Find the angle θmax at which the maximum radiation is emitted, in Ex. 11.3 (Fig. 11.13). Show that for ultra relativistic speeds ( υclose toc), θmax(1β)/2. What is the intensity of the radiation in this maximal direction (in the ultra relativistic case), in proportion to the same quantity for a particle instantaneously at rest? Give your answer in terms ofγ.

Short Answer

Expert verified

The value ofangle at which the power distribution is maximum at .

θm=arccos±1+15β213β

The value of ultrarelativistic speeds is θm=(1β)/2.

The ratio isη=148555γ8 .

Step by step solution

01

Write the given data from the question.

Consider themaximum angle of radiation emitted isθmax(1β)/2.

02

Determine the formulaof angle at which the power distribution is maximum, ultrarelativistic speeds and ratioη .

Write the formula of angle at which the power distribution is maximum.

cosθ=±1+15β23β…… (1)

Here,β is an ultrarelativistic speed.

Write the formula ofultrarelativistic speed.

cosθ=1+15β23β …… (2)

Here, β is an ultrarelativistic speed.

Write the formula ofη.

η=(45)512δ4 …… (3)

Here, δis intensity of the radiation.

03

Step 3:Determine thevalue of angle at which the power distribution is maximum, ultrarelativistic speeds and ratioη .

Determine the angle at which the power distribution is greatest first:

ddθdPdΩ=0ddθsin2θ(1βcosθ)5=0=2sinθcosθ(1βcosθ)55βsin3θ(1βcosθ)4(1βcosθ)10=02cosθ2βcos2θ=5βsin2θ

Solve further as

3βcos2θ+2cosθ5β=0

Determine the angle at which the power distribution is maximum.

For β0we expectθπ/2, so we want the RHS to tend to zero. We can only have that for the plus sign, so we have:

θm=arccos1+15β213β

We have β1 for the ultrarelativistic speeds, thereforeβ=1δ, δ<<1. Increase the cosine:

Determine the value of ultrarelativistic speeds.

Substitute(1δ) for β into equation (2).

cosθ=1+15(1δ)23(1δ)1+15(12δ)13(1δ)=1630δ13(1δ)13(1+δ)[41158δ1]

Solve further as

13(1+δ)[4(11516δ)1]=(1+δ)(154δ)114δ

But:

cosθm112θm2θm=δ/2=(1β)/2

Let's now compute the ratio. Starting with 11.69, the particle is immediately at rest.

|dPdΩ|rest=Sradr^r2=μ0q2a216π2csin2θ

Keep in mind that in this case, θ refers to the spherical angle rather than the angle closed with the velocity vector. So, in this instance, θπ/2, and:

|dPdΩ|rest=μ0q2a216π2c

The angle θm for the UR situation is what was discovered in the previous phase. The ratio is thus:

sin2θmθm2=δ2(1βcosθm)5=[1(1δ)(1+12δ2)]5(54)5δ5

Determine theη.

η=(dP/dΩ)UR(dP/dΩ)rest=sin2θm(1βcosθm)5=(45)512δ4

Now we need to determine theδ in terms ofγ:

γ=11β2=11(1δ)212δδ=12γ2

Determine the ratio is:

η=(45)512δ4=148555γ8

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