Tuesday 23 August 2016

homework and exercises - Is there a quick way of finding the kinetic energy on spherical coordinates?


Assume a particle in 3D euclidean space. Its kinetic energy: $$ T = \frac{1}{2}m\left(\dot x^2 + \dot y^2 + \dot z^2\right) $$


I need to change to spherical coordinates and find its kinetic energy: $$ T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right) $$


Its well known that: $$ x = r\sin\theta\cos\phi \\ y = r\sin\theta\sin\phi \\ z = r\cos\theta $$


A way of doing it is taking the time derivatives, arriving with $3+3+2=8$ different terms with some squares, then open it arriving at $6+6+3 = 12$ different terms majority of them with 4 sine or cossine multiplications. Then to cancel out some terms somehow to arrive in this neat $3$-term expression for kinetic energy in spherical coordinates. In short, a lot of work just to arrive in a simple expression.


Here is my question: Is there a shorter way? Or even better: is there an effortless way?



Answer



There is an effortless way, if you accept geometrical reasoning.


You know, that $T = \frac 1 2 m \vec v^2 = \frac 1 2 m \lvert \vec v \rvert^2$. Furthermore, spherical coordinates are orthogonal, therefore you can just write:



$$\lvert \vec v \rvert = \sqrt{v_\phi^2 + v_\theta^2 + v_r^2}$$


Geometrically, one easily finds: $v_r = \dot r$, $v_\theta = r \dot \theta$ and $v_\phi = r \sin(\theta) \dot \phi$.


And thus the result:


$$\lvert\vec v\rvert = \sqrt{\dot r^2 + r^2 \dot \theta^2 + r^2 \sin^2(\theta) \dot \phi^2}.$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...