# Constant velocity in a central force field

ilper
A body on a circular orbit in the field of a central force (satellites in gravity field of Earth; a charge in a magnetic field) is subjected to a force which is always perpendicular to its initial velocity v, hence in a time period dt it acquires an additional velocity dv, which is also perpendicular to v. Adding this two immediately gives a new velocity greater than the initial (a simple geometric property). As the velocity changes the body should radiate energy (surely for the charge in magnetic field). Does this energy release compensate exactly the increase in velocity so it stays constant?
If so, what radiates a body in gravitational field?

Delta2

DrStupid
hence in a time period dt it acquires an additional velocity dv, which is also perpendicular to v. Adding this two immediately gives a new velocity greater than the initial (a simple geometric property).

No, it doesn't. The velocity doesn't change in discrete steps but continuously. That results in constant speed:

##\frac{{dv^2 }}{{dt}} = 2 \cdot v \cdot \dot v = 0##

vanhees71, hutchphd and Dale
Mentor
Summary:: As the acquired under the force velocity dv is perpendicular to initial velocity v - geometrically adding of v and dv gives a new vector >v. But v must be constant.

Adding this two immediately gives a new velocity greater than the initial (a simple geometric property).
This is not correct. Remember that dv is not a finite length added at a discrete time. It is an infinitesimal length and as it is continuously added the direction of both v and dv change.

vanhees71
ilper
No, it doesn't. The velocity doesn't change in discrete steps but continuously. That results in constant speed:

##\frac{{dv^2 }}{{dt}} = 2 \cdot v \cdot \dot v = 0##
1. I don't agree that the velocity changes continuously. B.e. in an interaction with another charge a charge gains a photon from the source (the center of the force). So it gains Energy in finite steps - hence its velocity would also jump to a new value. Similarly it must be for gravity - exchange of gravitons.

2. Even if (adhering to classical physics) the velocity changes continuously it doesn't brake the law of adding two velocities as two vectors. E.g. an infinitesimal dv also will make 90 degrees to the initial velocity thus enhancing it. In order to change only the direction of the initial velocity even in infinitesimal dt, angle between v and dv will be infinitely close to 90 but not 90.

3. Even if one admits that in the limit as you say the reason why v does not change is that v increases continuously than by the same reasoning if the initial velocity is small (the body falls the to ground) than its velocity should also stay constant which is not the case indeed.

weirdoguy
Mentor
I don't agree that the velocity changes continuously. B.e. in an interaction with another charge a charge gains a photon from the source (the center of the force). So it gains Energy in finite steps
This is incorrect for a central force field, and @DrStupid is correct. First, this forum is classical, not quantum, but even in QM a central field doesn't work as you describe. In QM for a static central force field you are talking about virtual photons which do not behave at all like you described.

an infinitesimal dv also will make 90 degrees to the initial velocity thus enhancing it. In order to change only the direction of the initial velocity even in infinitesimal dt, angle between v and dv will be infinitely close to 90 but not 90.
This is simply incorrect. @DrStupid already posted the exceptionally simple math showing that it is incorrect. What part of his math did you not understand? Look at his equation. The conclusion is inescapable.

Even if one admits that in the limit as you say the reason why v does not change is that v increases continuously than by the same reasoning if the initial velocity is small (the body falls the to ground) than its velocity should also stay constant which is not the case indeed.
I am not sure what you are saying here. Can you write it mathematically?

vanhees71
In order to change only the direction of the initial velocity even in infinitesimal dt, angle between v and dv will be infinitely close to 90 but not 90.
It's very simple:
- If the angle is more than 90°, the speed decreases.
- If the angle is less than 90°, the speed increases.
- If the angle is exactly 90°, the speed remains unchanged.

Dale
Gold Member
2022 Award
It's very simple. If the speed stays constant that implies that its square is constant, i.e., you have
$$\vec{v}^2=\text{const}.$$
Now take the time derivative. With the product rule you find
$$\vec{v} \dot{\vec{v}}=0,$$
which means that the acceleration, ##\vec{a}=\dot{\vec{v}}## is always perpendicular to the velocity.

The most simple example for a force, where this is the case, is the motion of a charged particle in a magnetic field. There you have
$$m \dot{\vec{v}}=\frac{\vec{v}}{c} \times \vec{B} \; \Rightarrow \; \vec{v} \cdot \dot{\vec{v}}=0.$$
So a magnetic field doesn't change the speed but the direction of the particle's velocity.

E.g., it's easy to show that in a homogeneous magnetic field the particle moves with constant angular velocity (and thus with constant speed) around a circle. The speed doesn't change with time but the direction of the velocity does.

Dale and Delta2
ilper
This is incorrect for a central force field, and @DrStupid is correct. First, this forum is classical, not quantum, but even in QM a central field doesn't work as you describe. In QM for a static central force field you are talking about virtual photons which do not behave at all like you described.
I thought that the problem is classical that's why I post it here. I expected that the problem is solved by emitted energy from the body due to the Abraham Lorentz force.
Nevertheless the photons are virtual their energy is quantized. How are you conceiving the mechanism of transferring a force? What can influence a body without transfer of energy? Can it be a not finite amount but infinitely small and in the same time not zero? In order for velocity to change direction the body must receive influence and it is always going thought energy exchange in small but finite amounts.
This is simply incorrect. @DrStupid already posted the exceptionally simple math showing that it is incorrect. What part of his math did you not understand? Look at his equation. The conclusion is inescapable.
DrStupid with his simple math proves what he believes. He takes v=const silently, then the kinetic Energy is constant and it follows that velocity does not change.
I am not sure what you are saying here. Can you write it mathematically?
What has proven DrStupid mathematically? He says v changes continuously and that's why it can stay constant in value but change direction.
What I say is that according to classical physics v always changes continuously (not need to prove - in fact its a belief of classical physics(CP) - I don't deny it in the realm of CP). Nevertheless there are numerous problems where v changes value (in elementary textbooks).

weirdoguy
ilper
It's very simple:
- If the angle is more than 90°, the speed decreases.
- If the angle is less than 90°, the speed increases.
- If the angle is exactly 90°, the speed remains unchanged.
? Have do you drawn this geometrically?
In order v to stay constant dv must be less than 90. It can only stay unchanged if the angle is 90 when you add nothing.

2022 Award
? Have do you drawn this geometrically?
In order v to stay constant dv must be less than 90. It can only stay unchanged if the angle is 90 when you add nothing.
Not if you are adding infinitesimal perpendicular vectors (and if you are drawing it, you are drawing a finite vector). As @DrStupid showed, if you require constant velocity then ##\dot{\vec v}## must be perpendicular to ##\vec v##.

? Have do you drawn this geometrically?
Drawing finite vectors is just an approximation. You have to take the limit.

hutchphd
... force which is always perpendicular to its initial velocity v, hence in a time period dt it acquires an additional velocity dv, which is also perpendicular to v. Adding this two immediately gives a new velocity greater than the initial (a simple geometric property).
Apply time reversal symmetry to this construction, and you will get the opposite result, that the speed must be decreasing. This demonstrates that it's a flawed approach.

A better geometrical construction uses the vectors at two time points:
https://phys.libretexts.org/Bookshe...hree_Dimensions/4.05:_Uniform_Circular_Motion

jbriggs444 and hutchphd
Mentor
He takes v=const silently, then the kinetic Energy is constant and it follows that velocity does not change.
That is not at all what he did.

The speed is ##\sqrt{v^2}=\sqrt{\vec v \cdot \vec v}##. If ##\frac{d}{dt}\sqrt{v^2}=0## then ##\frac{d}{dt}v^2=0## and @DrStupid showed the condition for that to happen, namely that ##v \cdot \dot v = 0## or in other words if the acceleration is perpendicular to the velocity then the speed is constant. In direct contradiction to your multiple erroneous claims.

DrStupid
? Have do you drawn this geometrically?
In order v to stay constant dv must be less than 90. It can only stay unchanged if the angle is 90 when you add nothing.

Drawing the vector addition of a finite ##v## and an infinitesimal ##dv## is pretty much adding nothing. In order to draw something visible you need a finite ##\Delta v## and the angle between ##v## and ##\Delta v## actually is less than 90°. Just write the velocity as a taylor series

##v\left( {t + \Delta t} \right) = v\left( t \right) + \dot v\left( t \right) \cdot \Delta t + {\textstyle{1 \over 2}}\ddot v\left( t \right) \cdot \Delta t^2 + \cdots##

include the derivates of v for uniform circular motion, solve for the additional velocity

##\Delta v = v\left( {t + \Delta t} \right) - v\left( t \right) = \left[ { - \omega ^2 \cdot r\left( t \right) - {\textstyle{1 \over 2}}\omega ^3 \cdot v\left( t \right) \cdot \Delta t + \cdots } \right] \cdot \Delta t##

and you see a tangential second order term that is directed agains ##v##. The compensation for the supposed increase in speed is included in the tangential higher order terms. There is no physics involved. It's just math.

In the limit for infinitesimal ##dv## all higher order terms go to zero and only the centripetal acceleration

##\dot v = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta v}}{{\Delta t}} = - \omega ^2 \cdot r##

remains. The problem you are talking about does not exist.

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Dale
Gold Member
2022 Award
That is not at all what he did.

The speed is ##\sqrt{v^2}=\sqrt{\vec v \cdot \vec v}##. If ##\frac{d}{dt}\sqrt{v^2}=0## then ##\frac{d}{dt}v^2=0## and @DrStupid showed the condition for that to happen, namely that ##v \cdot \dot v = 0## or in other words if the acceleration is perpendicular to the velocity then the speed is constant. In direct contradiction to your multiple erroneous claims.
Of course, note that
$$v^2=\vec{v} \cdot \vec{v} \; \Rightarrow \; v \dot{v}=\vec{v} \cdot \dot{\vec{v}}.$$
If ##\vec{v} \neq 0 \Leftrightarrow v \neq 0## this implies that ##\dot{\vec{v}} \perp \vec{v}## (or ##\dot{\vec{v}}=0##).

ilper
Drawing the vector addition of a finite ##v## and an infinitesimal ##dv## is pretty much adding nothing. In order to draw something visible you need a finite ##\Delta v## and the angle between ##v## and ##\Delta v## actually is less than 90°. Just write the velocity as a taylor series

##v\left( {t + \Delta t} \right) = v\left( t \right) + \dot v\left( t \right) \cdot \Delta t + {\textstyle{1 \over 2}}\ddot v\left( t \right) \cdot \Delta t^2 + \cdots##

include the derivates of v for uniform circular motion, solve for the additional velocity

##\Delta v = v\left( {t + \Delta t} \right) - v\left( t \right) = \left[ { - \omega ^2 \cdot r\left( t \right) - {\textstyle{1 \over 2}}\omega ^3 \cdot v\left( t \right) \cdot \Delta t + \cdots } \right] \cdot \Delta t##

and you see a tangential second order term that is directed agains ##v##. The compensation for the supposed increase in speed is included in the tangential higher order terms. There is no physics involved. It's just math.

In the limit for infinitesimal ##dv## all higher order terms go to zero and only the centripetal acceleration

##\dot v = \mathop {\lim }\limits_{\Delta t \to 0} \frac{{\Delta v}}{{\Delta t}} = - \omega ^2 \cdot r##

remains. The problem you are talking about does not exist.
Does your considerations not hold true for every type of motion - even linear?

DrStupid
Does your considerations not hold true for every type of motion - even linear?

Yes, it does. But in linear motion there is no centripetal acceleration.

ilper
Yes, it does. But in linear motion there is no centripetal acceleration.
Then you prove that in the case of linear motion dv is second order of dt, even that is 0, as w is 0.
You accept that for finite Δv the angle is 90 (right?). Then making infinitesimal transition does not change this angle (even I think any other property of Δv).
As when Δv is unidirectional to v (0 grade) dv is also unidirectional to v (also 0 grade).
As when Δv is contra-directional to v (180 grade) dv is also contra-directional to v (also 180 grade).
Correct me if I am wrong.

weirdoguy
Mentor
You have already been corrected plenty, to no avail. This thread is closed.

weirdoguy