- #1
Kostik
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- TL;DR Summary
- How to show explicitly (analogous to the EM case) that the ratio (spin angular momentum density)/(energy density) for circularly-polarized gravitational plane wave ##= \pm 2/\omega##.
It is well understood that an infinite monochromatic, circularly-polarized electromagnetic plane wave has no angular momentum density. However, a finite monochromatic, circularly-polarized electromagnetic plane wave packet does have an angular momentum density, arising from effects at the border of the wave packet. Its ratio of spin angular momentum density to energy density is ##\pm 1/\omega##.
Here the electric and magnetic fields are, for example, a periodic function of ##\omega(t-z)##, taking ##c=1##. Thus one may take ##\mathbf{E_1} = e^{i\omega(t-z)}\mathbf{\hat{x}}##, ##\,\,\mathbf{E_2} = e^{i\omega(t-z)}\mathbf{\hat{y}}##, and define the circularly polarized waves ##\mathbf{E_\pm} = \mathbf{E_1} \pm i\mathbf{E_2}##, always taking the real part to obtain the physical wave.
This is clearly derived in
1. Ohanian, H. C. (2007). Classical Electrodynamics (2nd Editon).
2. Rohrlich, F. (2007) Classically Charged Particles (3rd Edition)
3. Ohanian, H. C. "What is spin?", Am. J. Phys. 54 (1985) 500–505 (https://physics.mcmaster.ca/phys3mm3/notes/whatisspin.pdf)
4. Gsponer, A. "What is spin?" (https://arxiv.org/abs/physics/0308027)
To find the spin angular momentum, one computes the total angular momentum $$\mathbf{J} = \mathbf{x} \times \frac{1}{4\pi}(\mathbf{E} \times \mathbf{B}) .$$ One writes ##\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}##, uses a couple of vector identities and the divergence theorem, and then obtains $$\mathbf{J} = \mathbf{L} + \mathbf{s}$$ where $$\mathbf{L} = \frac{1}{4\pi}\int \mathbf{x} \times E^n \nabla A^n\,dV$$ $$\mathbf{s} = \pm\frac{1}{4\pi}\int \mathbf{E} \times \mathbf{A}\,dV$$ integrated over the wave packet. The first integral (applied to a circularly polarized wave) is clearly identifiable as ordinary orbital angular momentum, while the second integral, which does not depend upon the coordinate ##\mathbf{x}##, is identifiable as spin angular momentum.
Gsponer remarks:
##\quad## "In principle, the method used here can be generalized to any spin. This requires to know the explicit form of the corresponding energy-momentum tensor from which the analogue of the Poynting vector can be derived and used to calculate the nonlocal part of the angular momentum. Unfortunately, there is no unique definition for this tensor, and there is much debate even for the spin 2 case which corresponds to gravitational radiation. Nevertheless, using Einstein’s energy-momentum pseudotensor, standard textbooks show how to derive the average rates of loss of angular momentum and energy of a point mass emitting gravitational waves in terms of its reduced mass quadrupole moments..."
I want to derive the ratio of spin angular momentum to energy density for the analogous gravitational wave case, which should be ##\pm 2/\omega##.
In Dirac's "General Theory of Relativity" (Ch. 34, p. 67) he shows that for a plane wave propagating in the ##z## direction, the Einstein pseudo-tensor has energy density ##{t_0}^0## and ##z##-momentum density ##{t_0}^3## both given by $${t_0}^0 = {t_0}^3 = (1/16𝜋) [ (u_{11})^2 + (u_{12})^2 ] .$$ Here, ##u_{𝜇𝜈}## is the derivative of the metric, ##dg_{𝜇𝜈}(𝜉)/d𝜉##, where ##g_{𝜇𝜈}## is a function of the single variable ##𝜉=𝜔(t-z)##.
I am at a loss as to how to proceed. I know "the explicit form of the corresponding energy-momentum tensor from which the analogue of the Poynting vector can be derived", but I do not know how to "calculate the nonlocal part of the angular momentum."
In order to get the spin term, I must somehow manipulate the total angular momentum ##\mathbf{x} \times \mathbf{S}## (##\mathbf{S}## is the momentum density, i.e., the Poynting vector) into a sum of terms ##\mathbf{L} + \mathbf{s}##, where the spin term ##\mathbf{s}## does not have a coordinate dependence on ##\mathbf{x}##. This probably requires some use of the divergence theorem, as in the electromagnetic case.
But I do not seem to have anything to work with, unlike in the electromagnetic case where we start by writing ##\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}##.
What I do know, from Weinberg's text ("Gravitation and Cosmology"), is that ##u_{11} + iu_{12}## transforms under a rotation by ##𝜃## according to the rule $$u'_{11} ∓ iu'_{12} = exp(±2i𝜃)(u_{11} ∓ iu_{12}) .$$ Weinberg refers to the "##±2##" which appears here as the helicity of the wave ##u_{11} ∓ iu_{12}##, which is also the ratio of the spin-angular momentum density to energy density. Of course, ##u_{11} ∓ iu_{12}## is a circularly polarized wave, so somehow, I think I should use these results.
Gsponer infers that the ratio of spin angular momentum density to energy density $$s_z/ℰ = ±2/𝜔$$ by examining the angular momentum and energy densities of a quadrupole (his equation (5)). However, I want to see how to mimic the direct derivation of the spin angular momentum integral (e.g., Ohanian, Rohrlich) in the case of general relativity, analogous to how it was done for electromagnetism.
Any hints or clues as to how to proceed?
Here the electric and magnetic fields are, for example, a periodic function of ##\omega(t-z)##, taking ##c=1##. Thus one may take ##\mathbf{E_1} = e^{i\omega(t-z)}\mathbf{\hat{x}}##, ##\,\,\mathbf{E_2} = e^{i\omega(t-z)}\mathbf{\hat{y}}##, and define the circularly polarized waves ##\mathbf{E_\pm} = \mathbf{E_1} \pm i\mathbf{E_2}##, always taking the real part to obtain the physical wave.
This is clearly derived in
1. Ohanian, H. C. (2007). Classical Electrodynamics (2nd Editon).
2. Rohrlich, F. (2007) Classically Charged Particles (3rd Edition)
3. Ohanian, H. C. "What is spin?", Am. J. Phys. 54 (1985) 500–505 (https://physics.mcmaster.ca/phys3mm3/notes/whatisspin.pdf)
4. Gsponer, A. "What is spin?" (https://arxiv.org/abs/physics/0308027)
To find the spin angular momentum, one computes the total angular momentum $$\mathbf{J} = \mathbf{x} \times \frac{1}{4\pi}(\mathbf{E} \times \mathbf{B}) .$$ One writes ##\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}##, uses a couple of vector identities and the divergence theorem, and then obtains $$\mathbf{J} = \mathbf{L} + \mathbf{s}$$ where $$\mathbf{L} = \frac{1}{4\pi}\int \mathbf{x} \times E^n \nabla A^n\,dV$$ $$\mathbf{s} = \pm\frac{1}{4\pi}\int \mathbf{E} \times \mathbf{A}\,dV$$ integrated over the wave packet. The first integral (applied to a circularly polarized wave) is clearly identifiable as ordinary orbital angular momentum, while the second integral, which does not depend upon the coordinate ##\mathbf{x}##, is identifiable as spin angular momentum.
Gsponer remarks:
##\quad## "In principle, the method used here can be generalized to any spin. This requires to know the explicit form of the corresponding energy-momentum tensor from which the analogue of the Poynting vector can be derived and used to calculate the nonlocal part of the angular momentum. Unfortunately, there is no unique definition for this tensor, and there is much debate even for the spin 2 case which corresponds to gravitational radiation. Nevertheless, using Einstein’s energy-momentum pseudotensor, standard textbooks show how to derive the average rates of loss of angular momentum and energy of a point mass emitting gravitational waves in terms of its reduced mass quadrupole moments..."
I want to derive the ratio of spin angular momentum to energy density for the analogous gravitational wave case, which should be ##\pm 2/\omega##.
In Dirac's "General Theory of Relativity" (Ch. 34, p. 67) he shows that for a plane wave propagating in the ##z## direction, the Einstein pseudo-tensor has energy density ##{t_0}^0## and ##z##-momentum density ##{t_0}^3## both given by $${t_0}^0 = {t_0}^3 = (1/16𝜋) [ (u_{11})^2 + (u_{12})^2 ] .$$ Here, ##u_{𝜇𝜈}## is the derivative of the metric, ##dg_{𝜇𝜈}(𝜉)/d𝜉##, where ##g_{𝜇𝜈}## is a function of the single variable ##𝜉=𝜔(t-z)##.
I am at a loss as to how to proceed. I know "the explicit form of the corresponding energy-momentum tensor from which the analogue of the Poynting vector can be derived", but I do not know how to "calculate the nonlocal part of the angular momentum."
In order to get the spin term, I must somehow manipulate the total angular momentum ##\mathbf{x} \times \mathbf{S}## (##\mathbf{S}## is the momentum density, i.e., the Poynting vector) into a sum of terms ##\mathbf{L} + \mathbf{s}##, where the spin term ##\mathbf{s}## does not have a coordinate dependence on ##\mathbf{x}##. This probably requires some use of the divergence theorem, as in the electromagnetic case.
But I do not seem to have anything to work with, unlike in the electromagnetic case where we start by writing ##\mathbf{B} = \mathbf{\nabla} \times \mathbf{A}##.
What I do know, from Weinberg's text ("Gravitation and Cosmology"), is that ##u_{11} + iu_{12}## transforms under a rotation by ##𝜃## according to the rule $$u'_{11} ∓ iu'_{12} = exp(±2i𝜃)(u_{11} ∓ iu_{12}) .$$ Weinberg refers to the "##±2##" which appears here as the helicity of the wave ##u_{11} ∓ iu_{12}##, which is also the ratio of the spin-angular momentum density to energy density. Of course, ##u_{11} ∓ iu_{12}## is a circularly polarized wave, so somehow, I think I should use these results.
Gsponer infers that the ratio of spin angular momentum density to energy density $$s_z/ℰ = ±2/𝜔$$ by examining the angular momentum and energy densities of a quadrupole (his equation (5)). However, I want to see how to mimic the direct derivation of the spin angular momentum integral (e.g., Ohanian, Rohrlich) in the case of general relativity, analogous to how it was done for electromagnetism.
Any hints or clues as to how to proceed?
