What is the origin of the reciprocity theorem in optics?

[Wiemster]

I just heard about the reciprocity theorem in optics which states that for two currents 1 and 2 and their corresponding fields

\int _V \vec{J_1} \cdot \vec{E_2} dV = \int _V \vec{J_2} \cdot \vec{E_1} dV

which basically comes down to the fact that you can interchange the source and detector in case of a current and an electric field and get the same result. I find this intriguing and was wondering about the origin of this phenomenon.

I saw the derivation and found that it comes down to the fact that the operator O that transforms between J and E

\vec{J}=\hat{O} \vec{E}

is Hermitian under the inner product

(\vec{E_1},\vec{E_2})=\int \vec{E_1} \cdot \vec{E_2} dV

which gives the result when one uses that for Hermitian operators (\vec{E_1},\hat{O} \vec{E_2})=(\hat{O} \vec{E_1}, \vec{E_2})[/tex].<br /> <br /> But I was wondering if there is something still more fundamental to say about this reciprocity theorem. It feels as though there must be some easy explanation for this symmetry of interchanging source and detector. Can somebody explain the origin of the theorem without mathematics perhaps?

 

[actionintegral]

Hi Wiemster,

For what it's worth, I just looked up reciprocity theorem and saw a reference to it in circuit theory. It says if you move a battery from one network branch to another, the same current appears in the original branch.

 

[Wiemster]

I found quite a few examples of recoprocity, the one you mention could be the same as the one I mentoned. It doesn't limit itself to lightwaves of course and could also be observed in a circuit where the electric field of the battery drives the current. The occurence of the theorem in other area's than EM is maybe also a consequence of the fact that an operator between two quantities is hermitian, so the quantities can be interchanged. If this is true there just might not be more to it than that and I am looking futile for a deeper explanation. It would be nice though to be able to explain the phenomena without mathematics...

 

[Hash]

A sort of recoprocity theorem is often used in antenna engeenering, under the name "equivalence theorem".
It states that the fields (E,H) in a region 1 can be expressed just as the radiated field by the electric and magnetic equivalent currents (J,M) of the exteriors fields in region 2, on a closed surface S delimiting the region 1 and 2.

A demonstration of this theorem is provided by the boundary conditions of the EM field, and integral version of the Maxwell equation. I could find reference if you want.

 

[marcusl]

All of these examples can be derived from each other, and all can be related ultimately back to the "Lorentz reciprocity theorem" relating E, H and J.

I'm puzzled with your comment that the operator is Hermitian because in circuit theory and microwave networks, the corresponding quantity is the impedance (or admittance or scattering) matrix, and it is always complex and symmetric, NOT Hermitian. Furthermore what you refer to as "inner products" aren't proper inner products in the usual derivations of these theorems because they don't employ complex conjugation--in other words, they don't give power but something called the "reaction", instead. Take a look at a comprehensive derivation like that in Harrington, Time-Harmonic Electromagnetic Fields, sec. 3.8 (1961, reissued 1987).

To discuss the physical intuition behind reciprocity, note that the reciprocity theorem(s) can always be related to the vector Green's theorem. The idea behind Green's theorem, and behind a Green function as well, is that a result or action at one point is related linearly to the source terms at other points and that the relation is mathematically symmetric. This is a familiar attribute of common Green Functions: they are always symmetric in the two space variables r and r'. So source and field can always be interchanged without changing the result, so long as the system is linear.

Note there are exceptions. In non-linear systems, the reciprocity theorems must be modified since symmetric matrices become non-symmetric tensors. This is the case in gyrotropic media like ferrites; reversing source and sink in a microwave circulator, for instance, gives different behavior (indeed this is by design and is what makes the device useful). Forms of the reciprocity theorem are restored, however, if the magnetic field is also reversed at the same time. Other applications where people worry about non-linear effects are plasmas and nuclear magnetic resonance.

 

[Wiemster]

what you refer to as "inner products" aren't proper inner products in the usual derivations of these theorems because they don't employ complex conjugation--in other words, they don't give power but something called the "reaction", instead. Take a look at a comprehensive derivation like that in Harrington, Time-Harmonic Electromagnetic Fields, sec. 3.8 (1961, reissued 1987).

I have seen a proper derivation as well, the hermiticity of the operator comment comes from wikipedia, so maybe that part needs some revision.

To discuss the physical intuition behind reciprocity, note that the reciprocity theorem(s) can always be related to the vector Green's theorem. The idea behind Green's theorem, and behind a Green function as well, is that a result or action at one point is related linearly to the source terms at other points and that the relation is mathematically symmetric. This is a familiar attribute of common Green Functions: they are always symmetric in the two space variables r and r'. So source and field can always be interchanged without changing the result, so long as the system is linear.

So it comes down to the linearity of the operator relating J and E

\vec{J} = \frac{1}{i \omega} \left[ \left( \nabla \times \frac{1}{\mu} \nabla \times \right) - \; \omega^2 \varepsilon \right] \vec{E} \equiv \hat{O}\vec{E}

and the fact that it is mathematically symmetric in r and r', can you explicitly show that for the above operator?

 

[actionintegral]

they are always symmetric in the two space variables r and r'. So source and field can always be interchanged without changing the result, so long as the system is linear.

I think that's the answer the original poster was looking for

 

[marcusl]

To facilitate my answer, can someone tell me how to insert or paste symbols and equations into the Response boxes on this thread? I'm afraid I'm hardly computer literate

 

[Wiemster]

Check: https://www.physicsforums.com/showthread.php?t=8997

 

[marcusl]

I have seen a proper derivation as well, the hermiticity of the operator comment comes from wikipedia, so maybe that part needs some revision.

I took a look at the Wikipedia site and it needs rework! It says their operator is Hermitian, but the next sentence says it isn't really, but then the article proceeds to use Hermiticity for their proof. For the record, the operator (or the corresponding scattering and impedance matrices in circuit theory) aren't Hermitian but rather complex symmetric.

So it comes down to the linearity of the operator relating J and E

[\vec{J} = \frac{1}{i \omega} \left[ \left( \nabla \times \frac{1}{\mu} \nabla \times \right) - \; \omega^2 \varepsilon \right] \vec{E} \equiv \hat{O}\vec{E}

and the fact that it is mathematically symmetric in r and r', can you explicitly show that for the above operator?

Well, that applies to the Green's theorem formulation. Concerning your equation above, I think it's the linearity of epsilon and mu that leads to the reciprocity. Doesn't sound profound but I'll try to make the case.

Write the quantity on the LHS below and expand to give the equation
\nabla \cdot (\vec{E_1} \times \vec{H_2} - \vec{E_2} \times \vec{H_1}) = \vec{H_2} \cdot \nabla \times \vec{E_1} - \vec{E_1} \cdot \nabla \times \vec{H_2} - \vec{H_1} \cdot \nabla \times \vec{E_2} - \vec{E_2} \cdot \nabla \times \vec{H_1}

Maxwell's equations give
\nabla \times \vec{E_1} = -i \omega \vec{\mu} \cdot vec{H_1}
\nabla \times \vec{H_1} = i \omega \vec{\varepsilon} \cdot vec{E_1} + vec{J_1},
and likewise for subscript 2. Epsilon and mu here are 2nd rank tensors to treat the general anisotropic case, mentioned but not treated in the Wikipedia article. They reduce to simple scalars for isotropic media, in which the two equations combine to give your equation. [Physics Forum's preview feature doesn't display equations, so I can't tell if the equations will come out right. I've never used Latex before!] Substitute into the first equation. For a passive linear medium, epsilon and mu are symmetric (NOT Hermitian!) so
\vec{\mu} \cdot \vec{H} = \vec{H} \cdot \vec{\mu}
and likewise for epsilon and E. This is key--the symmetry of the medium and of the equations leads to reciprocity of field and source.

Accordingly, the equation above simplifies to
\nabla \cdot (\vec{E_1} \times \vec{H_2} - \vec{E_2} \times \vec{H_1}) = \\<br /> -\vec{E_1} \cdot \vec{J_2} + \vec{E_2} \cdot \vec{J_1}.
Now integrate and evaluate. Various forms of the reciprocity theorem result, depending on the specifics of the problem. For media bounded by conductors, the left can be shown to vanish and your first equation
\int _V \vec{J_1} \cdot \vec{E_2} dV = \int _V \vec{J_2} \cdot \vec{E_1} dV
results.

Epsilon and mu aren't symmetric for gyrotropic media and so the reciprocity theorem above isn't valid for them, as I mentioned before; the message is that reciprocity arises from the symmetry and linearity of the medium and the linearity of Maxwell's equations.

 

[marcusl]

Sorry for the buggy reply above.

Moderator: Can you request the web gurus to fix the "Preview Post" feature so it again displays equations? Thanks!

Here is a better copy:

I have seen a proper derivation as well, the hermiticity of the operator comment comes from wikipedia, so maybe that part needs some revision.

I took a look at the Wikipedia site and it needs rework! It says their operator is Hermitian, but the next sentence says it isn't really, but then the article proceeds to use Hermiticity for their proof. For the record, the operator (or the corresponding scattering and impedance matrices in circuit theory) aren't Hermitian but rather complex symmetric.

So it comes down to the linearity of the operator relating J and E

[\vec{J} = \frac{1}{i \omega} \left[ \left( \nabla \times \frac{1}{\mu} \nabla \times \right) - \; \omega^2 \varepsilon \right] \vec{E} \equiv \hat{O}\vec{E}

and the fact that it is mathematically symmetric in r and r', can you explicitly show that for the above operator?

Well, that applies to the Green's theorem formulation. Concerning your equation above, I think it's the linearity of epsilon and mu that leads to the reciprocity. Doesn't sound profound but I'll try to make the case.

Write the quantity on the LHS below and expand to give the equation
\nabla \cdot (\vec{E_1} \times \vec{H_2} - \vec{E_2} \times \vec{H_1}) = \vec{H_2} \cdot \nabla \times \vec{E_1} - \vec{E_1} \cdot \nabla \times \vec{H_2} - \vec{H_1} \cdot \nabla \times \vec{E_2} - \vec{E_2} \cdot \nabla \times \vec{H_1}

Maxwell's equations give
\nabla \times \vec{E_1} = -i \omega \vec{\mu} \cdot \vec{H_1}
\nabla \times \vec{H_1} = i \omega \vec{\varepsilon} \cdot \vec{E_1} + \vec{J_1},
and likewise for subscript 2. Epsilon and mu here are 2nd rank tensors to treat the general anisotropic case, mentioned but not treated in the Wikipedia article. They reduce to simple scalars for isotropic media, in which the two equations combine to give your equation. [Physics Forum's preview feature doesn't display equations, so I can't tell if the equations will come out right. I've never used Latex before!] Substitute into the first equation. For a passive linear medium, epsilon and mu are symmetric (NOT Hermitian!) so
\vec{\mu} \cdot \vec{H} = \vec{H} \cdot \vec{\mu}
and likewise for epsilon and E. This is key--the symmetry of the medium and of the equations leads to reciprocity of field and source.

Accordingly, the equation above simplifies to
\nabla \cdot (\vec{E_1} \times \vec{H_2} - \vec{E_2} \times \vec{H_1}) = \\<br /> -\vec{E_1} \cdot \vec{J_2} + \vec{E_2} \cdot \vec{J_1}.
Now integrate and evaluate. Various forms of the reciprocity theorem result, depending on the specifics of the problem. For media bounded by conductors, your first equation
\int _V \vec{J_1} \cdot \vec{E_2} dV = \int _V \vec{J_2} \cdot \vec{E_1} dV
results.

Epsilon and mu aren't symmetric for gyrotropic media and so the reciprocity theorem above isn't valid for them, as I mentioned before; the message is that reciprocity arises from the symmetry and linearity of the medium and the linearity of Maxwell's equations.