0
$\begingroup$

Consider the dynamics of a particle P shown: Particle in 3D space with Radius r Newton's second law states that: $$\frac{d}{dt}(m\dot r) = \mathbf F$$

where, $\boldsymbol{r}$ is the position vector of the particle, $\boldsymbol{\dot r}$ is its velocity vector and $m$ is its mass. $\mathbf F$ is the total external force (vector) acting on P. Prove the following result regarding the existent of mechanical energy integral of motion of the system:

Theorem: For the dynamical system, if $F = -\nabla V (\boldsymbol{r}) $ i.e the force, $\mathbf F$, is the negative gradient of a scalar function $V(\mathbf r)$ of only the position vector, then the total mechanical energy of the system is conserved. Moreover, the function $V(\mathbf r)$ is nothing but the potential energy of the system.

Hints:

The above theorem states that if $F = -\nabla V (\boldsymbol{r}) $ then the mechanical energy is an integral of motion.

The gradient of a scalar function, $V$ is written as:

$\nabla V = \partial V / \partial \boldsymbol{r} $

Also, the gradient of a scalar quantity is a vector. Do not worry about writing it in terms of the components

Recall the mechanical energy is the sum of KE and PE. Mechanical energy is given by the standard form:

$$T=1/2 ∗ m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $$

where $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r}$ represents the dot product of $\boldsymbol{\dot r} $, which is the velocity vector, with itself (i.e. square of the speed: $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} = v^2$ )

Also, let $G = G(\mathbf r, t)$ be a scalar function of a vector, $r$, and a scalar, $t$. Then, the total derivative of $G$ is related to its partial derivatives by:

$$dG = \frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r + \frac{\partial G}{\partial t}\cdot dt$$

with $\frac{\partial G}{ \boldsymbol{\partial r}} $ identified as the gradient of G, which is a vector quantity, and $\frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r $ as the dot product between the gradient of $G$ and $d\boldsymbol{r}$

Do not assume that $V(\boldsymbol{r})$ is the potential energy as this should be a natural conclusion of the analysis.

To begin, take the time derivative of the kinetic energy. Recall that if $c = \boldsymbol{a} \cdot \boldsymbol{b} $ is the dot product of two vectors, then the time derivative of c is: $$\dot c = \boldsymbol{\dot a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{\dot b}$$

ASSUME THE PARTICLE'S MASS, $m$ is constant

Edit: Attempted solution: $$\frac {d}{dt}(m \dot r) = F $$ $$\frac {d}{dr}(m \dot r)\frac {dr}{dt} = F$$ $$d(mv)v = Fdr$$ $$\int mvdv = \int Fdr$$ $$\frac {1}{2}mv^2 + \int-Fdr = C $$

Seems to prove the above theorem, however this kind of seems to ignore the fact that the variables are vectors. How can a similar approach be applied to the vectors? And still derive the potential energy at the end?

Basically, I'm asking how the potential energy can be an outcome of $$\frac {d}{dt}(\frac{1}{2} m \dot r \cdot \dot r) $$ and $$\frac{d}{dt}(m\dot r) = \mathbf F$$

$\endgroup$
  • $\begingroup$ What is the question here? $\endgroup$ – probably_someone Sep 12 '19 at 13:24
  • $\begingroup$ Please note that this type of "solve this problem for me" question is off topic as per the homework policy. If there is some aspect of this question giving you trouble, try to ask a conceptual physics question, and show any relevant work. $\endgroup$ – JMac Sep 12 '19 at 13:28
  • $\begingroup$ ^question edited to hopefully address the concerns of @JMac $\endgroup$ – Clark Sep 12 '19 at 13:46
  • $\begingroup$ Why are you bolding some vectors and not others? $r$ and $\mathbf{r}$ are completely different things. $\endgroup$ – G. Smith Sep 12 '19 at 17:14
  • $\begingroup$ In $dG$, where did $c$ come from and what is it? $\endgroup$ – G. Smith Sep 12 '19 at 17:15
1
$\begingroup$

Multiply both sides by $\dot r$:

$m \ddot r \dot r = \frac{d}{dt}(\frac{1}{2}m \dot r^{2})$

$-\frac{\partial V } {\partial r} \frac{\partial r} {\partial t} = -\frac{\partial V}{\partial t}$

$\endgroup$
0
$\begingroup$

Instead of working in radial coordinates(which may be tricky because the unit vectors change with respect to time), I would suggest seeing the x,y,z equations of motion separately. So you have, $$m \ddot{x}= -\frac{\partial V}{\partial x}, \hspace{0.5mm}m \ddot{y}= -\frac{\partial V}{\partial y}, \hspace{0.5mm}m \ddot{z}= -\frac{\partial V}{\partial z}. $$ Now add these three equations and check for yourself that the following and the equations above imply the same thing i.e., $$\frac{d\Big(\frac{1}{2}m(\dot{x}^2 +\dot{y}^2+\dot{z}^2)+V(r)\Big)}{dt}=0. $$ So in some sense you could have integrated the equations of motion to get this.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.