c} \partial_t \chi - \Phi'=-\Phi \; \Rightarrow \; \Phi'=\Phi+\frac{1}{c} \partial_t \chi.$$
Thus with any scalar field ##\chi## the potentials
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi$$
describe the same physical situation as the original potentials. This is called gauge invariance of electrodynamics.

This means you have some freedom to choose your potentials conveniently. To see what's most convenient we use the inhomogeneous Maxwell equations. Plugging in the potentials gives
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A})-\frac{1}{c} \partial_t \left (-\frac{1}{c} \partial_t \vec{A}- \vec{\nabla} \Phi \right )=\frac{1}{c} \vec{j}.$$
The first expression can be simplified (in Cartesian coordinates!) to
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A},$$
i.e., the Ampere-Maxwell Law becomes
$$\Box \vec{A} + \vec{\nabla} \left (\frac{1}{c} \partial_t \Phi + \vec{\nabla} \cdot \vec{A} \right)=\frac{1}{c} \vec{j},$$
where
$$\Box=\frac{1}{c^2} \partial_t^2 -\Delta$$
is the D'Alembert operator. Obviously the above equations become most convenient, when we choose the potentials such that the parentheses vanish, because then the equation of motion for the vector potential separates into three wave equations for each component separately. We are free to so because of the gauge invariance, i.e., we can impose the constraint
$$\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,$$
which is a socalled gauge-fixing constraint. It's known as the Lorenz gauge (in the older literature somwhat injustly called the Lorentz gauge, mixing up the historical order, because it was the Danish physicist Ludvig Lorenz who found this particular gauge fixing rather than the Dutch physicist Hendrik A. Lorentz).

Now fortunately also the remaining equation becomes simple with the choice of the Lorenz gauge, because
$$-\vec{\nabla} \cdot \left (\frac{1}{c} \partial_t \vec{A}+\vec{\nabla} \Phi \right )=\rho.$$
Now we have because of the Lorenz-gauge condition
$$\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A}=-\frac{1}{c^2} \partial_t^2 \Phi,$$
and thus
$$\Box \Phi=\rho.$$
In the Lorenz gauge thus also the scalar potential separates from the vector potential and fulfills a simple inhomogeneous wave equation, which in the usual physical situations, where you like to calculate the emission of electromagnetic waves from given charge-current distributions, is solved by the retarded Green's function, leading to the retarded potentials.
I appreciate you for your detailed comment.
I am an engineering student so, it took me a while to catch up heavyside Lorentz Unit, but wow. This is indeed beautiful.

Ah, and stupid me.. I should've not used inverse del operator..

I have one more question.

1.

Now for a given field configuration E→" role="presentation" style="display: inline-block; line-height: 0; font-size: 14.69px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: 'PT Sans', san-serif; position: relative; background-color: rgb(255, 255, 255);">⃗EE→ and B→" role="presentation" style="display: inline-block; line-height: 0; font-size: 14.69px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; font-family: 'PT Sans', san-serif; position: relative; background-color: rgb(255, 255, 255);">⃗BB→ these potentials are not unique. You can add a gradient to the vector potential without changing the magnetic field, i.e.,Isn't B uniquely determined as we know that divergence of B is 0 and curl of A is a non-zero vector field B. Did you mean to say vector potential A here? or because A is inderminate so B is also not uniquely determined?

Thank you!
 
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  • #7
The physical situation is uniquely described by the electromagnetic field with components ##\vec{E}## and ##\vec{B}##. For this given physical field configuration the scalar and vector potential are only defined up to a gauge transformation given above. That's called the "gauge invariance" of electrodynamics.
 

Related to [electromagnetics] Lorentz Gauge

1. What is the Lorentz gauge in electromagnetics?

The Lorentz gauge is a mathematical constraint used in electromagnetics to simplify the equations that describe the behavior of electromagnetic fields. It is named after physicist Hendrik Lorentz and is one of the four main gauge choices used in electromagnetism, along with the Coulomb gauge, the radiation gauge, and the Lorenz gauge.

2. How does the Lorentz gauge differ from other gauge choices?

The Lorentz gauge is unique in that it ensures that the electric and magnetic fields in a given system satisfy the wave equation, which is a fundamental equation in electromagnetism. This makes it a convenient choice for solving problems involving electromagnetic waves.

3. What are the benefits of using the Lorentz gauge in electromagnetics?

One of the main benefits of using the Lorentz gauge is that it simplifies the equations used to describe electromagnetic fields, making them easier to solve and analyze. It also ensures that the solutions obtained are physically meaningful and satisfy the wave equation, providing a more accurate representation of the behavior of electromagnetic fields in a given system.

4. Are there any limitations to using the Lorentz gauge?

While the Lorentz gauge is a useful tool in electromagnetics, it does have some limitations. It cannot be applied in cases where charges or currents are not present, such as in empty space. It also does not account for any sources of radiation, so it may not be suitable for studying systems that emit electromagnetic waves.

5. How is the Lorentz gauge used in practical applications?

The Lorentz gauge is commonly used in practical applications, such as in the design and analysis of electromagnetic devices, antennas, and communication systems. It is also used in theoretical studies of electromagnetism, such as in quantum electrodynamics and the study of fundamental particles.

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