4
$\begingroup$

For a conservative force $\vec{F}(x,y)$ we have $$\vec{F}(x,y) = - \vec{\nabla}\phi(x,y)$$ where $\phi(x,y)$ is a potential function for $\vec{F}(x,y)$.

Again, Newton's second law states that $$ \vec{F}(t) = \frac{d \vec{P}(t)}{dt}$$ where $\vec{P}(t)$ is the momentum.

How these two equations are compatible; i.e, for a conservative force $\vec{F}(x,y)$ how can we write the Newton's second law?

Edit

For a conservative force $\vec{F}(x,y)$ how can we write the Newton's second law?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ A particle at position $\vec{x}(t)$ at time $t$ will experience a force $F(t)=-\vec{\nabla}\phi(\vec{x}(t))$. $\endgroup$
    – lemon
    Commented Oct 2, 2014 at 11:13
  • $\begingroup$ I don't understand what you are asking...in what sense do you think there is a problem of "compatibility" if you write $\dot{p}=-\nabla\phi$? $\endgroup$
    – yuggib
    Commented Oct 2, 2014 at 11:37
  • $\begingroup$ @yugib: I have edited the question. $\endgroup$
    – rainman
    Commented Oct 2, 2014 at 11:39
  • $\begingroup$ You edited the question so that the last line is exactly repeated. Why do you expect that this will clarify the question? $\endgroup$
    – Jim
    Commented Oct 2, 2014 at 13:40
  • $\begingroup$ @Jim: Yes, you are right. I should have posed the question in a different way as it clarifies the query. $\endgroup$
    – rainman
    Commented Oct 2, 2014 at 14:35

1 Answer 1

Reset to default
2
$\begingroup$

What @lemon said is right. While $\bf F = -\nabla \phi\left({\bf r}\right)$ suggests that $\bf F$ depends on $\bf r$, it ultimately depends on time through ${\bf r} = {\bf r}\left(t\right)$. That is, ${\bf F} = -\nabla \phi\left({\bf r}\left(t\right)\right)$. So, there is no inconsistency in writing Newton's second law as (a dot represents a time derivative) $$ {\bf F} = m \ddot{\bf r} = -\nabla \phi $$ In fact, we don't have to restrict ourselves to a conservative force. $\bf F$ can depend on position, velocity, and time explicitly, ${\bf F} = {\bf F}\left({\bf r}\left(t\right),\dot{\bf r}\left(t\right),t\right)$, since you will end up with a perfectly consistent system of differential equations $$ \begin{eqnarray} \ddot{x} &=& F_x\left(x,y,z,\dot{x},\dot{y},\dot{z},t\right)/m \\ \ddot{y} &=& F_y\left(x,y,z,\dot{x},\dot{y},\dot{z},t\right)/m \\ \ddot{z} &=& F_z\left(x,y,z,\dot{x},\dot{y},\dot{z},t\right)/m \\ \end{eqnarray} $$ To work a simple example, consider the potential energy $$ \phi\left(x\right) = \frac{1}{2} k x^2. $$ Then $$ {\bf F} = -\nabla\left(\frac{1}{2} k x^2\right) = -k x \ \hat{\bf x} = m \left(\ddot{x} \ \hat{\bf x} + \ddot{y} \ \hat{\bf y} + \ddot{z} \ \hat{\bf z}\right), $$ or $$ \begin{eqnarray} m \ddot{x} = -kx &\Rightarrow& x\left(t\right) = x\left(0\right) \cos\left(\omega t \right) + \frac{\dot{x}\left(0\right)}{\omega} \sin\left(\omega t\right), \omega = \sqrt{k/m} \\ m \ddot{y} = 0 &\Rightarrow& y\left(t\right) = {\dot y}\left(0\right) t + y\left(0\right)\\ m \ddot{z} = 0 &\Rightarrow& z\left(t\right) = {\dot z}\left(0\right) t + z\left(0\right)\\ \end{eqnarray} $$

Improve this answer
$\endgroup$

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