Goldstein: derivation of work-energy theorem

Question

I am reading "Classical Mechanics-Third Edition; Herbert Goldstein, Charles P. Poole, John L. Safko" and in the first chapter I came across the work-energy theorem (paraphrased) as follows:

The work done on a particle by a force $F$ from point 1 to point 2 is given by: $$ W_{12} = \int_{1}^{2}F\cdot ds \tag{1.29} $$ Since $F = \dfrac{dp}{dt}$, we can write this as: $$ W_{12} = \int_{1}^{2}\left(\dfrac{dp}{dt}\right).ds $$ Also, since $p = mv$ and $ds = vdt$: $$ W_{12} = \int_{1}^{2}\left(\dfrac{d(mv)}{dt}\right).vdt $$ Now, we take the case of unchanging mass and we get $$ \begin{align} W_{12} &= m\int_{1}^{2}\left(\dfrac{dv}{dt}\right).vdt\\ &= \dfrac{m}{2}\int_{1}^{2}\dfrac{d}{dt}\left(v^2\right)dt\\ &= m\int_{1}^{2}\left(\dfrac{d}{dt}v.v\right)dt\\ &= m\int_{1}^{2}\left(v_2^2 -v_1^2\right) \end{align} $$ Since $\dfrac{m}{2}v^2$ is the kinetic energy (denoted by $T$): $$ W_{12} = T_2 - T_1 $$ I am not comfortable with how that factor of $\dfrac{1}{2}$ magically appears when going from $m\int_{1}^{2}\left(\dfrac{dv}{dt}\right).vdt$ to $\dfrac{m}{2}\int_{1}^{2}\dfrac{d}{dt}\left(v^2\right)dt$. Could someone help with this? I have a feeling that it has something to do with vector calculus but I don't know how.

As said in answers: $$ \begin{align} \int_{1}^{2}\left(\dfrac{mdv}{dt}.v\right)dt &= m\int_{1}^{2}\left(\dfrac{dv.v}{dt}\right)dt\\ &= m\int_{1}^{2}\left(\dfrac{v.dv}{dt}\right)dt\\ &= m\int_{1}^{2}\left(\dfrac{\dfrac{1}{2}dv^2}{dt}\right)dt\\ &= \dfrac{1}{2}m\int_{1}^{2}\left(\dfrac{dv^2}{dt}\right)dt\\ &= \dfrac{1}{2}mv^2\\ \end{align} $$

Also, the entire derivation is done under the assumption that $dm=0$. Does this mean that when $dm\neq 0$, we can not compare work done with change in kinetic energy?

Answer 1

I am not comfortable with how that factor of 1/2 magically appears... Could someone help with this?

It does not "magically" appear. It is just from the basic rules of differential calculus. In particular: $$ \frac{d}{dv} v^2 = 2v\;. $$

The $2$ on the RHS of the above equation can move to the LHS and become a 1/2 (this is just basic algebra). Also moving the $dv$ from one side to the other gives: $$ \frac{1}{2}d(v^2) = vdv $$

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The vector calculus analog is (again just the product rule): $$ d(\vec v \cdot \vec v) = 2 \vec v \cdot d\vec v $$

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Also, the entire derivation is done under the assumption that $dm=0$. Does this mean that when $dm\neq 0$, we can not compare work done with change in kinetic energy[?]

No, it does not necessarily. You can use the Hamiltonian equations of motion to prove more generally that when $H(x,p) = T(p) + U(x)$ the work-KE theorem holds: $$ \int \vec F \cdot \vec dx $$ $$ = \int \frac{d\vec p}{dt} \cdot \vec dx = \int d\vec p \cdot \vec v $$ $$ =\int d\vec p \cdot \frac{\partial H}{\partial \vec p} = \int d\vec p \cdot \frac{\partial T}{\partial \vec p} $$ $$ =\Delta T $$

Answer 2

The presentation in Goldstein is unnecessarily convoluted.

Let me first address the appearance of the factor $\tfrac{1}{2}$

Take the following integral of velocity, from starting velocity $v_0$ to end velocity $v$:

$$ \int_{v_0}^v v \ dv \tag{1} $$

As we know: if we have a function $f(x) = x$ then we have the following for the primitive $F$ of that function:

$$ F(x) = \tfrac{1}{2}x^2 \tag{2} $$

So we have for the integral of velocity, from starting velocity $v_0$ to end velocity $v$

$$ \int_{v_0}^v v \ dv = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{3} $$

That is the reason why the factor $\tfrac{1}{2}$ appears.

The derivation in Goldstein is a struggle.

The natural starting point is to define the integral of force with respect to position coordinate, from starting point $s_0$ to end point $s$

$$ \int_{s_0}^s F \ ds \tag{4} $$

In Goldstein the integration is first stated from start point '1' to end point '2', but in the next expression those integration limits are dropped. Throughout the presentation the steps are haphazard.

For the series of steps that lead from (4) the natural starting point, to (1) see derivation of the Work-Energy theorem.

Answer 3

First off, when $\mathrm dm\ne0,$ the key is that you have to figure out what the “missing” mass went off to do. That mass generally represents some sort of momentum flow out of the system! You have to quantify this and use conservation of momentum to get a good result. A good example of this is derivations of the rocket equation.

Now for the factor of $\frac12$, it has nothing to do with vector calculus whatsoever. It's universal. Just has to do with the product rule on a commutative product. So you have two options. You can either do a top down derivation:$$ \frac{\mathrm d\phantom t}{\mathrm dt} \left(v\cdot v\right) = \frac{\mathrm dv}{\mathrm dt}\cdot v + v\cdot \frac{\mathrm dv}{\mathrm dt}\\= 2 v \cdot \frac{\mathrm dv}{\mathrm dt},$$divide both sides by two and you are done. There is no difference in this derivation between the dot product of vectors and the ordinary product of real numbers or even its extension to complex numbers. As long as it obeys that first product rule and is commutative, you get this result.

Or, you can approach it bottom-up. This is maybe more fun and is called “integration by parts.” Start from this: $$ I(\tau) = \int_0^\tau \mathrm dt~v(t)\cdot \frac{\mathrm dv}{\mathrm dt}.$$ When we integrate by parts we go up and down the integral/derivative ladder. In this case we see something that we naïvely want to raise, $\frac{\mathrm dv}{\mathrm dt}\to v$, and we want to lower the remaining $v(t)$. So then we get, $$ I(\tau) = v\cdot v\Big|^\tau_0- \int_0^\tau \mathrm dt~\frac{\mathrm dv}{\mathrm dt}\cdot v(t). $$ At this point it is tempting to despair! We wanted to reduce complexity, it looks like we have only increased it. But commutativity is what saves us: for we can immediately recognize that we have written: $$I(\tau) = v\cdot v\Big|^\tau_0-I(\tau),\\ 2 I(\tau) = v\cdot v\Big|^\tau_0,\\ I(\tau) = \frac12 v\cdot v\Big|^\tau_0.$$ Again, I didn't say anything about vectors, I just said that I needed a notion of product for which integration by parts works. It works for the scalar product and for the dot product, and even for the cross product, although that is not commutative but anticommutative and so you get a tautology $0=0$ at the end. But this is a very general result, it does not depend on the specifics of product.