Physics Asked on December 4, 2021
I think for radio antenna transmitters, if the AC current’s frequency is at the frequency of radiowaves, EM waves in the radio frequency will be emitted? Then the frequency has something to do with the way the photons were generated ?
Anyways I have not been able to make the connection between the interpretation of photons as packets of energy and EM radiation, if there’s a difference at all.
The Poynting vector gives the energy flux per unit of area and time. Suppose $Sigma$ is a closed surface that encloses an antenna. Then the total power radiated by the antenna is
$$P = int_{Sigma}mathbf Scdothat{mathbf n} text dSigma$$
where $mathbf S$ is the Poynting vector. Assuming that the antenna is transmitting at a frequency $nu$, each emitted photon has an energy of
$$E_nu = hnu$$
Knowing the power $P$ of the antenna and the frequency $nu$ we can estimate the photon flux $dot N$ by means of the equation
$$P = E_nudot N,$$
which is just a statement of the law of energy conservation.
More generally, if an antenna is transmitting with a spectral power density $pi(nu)$, we would have a spectral photon flux $dot n(nu)$ given by
$$dot n(nu)=frac{pi(nu)}{hnu}.$$
The total photon flux is then given by the integral over all frequencies, viz.
$$dot N = int_0^inftyfrac{pi(nu)}{hnu} text dnu.$$
Indeed, for an antenna transmitting at a pure frequency $nu^*$ we would have $pi(nu) = P^*delta(nu-nu^*)$, which yields
$$dot N = frac{P^*}{hnu^*}.$$
Answered by Phoenix87 on December 4, 2021
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