# Relative velocity and lenticity

Consider a particle P in uniform motion. Suppose two inertial observers observe its motion. Observer K is stationary relative to the ground, and observer L is in uniform motion in the same direction as P but at a different rate.

(A) Say the spatial position of P relative to K is x(P, K), the spatial position of P relative to L is x(P, L), and the spatial position of L relative to K is x(L, K). The relation between these observations is

$\mathbf{x}(P,K)=\mathbf{x}(P,L)&space;+&space;\mathbf{x}(L,K)$

Differentiate these spatial positions with respect to time t to get time velocities (since time is the independent variable)

$\frac{d\mathbf{x}(P,K)}{dt}=\frac{d\mathbf{x}(P,L)}{dt}+\frac{d\mathbf{x}(L,K)}{dt}$

Differentiate these velocities v with respect to time to get accelerations

$\frac{d\mathbf{v}(P,K)}{dt}=\frac{d\mathbf{v}(P,L)}{dt}+\frac{d\mathbf{v}(L,K)}{dt}=\frac{d\mathbf{v}(P,L)}{dt}$

Since the acceleration of L relative to K is zero, the acceleration of the particle P measured by both inertial observers is equal.

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(B) Now say the temporal position of P relative to K is t(P, K), the temporal position of P relative to L is t(P, L), and the temporal position of L relative to K is t(L, K). The relation between these observations is

$\mathbf{t}(P,K)=\mathbf{t}(P,L)&space;+&space;\mathbf{t}(L,K)$

Differentiate these temporal positions with respect to distance s to get space lenticities (since space is the independent variable)

$\frac{d\mathbf{t}(P,K)}{ds}=\frac{d\mathbf{t}(P,L)}{ds}+\frac{d\mathbf{t}(L,K)}{ds}$

Differentiate these lenticities w with respect to distance to get relentations

$\frac{d\mathbf{w}(P,K)}{ds}=\frac{d\mathbf{w}(P,L)}{ds}+\frac{d\mathbf{w}(L,K)}{ds}=\frac{d\mathbf{w}(P,L)}{ds}$

Since the relentation of L relative to K is zero, the relentation of the particle P measured by both inertial observers is equal. Note the symmetry with (A).

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(C) Say the spatial position of P relative to K is x(P, K), the position of P relative to L is s(P, L), and the position of L relative to K is x(L, K). The relation between these observations is

$\mathbf{x}(P,K)=\mathbf{x}(P,L)&space;+&space;\mathbf{x}(L,K)$

Differentiate time t with respect to these spatial positions to get time lenticities (since time is the independent variable)

$\frac{dt}{d\mathbf{x}(P,K)}=\frac{dt}{d\mathbf{x}(P,L)}&space;\boxplus&space;\frac{dt}{d\mathbf{x}(L,K)}$

with the boxplus representing harmonic addition. Why not arithmetic addition? Because, apart from division by zero,

$\frac{dt}{d\mathbf{x}(P,L)}&space;=&space;\left&space;(\frac{d\mathbf{x}(P,L)}{dt}&space;\right&space;)^{-1}&space;\;&space;\mathrm{and}\;&space;\:&space;\frac{dt}{d\mathbf{x}(L,K)}&space;=&space;\left&space;(\frac{d\mathbf{x}(L,K)}{dt}&space;\right&space;)^{-1}$

Differentiate time with respect to these lenticities w to get relentation

$\frac{dt}{d\mathbf{w}(P,K)}=\frac{dt}{d\mathbf{w}(P,L)}&space;\boxplus&space;\frac{dt}{d\mathbf{w}(L,K)}=\frac{dt}{d\mathbf{w}(P,L)}$

Since the relentation of L relative to K is infinite, the relentation of the particle P measured by both inertial observers is equal.

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(D) Say the temporal position of P relative to K is t(P, K), the temporal position of P relative to L is t(P, L), and the temporal position of L relative to K is t(L, K). The relation between these observations is

$\mathbf{t}(P,K)=\mathbf{t}(P,L)&space;+&space;\mathbf{t}(L,K)$

Differentiate distance s with respect to these temporal positions to get space velocities (since space is the independent variable)

$\frac{ds}{d\mathbf{t}(P,K)}=\frac{ds}{d\mathbf{t}(P,L)}&space;\boxplus&space;\frac{ds}{d\mathbf{t}(L,K)}$

with the boxplus representing harmonic addition. Why not arithmetic addition? Because, apart from division by zero,

$\frac{ds}{d\mathbf{t}(P,L)}&space;=&space;\left&space;(\frac{d\mathbf{t}(P,L)}{ds}&space;\right&space;)^{-1}&space;\;&space;\mathrm{and}\;&space;\:&space;\frac{ds}{d\mathbf{t}(L,K)}&space;=&space;\left&space;(\frac{d\mathbf{t}(L,K)}{ds}&space;\right&space;)^{-1}$

Differentiate distance with respect to these velocities v to get accelerations

$\frac{ds}{d\mathbf{v}(P,K)}=\frac{ds}{d\mathbf{v}(P,L)}&space;\boxplus&space;\frac{ds}{d\mathbf{v}(L,K)}=\frac{ds}{d\mathbf{v}(P,L)}$

Since the acceleration of L relative to K is infinite, the acceleration of the particle P measured by both inertial observers is equal. Note the symmetry with (C).