1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relationship of Gradients to Work, PE and KE

  1. Oct 4, 2012 #1
    I am having a hard time relating work, potential energy and kinetic energy with my knowledge of calculus. I have a feeling that it somehow relates to gradients, is that correct? If so; how would you relate gradients to those concepts? And why is it that now derivatives are being taken with respect to position as opposed to time? If position doesn't change with respect to time, couldn't I just specify that? I can do all the calculations because I have my notes with some equations, but I want to understand those concepts in a more general setting.

    Also, how much calculus is needed to achieve an understanding of all of classical mechanics in a general fashion? I din't know any calculus when I first learned about velocity, acceleration, forces, etc (high school). And now I realize how much clearer your perspective of physics can be with an intuitive knowledge of calculus. But I've only been as far as gradients. Should I take a look at double integrals, curl, gradient fields and the like? Am I gonna run into those later in my undergraduate classical mechanics course? Which, by the way, is completely based on calculus.

  2. jcsd
  3. Oct 5, 2012 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    For classical mechanics the most important part of calculus are vectors and derivatives. Concerning vector analysis it's enough to know the gradient to begin with.

    Suppose you consider a problem, where a point particle moves in some force field (e.g., a planet in the gravitational field of the sun). Then the force is a function of position of the particle, and you have to solve Newton's equation of motion, which is of the form
    [tex]\frac{\mathrm{d}^2 \vec{x}}{\mathrm{d} t^2}=\vec{F}(\vec{x}).[/tex]
    I also assumed that [itex]\vec{F}[/itex] doesn't depend explicitly on time.

    Now many forces can be written as the gradient of a potential, i.e.,
    [tex]\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).[/tex]
    The gradient in Cartesian coordinates is given by the partial derivatives, i.e.,
    \partial_x V \\ \partial_y V \\ \partial_z V
    Consider [itex]\vec{x}(t)[/itex] to be a solution of Newton's equation of motion. The time derivative of the potential, when you plug in this solution reads (according to the chain rule for functions of more than one variable):
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} V[\vec{x}(t)]=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \cdot \vec{\nabla} V[\vec{x}(t)]=-\vec{v}(t) \cdot \vec{F}[\vec{x}(t)].[/tex]
    Here, I've used that [itex]\vec{v}=\mathrm{d} \vec{x}/\mathrm{d} t[/itex] is the velocity of the particle along its trajectory.

    Now consider the following derivative
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \vec{v}^2(t)=2 \vec{v} \cdot \frac{\mathrm{d} \vec{v}}{\mathrm{d} t}=2 \vec{v} \cdot \frac{\mathrm{d}^2 \vec{x}}{\mathrm{d} t^2}.[/tex]
    Now, multiplying the equation of motion with [itex]\vec{v}[/itex] and using our two derivatives calculated above we find
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right )=-\frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}).[/tex]
    Bringing both sides of the equation together gives
    [tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 + \frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x}) \right )=0[/tex]
    along the trajectory of the particle.

    This equation tells us that the quantity
    [tex]E=\frac{m}{2} \vec{v}^2 +V(\vec{x})[/tex]
    is constant along the trajectory of the particle. This quantity is called total energy, and the first term is the kinetic (because it contains the velocity) and the second term is the potential energy (because it derives from the scalar potential of the force).

    Now, the work done on the particle when moving is defined by
    [tex]W=\int_{t_1}^{t_2} \mathrm{d} t \; \vec{v}(t) \cdot \vec{F}[\vec{x}(t)].[/tex]
    Comparing the integrand with the above derivative of the potential gives
    [tex]W=-\int_{t_1}^{t_2} \mathrm{d t} \; \frac{\mathrm{d}}{\mathrm{d} t} V[\vec{x}(t)]=-\{V[\vec{x}(t_2)]-V[\vec{x}(t_1)] \}.[/tex]
    Thus, for a force which is derivable as the gradient of a scalar potential, the work done is given by the potential difference between the endpoints of the trajectory (up to a sign).

    I hope this helps to understand the interrelations between the various quantities related to energy.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook