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Velocity

Consider a particle P described by a repetitive chain  \Psi of historically ordered space-time events  \mathsf{\Omega}. Let these events be characterized by their position  \bar{r} and time of occurrence  t. Then we write

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

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 these events is \Delta t = t_{\it{f}} - t_{\it{i}}. Then a velocity vector is defined by the ordered set of three numbers

\overline{\mathsf{v}} \equiv \dfrac{\Delta \overline{r}}{\Delta t}

The speed of P is defined by the norm of the velocity

\mathsf{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  \mathbf{\Theta}. To determine the velocity first measure the elapsed time between events \mathsf{\Omega}_{\it{i}} and \mathsf{\Omega}_{ \it{f}} as

\Delta t = \left(k-j \right) \widehat{\tau}^{\, \mathbf{\Theta}} = \left(f - i \right) \widehat{\tau}^{\, \mathsf{P}} = \dfrac{f - i}{\nu}

where  \left( k-j \right) is the number of clock cycles between initial and final events,  \hat{\tau} is a period and  \nu is the frequency of P. Then make three length measurements along the spatial axes that are noted by  \ell_{x},  \ell_{y} and  \ell_{ z}. Combine these measurements to obtain the observed separation vector  \Delta \bar{r}  =  \left( \ell_{x},  \,  \ell_{y},  \, \ell_{z}  \right) between events. This separation is due to a sum of displacements that may be written as

\Delta \bar{r} = \Delta \bar{r} \! \left(\mathsf{\Omega}_{i} \right) +  \Delta \bar{r} \! \left( \mathsf{\Omega}_{i+1} \right) + \; \ldots \; +  \Delta \bar{r} \! \left( \mathsf{\Omega}_{f} \right)

where \Delta \bar{r} \! \left(\mathsf{\Omega} \right) notes the displacement of P during one complete orbital cycle  \mathsf{\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} \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

\mathsf{v} = \left\| \dfrac{\Delta \bar{r}}{\Delta t} \right\| = \dfrac{\Delta r}{\Delta t} = \dfrac{\left(f-i \right) \lambda}{\left(f-i \right) \widehat{\tau}} = \dfrac{\lambda}{\widehat{\tau}} = \nu \lambda

If P contains many quarks we can use Planck’s postulate to substitute the mechanical energy  E for the frequency to obtain

\mathsf{v} = \nu \lambda = \left( \dfrac{E}{h} \rule{0px}{14px} \right) \lambda

And if the frame of reference is inertial then de Broglie’s postulate can be used to replace the wavelength with the momentum  p to obtain

\mathsf{v} = \left( \dfrac{E}{h} \rule{0px}{14px} \right) \! \left( \dfrac{h}{p} \rule{0px}{14px} \right) =\dfrac{E}{p}

The Speed of a Photon

If the particle under consideration is a photon  \boldsymbol{\gamma} then its mass  m is zero, and its energy is given by  E=cp. So the speed of a photon is

\mathsf{v} \! \left( \boldsymbol{\gamma} \right) = \dfrac{E}{p}= \dfrac{cp}{p} = c

For photons  \lambda \nu = c. And for this reason the constant  c is usually called the speed of light. Thus the wavenumber  \kappa and other photon characteristics are all related as

\kappa = \dfrac{2 \pi}{\lambda} = \dfrac{2\pi}{h} p = \dfrac{2\pi}{hc} E

The Speed of a Newtonian Particle

If P is Newtonian then m \ne 0 and the particle is presumably in dynamic equilibrium with its environment. For this case, the mechanical energy is given by E = 2K where  K is the kinetic energy. So for Newtonian particles

\mathsf{v} = \dfrac{E}{p} = \dfrac{2K}{p}

But recall that kinetic energy is defined by K \equiv p^{ 2} / \, 2m. So for Newtonian particles

\mathsf{v} =  \dfrac{2K}{p} = \dfrac{2}{p} \left( \dfrac{p^{2}}{2m} \right) = \dfrac{p}{m}

This equation can be rearranged to obtain the utterly conventional relationship

p = m \mathsf{v}

The term momentum is the modern English word used for translating the phrase; quantity of motion.1Isaac Newton, Mathematical Principles of Natural Philosophy, page 639. Translated by Andrew Motte and Florian Cajori. University of California Press, 1934. So the foregoing relationship was stated by Sir Isaac Newton when he wrote2Isaac Newton, Mathematical Principles of Natural Philosophy, page 404. Translated by I. Bernard Cohen and Anne Whitman. University of California Press, 1999.

An icon indicating a quotation.Quantity of motion is a measure of motion that arises from the velocity and the quantity of matter jointly.

β€” Sir Isaac Newton

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  \gamma which can now be expressed as

\gamma \equiv \dfrac{1}{\sqrt{\; 1 - \left(p/mc \right)^{2} \; }} = \dfrac{1}{\sqrt{\; 1 - \left( \mathsf{v}/c \right)^{2} \; }}

And for Newtonian particles the kinetic energy can be written as

K \equiv \dfrac{\; p^{2}}{2m} = \frac{1}{2} m \mathsf{v}^{2}

Bidang, Iban people. Sarawak 20th century, 48 x 107 cm. Nabau motif. From the Teo Family collection, Kuching. Photograph by D Dunlop.
References
1Isaac Newton, Mathematical Principles of Natural Philosophy, page 639. Translated by Andrew Motte and Florian Cajori. University of California Press, 1934.
2Isaac Newton, Mathematical Principles of Natural Philosophy, page 404. Translated by I. Bernard Cohen and Anne Whitman. University of California Press, 1999.