# Integrals of Power

This question has two parts: a question about a specific expression and a question about a family of expressions. In what follows I will omit all constants of integration.

If we integrate force with respect to time we get momentum: $\int F \, \operatorname{d}\!t = mv$. If an object accelerates along a line then we can find its acceleration at any given point and write force as a function of distance. Doing so, and integrating gives kinetic energy (or work done): $\int F \, \operatorname{d}\!x = \frac{1}{2}mv^2$.

I decided to try the same thing with power using the definition $P = Fv$. When I integrated with respect to time I got, as expected, $\frac{1}{2}mv^2$ which is work done. If an object accelerates along a line then we can measure the power at regular distances and write power as a function of distance. Then we may integrate power with respect to distance: $$\int P \, \operatorname{d}\!x = \int mav \, \operatorname{d}\!x = m\int\frac{\operatorname{d}\!v}{\operatorname{d}\!t}v\, \operatorname{d}\!x = m\int\frac{\operatorname{d}\!x}{\operatorname{d}\!t}\frac{\operatorname{d}\!v}{\operatorname{d}\!x}v\,\operatorname{d}\!x = m\int v^2\operatorname{d}\!v = \tfrac{1}{3}mv^3$$

Here is the first question: What is $\frac{1}{3}mv^3$? Does it have a name? How can I understand it conceptually?

Here is the second question: Momentum is $mv$, kinetic energy is $\frac{1}{2}mv^2$, this new quantity $\frac{1}{3}mv^3$ has turned up. It seems that $\frac{1}{k}mv^k$ for positive whole numbers $k$ plays a role. These can all be combined into one power series. Provided $v$ is very small, we have

$$mv + \tfrac{1}{2}mv^2 + \tfrac{1}{3}mv^3 + \cdots + \tfrac{1}{k}mv^k + \cdots \equiv m\ln\left(\frac{1}{1-v}\right)$$

Do the expressions $\frac{1}{k}mv^k$ have any general meaning; and does this logarithmic expression have any use? (I realise that the units in the sum are all different.)

• @BMS, He mentioned that he knows they're all in different units, but he's wondering if that expression demonstrates any physical relationship as far as motion is concerned in classical mechanics - in which case I am not sure. Having a number of various units, I can't say that I think so. However he does pose an interesting question. – Signus Sep 19 '13 at 17:21
• @Signus Indeed he did; missed that. As far is attempting to interpret $\frac{1}{3}mv^3$, you may want to try interpreting the quantity $P\,dx$. Power is strongly linked with a time derivative, so this may be a bit difficult, though interesting! – BMS Sep 19 '13 at 17:47
• @BMS, that it is. When I think of power I begin to think of impulse because it is the average force over time $J = \int_{{t}_{1}}^{{t}_{2}} F dt$. – Signus Sep 19 '13 at 18:09
• @Signus Your edit was incorrect: the integral of power w.r.t. time is work done, as I originally said. Power is measured in J/s. Power x Time = Work. – Fly by Night Sep 19 '13 at 19:20
• @FlybyNight, Indeed, I was trying to explore the possible relation with impulse and power to evaluate $\frac{1}{3}mv^{3}$ considering that $J = \frac{N}{s}$ and $W = \frac{N*m}{s}$ – Signus Sep 19 '13 at 19:30

Comments to the question (v2): OP's observation can be summarized as follows.

1. In $D$ dimensions: Let $g=g(v^2)$ be a function of the square $v^2$ of the speed $v=|{\bf v}|$. Let $G=G(v^2)$ be a primitive integral (aka. antiderivative or indefinite integral) of $g$, i.e. $G^{\prime}(v^2) = g(v^2)$. Then in analogy with the work-energy theorem for a non-relativistic particle with mass $m$, one has $$\int_{{\bf r}_i}^{{\bf r}_f} \! g(v^2)~ {\bf F}\cdot {\rm d}{\bf r} ~=~\int_{t_i}^{t_f} \! g(v^2)~ m\frac{{\rm d}{\bf v}}{{\rm d}t}\cdot{\bf v}~ {\rm d}t ~=~\frac{m}{2}\int_{t_i}^{t_f} \! g(v^2)~ \frac{{\rm d}(v^2)}{{\rm d}t}{\rm d}t$$ $$~=~\frac{m}{2}\int_{v_i^2}^{v_f^2} \! g(v^2) ~ {\rm d}(v^2) ~=~\frac{m}{2} G(v_f^2) -\frac{m}{2} G(v_i^2).$$ The standard work-energy theorem corresponds to $g(v^2)=1$ and $G(v^2)=v^2$. OP's first non-standard example corresponds to $g(v^2)=v$ and $G(v^2)=\frac{2}{3}v^3$.

2. In $D=1$ dimension: We use boldface to denote signed 1D quantities, i.e. vectors in 1D. Let $h=h({\bf v})$ be a function of the 1D velocity ${\bf v}$. Let $H=H({\bf v})$ be a primitive integral of ${\bf v}h({\bf v})$, i.e. $H^{\prime}({\bf v}) = {\bf v}h({\bf v})$. Then the generalized work-energy theorem read $$\int_{{\bf x}_i}^{{\bf x}_f} \! h({\bf v})~{\bf F}~{\rm d}{\bf x} ~=~\int_{t_i}^{t_f} \! h({\bf v})~ m\frac{{\rm d}{\bf v}}{{\rm d}t} {\bf v} ~{\rm d}t$$ $$~=~m\int_{{\bf v}_i}^{{\bf v}_f} \! H^{\prime}({\bf v})~{\rm d}{\bf v} ~=~m H({\bf v}_i)-m H({\bf v}_f).$$ The standard work-energy theorem corresponds to $h({\bf v})=1$ and $H({\bf v})=\frac{1}{2}{\bf v}^2$. OP's first non-standard example corresponds to $h({\bf v})={\bf v}$ and $H({\bf v})=\frac{1}{3}{\bf v}^3$.

I don't know if the above generalizations of the work-energy theorem have a name, though. Usually these generalizations are not needed in order to solve a physical problem. I speculate that similar observations might be useful in integrable systems to keep track of an infinite hierarchy of higher conservation laws.

• Why have a function of $v^2$ and not just $v$? – Fly by Night Sep 19 '13 at 18:00
• @Fly by Night: A function of $v^2$ can easily be converted to a function of speed $v\geq 0$, and vice-versa, so there is no loss of generality. Moreover, it makes the formula simpler. Plus it eliminates any confusion between the use of speed $v$ versus velocity ${\bf v}$. – Qmechanic Sep 19 '13 at 18:10
• The fact that the expression involves $v^3$ means that it is sign dependent, and hence direction dependent. Just as momentum ($mv$) is direction dependent and is a vector in higher dimensions and kinetic energy ($\frac{1}{2}mv^2$) is not direction dependent and is a scalar in higher dimensions, it seems that $\frac{1}{3}mv^3$ might be more than a scalar, perhaps a vector and perhaps a tensor. – Fly by Night Sep 20 '13 at 18:07
• @Fly by Night: I agree that in 1D there exists a signed version, cf. new update. I leave it to you and others to make sense of such a construction in higher dimensions. – Qmechanic Sep 20 '13 at 19:42
• @Fly by Night: Please note that the answer only talks about ${\bf v}^3$ in 1D. So the velocity ${\bf v}\in \mathbb{R}$, and the cube ${\bf v}^3\in \mathbb{R}$ makes sense. – Qmechanic Sep 20 '13 at 21:39