Orbital and Spin Angular momentum of light derivation

In summary, the problem is asking to show that the angular momentum carried by an electromagnetic field is equal to the integral of the vector potential in radiation gauge multiplied by the vector of the electric field and the operator for the angular momentum, using partial integration. The assumptions made are that the field extension is finite and that the subscripts in the equation represent a sum. The solution gets stuck when trying to expand the vector potential using a vector identity. The issue may be related to the use of partial integration, which can be an alternative term for integration by parts or integrating a partial derivative. The exact integration or partial derivative to be used is unclear in this case.
  • #1
calculo2718
25
1

Homework Statement


The momentum carried by an electromagnetic field is [;\vec{P}(\vec{x}, t) = \frac{1}{4\pi c} \int d\vec{x}\vec{E}(\vec{x},t) \times \vec{B}(\vec{x},t);]

show that for a finite field extension
[;\vec{J}(\vec{x}, t) = \frac{1}{4\pi c}\int -i A_i(\vec{x},t)\vec{L}E_i(\vec{x},t) + \vec{E}(\vec{x},t) \times \vec{A}(\vec{x},t) d\vec{x};] where [;\vec{J};] is the angular momentum, [;L = -ir \times \nabla;], and [;\vec{A};] is the vector potential in radiation gauge, i.e. [;A^0 = 0;] and [;\nabla \cdot \vec{A} = 0;] (hint: using partial integration)

Homework Equations


coulomb gauge [;\vec{B} = \nabla \times \vec{A};]

The Attempt at a Solution


There are some things in the wording in the problem that I do not understand so I am making some assumptions that may not even be true.

1) Finite field extension means that the field does NOT go to infinite so [;\nabla \cdot \vec{E} = 0;]
2) The subscripts in the [;-i A_i(\vec{x},t)\vec{L}E_i(\vec{x},t);] represents a sum.
3) I have no clue what partial integration is.
[;\int \vec{x} \times (\vec{E} \times \vec{B}) d\vec{x};]
[;\int \vec{x} \times (\vec{E} \times (\nabla \times \vec{A})) d\vec{x};]
using triple product expansion
[;\int \vec{x} \times (\nabla(\vec{E} \cdot \vec{A}) - \vec{A}(\vec{E}\cdot \nabla)) d\vec{x};]
distributing the [;\vec{x} \times;]
[;\int \vec{x} \times \nabla(\vec{E} \cdot \vec{A}) - \vec{x} \times \vec{A}(\vec{E}\cdot \nabla) d\vec{x};]

This is where I get stuck. If I assume that [;\nabla \cdot \vec{E} = 0;], I have no idea where to get the [;\vec{E} \times \vec{A};] term in the answer I am supposed to get. If I don't assume that, I try to expand my [;\nabla(\vec{E} \cdot \vec{A});] term using some vector identity the [;\nabla \cdot \vec{E};] cancels anyway and I simply get the equation in my second line of work.

My gut tells me that I missing something having to do with this "partial integration" business. I have googled this and I get it as an alternative term to integration by parts or as integrating a partial derivative. If I am to integrate by parts, what am I integrating by parts? I don't have a product of two functions I have different vector products of different vector fields. If I am supposed to integrate some partial derivative, which one? I have so many!

please help, thanks!
 
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  • #2
Your TeX formulas get displayed properly if you enclose them in $$ or ## (inline mode). Currently they are hard to read.
 
  • #3
mfb said:
Your TeX formulas get displayed properly if you enclose them in $$ or ## (inline mode). Currently they are hard to read.
I can't seem to edit the post so I am re-posting here.

Homework Statement


The momentum carried by an electromagnetic field is $$\vec{P}(\vec{x}, t) = \frac{1}{4\pi c} \int d\vec{x}\vec{E}(\vec{x},t) \times \vec{B}(\vec{x},t)$$

show that for a finite field extension
$$\vec{J}(\vec{x}, t) = \frac{1}{4\pi c}\int -i A_i(\vec{x},t)\vec{L}E_i(\vec{x},t) + \vec{E}(\vec{x},t) \times \vec{A}(\vec{x},t) d\vec{x}] where $\vec{J}$$ is the angular momentum, $$L = -ir \times \nabla$, and [\vec{A}$$ is the vector potential in radiation gauge, i.e. $$A^0 = 0$$ and $$\nabla \cdot \vec{A} = 0$$ (hint: using partial integration)

Homework Equations


coulomb gauge $$\vec{B} = \nabla \times \vec{A}$$

The Attempt at a Solution


There are some things in the wording in the problem that I do not understand so I am making some assumptions that may not even be true.

1) Finite field extension means that the field does NOT go to infinite so $$\nabla \cdot \vec{E} = 0$$
2) The subscripts in the $$-i A_i(\vec{x},t)\vec{L}E_i(\vec{x},t)$$ represents a sum.
3) I have no clue what partial integration is.
$$\int \vec{x} \times (\vec{E} \times \vec{B}) d\vec{x}$$
$$\int \vec{x} \times (\vec{E} \times (\nabla \times \vec{A})) d\vec{x}$$
using triple product expansion
$$\int \vec{x} \times (\nabla(\vec{E} \cdot \vec{A}) - \vec{A}(\vec{E}\cdot \nabla)) d\vec{x}$$
distributing the $$\hat{x}\times$$
$$\int \vec{x} \times \nabla(\vec{E} \cdot \vec{A}) - \vec{x} \times \vec{A}(\vec{E}\cdot \nabla) d\vec{x}$$

This is where I get stuck. If I assume that $$\nabla \cdot \vec{E} = 0$$, I have no idea where to get the $$\vec{E} \times \vec{A}$$ term in the answer I am supposed to get. If I don't assume that, I try to expand my $$\nabla(\vec{E} \cdot \vec{A})$$ term using some vector identity the $$\nabla \cdot \vec{E}$$ cancels anyway and I simply get the equation in my second line of work.

My gut tells me that I missing something having to do with this "partial integration" business. I have googled this and I get it as an alternative term to integration by parts or as integrating a partial derivative. If I am to integrate by parts, what am I integrating by parts? I don't have a product of two functions I have different vector products of different vector fields. If I am supposed to integrate some partial derivative, which one? I have so many!

please help, thanks!
 

FAQ: Orbital and Spin Angular momentum of light derivation

1. What is the concept of orbital and spin angular momentum of light?

The orbital and spin angular momentum of light refers to the two types of angular momentum that can be associated with a beam of light. The orbital angular momentum is related to the spatial distribution of the beam, while the spin angular momentum is related to the polarization of the light.

2. How is the orbital and spin angular momentum of light derived?

The derivation of the orbital and spin angular momentum of light involves using the Maxwell's equations and the properties of the electromagnetic field. By analyzing the electromagnetic field in terms of its electric and magnetic components, the expressions for the orbital and spin angular momentum can be obtained.

3. What are the applications of orbital and spin angular momentum of light?

The orbital and spin angular momentum of light have various applications in optics and photonics. It has been used in the development of new techniques for optical communications, imaging, and manipulation of particles at the nanoscale level.

4. Can light possess both orbital and spin angular momentum?

Yes, light can possess both orbital and spin angular momentum at the same time. In fact, most light beams have a combination of both types of angular momentum, with varying magnitudes and directions.

5. How does the orbital and spin angular momentum of light affect its propagation through different media?

The orbital and spin angular momentum of light can be affected by the properties of the medium it is propagating through, such as its refractive index and polarization. This can lead to changes in the magnitude and direction of the orbital and spin angular momentum, which can have implications for the behavior of light in different environments.

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