To find the number of micro states in a statistical system for N number of particles in a fixed volume and energy within the range E and $E+\delta{E}$, we require the knowledge on how to find the volume for N dimensional hypersphere in momentum space.
In the simplest way, the problem is to find the volume of an N-dimensional hypersphere. If we try to find the general formula by mathematical induction, using the equation in Cartesian form, it will produce a complex equation.
Instead there is a clever trick to simplify the procedure based on the concept of infinity. The equation of a sphere in Cartesian form is $$ x_1^2+x_2^2+x_3^2+...+x_N^2 = r^2$$ where 'r' is the radius of the sphere.
In general, volume of an n-dimensional sphere should be proportional to $r^N$ and its surface is proportional to $r^{N-1}$. So, $$V \propto r^N\\ S \propto r^{N-1} $$ or $$ S = A r^{N-1}$$ where A is the proportionality constant that is equal to the surface area of a unit sphere. If we could find somehow this value of A in n-dimension, it is then a simple task to proceed to the volume. The special characteristic of the surface of a sphere is that, once we know the surface, the volume is just a simple line integral of the surface value along 'r'.
$$ dV = S dr = A r^{N-1} dr $$
The generalization of the equation of sphere in Cartesian form is a straight task. To connect all these things, we use the property of the infinity given as,
Integral of a function over the infinite volume is the same independent of whether the integral is done with Cartesian cubic volume element (cubic volume extending to infinity) or spherical volume element (spherical volume extending to infinity), the result is the same. This is not a completely new one because we use this same fact to find the integral of $$\int_{-\infty}^{\infty}e^{-\alpha{x^2}}\,dx = \sqrt{\frac{\pi}{\alpha}}$$
Using the same integral we have, $$ I = \int_{-\infty}^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} \,dx_1\,dx_2\,...\,dx_N = \pi^{N/2} $$ Now,the same function integrated over spherical volume element gives, $$ I = \int_0^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} A r^{N-1}dr = \int_0^{\infty} A e^{-{r^2}} r^{N-1} dr $$ Substituting $y = r^2 $ we have $$I = \int_0^{\infty} A e^{-y} y^{\frac{N-1}{2}} \,\frac{1}{2y^{1/2}}\,dy = \int_0^{\infty} A \frac{1}{2} e^{-y} y^{\frac{N}{2}-1}dy = \frac{1}{2} A \Gamma(N/2)$$ then we have, by equating the value of I in two ways, $$ A = \frac{2\pi^{N/2}}{\Gamma(N/2)}$$ Now, the volume is evaluated to be $$V = \int _0^R \,S \,dr = \int_0^R \,A r^{N-1}\,dr \\ V = \int_0^R\,\frac{2\pi^{N/2}}{\Gamma(N/2)} r^{N-1} \, dr = \frac{\pi^{N/2}}{\Gamma(N/2)} \frac{R^{N}}{N/2} = \frac{\pi^{N/2} R^N}{\Gamma(\frac{N}{2}+1)}$$ where we used the fact $$\frac{N}{2} \Gamma(\frac{N}{2}) = \Gamma(\frac{N}{2}+1)$$
Thus we can determine the volume and so the surface area of an n-dimensional hypersphere in a simple form.
In the simplest way, the problem is to find the volume of an N-dimensional hypersphere. If we try to find the general formula by mathematical induction, using the equation in Cartesian form, it will produce a complex equation.
Instead there is a clever trick to simplify the procedure based on the concept of infinity. The equation of a sphere in Cartesian form is $$ x_1^2+x_2^2+x_3^2+...+x_N^2 = r^2$$ where 'r' is the radius of the sphere.
In general, volume of an n-dimensional sphere should be proportional to $r^N$ and its surface is proportional to $r^{N-1}$. So, $$V \propto r^N\\ S \propto r^{N-1} $$ or $$ S = A r^{N-1}$$ where A is the proportionality constant that is equal to the surface area of a unit sphere. If we could find somehow this value of A in n-dimension, it is then a simple task to proceed to the volume. The special characteristic of the surface of a sphere is that, once we know the surface, the volume is just a simple line integral of the surface value along 'r'.
$$ dV = S dr = A r^{N-1} dr $$
The generalization of the equation of sphere in Cartesian form is a straight task. To connect all these things, we use the property of the infinity given as,
Integral of a function over the infinite volume is the same independent of whether the integral is done with Cartesian cubic volume element (cubic volume extending to infinity) or spherical volume element (spherical volume extending to infinity), the result is the same. This is not a completely new one because we use this same fact to find the integral of $$\int_{-\infty}^{\infty}e^{-\alpha{x^2}}\,dx = \sqrt{\frac{\pi}{\alpha}}$$
Using the same integral we have, $$ I = \int_{-\infty}^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} \,dx_1\,dx_2\,...\,dx_N = \pi^{N/2} $$ Now,the same function integrated over spherical volume element gives, $$ I = \int_0^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} A r^{N-1}dr = \int_0^{\infty} A e^{-{r^2}} r^{N-1} dr $$ Substituting $y = r^2 $ we have $$I = \int_0^{\infty} A e^{-y} y^{\frac{N-1}{2}} \,\frac{1}{2y^{1/2}}\,dy = \int_0^{\infty} A \frac{1}{2} e^{-y} y^{\frac{N}{2}-1}dy = \frac{1}{2} A \Gamma(N/2)$$ then we have, by equating the value of I in two ways, $$ A = \frac{2\pi^{N/2}}{\Gamma(N/2)}$$ Now, the volume is evaluated to be $$V = \int _0^R \,S \,dr = \int_0^R \,A r^{N-1}\,dr \\ V = \int_0^R\,\frac{2\pi^{N/2}}{\Gamma(N/2)} r^{N-1} \, dr = \frac{\pi^{N/2}}{\Gamma(N/2)} \frac{R^{N}}{N/2} = \frac{\pi^{N/2} R^N}{\Gamma(\frac{N}{2}+1)}$$ where we used the fact $$\frac{N}{2} \Gamma(\frac{N}{2}) = \Gamma(\frac{N}{2}+1)$$
Thus we can determine the volume and so the surface area of an n-dimensional hypersphere in a simple form.
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