Is the Pythagorean Theorem true for infinitesimal leg?

Question

In Newtonian Physics (Classical Mechanics) a body rotates around another (massive) body with constant circular speed ω (stationary trajectory) eternally. The linear velocity v is perpendicular to the radius and the force F is along radius and hence perpendicular to v. F induces acceleration 'a' (F=ma) which in time dt induces dv along the force. E.g. v and dv are strictly perpendicular. This two are added like vectors and the sum is again v, but not in the same direction as v but a bit tilted toward the second (the massive) body. This seems to be very much in conflict with the Pythagorean Theorem (PT). Physicists insist that this is due to the infinitesimal character of dv, but this nevertheless does not violate PT. E.g. adding a perpendicular dv to v leads to the same magnitude of v (not $\sqrt{v^{2}+\left( dv\right) ^{2}}$) but this is not violation of PT. I am in total misunderstanding. In fact they insist that this happens because while dv is infiniesimal than $\left( dv\right) ^{2}=0$. I don’t believe this is also true. So the question is: Does PT hold for infinitesimal leg or not and how is it possible v+dv to equal v and simultaneously PT to be true?

Answer 1

$dv$ is not even a number in math, this is another kind of object, so physicists clearly abuse some natural ways of understanding and visualising the differential. In math, expressions like $v^2 + (dv)^2$ or $v + dv$ usually make no sense.

Answer 2

Expanding @Intelligenti pauca's comment, the power series for $\sqrt{x^2+y^2}$ in terms of $y$ centered at $0$ starts as $$\lvert x\rvert + \frac{y^2}{2\lvert x\rvert} + \frac{y^4}{8 \lvert x\rvert^3} + \frac{y^6}{16\lvert x\rvert^5} +\cdots$$ So, to first order with respect to $y$, $\sqrt{x^2+y^2}\approx \lvert x\rvert$.

Before seeing how this interacts with calculating $\frac{d\lvert \mathrm{v}\rvert}{dt}$, let's calculate the derivative directly so we can believe it's zero at least mathematically. Write $v_x$ and $v_y$ for the two components of $\mathrm{v}$. Taking derivatives, we get $$\frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\sqrt{v_x^2+v_y^2}}{dt} = \frac{1}{\sqrt{v_x^2+v_y^2}}\left(v_x\frac{d v_x}{dt} + v_y\frac{d v_y}{dt}\right) = \frac{1}{\lvert \mathrm{v}\rvert}\left(v_x\frac{d v_x}{dt} + v_y\frac{d v_y}{dt}\right)$$ When we evaluate at the $t$ where $v_x=v$ and $v_y=0$, which by physics is when $\frac{d v_x}{dt}=0$ (non-rigorously, this is that "$\mathrm{v}$ and $d\mathrm{v}$ are orthogonal"), then we can see the derivative evaluates to $0$.

Now let's use the power series to compute the derivative. Substituting in $v_x$ and $v_y$, we get $$\lvert\mathrm{v}\rvert = \lvert v_x\rvert + \frac{v_y^2}{2\lvert v_x\rvert} + \cdots$$ Taking the derivative with respect to $t$, we get $$ \frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\lvert v_x\rvert}{dt}+\left(-\frac{v_y^2}{2v_x^2}\frac{d\lvert v_x\rvert}{dt} + \frac{v_y}{\lvert v_x\rvert}\frac{d v_y}{dt} \right) + \cdots $$ This is where we can see that the fact that using $\sqrt{x^2+y^2}$ to first order is all what matters: since $v_y=0$, everything from the second term onward is zero and we are left with $\frac{d\lvert\mathrm{v}\rvert}{dt} = \frac{d\lvert v_x\rvert}{dt}$.

In short: when in doubt, try to eliminate differentials and instead use derivatives directly. Most of the time statements like "$dv^2=0$" represent manipulation of a power series, in this case that second-order terms and above don't contribute to the first derivative.