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We all know acceleration is the time-derivative of velocity which in turn is the time-derivative of position. Vice versa: position is the integration of velocity and velocity itself is the integration of acceleration. Just out of fun I did the following derivation which ends up equating velocity to position! And I don't know why.

$$ \begin{array}{r c l} \begin{array}{r c l} \mathbf{a}(\mathbf{r},t) = \displaystyle\frac{\mathrm{d}}{\mathrm{d}t} \left[\mathbf{v}(\mathbf{r},t)\right] \\ \mathbf{a}(\mathbf{r},t) = \mathbf{a}(\mathbf{r}) = \mathbf{a}(t) \end{array} &\implies& \begin{array}{r c l} a(r_x,t) &=& \displaystyle\frac{\mathrm{d}}{\mathrm{d}t} \left[v(r_x,t)\right] &(1)\\ a(r_x) &=& \displaystyle\frac{\mathrm{d}}{\mathrm{d}t} \left[v(r_x,t)\right] &(2)\\ a(r_x) &=& \displaystyle\frac{\mathrm{d}v(r_x,t)}{\mathrm{d}r_x}\frac{\mathrm{d}r_x(t)}{\mathrm{d}t} &(3)\\ a(r_x) &=& \displaystyle\frac{\mathrm{d}v(r_x)}{\mathrm{d}r_x}\, v(t) &(4)\\ a(r_x)\,\mathrm{d}r_x &=& v(t)\,\mathrm{d}v(r_x) =v(t)\,\mathrm{d}v(r_x,t) =v(t)\,\mathrm{d}v(t) &(5)\\ a(r_x)\,\mathrm{d}r_x &=& v(t)\,\mathrm{d}v(t) &(6)\\ \displaystyle\int a(r_x)\,\mathrm{d}r_x &=& \displaystyle \int v(t)\,\mathrm{d}v(t) &(7)\\ v(r_x) &=& r_x(t) &(8)\\ \end{array} \end{array} $$

I suspect the error occurs around line (6) to (8). If integrating velocity w.r.t the change in velocity is not the position, then what else could that be?

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    $\begingroup$ $\int v\, dv = v^2/2 + C$ $\endgroup$
    – d_b
    Commented Jan 9, 2023 at 5:01
  • $\begingroup$ Also issue: a dt = v. But a dr =/= v $\endgroup$
    – planetmaker
    Commented Jan 9, 2023 at 5:01

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It is not clear what you mean by $a(r_x, t)$. If you are describing the acceleration of an object you would have either $a(r_x)$ or $a(t)$. Saying that the acceleration is a function if both implies you have something like a fluid.

If $v$ is truly a function of two variables, then step 2 to 3 is an incorrect application of the chain rule. What it should actually say is: $\frac{dv}{dt}=\frac{\partial v}{\partial t} + \frac{\partial v}{\partial r}\frac{dr}{dt}$

The left hand side of step 7 to 8 is incorrect. The integral of acceleration is only the velocity for the integral with respect to $t:$ $$\int a(t) dt =\Delta v \neq \int a(r) dr$$

The right hand side is also in error. $$\int v dv =\Delta (\frac{v^2}{2})$$

Overall, there seems to be a big confusion about what it means for a function to be a function with respect to two arguments.

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  • $\begingroup$ I was thinking since position is a function of time and that time is one directional, at any specific time the object will have a corresponding position and along with it are the derivatives. If I have all the data in hand, I can plot a 3D line chart with acceleration, position and time each represented by an axis. I can then point to the same acceleration either by position or by time hence making acceleration a function of either position or time.... or both. $\endgroup$
    – KMC
    Commented Jan 9, 2023 at 6:56
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    $\begingroup$ @KMC Yes, so what you really mean is $a=a(r)=a(r(t))$ or $a=a(t)=a(t(r)).$ But writing $a=a(r,t)$ implies you have have two independent parameters you can vary. Furthermore, you have to be careful with $a(r),$ because likely the acceleration is not defined for all positions, only positions on some one-dimensional path $r \in R \subset \mathbb{R}.$ $\endgroup$
    – Jbag1212
    Commented Jan 9, 2023 at 7:58

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