show $ -\int \frac{d^{3}p}{(2\pi)^3}p\frac{\partial f(p,t)}{\partial p}=3F$

Question

Hi I am stuck with an integral problem trying to show $$ -\int \frac{d^{3}p}{(2\pi)^3}p\frac{\partial f(p,t)}{\partial p}=3F$$ where $F=\frac{1}{(2\pi)^3}\int d^{3}pf(p,t)$ I have read in many books that simply integrated by parts to get the solution, but I just dont get it can anyone help?

Answer 1

Let me translate the physics type of notation back to calculus: assume your configuration space is just $\mathbb{R}^3$, and $p = \langle x,y,z\rangle $ is a vector in $\mathbb{R}^3$, also the function $f$ is smooth and the product with $x$, $y$,or $z$ vanishes when $\|p\| = \sqrt{x^2+y^2+z^2}\to \infty$. Since there is no time here we treat $t$ as a constant. Then the integral can be rewritten in mathematical rigor to avoid vagueness: $$ \begin{aligned} -\int \frac{d^{3}p}{(2\pi)^3}p\frac{\partial f(p,t)}{\partial p} &= -\int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} \nabla_p f\cdot p \,dxdydz \\ &= -\int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} \langle \partial_x f, \partial_y f, \partial_z f \rangle\cdot \langle x,y,z\rangle \,dxdydz \\ &= -\int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} (x \, \partial_x f + y\, \partial_y + z\, \partial_z f)\,dxdydz \end{aligned} $$ Split the integrands into three parts: $$ \begin{aligned} &\int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} (x\, \partial_x f + y \, \partial_y + z\, \partial_z f)\,dxdydz \\ =&\int_{\mathbb{R}^2}(\int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} x \partial_x f\,dx)dydz + \int_{\mathbb{R}^2}(\int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} y \partial_y f\,dy)dxdz + \int_{\mathbb{R}^2}(\int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} z \partial_z f\,dz)dxdy \end{aligned} $$ For each integral: let $u= x$, $dv = \partial_x f\,dx$, then by integral by parts formula $\displaystyle \int_{\mathbb{R}^1}u dv = - \int_{\mathbb{R}^1}v du$, if $uv$ goes to zero at infinity. $$ \int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} x \partial_x f\,dx = - \int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} f\,dx $$ Hence: $$ \int_{\mathbb{R}^2}(\int_{\mathbb{R}^1}\frac{1}{(2\pi)^3} x \partial_x f\,dx)dydz = - \int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} f\,dx dydz $$ for we assumed $xf(x,y,z,t) \to 0$ when $x\to \infty$. Do the same thing for other two terms: $$ \int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} (x\, \partial_x f + y \, \partial_y + z\, \partial_z f)\,dxdydz = -3\int_{\mathbb{R}^3}\frac{1}{(2\pi)^3} f\,dx dydz $$ Therefore: $$ -\int \frac{d^{3}p}{(2\pi)^3}p\frac{\partial f(p,t)}{\partial p} = \frac{3}{(2\pi)^3} \int_{\mathbb{R}^3}f\,dx dydz = \frac{3}{(2\pi)^3} \int f(p,t) \,d^3p = 3F. $$ Also the last remark is that: the configuration space needs not to be $\mathbb{R}^3$, it may be a manifold, $p$ is a tangent vector, momentum is cotangent. And the formula is still valid.