What conditions are required for the derivative of kinetic energy to be F.v?

Question

In Ch. 1 Derivation 1 of Goldstein's mechanics, we have

Show that for a single particle with constant mass the equation of motion implies $$ \frac{dT}{dt} = \vec{F}\cdot\vec{v} $$

The first step seems straightforward $$ \frac{dT}{dt} = mv\dot{v} $$

But $$ \vec{F}\cdot\vec{v} = mv\dot{v}\cos\theta $$ where $\theta$ is the angle between force and velocity, so the relation that I'm trying to prove only holds if $\vec{F}\,||\,\vec{v}$. I come up against this same parallel requirement in the second part of the derivation trying to show $\frac{d(mT)}{dt} = \vec{F} \cdot \vec{p}$.

In my mind, counterexamples come to mind in the form of circular motion, where $\cos\theta = 0$ or a force against the velocity where $\cos\theta = -1$. Is there something that I'm missing in the problem or derivation here that causes the time derivative of kinetic energy to always be $\vec{F} \cdot \vec{v}$?

Just as a note, there do seem to be quite a few questions about this derivation around that I've found, but none of them seem to address the issue that I'm having.

I just thought of instead doing the derivation using $$ \frac{dT}{dt} = \frac{d}{dt} \frac{m}{2} \vec{v} \cdot \vec{v} = m \dot{\vec{v}} \cdot \vec{v} = \vec{F} \cdot \vec{v} $$ that the result seems to just pop out. I'm not sure of where any meaning about the angle comes out of using $v^2 = \vec{v} \cdot \vec{v}$. It does mean that the direction of $\vec{v}$ doesn't matter, but it isn't clear to me how that affects the dot product $\vec{F} \cdot \vec{v}$.

Out of curiosity, I also came up with the following that I think should also be valid starting with the definition of work: $$ dW = \vec{F} \cdot d\vec{l} = \vec{F} \cdot \vec{v} dt \implies \frac{dW}{dt} = \vec{F} \cdot \vec{v} $$ and by work-energy, $\frac{dW}{dt} = \frac{dT}{dt}$ (throwing away the delta since we're looking at a time derivative), so $\frac{dT}{dt} = \vec{F} \cdot \vec{v}$.

In addition to the accepted answer, I'll also point out that the correct calculation of $\frac{dv}{dt}$ seems to be $\frac{\vec{v} \cdot \dot{\vec{v}}}{v}$, which changes $\frac{dT}{dt}$ to the expected result.

Answer 1

The first step seems straightforward $$ \frac{dT}{dt} = mv\dot{v} $$

To understand your error one needs to look at how speed and velocity are related, $\vec v = v \,\hat v$, where $\vec v$ is the velocity, $\hat v$ is the unit vector in the direction of the velocity and $v$ is the speed.

You started out with the speed squared $v^2$ and differentiated with respect to time to get $(2)\,v\,\dfrac {dv}{dt}$.

In the other part of the derivation you had a term $\dfrac {d\vec v}{dt}\cdot \vec v$

If $\vec v = v \,\hat v$ then $\dfrac {d\vec v}{dt} = \dfrac{dv}{dt} \hat v + v \dfrac{d\hat v}{dt}$

This means that $\dfrac {d\vec v}{dt} \ne \dfrac{dv}{dt} \hat v \Rightarrow \left |\dfrac {d\vec v}{dt}\right |\ne \dfrac{dv}{dt} $ unless $\dfrac{d\hat v}{dt} =0$ which is only true if the direction of the velocity does not change.

Your anomaly is due to assuming that $\left |\dfrac {d\vec v}{dt}\right |$ and $\dfrac{dv}{dt} $ are equal.

The kinetic energy should be written as $\frac 1 2 m (\vec v \cdot \vec v)$ and differentiate with respect to time and then you can compare like $\dfrac {d\vec v}{dt}$ on the left hand side with like $\dfrac {d\vec v}{dt}$ on the right hand side.

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Put another way.

The magnitude of the change in the velocity $|(\vec v + \Delta \vec v) - \vec v| = |\Delta \vec v|$

is not equal to

the change in the speed $|\vec v + \Delta v| - |\vec v|$

unless the direction of the velocity does not change.

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Update in response to a comment from @danielunderwood.

Look at where kinetic energy $T = \frac 12 m v^2$ comes from.

A force $\vec F(\vec r)$ is applied to a mass $m$ and using Newton's second law $\vec F = \dfrac {d\vec p}{dt} = m \dfrac{d \vec v}{dt}$.

The work done when this force is displaced by $d\vec r$ is $\vec F \cdot d\vec r = m \dfrac{d \vec v}{dt} \cdot d\vec r = m \vec v \cdot d\vec v$ and this is the change in the kinetic energy $dT$.

$dT = m \vec v \cdot d\vec v = \dfrac 1 2 m \,{d(\vec v \cdot \vec v)} =\dfrac 1 2 m \,{d(v^2)}$

Integration with the mass starting from rest gives an expression of the kinetic energy of a mass $m$ moving with a speed $v$.

You have gone in the reverse direction in one of your derivations and ignored the fact that the velocity is a vector and $v^2$ is $\vec v \cdot \vec v$.

Answer 2

The dot product between two vectors is $\vec a \cdot \vec b = |\vec a|~|\vec b|~\cos\theta_{ab},$ where $\theta_{ab}$ is the angle between $\vec a$ and $\vec b.$

Your derivation involving $\frac12 m \vec v\cdot\vec v$ is indeed correct and your derivation involving $\frac12 mv^2$ is indeed limited to being only one-dimensional. The problem is that in three dimensions, $v = |\vec v|$ but your "derivation" implicitly assumes that $\frac{d}{dt}|\vec v| = \left|\frac{d\vec v}{dt}\right|$, which is not true in general.

However dot products and cross products both obey a form of the product rule; we might say in more advanced mathematics that the dot product is embodied by a "metric tensor" $g_{\bullet\bullet}$ such that the inner product of two vectors $u^\bullet$ and $v^\bullet$ is $g_{ab}~u^a~v^b$ and the derivative is given by $\dot g_{ab}~u^a~v^b + g_{ab}~\dot u^a~v^b + g_{ab}~u^a~\dot v^b$ and all we need is that this tensor remains constant over time, $\dot g_{ab}=0,$ to understand your result. Similarly a cross product in 3D comes from a [0, 3]-valence orientation tensor $\epsilon_{abc}$ and if that's constant with respect to time then you get a normal product rule for the cross product.

And then you can see what you need to not get these, like time-varying coordinates.