Work from a perpendicular force?

Question

I am deeply concerned about a problem I have in past time even not noticed and slid past it without paying any attention, accepting what books write without the slightest idea that this can be a problem.

According to Newton’s second law if a force $F $ is applied to a body (pay attention! – there is no any option weather it is moving or not) it gets immediately accelerated which means that its speed $v $ changes in the direction of the force immediately after $F$ is applied e.g. after $dt $ – $v\rightarrow v+dv$

That would mean that if $F$ is perpendicular to $v $ the speed $dv$ is also perpendicular to $v$-> which means that $v $ increases in magnitude (not only change of direction) and gets $v+dv $ (or Euclidean geometry is not true in the limit of infinitesimal values which is absurd in the Newtonian space). But this means that work is done by the force and the kinetic energy increases.

If someone insists that nevertheless adding infinitesimal $dv $ doesn’t change the magnitude of $v $ e.g $v+dv=v$ then the same would apply even when $v $ and $dv $ are collinear e.g. adding $dv $ to $v $ doesn’t change the magnitude of $v $ at all.

In textbooks it is stated that $F $ (when perpendicular to $v$) only changes the direction of $v $ which means kinetic energy doesn’t change.

What is wrong in my considerations?

Answer 1

In textbooks it is stated that 𝐹 (when perpendicular to 𝑣) only changes the direction of 𝑣 which means kinetic energy doesn’t change.

Yes, the textbooks are correct. Since $KE=\frac{m}{2}\vec v^2$ if we change $\vec v$ infinitesimally then we get $$dKE= KE(\vec v +d \vec {v}) -KE(\vec v)$$ $$=\frac{m}{2}\left( \vec v^2 + 2 \vec v \cdot d\vec {v} +d \vec {v}^2\right) -\frac{m}{2} \vec v^2$$ The last term in the parentheses drops out because it is a product of infinitesimals, and we get $$dKE= m \ \vec v \cdot d\vec {v}$$ Thus the change in KE can be 0 if $\vec v =0$ or if $\vec v$ is perpendicular to $d\vec {v }$

If someone insists that nevertheless adding infinitesimal 𝑑𝑣 doesn’t change the magnitude of 𝑣 e.g 𝑣+𝑑𝑣=𝑣 then the same would apply even when 𝑣 and 𝑑𝑣 are collinear e.g. adding 𝑑𝑣 to 𝑣 doesn’t change the magnitude of 𝑣 at all.

This does not follow. A similar approach as above can be used here, although the calculation is more difficult. Since $|\vec v|=\sqrt{\vec v^2}=\sqrt{\vec v \cdot \vec v}$ we can write: $$d|\vec v|=|\vec v + d\vec v|-|\vec v|$$ $$ = \sqrt{\left(\vec v + d \vec v \right)^2} - \sqrt{ \vec v^2}$$ as before the products of infinitesimals drop out and we have $$ = \sqrt{\vec v^2 + 2 \vec v \cdot d\vec v} - \sqrt{\vec v^2}$$ we can series expand the first radical to get $$ = \sqrt{\vec v^2} + \frac{\vec v \cdot d\vec v}{\sqrt{\vec v^2}} - \sqrt{\vec v^2}$$ $$d|\vec v|=\hat v \cdot d\vec v$$ So again, $d|\vec v|$ can be zero if $d\vec v$ is perpendicular to $\vec v$. There is no contradiction to say that $d|\vec v|=0$ in some situations and not in others. That is a direct consequence of the math, colinear is fundamentally different from perpendicular.

That would mean that if 𝐹 is perpendicular to 𝑣 the speed 𝑑𝑣 is also perpendicular to 𝑣-> which means that 𝑣 increases in magnitude (not only change of direction) and gets 𝑣+𝑑𝑣 (or Euclidean geometry is not true in the limit of infinitesimal values which is absurd in the Newtonian space).

Note, in all of the above derivations the vector $\vec v+ d\vec v$ is obtained from the standard Euclidean vector addition. $\vec v$ is infinitesimally different from $\vec v + d\vec v$ in the usual way. However, your statements about $dKE$ and $d|\vec v|$ simply do not follow from the Euclidean vector addition of $\vec v + d\vec v$.

Edit: for more information on how infinitesimals, differentials and so forth work see: https://people.math.wisc.edu/~keisler/foundations.pdf especially p. 34.

In the above I have made a slight abuse of notation by writing $$dy=y(x+dx)-y(x)$$ instead of the more complete and correct $$dy=\text{st}\left( \frac{y(x+dx)-y(x)}{dx} \right) dx$$ Dividing by $dx$, taking the standard part, $\text{st}$, and multiplying by $dx$ is what removes the products of infinitesimals.

I thought that it was not a major abuse of notation since in other contexts $dx\ne 0$ and $dx^2=0$ is a defining property of infinitesimals, but from some comments it seems these basic properties of infinitesimals and differentials may not be understood by all readers.

Answer 2

I think there are a few points of confusion.

First, if a force $\vec{F}$ is applied to a mass, m, then regardless of the reference frame (whether or not you are moving in some frame of reference) $\vec{F}=m\vec{a}$. In other words the net force is proportional and parallel to the acceleration, not perpendicular. Remember, if things are moving in the same direction they are parallel, if they move at a right angle they are perpendicular.

Note also that $\vec{v}(t)=\int_0^t\vec{a}(t')dt'+\vec{v}(0)$.

So your statement about the force being perpendicular to dv (or a(t)dt) completely depends on the problem.

For instance, for a satellite in circular orbit the direction of motion is in the $\hat{\phi}$ direction whereas the centripetal force that is responsible for keeping the satellite in it circular motion is along the $\hat{r}$ direction.

The equation for work done by a force is $W=\int \vec{F}\cdot d\vec{x}$

So the work done by the centripetal force is $W=\int (F\hat{r})\cdot(r\omega \hat{\phi}dt)$ and since $\hat{r}\cdot \hat{\phi}=0$, no work is done by the centripetal force to keep the satellite in orbit... but it is doing something. It is changing the vector direction of the satellite!

How can a force change the direction while not doing work? another way to think about it is that work is the change in the kinetic energy. $W=\Delta K=K_f-K_i=1/2mv_f^2-1/2mv_i^2$. Since the speed (scalar) does not change $W=\Delta K=0$

Maybe a more helpful example is how adding a vector does not necessarily change the magnitude of the original vector.

consider a circle described by $\vec{r}=R\{cos(\theta),sin(\theta)\}$. If you imagine moving along the circle, there is some vector that takes you from where you are to where you want to go. Say you are at $\vec{A}=\{1,0\}$ and you want to get to $\vec{B}=\{0,1\}$. The vector you need is $\vec{C}=\{-1,1\}$ so that $\vec{B}=\vec{A}+\vec{C}$. Note that the length of the vector A and B are both 1, but you definitely did add a non-trivial vector, C. This works for any two point on a circle.

Answer 3

which means that its speed v increases in the direction of the force immediately after F is applied

No. Imagine uniform circular motion for an object moving at speed $v$ in a circle with radius $r$. The acceleration will always be $a=\dfrac {v^2}r$ but the speed will remain the same.

Second, $\bf \vec v + \delta \vec v$ does not necessarily change the magnitude. Consider two vectors ${\bf \vec a} = \left(1,0 \right)$ and ${\bf \vec b} = \left( -1,1\right)$. Then ${\bf \vec a + \vec b} =\left( 0,1\right)$, and yet $\| \bf \vec a \| = \| \vec a + \vec b \| = \mathrm 1$.