Velocity
//Bidang//, Iban people. Sarawak 20th century, 48 x 107 cm. Nabau motif. From the Teo Family collection, Kuching. Photograph by D Dunlop.
Bidang, Iban people. Sarawak 20th century, 48 x 107 cm. Nabau motif. From the Teo Family collection, Kuching. Photograph by D Dunlop.

Consider a particle P described by a repetitive chain of historically ordered space-time events

$\Psi \left( \bar{r}, t \right) ^{\sf{P}} = \left( \sf{\Omega}_{1} , \sf{\Omega}_{2} \ldots \, \sf{\Omega}_{\it{i}} \, \ldots \, \sf{\Omega}_{\it{f}} \, \ldots \right)$

which are characterized by their position $\bar{r}$ and time of occurrence $\, t \,$. The separation between some arbitrary initial and final pair of events is given by

$\Delta \overline{r} = \overline{r}_{\it{f}} - \overline{r}_{\it{i}}$

and the elapsed time between events is

$\Delta t = t_{\it{f}} - t_{\it{i}}$

Definition: the velocity is the ordered set of three numbers

$\begin{align} \overline{\sf{v}} \equiv \frac{ \Delta \overline{r} }{ \Delta t } \end{align}$

Definition: the speed of P is the norm of the velocity

$\sf{v} \equiv \left\| \, \overline{\sf{v}} \, \right\|$

Consider measuring this quantity based on observations of length and time. Recall that length is defined only for particles that are at least as big as atoms. And we also presume that P is compared to a frame of reference that includes a calibrated rod and a calibrated clock. To determine the velocity first measure the elapsed time between events $\sf{\Omega}_{\it{i}}$ and $\sf{\Omega}_{ \it{f}}$ as

$\begin{align} \Delta t = \left( k-j \right) \hat{\tau}^{ \mathbf{\Theta}} = \left( f - i \right) \hat{\tau} ^{\sf{P}} = \frac{ f - i }{ \nu } \end{align}$

where $k-j$ is the number of clock cycles between events, $\; \hat{\tau}$ is a period and $\nu$ is the frequency of P. Then make three length measurements $\ell _{x}$ $\ell _{y}$ and $\ell _{ z}$ along the spatial axes. Combine them to obtain the separation vector $\Delta \bar{r} = \left( \ell _{x}, \, \ell _{y}, \, \ell _{z} \right)$ between events. This separation is due to a sum of displacements

$\Delta \bar{r} = \Delta \bar{r}^{ \, \sf{\Omega}}_{i} + \Delta \bar{r}^{ \, \sf{\Omega}}_{i+1} + \, \ldots \, + \Delta \bar{r}^{ \, \sf{\Omega}}_{f}$

where $\Delta \bar{r}^{ \, \sf{\Omega}}$ notes the displacement of P during each orbital cycle $\sf{\Omega}$. As discussed earlier atomic cycles are separated from each other by one wavelength $\, \lambda \,$. And if measurements are not too disruptive so that $\lambda$ is constant, then the distance between initial and final events is

$\Delta r \equiv \left\| \, \Delta \bar{r} \vphantom{\ell_{y}} \, \right\| = \left\| \, \left( \ell _{x}, \, \ell _{y}, \, \ell _{z} \right) \, \right\| = \left( f- i \right) \lambda$

Combining these observations gives the measured speed of P as

$\begin{align} \sf{v} = \left\| \, \frac{ \rm{\Delta} \mathit{\bar{r}}}{ \rm{\Delta} \mathit{t} } \, \right\| = \frac{ \rm{\Delta} \mathit{r} }{ \rm{\Delta} \mathit{t}} = \mathit{ \frac{ \left( f- i \right) \, \lambda}{ \left( f- i \right) \, \hat{\tau}} } = \nu \lambda \end{align}$

If the particle contains lots of quarks we can use Planck's postulate $E = h \nu$ to express the frequency in terms of the mechanical energy $E$ as

$\begin{align} \sf{v} = \frac{ \it{E} }{\it{h}} \lambda \end{align}$

And if the frame of reference is inertial then de Broglie's postulate $\lambda = h / p$ can be used to replace the wavelength with the momentum $p$ to obtain

$\begin{align} \sf{v} = \left( \frac{\it{E}}{\it{h}} \right) \left( \frac{\it{h}}{\mathit{p}} \right) = \frac{ \mathit{E} }{ \mathit{p} } \end{align}$

Now if the particle under consideration is a photon then its mass $m$ is zero, and its energy is $E=cp$. So the speed of a photon is

$\begin{align} \sf{v} = \frac{ \mathit{E} }{ \mathit{p} }= \frac{ \mathit{cp} }{ \mathit{p} }= \mathit{c} \end{align}$

and for this reason the constant $c$ is usually called the speed of light. For photons $\lambda \nu = c$. Then in terms of the wavenumber

$\begin{align} \kappa = \frac{1}{\lambda} = \frac{\nu}{c} = \frac{E}{hc} \end{align}$

Velocity of a Newtonian Particle

If a particle is Newtonian then it is presumably in dynamic equilibrium where its mechanical energy is $E = 2K$ and $K$ is its kinetic energy. But recall that kinetic energy is defined by $K \equiv p^{ 2} / \,2m$ so that for Newtonian particles

$\begin{align} \sf{v} = \frac{\mathit{E}}{\mathit{p}} = \frac{ \mathrm{2}\mathit{K} }{ \mathit{p} } = \left( \frac{ \mathrm{2} }{ \mathit{p} } \right) \left( \frac{ \mathit{p}^{ \mathrm{2}}}{\mathrm{2}\mathit{m}} \right) = \frac{\mathit{p} }{ \mathit{m} } \end{align}$

The term momentum is the modern English word used for translating the phrase "quantity of motion"1. So the foregoing relationship, which can be rearranged as

$p   = m \sf{v}$

was stated by Isaac NewtonXlink.png when he wrote

"Quantity of motion is a measure of motion that arises from the velocity and the quantity of matter jointly."2

This direct proportionality between speed and momentum is traditional and simple. It can be used to eliminate the momentum in some previously defined quantities such as the Lorentz factor which can now be expressed as

$\begin{align} \gamma \equiv \frac{ 1 }{ \sqrt{ \ 1 - \left( p/mc \right)^{2} \ \vphantom{{\left( p/mc \right)^{2}}^{2}} } } = \frac{ 1 }{ \sqrt{ \ 1 - \left( \sf{v}/\mathit{c} \right)^{2} \ \vphantom{{\left( p/mc \right)^{2}}^{2}} } } \end{align}$

And for Newtonian particles the kinetic energy can be written as

$\begin{align} K \equiv \frac{\, p^{ 2} }{ 2m } = \small{\frac{1}{2}} \normalsize{ m \sf{v}^{\mathrm{2}} } \end{align}$

Right.png Next step: acceleration.
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