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

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.

• How much does the v perp component change T? – mmesser314 Jul 31 '17 at 3:24
• @mmesser314 it should change it in the same way that the parallel component would due to the square, correct? – danielunderwood Jul 31 '17 at 3:51
• If the parallel component is in the direction of F, the force increases the speed, and hence increases T. If they are opposed, the force decreases the speed and T. The perpendicular component does neither. It changes only the direction of v. Think of an electron in a uniform B field, where F = qv x B. You will need to show this. – mmesser314 Jul 31 '17 at 6:25
• @mmesser314 Could that also be argued by forces perpendicular to velocity doing no work and the work-energy theorem? – danielunderwood Jul 31 '17 at 16:11

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.

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.

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 $\vec v \cdot \vec v$.

• So from what you're saying, it looks like I messed up saying $\frac{d}{dt}v^2 = 2v\dot{v}$? I understand that the vector time derivative needs a time derivative of its basis vectors, but I don't seem to understand why this would be needed to handle the time derivative of $v$ since it's a scalar function. Does the magnitude of a vector not work like a scalar function? I feel like I'm missing something rather obvious, but I'm not sure what it is. – danielunderwood Aug 2 '17 at 2:45
• @danielunderwood I have updated my answer to try and explain better your error. – Farcher Aug 2 '17 at 7:28

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.