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- The ratio of energy density to angular momentum density in a plane wave of monochromatic, circularly polarized electromagnetic radiation is ##\pm 1/\omega##. It should be ##\pm 2/\omega## for a gravitational wave. But, using the known formulae for ##\dot{E}## and ##\dot{\textbf{M}}## for a radiating system (EM and gravitational), one obtains ##\dot{M}/\dot{E} = 1/\omega##.
Background: For electric dipole radiation, the energy and angular momentum lost by radiation from a system of charges by radiation is given by:
$$\dot{E}_{dip} = -\frac{2}{3c^3} \ddot{\textbf{d}}^2$$ $$\overline{ \dot{\textbf{M}}_{dip} } = -\frac{2}{3c^3}\overline{\dot{\textbf{d}} \times \ddot{\textbf{d}} }$$ Here ##\textbf{d}## is the dipole moment of the system, and the overbar denotes a time-average. These formulae are in Landau & Lifshitz Vol. 2, p. 175, eq. (67.8) and p. 206, eq. (75.7), and other texts as well.
For electric quadrupole radiation, the formulae are:
$$\dot{E}_{quad} = -\frac{1}{180c^5} \dddot{Q}_{kl}\dddot{Q}_{kl}$$ $$\overline{ (\dot{M}_{quad})_i } = -\frac{1}{90c^5} \epsilon_{ijk}\, \overline{\ddot{Q}_{jl} \dddot{Q}_{kl}}$$ Here, ##Q_{kl}## is the electric quadrupole moment, ##\epsilon_{ijk}## the antisymmetric Levi-Civita symbol. Repeated indices are summed over. Note the similarity to the dipole case, which can also be written $$\dot{E}_{dip} = -\frac{2}{3c^3} \ddot{d_k} \ddot{d_k} \quad , \quad\quad \overline{ (\dot{M}_{dip})_i } = -\frac{2}{3c^3} \epsilon_{ijk}\, \overline{\dot{d_j} \ddot{d_k}}$$ The first equation (for energy) is a standard result; e.g., Landau & Lifshitz p. 189, eq. (71.5). The second equation is done by Landau & Lifshitz for a gravitating system where ##Q_{kl}## is the mass quadrupole moment, but the derivation is identical: see p. 357, eq. (3).
Question: I am trying to show that for monochromatic, circularly-polarized dipole radiation, ##M/E = \pm 1/\omega##, but for monochromatic, circularly-polarized quadrupole radiation, ##M/E = \pm 2/\omega##. Here ##M/E## means the ratio of the density of energy to the density of angular momentum carried by the radiation field.
Obviously, the rate of change of energy (and angular momentum) of the system is proportional to the energy density in the radiation field. Hence, ##M_i/E## of the radiation field is equal to ##\dot{M}_i/\dot{E}## of the system.
Example: Take the case of a charge ##q## in a circular orbit in the ##x##-##y## plane given by $$\textbf{r} = r_0 \cos\omega t \, \hat{\bf{x}} + r_0 \sin\omega t \, \hat{\bf{y}}.$$ Then ##\textbf{d} = q\textbf{r}## and $$\dot{\textbf{d}} = q\omega (-r_0 \sin\omega t \, \hat{\bf{x}} + r_0 \cos\omega t \, \hat{\bf{y}}) \quad , \quad\quad \ddot{\textbf{d}} = -\omega^2 \textbf{d}$$ and one easily finds (for dipole radiation) that ##M_z/E = \pm 1/\omega##.
Now, if you carry out the same calculation for the quadrupole radiation (details omitted), you also obtain ##M_z/E = \pm 1/\omega##. This means that for gravitational radiation, for which the formulas are essentially the same: $$\dot{E}_{quad} = -\frac{G}{45c^5} \dddot{Q}_{kl}\dddot{Q}_{kl} \quad\quad(*)$$ $$\overline{ (\dot{M}_{quad})_i } = -\frac{2G}{45c^5} \epsilon_{ijk}\, \overline{\ddot{Q}_{jl} \dddot{Q}_{kl}} \quad\quad(*)$$ one also obtains ##M_z/E = \pm 1/\omega##. However, this simply isn't true. It's known that the gravitational field is spin 2 and the correct result is ##M/E = \pm 2/\omega##.
There is no question about the validity of the formulae ##(*)## for quadrupole radiation. The first one is in Landau & Lifshitz and Ohanian & Ruffini and many other texts; the second one is in Landau and MTW. The paradox is that the quadrupole formulas (except for constants) are the same for electromagnetic radiation and gravitational radiation. But for EM the result should be ##M_z/E = 1/\omega## while for gravity it should be ##M_z/E = 2/\omega##.
If you prefer to replace the system with two charges ##q## (or two masses ##m##) connected by a rigid massless rod (thus, there is no electric dipole radiation), nothing changes -- the quadrupole moments are simply multiplied by 2, and the result is the same.
Can someone help to identify the mistake I am overlooking?
$$\dot{E}_{dip} = -\frac{2}{3c^3} \ddot{\textbf{d}}^2$$ $$\overline{ \dot{\textbf{M}}_{dip} } = -\frac{2}{3c^3}\overline{\dot{\textbf{d}} \times \ddot{\textbf{d}} }$$ Here ##\textbf{d}## is the dipole moment of the system, and the overbar denotes a time-average. These formulae are in Landau & Lifshitz Vol. 2, p. 175, eq. (67.8) and p. 206, eq. (75.7), and other texts as well.
For electric quadrupole radiation, the formulae are:
$$\dot{E}_{quad} = -\frac{1}{180c^5} \dddot{Q}_{kl}\dddot{Q}_{kl}$$ $$\overline{ (\dot{M}_{quad})_i } = -\frac{1}{90c^5} \epsilon_{ijk}\, \overline{\ddot{Q}_{jl} \dddot{Q}_{kl}}$$ Here, ##Q_{kl}## is the electric quadrupole moment, ##\epsilon_{ijk}## the antisymmetric Levi-Civita symbol. Repeated indices are summed over. Note the similarity to the dipole case, which can also be written $$\dot{E}_{dip} = -\frac{2}{3c^3} \ddot{d_k} \ddot{d_k} \quad , \quad\quad \overline{ (\dot{M}_{dip})_i } = -\frac{2}{3c^3} \epsilon_{ijk}\, \overline{\dot{d_j} \ddot{d_k}}$$ The first equation (for energy) is a standard result; e.g., Landau & Lifshitz p. 189, eq. (71.5). The second equation is done by Landau & Lifshitz for a gravitating system where ##Q_{kl}## is the mass quadrupole moment, but the derivation is identical: see p. 357, eq. (3).
Question: I am trying to show that for monochromatic, circularly-polarized dipole radiation, ##M/E = \pm 1/\omega##, but for monochromatic, circularly-polarized quadrupole radiation, ##M/E = \pm 2/\omega##. Here ##M/E## means the ratio of the density of energy to the density of angular momentum carried by the radiation field.
Obviously, the rate of change of energy (and angular momentum) of the system is proportional to the energy density in the radiation field. Hence, ##M_i/E## of the radiation field is equal to ##\dot{M}_i/\dot{E}## of the system.
Example: Take the case of a charge ##q## in a circular orbit in the ##x##-##y## plane given by $$\textbf{r} = r_0 \cos\omega t \, \hat{\bf{x}} + r_0 \sin\omega t \, \hat{\bf{y}}.$$ Then ##\textbf{d} = q\textbf{r}## and $$\dot{\textbf{d}} = q\omega (-r_0 \sin\omega t \, \hat{\bf{x}} + r_0 \cos\omega t \, \hat{\bf{y}}) \quad , \quad\quad \ddot{\textbf{d}} = -\omega^2 \textbf{d}$$ and one easily finds (for dipole radiation) that ##M_z/E = \pm 1/\omega##.
Now, if you carry out the same calculation for the quadrupole radiation (details omitted), you also obtain ##M_z/E = \pm 1/\omega##. This means that for gravitational radiation, for which the formulas are essentially the same: $$\dot{E}_{quad} = -\frac{G}{45c^5} \dddot{Q}_{kl}\dddot{Q}_{kl} \quad\quad(*)$$ $$\overline{ (\dot{M}_{quad})_i } = -\frac{2G}{45c^5} \epsilon_{ijk}\, \overline{\ddot{Q}_{jl} \dddot{Q}_{kl}} \quad\quad(*)$$ one also obtains ##M_z/E = \pm 1/\omega##. However, this simply isn't true. It's known that the gravitational field is spin 2 and the correct result is ##M/E = \pm 2/\omega##.
There is no question about the validity of the formulae ##(*)## for quadrupole radiation. The first one is in Landau & Lifshitz and Ohanian & Ruffini and many other texts; the second one is in Landau and MTW. The paradox is that the quadrupole formulas (except for constants) are the same for electromagnetic radiation and gravitational radiation. But for EM the result should be ##M_z/E = 1/\omega## while for gravity it should be ##M_z/E = 2/\omega##.
If you prefer to replace the system with two charges ##q## (or two masses ##m##) connected by a rigid massless rod (thus, there is no electric dipole radiation), nothing changes -- the quadrupole moments are simply multiplied by 2, and the result is the same.
Can someone help to identify the mistake I am overlooking?
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