Question

help me

Answer 1

question?

Answer 2

take a look the piciture attached below

Answer 3

i'm trying to derive 12.8 from 12.7.,

Answer 4

i'm trying to derive equations 12.8 from 12.7.,

Answer 5

Im not sure what kind of physics class your in but im in physics 1 and not even sure what a few of those symbols even mean.

Answer 6

sorry this is for 4th year bachelor student in the university

Answer 7

To go from 12.7 to 12.8 is a non trivial exercise, the main task is to calculate the electric and magnetic fields generated by the oscillating dipole, then you can compute the poynting vector, integrate over all directions and take a time average to find the average power radiated.
The Feynman lectures discuss this kind of stuff somewhere, not sure if he gives all the details, from memory I can't think of a good reference for this problem.

Answer 8

let me show you what i've done

Answer 9

wait a minute :)

Answer 10

\[d = e X_0 \cos (\omega_0 t) \]

\[\frac{ d }{ dx } (d) = \frac{ d }{ dx } \left( e X_0 cow (\omega_0 t) \right) = -e \omega_0 X_0 \sin (\omega_0 t)\]
\[\frac{ d^2 }{ dx } (d) = \frac{ d }{ dx } \left( -e \omega_0 X_0 \sin (\omega_0 t) \right) = -e \omega_0^2 X_0 \cos (\omega_0 t)\]
\[\bar S = \frac{ 2 \bar {{\frac{ d^2 (d)}{ dx }}^2} }{ 3 c^3 } = \frac{ 2 (-e \omega_0^2 X_0 \cos \omega_0t)^2 }{ 3 c^3 } = \frac{ 2 e^2 \omega_0^4 X_0^2 \cos^2(\omega_0t) }{ 3c^3 } \]
at t =0
\[\bar S = \frac{ 2e^2 \omega_0^4 X_0^2 \cos^2(\omega_0 (0)) }{ 3c^3 } = \frac{ 2 e^2 \omega_0^4 X_0^2 }{ 3c^3 }\]

Answer 11

@ProfBrainstorm :)

Answer 12

@cwrw238

Answer 13

based on what i've done.,
\[\bar S = \frac{ 2 e^2 \omega_0^4 X_0^2 }{ 3 c^2 }\]
but based on the book which i read, it is
\[\bar S = \frac{ e^2 \omega_0^4 X_0^2 }{ 3 c^2 }\]
PLEASE HELP ME TO SOLVE THIS

Answer 14

You've not treated the averaging correctly.
Rather than just taking the value of S at t=0, you need to take an average value of S over time.
The time average of cos^2(omega t) is one half, so that's where your unwanted factor of 2 goes.

Answer 15

When I first read your question I thought that you wanted to derive equation 12.8 rigorously from maxwell's equations : )

Answer 16

ahhh i see

Answer 17

so you said that the time average of this term is 1/2?

Answer 18

yes i think so, just from memory

Answer 19

thank you @ProfBrainstorm

Answer 20

you're welcome - btw i checked, the average value of cos squared is 1/2 as i thought.