Fictitious forces and their associated energies

[phantomvommand]

TL;DR Summary

Why certain forces like centrifugal forces have an associated centrifugal potential energy.

I understand the idea of a fictitious force. What I am confused about is the energy/potential associated with it.
For example, if a cylinder of water is rotating, there apparently exists this centrifugal potential energy, which is obtained by integrating mrw^2 dr. Why is it that the “centripetal potential energy” does not exist, and thus does not add an an additional 1/2 mr^2w^2? Furthermore, does there exist a coriolis potential energy?
What exactly is the reason why some forces are associated with an energy, while others are not?
Thank you.

 

[etotheipi]

It is nothing more than mathematical convenience. In a general non-inertial frame the second law of motion for a particle $P$ is modified to $\mathbf{F} = m[\mathbf{A} + \dot{\boldsymbol{\Omega}} \times \boldsymbol{x}' + 2\boldsymbol{\Omega} \times \mathbf{v}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{x}') + \mathbf{a}']$, where $\boldsymbol{x}' = \overrightarrow{O'P}$, $\mathbf{v}' = (\mathrm{d}\boldsymbol{x}'/ \mathrm{d}t) \big{|}_{O'x'y'z'}$ and $\mathbf{a}' = (\mathrm{d}\mathbf{v}'/ \mathrm{d}t) \big{|}_{O'x'y'z'}$ are the position, velocity and acceleration of the particle with respect to the non-inertial frame $O'x'y'z'$, and $\mathbf{A}$ and $\boldsymbol{\Omega}$ are the linear acceleration and angular velocity of $O'x'y'z'$ with respect to $Oxyz$.

Now suppose that the angular acceleration of $O'x'y'z'$ relative to $Oxyz$ is zero , i.e. $\dot{\boldsymbol{\Omega}} = 0$, and further that $O'$ does not accelerate with respect to $O$, i.e. $\mathbf{A} = 0$. Consider the resulting equation of motion in the non-inertial frame, and then take the scalar product of both sides with $\mathbf{v}'$,\begin{align*}
\mathbf{F} &= m[2\boldsymbol{\Omega} \times \mathbf{v}' + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{x}') + \mathbf{a}'] \\

\implies \mathbf{v}' \cdot \mathbf{F} &= m[\mathbf{v}' \cdot \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{x}') + \mathbf{v}' \cdot \mathbf{a}'] \quad \quad \quad (1)
\end{align*}since $\mathbf{v}' \cdot (\boldsymbol{\Omega} \times \mathbf{v}') = 0$. By a vector identity we can re-write $\boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{x}') = (\boldsymbol{\Omega} \cdot \boldsymbol{x}') \boldsymbol{\Omega} - \Omega^2 \boldsymbol{x}'$, hence\begin{align*}

\mathbf{v}' \cdot \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{x}') &= (\boldsymbol{\Omega} \cdot \mathbf{v}')(\boldsymbol{\Omega} \cdot \boldsymbol{x}') - \Omega^2 \mathbf{v}' \cdot \boldsymbol{x}' \\

&= \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{1}{2} (\boldsymbol{\Omega} \cdot \boldsymbol{x}')^2 - \frac{1}{2} \Omega^2 \boldsymbol{x}' \cdot \boldsymbol{x}' \right]\end{align*}We can also re-write $m \mathbf{v}' \cdot \mathbf{a}' = \frac{m}{2} \frac{\mathrm{d}}{\mathrm{d}t} \left[ \mathbf{v}' \cdot \mathbf{v}' \right] = \mathrm{d}T / \mathrm{d}t$. Thus equation $(1)$ is equivalent to\begin{align*}

\mathbf{v}' \cdot \mathbf{F} = \frac{\mathrm{d}}{\mathrm{d}t} \left[ \frac{m}{2} \mathbf{v}' \cdot \mathbf{v}' + \left( \frac{m}{2} (\boldsymbol{\Omega} \cdot \boldsymbol{x}')^2 - \frac{m}{2} \Omega^2 \boldsymbol{x}' \cdot \boldsymbol{x}' \right) \right] \end{align*}Further, notice that $\frac{m}{2} (\boldsymbol{\Omega} \cdot \boldsymbol{x}')^2 - \frac{m}{2} \Omega^2 \boldsymbol{x}' \cdot \boldsymbol{x}' = \frac{-m}{2}\Omega^2 \left( |\boldsymbol{x}'|^2 - (|\boldsymbol{x}'| \cos{\theta})^2 \right)=\frac{-1}{2} \Omega^2 md^2$, where $d = \sqrt{|\boldsymbol{x}'|^2 - (|\boldsymbol{x}'| \cos{\theta})^2}$ is the perpendicular distance of the particle from the axis ($O', \boldsymbol{\Omega}$). And since $I := md^2$ is the moment of inertia of the particle about this axis, the equation may be re-written in the simple form\begin{align*}
\mathbf{v}' \cdot \mathbf{F} = \frac{\mathrm{d}}{\mathrm{d}t} \left[ T - \frac{1}{2}I \Omega^2 \right]
\end{align*}And the $\varphi_R := \frac{-1}{2}I \Omega^2$ term can be identified as a "centrifugal potential energy" term. Further, suppose that the real force $\mathbf{F}$ can be expressed as $\mathbf{F} = - \nabla_{\boldsymbol{x}'} U$. In such cases, $\mathbf{v}' \cdot \mathbf{F} = - \mathbf{v}' \cdot \nabla_{\boldsymbol{x}'} U = - \mathrm{d}U/\mathrm{d}t$. Thus, integrating and re-arranging, you'd obtain\begin{align*}
T + \left[\varphi_R + U\right] := T + V = E

\end{align*}where $E$ is the constant of integration, and $V := \varphi_R + U$ is the effective potential energy of the system.

 

[A.T.]

Why certain forces like centrifugal forces have an associated centrifugal potential energy.

You can associate a potential with a force field, that depends only on position and on conserved object properties (like mass or charge), but not on velocity etc.

Why is it that the “centripetal potential energy” does not exist, ... does there exist a coriolis potential energy?

They both depend on velocity.

 

[etotheipi]

Why is it that the “centripetal potential energy” does not exist... does there exist a coriolis potential energy?

They both depend on velocity.

I don't think this is correct. For example centripetal forces can often be derived from potential functions, e.g. consider a particle orbiting in a central gravitational field, or a central electric field, etc; none of these forces depend on velocity.

Also, the Coriolis force cannot do work, so there is no need for a potential for it in the first place anyway.

 

[A.T.]

For example centripetal forces can often be derived from potential functions, e.g. consider a particle orbiting in a central gravitational field, or a central electric field, etc; none of these forces depend on velocity.

Often is not always. In general, a centripetal force doesn't have to be a single conservative force. For example, in a rotating frame the Coriolis force can act as a centripetal force.

 

[etotheipi]

Sure, but it's still not true to say that centripetal forces depend on velocity, in the general case.

Aside; I actually quite dislike the terminology 'centripetal force', precisely because by centripetal force one usually means the component of a sum of a specified set of forces toward the instantaneous centre of curvature, $\sum \mathbf{F}_i \cdot \mathbf{e}_{n}$. And of course some of those in the sum might be conservative, and some of them not.

 

[A.T.]

Sure, but it's still not true to say that centripetal forces depend on velocity, in the general case.

The most general definition of "centripetal force" does depend on velocity:
https://en.wikipedia.org/wiki/Centripetal_force#Formulae
So you cannot define a general "potential of centripetal force", just for special cases.

 

[etotheipi]

The most general definition of "centripetal force" does depend on velocity:
https://en.wikipedia.org/wiki/Centripetal_force#Formulae

You misunderstand what it means to say a force depends on velocity. The most general force defined on a system of particles is a function $\mathbf{F}: \mathbf{R}^{3n} \times \mathbf{R}^{3n} \times \mathbf{R} \longrightarrow \mathbf{R}^{3n}$ whose argument is the $(6n+1)$-tuple $(\mathbf{x}, \dot{\mathbf{x}}, t)$. This function is defined such that $m\ddot{\mathbf{x}} = \mathbf{F}( \mathbf{x}, \dot{\mathbf{x}}, t)$ is safisfied, and the functional form of $\mathbf{F}$ indeed constitutes a definition of the system. A force depends on velocity if it is a non-trivial function of velocity!

It is possible to determine a further kinematic relation between $\ddot{\mathbf{x}}_i$ and $\dot{\mathbf{x}}_i$, namely that $(\ddot{\mathbf{x}}_i)_n = (\dot{\mathbf{x}}_i^2)/\rho$. But this relation does not mean that the force depends on velocity.

Again, consider the example of a gravitational force, which may be considered a centripetal force. It is merely a function of position co-ordinates, and not a function of velocities, i.e. $\mathbf{F}_{12} = (\gamma \mathbf{e}_{12})/|\mathbf{x}_1 - \mathbf{x}_2|^2$. Even though you can relate it to velocity by means of Newton's equation, it does not depend on velocity!

A force that does depend on velocity, for instance, is the magnetic force $\mathbf{F}_m = q\dot{\mathbf{x}} \times \mathbf{B}$.

 

[A.T.]

Again, consider the example of a gravitational force, which may be considered a centripetal force.

The OP (if I understand the question correctly) asked why there is no "centripetal force potential" based on the general concept of the centripetal force, not on some special cases of centripetal force.

 

[etotheipi]

I was responding to your claim in post #3, which is incorrect, firstly because centripetal forces need not have any velocity dependence and secondly because you can of course also have forces/vector fields depending on only position co-ordinates which are not conservative.

In any case I already said at the end of post #2 that if the real forces are derivable from a scalar field then you may define a potential in the usual fashion in order to simplify the problem again to energy methods.

 

[hutchphd]

Centripetal force is a descriptive term, not a definition. As soon as you define an origin you can define centripetal force as any force toward that origin. No velocity need be invoked. In fact the origin does not have to be the canter of rotation (although...)
I believe Centrifugal Force is very different and defines the particular pseudo-force associated with a rotating coordinate system. Its value is determined by that coordinate transformation parameterized by the rotation speed
As usual I believe the disagreement here is largely semantic but think @etotheipi correct.

 

[A.T.]

... you can of course also have forces/vector fields depending on only position co-ordinates which are not conservative...

That is true, mathematically. But are there physical examples?

 

[etotheipi]

That is true, mathematically. But are there physical examples?

What about a particle moving in a non-conservative electric field?

 

[A.T.]

Centripetal force is a descriptive term, not a definition.

Yes, and the term is not even used consistently. Which is maybe a reason why we usually don't use "centripetal potential", but rather explicitly state the force, like "gravitational potential".

 

[A.T.]

What about a particle moving in a non-conservative electric field?

Time dependent?

 

[hutchphd]

What's a "physical" field? For instance the drag force on a "unit" swimmer in a steady rotating current could be represented as a field...

 

[etotheipi]

Time dependent?

Fair enough, but what about if we also impose that $\partial \mathbf{B} / \partial t$ is constant? I'm sure there are also other less fundamental examples.