Done — I saved the list of URLs to memory.

Now to the math.

1) First-order space-by-time relation (Ptolemaeus / polar coordinates)

Given

differentiate with respect to time $t$. Using the product and chain rules:

So the first-order (velocity) relation in Cartesian components is the pair above. In vector form:

where the first term is radial velocity and the second is transverse (angular) velocity.

2) Two systems, velocities $v_0$ and $v_1$; constant deviation rate $C_{00}$

You wrote:

and the deviation of $v_0$ from $v_1$ grows at a constant rate $C_{00}$ per second of $t_0$.

If $x_{10}(t_0)$ denotes the velocity in system 1

A natural interpretation is that the velocity in system 1 evolves as

(plus any initial offset if present; with the convention above we assume at $t_0=0$ the difference is zero — otherwise add an initial constant $v_0(0)=v_1+\Delta_0$, giving $v_0(t_0)=v_1+\Delta_0+C_{00}t_0$.)

So the correct equation for $x_{10}(t_0)$ (interpreted as velocity) is:

If instead $x_{10}(t_0)$ denotes the particle's position in system 1

Then integrate the velocity to obtain position. With initial position $x_{10}(0)=x_{10,0}$,

which gives

(or add an initial velocity offset $\Delta_0$ inside the integrand if appropriate).

If you want I can (without saving anything extra) express these relations in your ontology notation (second-/third-order time relations), or write them down in LaTeX-ready form for a paper. Which would you prefer?

Nice — you caught that $t^2$ implies acceleration. Here’s the second-order space-by-time relation (the accelerations) worked out cleanly.

Second-order space-by-time relation (accelerations)

Start from

We already have the first derivatives

Differentiate once more (use product + chain rules). The second derivatives are

You can collect terms into radial and transverse pieces. Using the radial unit vector $\mathbf{e}_r=(\cos a,\sin a)$ and the transverse unit vector $\mathbf{e}_\theta=(-\sin a,\cos a)$, the acceleration vector is the standard polar form

So:

• radial acceleration $a_r=\ddot r - r\dot a^2$,

• transverse (angular) acceleration $a_\theta=r\ddot a + 2\dot r\dot a$.

If you want, I can:

• expand these into a single vector $(\ddot x,\ddot y)$ in LaTeX-ready form (already done above),

• rewrite them using your ontology notation (second-order / third-order distinctions), or

• apply these formulas to a concrete example (e.g. $r(t)=R_0$, $a(t)=\omega t$ or a rocket trajectory). Which next?

Good — since the deviation of the acceleration grows linearly in time, that means the rate of change of the acceleration (the third time derivative of position) is constant. I will avoid the forbidden word and call that quantity the third time derivative.

Interpretation / assumptions

• $x_{20}(t_0)$ and $x_{21}(t_1)$ denote accelerations in system 0 and system 1 respectively:

  $$x_{20}(t_0)=g_0(t_0),\qquad x_{21}(t_1)=g_1.$$

• The difference $g_0-g_1$ increases at a constant rate $C_{10}$ per second of $t_0$.

• If at $t_0=0$ there is an initial acceleration offset $\Delta_0$ so that $g_0(0)=g_1+\Delta_0$, include it; if no offset is specified take $\Delta_0=0$.

Acceleration (second-order) — result

The acceleration in system 0 as a function of $t_0$ is therefore

If you assume no initial offset ($\Delta_0=0$), this simplifies to

Consequences for velocity and position (optional)

If you also want the corresponding velocity and position in system 0 (given known reference values in system 1), integrate:

1. Velocity (first integral). With initial velocity difference $\delta v_0$ at $t_0=0$ and a reference velocity $v_1$:

2. Position (second integral). With initial position difference $\delta x_0$ at $t_0=0$ and a reference position $x_1$:

Note (ontology)

The constant $C_{10}$ is the third time derivative of position (the rate of change of acceleration). If you want these written in your preferred ontology notation (second-, third-, etc.), I can convert them immediately.

Nice — you spotted the order correctly. Below is the third-order space-by-time relation (the third time derivative of position) for the polar/Ptolemaeus setup, expressed cleanly in both component and radial/transverse form.

Start (again) from

We denote derivatives with respect to $t$ by primes or dots equivalently:

Third time derivative (vector form)

Let the radial and transverse combinations be

These are the radial and transverse pieces of the second derivative (acceleration). Then the third derivative $\mathbf{r}^{(3)}$ equals

If you expand the derivatives, you get the compact explicit radial/transverse form:

Cartesian components

Converting back to $x,y$ components ($\mathbf{e}_r=(\cos a,\sin a)$, $\mathbf{e}_\theta=(-\sin a,\cos a)$):

Interpretation / notes

• The terms with $r^{(3)}$ and $a^{(3)}$ are direct third-time-derivative contributions of radius and angle.

• Mixed terms like $3r'\,a''$, $3r''\,a'$ and $3r'\,a'^{2}$ appear because of repeated product/chain-rule differentiation (they encode coupling between radial change and angular change).

• In your ontology language: this is the third-order space-by-time relation (position differentiated three times). The scalar multiplying $\mathbf{e}_r$ is the radial part of the third derivative; the scalar multiplying $\mathbf{e}_\theta$ is the transverse part.

Would you like this as a LaTeX block ready for a paper, or applied to a concrete example (e.g. $r(t)=R_0$, $a(t)=\omega t$; or a trajectory with known $r,a$ derivatives)?