Question 11.27: In Section 11.2.1 we calculated the energy per unit time rad...

The momentum flux density is (minus) the Maxwell stress tensor (Section 8.2.3),

T_{i j}=\epsilon_{0}\left(E_{i} E_{j}-\frac{1}{2} \delta_{i j} E^{2}\right)+\frac{1}{\mu_{0}}\left(B_{i} B_{j}-\frac{1}{2} \delta_{i j} B^{2}\right),

(Eq. 8.17) so the momentum radiated per unit (retarded) time is

T_{i j} \equiv \epsilon_{0}\left(E_{i} E_{j}-\frac{1}{2} \delta_{i j} E^{2}\right)+\frac{1}{\mu_{0}}\left(B_{i} B_{j}-\frac{1}{2} \delta_{i j} B^{2}\right)                                (8.17)

\frac{d p }{d t_{r}}=-\oint \frac{1}{\left(\partial t_{r} / \partial t\right)} \overleftrightarrow{T} \cdot d A,

where (Eq. 10.78)

\frac{\partial t_{r}}{\partial t}=\frac{ᴫ c}{ ᴫ \cdot u }                       (10.78)

\frac{\partial t_{r}}{\partial t}=\frac{ᴫ c}{ ᴫ \cdot u }=\left(1-\frac{\hat{ᴫ } \cdot v }{c}\right)^{-1}.

(This factor is 1, when v = 0; it can be ignored in deriving the Larmor formula, but it does contribute to the momentum radiated.) The integration is over a large spherical surface centered on the charge:

As in the case of the power radiated, only the radiation fields contribute (Eq. 11.66):

E _{ rad }=\frac{q}{4 \pi \epsilon_{0}} \frac{ᴫ}{(ᴫ \cdot u )^{3}}[ ᴫ \times( u \times a )]                                    (11.66)

E _{ rad }=\frac{q}{4 \pi \epsilon_{0}} \frac{ ᴫ }{( ᴫ \cdot u )^{3}}[ ᴫ \times( u \times a )], \quad B _{ rad }=\frac{1}{c} \hat{ᴫ } \times E _{ rad }.

Thus

(note that B · dA = 0, and, for radiation fields, E \cdot d A =0,\left(1 / \mu_{0}\right) B^{2}=\left(1 / \mu_{0} c^{2}\right)(\hat{ ᴫ } \times E ) \cdot(\hat{ ᴫ } \times E )=\left.\epsilon_{0} \hat{ᴫ } \cdot[ E \times(\hat{ ᴫ } \times E )]=\epsilon_{0} \hat{ᴫ } \cdot\left[\hat{ ᴫ } E^{2}- E (\hat{ ᴫ } \cdot E )\right]=\epsilon_{0} E^{2}\right). So

\frac{d p }{d t_{r}}=\epsilon_{0} \oint\left(\frac{ ᴫ \cdot u }{ᴫ c }\right) E^{2} d A =\frac{q^{2}}{16 \pi^{2} \epsilon_{0} c} \oint \frac{ᴫ}{( ᴫ \cdot u )^{5}}[ ᴫ \times( u \times a )]^{2} d A.

We expand the integrand to first order in \beta \equiv v / c: u =c \hat{ ᴫ }- v =c(\hat{ ᴫ}- \beta ), ᴫ \cdot u =c ᴫ (1-\hat{ ᴫ } \cdot \beta ),

\frac{1}{( ᴫ \cdot u )^{5}}=\frac{1}{ᴫ{ }^{5} c^{5}}[1+5(\hat{ ᴫ } \cdot \beta )].

ᴫ \times( u \times a )= u ( ᴫ \cdot a )- a ( ᴫ \cdot u )=ᴫ c(\hat{ ᴫ }- \beta )(\hat{ᴫ } \cdot a )- a ᴫ c(1-\hat{ ᴫ } \cdot \beta )=ᴫ c[\hat{ ᴫ}(\hat{ ᴫ } \cdot a )- a + a (\hat{ ᴫ } \cdot \beta )- \beta (\hat{ᴫ} \cdot a )].

+2( a \cdot \beta )(\hat{ᴫ } \cdot a )]=ᴫ^{2} c^{2}\left[a^{2}-(\hat{ᴫ } \cdot a )^{2}-2 a^{2}(\hat{ᴫ } \cdot \beta )+2(\hat{ ᴫ } \cdot a )( a \cdot \beta )\right].

=\frac{1}{ᴫ^{2} c^{3}}\left[(\hat{ᴫ} \times a )^{2}+\hat{ ᴫ } \cdot\left[3 a^{2} \beta +2( a \cdot \beta ) a \right]-5(\hat{ ᴫ } \cdot \beta )(\hat{ ᴫ } \cdot a )^{2}\right].

To integrate the first term we set the polar axis along a, so (\hat{ ᴫ } \times a )^{2}=a^{2} \sin ^{2} \theta \text {, while } \hat{ ᴫ }=\sin \theta \cos \phi \hat{ x }+\sin \theta \sin \phi \hat{ y }+\cos \theta \hat{ z }. The \phi integral kills the \hat{ x } \text { and } \hat{ y } components, leaving

\oint \frac{1}{ᴫ^{2}}(\hat{ ᴫ } \times a )^{2} d A =2 \pi a^{2} \hat{ z } \int_{0}^{\pi} \sin ^{3} \theta \cos \theta d \theta=0.

[Note that if v = 0 then  d p / d t_{r}= 0—a particle instantaneously at rest radiates no momentum. That’s why we had to carry the expansion to first order in \beta , whereas in deriving the Larmor formula we could afford to set v=0.] The second term is of the form (\hat{ᴫ} \cdot g ) , for a constant vector g. Setting the polar axis along g, so \hat{ ᴫ} \cdot g =g \cos \theta , the x and y components again vanish, leaving 

\oint \frac{1}{ᴫ^{2}}(\hat{ᴫ} \cdot g ) d A =2 \pi g \hat{ z } \int_{0}^{\pi} \cos ^{2} \theta \sin \theta d \theta=\frac{4 \pi}{3} g.

The last term involves (\hat{ ᴫ} \cdot \beta )(\hat{ ᴫ} \cdot a )^{2} ; this time we orient the polar axis along a and let v lie in the xz plane:

\beta =\beta_{x} \hat{ x }+\beta_{z} \hat{ z }, \text { so }(\hat{ ᴫ } \cdot \beta )=\beta_{x} \sin \theta \cos \phi+\beta_{z} \cos \theta . Then

=a^{2}\left(\beta_{x} \hat{ x } \int \sin ^{3} \theta \cos ^{2} \theta \cos ^{2} \phi d \theta d \phi+\beta_{z} \hat{ z } \int \cos ^{4} \theta \sin \theta d \theta d \phi\right)=\frac{4 \pi}{15} a^{2}\left(\beta_{x} \hat{ x }+3 \beta_{z} \hat{ z }\right)=\frac{4 \pi}{15}\left(a^{2} \beta +2( a \cdot \beta ) a \right).

Putting all this together,

\frac{d p }{d t_{r}}=\frac{q^{2}}{16 \pi^{2} \epsilon_{0}} \frac{1}{c^{4}}\left\{\frac{4 \pi}{3}\left[2 a ( a \cdot \beta )+3 a^{2} \beta \right]-5 \frac{4 \pi}{15}\left[a^{2} \beta +2( a \cdot \beta ) a \right]\right\}=\frac{\mu_{0} q^{2}}{6 \pi c^{3}} a^{2} v.

The angular momentum radiated is

\frac{d L }{d t_{r}}=-\oint \frac{1}{\left(\partial t_{r} / \partial t\right)}( ᴫ \times \overleftrightarrow{T}) \cdot d A.

Because of the “extra” ᴫ in the integrand, it seems at first glance that the radiation fields alone will produce a result that grows without limit (as ᴫ → ∞); however, the coe!cient of this term is precisely zero:  (ᴫ \times \overleftrightarrow{T}) . d A =-\epsilon_{0} E^{2}( ᴫ \times d A )=0 (for radiation fields). The finite contribution comes from the cross terms, in which one field (\operatorname{in} \overleftrightarrow{T}) is a radiation field and the other a Coulomb field \left( E _{\text {coul }} \equiv \frac{q}{4 \pi \epsilon_{0}} \frac{ᴫ}{( ᴫ \cdot u )^{3}}\left(c^{2}-v^{2}\right) u \right) :

T_{i j}=\epsilon_{0}\left(E_{i}^{(r)} E_{j}^{(c)}+E_{i}^{(c)} E_{j}^{(r)}-\delta_{i j} E ^{(r)} \cdot E ^{(c)}\right)+\frac{1}{\mu_{0}}\left(B_{i}^{(r)} B_{j}^{(c)}+B_{i}^{(c)} B_{j}^{(r)}-\delta_{i j} B ^{(r)} \cdot B ^{(c)}\right).

This time

\overleftrightarrow{T} \cdot d A =\epsilon_{0}\left[ E ^{(r)}\left( E ^{(c)} \cdot d A \right)-\left( E ^{(r)} \cdot E ^{(c)}\right) d A \right]-\frac{1}{\mu_{0}}\left( B ^{(r)} \cdot B ^{(c)}\right) d A,

so

( ᴫ \times \overleftrightarrow{T}) \cdot d A =\epsilon_{0}\left( ᴫ \times E ^{(r)}\right)\left( E ^{(c)} \cdot d A \right).

Thus

\frac{d L }{d t_{r}}=-\frac{q^{2} c}{16 \pi^{2} \epsilon_{0} \gamma^{2}} \oint \frac{ᴫ}{( ᴫ \cdot u )^{5}}\{ᴫ \times[ ᴫ \times( u \times a )]\} u \cdot d A , \text { where } \gamma \equiv \frac{1}{\sqrt{1-(v / c)^{2}}}.

Now, \{ᴫ \times[ ᴫ \times( u \times a )]\}= ᴫ [ ᴫ \cdot( u \times a )]-( u \times a ) ᴫ^{2} , and ( u \times a )=c(\hat{ ᴫ}- \beta ) \times a =c[(\hat{ᴫ } \times a )-( \beta \times a )],

\hat{ ᴫ } \cdot( u \times a )=-c \hat{ ᴫ } \cdot( \beta \times a ), \text { so }\{ ᴫ \times[ ᴫ \times( u \times a )]\}=- ᴫ ^{2} c\{(\hat{ ᴫ } \times a )-( \beta \times a )+\hat{ ᴫ }[\hat{ ᴫ } \cdot( \beta \times a )]\}.

Meanwhile u \cdot d A =( u \cdot \hat{ᴫ }) ᴫ^{2} d \Omega \text {, where } d \Omega \equiv \sin \theta d \theta d \phi \text {. Expanding to first order in } \beta,

The first integral is

– a \times \int \hat{ ᴫ } \sin \theta d \theta d \phi=- a \times \int(\sin \theta \cos \phi \hat{ x }+\sin \theta \sin \phi \hat{ y }+\cos \theta \hat{ z }) \sin \theta d \theta d \phi= 0,

the second is

-( \beta \times a ) \int \sin \theta d \theta d \phi=-4 \pi( \beta \times a ),

for the third we set the polar axis along g \equiv \beta \times a and get

g \int \hat{ᴫ} \cos \theta \sin \theta d \theta d \phi=2 \pi g \hat{ z } \int \cos ^{2} \theta \sin \theta d \theta=\frac{4 \pi}{3} g =\frac{4 \pi}{3}( \beta \times a ),

and for the last we put the polar axis along a and let \beta =\beta_{x} \hat{ x }+\beta_{z} \hat{ z } lie in the xz plane:

=-4 a \beta_{x} \hat{ y } \int \cos ^{2} \phi \sin ^{3} \theta d \theta d \phi=-\frac{16 \pi}{3} a \beta_{x} \hat{ y }=\frac{16 \pi}{3}( \beta \times a ).

Putting this all together, we conclude