Question
A star, radius $R$, emits isotropically with a black body spectrum with temperature T.
I want to show that the radiation pressure at a distance $d>R$ from the centre of the star is $P_{rad}=\large{\frac{4\sigma T^4}{3c}}\left[1-\left(1-\frac{R^2}{d^2}\right)^{\frac{3}{2}}\right]$.
Background information:
From definitions here is some background information on formula that I believe are relevant:
For a particular radiation field, $dE_v(\theta,\phi)= I_v(\theta,\phi)\ dv\ dAcos\theta\ d\Omega\ dt$, where $d\Omega= sin\theta\ d\theta\ d\phi$ is the solid angle and $dA$ is perpendicular to flow of energy.
We can express flux, $F$ in therms of $I_v$, the specific intensity, by integrating $dE_v(\theta,\phi)$ over a solid angle $d\Omega$.
Therefore, $dE_v(\theta,\phi)= \oint dE_v(\theta,\phi)\ d\Omega = \oint I_v(\theta,\phi)\ dv\ dAcos\theta\ d\Omega\ dt$
and so,
$F_v = \oint I_v\ cos\theta\ d\Omega = \int_{0}^{2\pi} \int_{0}^{\pi} I_v\ cos\theta\ sin\theta\ d\theta\ d\phi$.
For an isotropic black body, $I_v$ does not depend on $\theta,\phi$ and so $F_v=0$.
At the surface of a black body, the outward flux is $F=\int_{0}^{\infty}F_vdv=\pi \int_{0}^{\infty}B_v(T)dv =\sigma T^4$.
The radiation pressure for isotropic $I_v$ is $P_{rad}= \frac{2}{c} \int_{0}^{\infty} \int_{hemisphere} I_v\ cos^2\theta\ d\Omega\ dv = \frac{4\pi}{3c} \int_{0}^{\infty} I_vdv$
and specifically, for isotropic black body radiation, $I_v=B_v(T)$ giving $P_{rad}=\frac{4\sigma T^4}{3c}$.
Method:
I have been told to define the solid angle $d\Omega$ of the system and use trigonometry in which I must use the known distances of the system (the stars radius $R$, the distance to the point $d>R$) to express the defining angles in terms of distances.
i.e. I believe we are required to use $sin\theta$ or $cos\theta$ with the side opposite to theta being the radius and the distance from the center $d$ being the hypotenuse of the triangle (I am not sure what the adjacent side is equal to).
I believe that to obtain the correct final answer, a combination of integration using u substitution and trigonometry is necessary.
What I have tried:
I am able to derive the first part of the answer, $P_{rad}=\large{\frac{4\sigma T^4}{3c}}$ through integrating $P_{rad}= \frac{2}{c} \int I_v\ cos^2\theta\ sin\theta\ d\theta\ dv$ with $u=cos\theta$ and $du= -sin\theta\ d\theta$ with the integral $\int_{1}^{-1} -u^2du= -\int_{-1}^{1} -u^2du$ giving $P_{rad} = \frac{4}{3c}\int I_v\ dv$ and since $F= \int_{0}^{\infty} I_v\ dv = \pi \int_{0}^{\infty} I_v\ dv = \sigma T^4$ then $P_{rad} = \frac{4\pi}{3c} \int I_v\ dv = \frac{4\sigma T^4}{3c}$.
My problem lies in the fact that I cannot correctly determine the terms within the brackets as well to give the full expected result.
Best Answer
The terms in the brackets are there because the star only subtends a finite angle as seen by the observer. If we drew a triangle with vertices at the observer, the center of the star, and a point where the observer's line of sight is tangent to the star, it would be a right triangle with hypotenuse $d$ and leg $R$, thus the angle the star subtends from center to limb is $\theta=\sin^{-1}\frac Rd$. So your integral is $$\begin{align}P&=\int_0^{2\pi}\int_0^{\sin^{-1}\frac Rd}\frac{2\sigma T^4}{\pi c}\cos^2\theta\sin\theta\,d\theta\,d\phi=-\left.\frac{4\sigma T^4}{3c}\cos^3\theta\right|_0^{\sin^{-1}\frac Rd}\\ &=\frac{4\sigma T^4}{3c}\left[1-\cos^3\left(\sin^{-1}\frac Rd\right)\right]=\frac{4\sigma T^4}{3c}\left[1-\left(1-\frac{R^2}{d^2}\right)^{3/2}\right]\end{align}$$ I think this formula is only right for pressure on a perfectly reflecting surface. For a perfectly absorbing surface I think the answer should only be half as big.
EDIT: Some derivations. If we imagine a little area $A$ in thermal equilibrium with its surroundings at temperature $T$ and also it can only exchange energy through radiation and it absorbs all radiation that hits it and that the upward normal to its surface is the $z$-direction, then the power incident on its surface is $$\text{power}=\int_0^{2\pi}\int_0^{\pi/2}AI_0\cos\theta\sin\theta\,d\theta\,d\phi$$ The $AI_0\cos\theta$ factor is the energy incoming pre unit solid angle per unit time. The $\cos\theta$ is there because of the inclination of the source to the surface. The $\sin\theta\,d\theta\,d\phi$ is the element of solid angle. That $\sin\theta$ is there because that's the radius of a small circle on the unit sphere. So the $\phi$ integral is easy with result $2\pi$ and to do the $\theta$ integral we let $u=\cos\theta$ so $du=-\sin\theta\,d\theta$ and $$\int\cos\theta\sin\theta\,d\theta=\int-u\,du=-\frac12u^2+C=-\frac12\cos^2\theta+C$$ And so $$\text{power}=2\pi AI_0\left[-\frac12\cos^2\theta\right]_0^{\pi/2}=\pi AI_0=\sigma AT^4$$ Which is the Stefan-Boltzmann law. The momentum incoming per unit area per unit solid angle per unit time is thus $$p_z=-\frac{I_0}c\cos\theta=-\frac{\sigma T^4}{\pi c}\cos\theta$$ Because we know that electromagnetic radiation carries momentum proportional to its energy: $E=pc$ and the factor of $-\cos\theta$ is there because we only need the $z$-component of momentum, the other components will cancel out due to symmetry. Notice how we used the Stefan-Boltzmann law to find out what that unknown energy per unit area per unit solid angle per unit time $I_0$ was. So to get pressure, which is force per unit area and so is momentum incoming per unit area per unit time we just have to integrate this over solid angle: $$P=\int_0^{2\pi}\int_0^{\theta_{\max}}\frac{\sigma T^4}{\pi c}\cos^2\theta\sin\theta\,d\theta\,d\phi$$ The minus sign went away because we are going to consider force to be positive down. Again, the $\phi$ integral is easy and we let $u=\cos\theta$ to get $$\int\cos^2\theta\sin\theta\,d\theta=\int-u^2\,du=-\frac13u^3+C=-\frac13\cos^3\theta+C$$ So we have $$P=2\pi\frac{\sigma T^4}{\pi c}\left[-\frac13\cos^3\theta\right]_0^{\theta_{\max}}=\frac{2\sigma T^4}c\left[-\frac13\cos^3\theta_{\max}+\frac13\right]=\frac{2\sigma T^4}{3c}\left[1-\cos^3\theta_{\max}\right]$$ As I said above, this is for a perfectly absorbing surface. If it's perfectly reflecting, the pressure will be twice as great as the momentum change of the incident radiation is doubled.
Just draw a picture of the observer, the star, and the triangle I described earlier. You will see that $\cos\theta_{\max}=\sqrt{1-\frac{R^2}{d^2}}$ by the Pythagorean theorem. I don't have the energy to compose a graph of this for you right now.
EDIT: Here is the picture I was encouraging you to draw. The observer is at the leftmost vertex of the red triangle, the center of the star at its lower right corner, and the stellar limb at its top corner. The angle at the top must be a right angle because there the line of sight is tangent to the star, represented by the blue circle. We were given the distance to the center $d$ and the radius of the star $R$, so trigonometry tells us that $\sin\theta_{\max}=\frac Rd$ and then some more trigonometry tells us that $\cos\theta_{\max}=\frac{\sqrt{d^2-R^2}}{d}$.
