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Okay y'all. I cannot be the the first person to notice that $$ \frac{dE}{dv} = p$$ (or, alternatively, that $\int pdv = E$). I futzed around with the equations a little bit to see if replacing velocity with it's definition would yield anything fruitful and I came up with this thing that may or may not be intelligible:

$$ \int pd (\frac{dx}{dt} ) = \frac{1}{2}mv^2 = E$$

So ultimately, I am confused. Why on earth would it be the case the momentum is the derivative of kinetic energy with respect to velocity? Or, maybe the better question is, is there a physical meaning to the fact that the derivative of kinetic energy with respect to velocity is momentum (or that the integral of momentum over velocity is kinetic energy)?

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    $\begingroup$ Hint: Kinetic energy is work done against the force $ma\,.$ Please futz around a bit more. $\endgroup$
    – Kurt G.
    Commented Jun 15, 2023 at 14:04
  • $\begingroup$ @Kurt Are you thinking of potential energy, i.e. $\int_C \frac{\partial p}{\partial t} dx$? $\endgroup$ Commented Jun 15, 2023 at 16:05
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    $\begingroup$ No. Kinetic energy. When we accelerate a body of mass $m$ we are working against its inertial force $ma\,.$ When this happens over the distance $dx$ we do the work $$dE=m\,a\,dx=m\frac{dv}{dt}dx=mv\,dv\,.$$ Now integrate and you get $E=mv^2/2\,.$ I am sure this was mentioned here on PSE before. $\endgroup$
    – Kurt G.
    Commented Jun 15, 2023 at 16:10
  • $\begingroup$ @JAlex Please don't put your integration variable in your integration bounds! (or leave comments consisting only of unexplained formulae...) $\endgroup$ Commented Jun 16, 2023 at 12:31
  • $\begingroup$ @preferred_anon point taken. tx. $\endgroup$
    – JAlex
    Commented Jun 16, 2023 at 14:58

3 Answers 3

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I cannot be the the first person to notice that $ \frac{dE}{dv} = p$ (or, alternatively, that $\int pdv = E$).

The Hamiltonian equations of motion are: $$ \frac{\partial H}{\partial \vec p} = \frac{d\vec x}{dt}=\vec v \tag{1} $$ and $$ \frac{\partial H}{\partial \vec q} = -\frac{d\vec p}{dt} \tag{2}\;, $$ where $H$ is the energy, which usually corresponds to the total mechanical energy, which is the kinetic energy plus the potential energy.

You seem to be using the symbol $E$ for kinetic energy. I will use the symbol $T$ for kinetic energy, since $E$ is often reserved for total energy.

Why on earth would it be the case the momentum is the derivative of kinetic energy with respect to velocity?

It is not always, but often. What is true is: The velocity is the derivative of the energy with respect to the momentum: $$ \vec v = \frac{\partial H}{\partial \vec p}\;. $$

In the case where the momentum $\vec p$ is equal to the mass $m$ times the velocity $\vec v$ (which is not always the case, but is often the case): $$ v_i = \frac{\partial H}{\partial p_i} = \sum_j\frac{\partial H}{\partial \vec v_j}\frac{\partial v_j}{\partial p_i} = \frac{\partial H}{\partial \vec v_i}\frac{1}{m}\;. $$ Or, moving the $m$ to the other side, in this case where $\vec p = m\vec v$, we see: $$ \vec p = \frac{\partial H}{\partial \vec v} $$

Now, assuming the potential energy $U$ doesn't depend on the velocity: $$ H(\vec x, \vec p) = T(\vec p) + U(\vec x)\;. $$

So, we can further write: $$ \vec p = \frac{\partial H}{\partial \vec v}=\frac{\partial T}{\partial \vec v} $$

Or, maybe the better question is, is there a physical meaning to the fact that the derivative of kinetic energy with respect to velocity is momentum (or that the integral of momentum over velocity is kinetic energy)?

As mentioned above, this is not an unqualified fact. It is sometimes/often true.

You may be aware of the work-kinetic-energy theorem. This says that the work done by the net force is equal to the change in kinetic energy: $$ \delta T = \int d\vec x \cdot \vec F = \int d\vec x \cdot \frac{d\vec p}{dt} = \int \vec v\cdot d\vec p\;. $$

If we can write $\vec p = m\vec v$ then we again arrive at your conclusion: $$ \delta T = \int \vec p \cdot d\vec v $$

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  • $\begingroup$ Hot. That makes complete sense. Upvote for you. Also, how do you get the (1) to the right of your equations? $\endgroup$ Commented Jun 15, 2023 at 19:30
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    $\begingroup$ @AdamGluntz, if you edit the answer, you can see the \tag{1} or \tag{2} inside the math block. $\endgroup$
    – cjm
    Commented Jun 15, 2023 at 22:24
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    $\begingroup$ Or even \tag1 \tag2 etc. (for equations labelled with a single character) if you can't be bothered with the parentheses. $\endgroup$
    – joseph h
    Commented Jun 17, 2023 at 3:04
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To go from $p$ to $p+dp$ requires an impulse of

$$dp = F dt$$

In that time, the particle travels:

$$ dx = v dt $$

which requires work:

$$ dE = F dx = F v dt = F v \frac{dp}F = v dp$$

So:

$$ \frac{dE}{dp} = v $$

or

$$ E = \int dE = \int v dp = \int \frac p m dp = \frac{p^2}{2m} $$

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    $\begingroup$ I think the @Adam wanted more of a physical significance part rather than the mathematics of it $\endgroup$ Commented Jun 16, 2023 at 5:47
  • $\begingroup$ @HarshdeepChhabra going from $F dx$ to $v dp$ is the physical significance. $\endgroup$
    – JEB
    Commented Jun 16, 2023 at 15:17
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Now here the energy (Kinetic) is directly dependent on the velocity. So if you see momentum as a analog of velocity, since it's just a constant (classically) times the velocity. This is saying that as your energy is increasing with respect to velocity it is directly proportional to how fast you're going. Similarly if you take the harmonic oscillator equation $$E = \frac{1}{2} kx^2$$ If you take $dE/dx=kx$ is a quantity similar to momentum which is analog to position and the energy increases as per the position

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