# Do bigger stars emit higher amplitude waves?

So I’ve been reading about luminosity and I get the concept of the size and temperature determining the intrinsic brightness and total energy emission of the star. I also understand how the further away the star is, the more the brightness will decrease since the waves spread out more meaning less will hit your eye and that the intensity of light is equal to the amplitude squared.

The thing I’m just wondering is if the temperature of the star determines the wavelength of the waves. Does the number of photons and the size of the star determine the amplitude of the waves? So if you had a red giant and a red dwarf star the red giant would be emitting higher amplitude waves? Or is it that they both emit electromagnetic waves with the same amplitude hit because the red giant is bigger it emits more waves and thus more total energy.

• @Qmechanic Thanks so much for your response. I’m only in year nine but I have a strong interest for astronomy which is why I know a lot about it. I think I understand most of your answer and get the general idea behind it. I’ve been trying to learn as much about astronomy as I can while only using basic maths but I think I’ve reached a point where if I wish to know more I’m going to have to learn some more maths. So far I’m familiar with the basic. energy, force, time, mass, work done and some basic luminosity equations but I don’t know much more than that – Spacegamer 383 May 5 '18 at 13:26

The simple answer is yes, providing you understand the limitations of a classical treatment and consider the radiation field at a fixed distance from the star.

In classical terms, the amplitude of the electromagnetic waves is proportional to the square root of the flux of energy carried (power per unit area) at any point in space. It doesn't have anything to do with the wavelength of the radiation, but clearly as we move further away from a star, the power per unit area we measure, and therefore the electric field amplitude decrease. Where the temperature of the star does come into it is that hot stars emit more power for a given size of star.

The details

The time-averaged power per unit area carried by a plane-polarised electromagnetic wave is given by $$\langle S \rangle = \frac{1}{2} \epsilon_0 c E_0^2,$$ where $E_0$ is the electric field amplitude of the wave.

In general, the light from a star is unpolarised (there are special exceptions) and can be modelled as consisting of an equal mixture of light in two perpendicular polarisation states with a random, time-dependent phase between them.

The time-averaged power per unit area from this wave will be $$\langle S \rangle = \frac{1}{2}\epsilon_0 c E_0^2 + \frac{1}{2}\epsilon_0 c E_0^2 = \epsilon_0 c E_0^2,$$ where now $E_0$ is the electric field amplitude of each of the two perpendicular polarisation components, and $$E_0 = \sqrt{\frac{\langle S \rangle}{\epsilon_0 c}}.$$ This would also be the rms value of the total E-field, since the two polarisations are at right angles.

Some examples

The "solar constant" (the average power per unit area incident at the top of the Earth's atmosphere) is about 1360 W m$^{-2}$. Thus the electric field amplitude is 716 V/m.

You can extend the former calculation to stars of different temperature by saying that the electric field amplitude was related to the luminosity of the star and the distance from the star that you measured the field by $$E_0 = 716 \left(\frac{L}{L_{\odot}}\right)^{1/2} \left(\frac{d}{{\rm 1\ au}}\right)^{-1} \ {\rm V/m},$$ where $L_{\odot}$ is the solar luminosity and the distance $d$ is in astronomical units.

The luminosity $L$ then depends on both the temperature of the star and its radius, through the Stefan-Boltzmann law. $$E_0 = 716 \left(\frac{T}{T_{\odot}}\right)^{2} \left(\frac{R}{R_{\odot}}\right)\ {\rm V/m},$$ where $T$ is the "effective temperature" of the stellar photosphere.

On the other hand if we were just to measure the radiation at the solar surface, the time-averaged power per unit area is approximately $\sigma T^4$, where $\sigma$ is the Stefan-Boltzmann constant. For the Sun, where $T_{\odot} \sim 5760$K, $\langle S \rangle = 6.2\times 10^{7}$ W m$^{-2}$ and $E_0 = 1.5 \times 10^{5}$ V/m.

The extension of this calculation to the surfaces of stars with different temperatures would be $$E_0 = 1.5\times 10^{5} \left(\frac{T}{T_{\odot}}\right)^{2} \ {\rm V/m}$$