Skip to main content
SearchLoginLogin or Signup

Centrifugal forces from seven perspectives

Published onApr 09, 2024
Centrifugal forces from seven perspectives
·

Abstract

Everyone has experienced centrifugal forces as exciting and possibly scary. From a physical point of view, they are ficticious accelerations, as they appear to be related to accelerating frames of reference and don’t depend on the test particle’s mass, and from a relativists point of view, they are gravitational or inertial, with exactly the same properties. All students have used the iconic result that the centrifugal acceleration aa is given by υ2/r\upsilon^2/r, and I’d like to illustrate different ways to think about it.

Layperson’s perspective

Imainge a particle on a circular trajectory of radius rr and velocity υ\upsilon. Over the course of half an orbit which is covered in time Δt=πr/υ\Delta t = \pi r/\upsilon, the velocity changes by Δυ=2υ\Delta\upsilon = 2\upsilon from +υ+\upsilon to υ-\upsilon. Under the approximation that the acceleration is constant, one arrives at aΔυ/Δt=2/π×υ2/ra\simeq \Delta\upsilon/\Delta t = 2/\pi\times\upsilon^2/r. Already now one can recognise the iconic a=υ2/ra = \upsilon^2/r, and the nasty prefactor 2/π12/\pi \neq 1 is a remnant of the questionable approximation of constant acceleration.

Newton’s perspective

A remedy would be to carry out the limit Δt0\Delta t\rightarrow0 and to compute the acceleration through a proper differentiation: If the particle moves in the (xy)(xy)-plane on a circle of radius rr, its velocity is given by x˙=υsin(ωt)\dot{x} = \upsilon\sin(\omega t) so that the acceleration becomes a=x¨=υωcos(ωt)υωa = \ddot{x} = \upsilon\omega\cos(\omega t) \simeq \upsilon\omega at t0t \rightarrow 0. Effectively, we’re ignoring the change of y˙=υcos(ωt)\dot{y} = \upsilon\cos(\omega t) as y¨=υωsin(ωt)\ddot{y} = \upsilon\omega\sin(\omega t) because the sine is just 0 at t=0t=0. Then, the instantaneous value of the centrifugal acceleration is a=υω=ω2r=υ2/ra = \upsilon\omega = \omega^2 r = \upsilon^2/r. Due to symmetry, the chosen instant in time t=0t=0 is no particular choice, so the result needs to be valid in generality.

Lagrange’s perspective

A classical, Lagrangian approach for circular motion on the (xy)(xy)-plane would entail to write down a kinetic term m/2(x˙2+y˙2)m/2\:(\dot{x}^2+\dot{y}^2) and the addition of a Lagrange-multiplier to enforce the boundary condition that x2+y2x^2+y^2 always stays equal to r2r^2:

L=m2(x˙2+y˙2)λ2(x2+y2r2).(1)\mathcal{L} = \frac{m}{2}(\dot{x}^2+\dot{y}^2) - \frac{\lambda}{2}(x^2+y^2-r^2). \tag{1}

For convenience, I’ve chosen λ/2\lambda/2 to be the Lagrange multiplier. Interestingly, the term λr2/2\lambda r^2/2 can be dropped because it is a total time derivative and would not matter in the actual equation of motion, leaving the Lagrange-function of two harmonic osillators,

L=m2(x˙2+y˙2)λ2(x2+y2),(2)\mathcal{L} = \frac{m}{2}(\dot{x}^2+\dot{y}^2) - \frac{\lambda}{2}(x^2+y^2), \tag{2}

trading the constraint for a potential. Variation of the Lagrange-function by application of the Euler-Lagrange equations leads directly to the equations of motion

mx¨=λxandmy¨=λy(3)m\ddot{x} = -\lambda x \quad\text{and}\quad m\ddot{y} = -\lambda y \tag{3}

which remain coupled through the boundary condition x2+y2=r2x^2+y^2=r^2. You can see that in the following step: Choosing x(t)=rcos(ωt)x(t) = r\cos(\omega t) with ω2=λ/m\omega^2 = \lambda/m forces y(t)y(t) to be rsin(ωt)r\sin(\omega t), by virtue of x2+y2=r2(cos2(ωt)+sin2(ωt))=r2x^2+y^2 = r^2\left(\cos^2(\omega t) + \sin^2(\omega t)\right) = r^2.

The Lagrange-multiplier was fixed by the condition λ=mω2\lambda = m\omega^2: Substitution into the Lagrange-function and getting rid of the factor mm by invoking mechanical similarity (the Lagrange-function is only ever defined up to an affine transform LαL+β\mathcal{L}\rightarrow\alpha\mathcal{L}+\beta with two constants α,β\alpha,\beta) one arrives at

L=12(x˙2+y˙2)ω22(x2+y2).(4)\mathcal{L} = \frac{1}{2}(\dot{x}^2+\dot{y}^2) - \frac{\omega^2}{2}(x^2+y^2). \tag{4}

which has replaced the boundary condition completely by the harmonic, centrifugal potential Φ=ω2/2(x2+y2)\Phi = \omega^2/2\:(x^2+y^2). The acceleration would be a2=x¨2+y¨2=r2ω4a^2 = \ddot{x}^2 + \ddot{y}^2 = r^2\omega^4, while the velocity remains υ2=x˙2+y˙2=r2ω2\upsilon^2 = \dot{x}^2+\dot{y}^2 = r^2\omega^2, leading again to a=rω2=υ2/ra = r\omega^2 = \upsilon^2/r. The fact that the mass mm drops out is a signature of the centrifugal force being inertial.

Jacobi’s and d’Alembert’s perspective

Jacobi was one of the first to think about the relation between the average kinetic and potential energies in classial systems, what is today known as the virial theorem. The harmonic oscillator has a peculiar property in this respect as its average kinetic energy T\langle T\rangle and potential energy Φ\langle\Phi\rangle are identical:

T=12x˙2+y˙2=r2ω22=ω22x2+y2=Φ(5)\langle T\rangle = \frac{1}{2}\langle \dot{x}^2+\dot{y}^2\rangle = \frac{r^2\omega^2}{2} = \frac{\omega^2}{2}\langle x^2+y^2\rangle = \langle\Phi\rangle \tag{5}

This is almost a triviality in this context because the energies are not only equal on average, but, due to the boundary condition, equal at every instant:

T=12(x˙2+y˙2)=r2ω22=ω22(x2+y2)=Φ,(6)T = \frac{1}{2}\left(\dot{x}^2+\dot{y}^2\right) = \frac{r^2\omega^2}{2} = \frac{\omega^2}{2}\left(x^2+y^2\right) = \Phi, \tag{6}

actually implying L0\mathcal{L} \equiv 0 with no further implication within classical mechanics. As the total energy is made up from T+ΦT+\Phi, both TT and Φ\Phi must be conserved individually.

The particular shape of the parabolic, central potential Φ=ω2r2/2\Phi = \omega^2r^2/2, here in polar coordinates, makes sure that the acceleration a=Φ/r=ω2r=υ2/ra = \partial\Phi/\partial r = \omega^2 r = \upsilon^2/r is always radial and perpendicular to the velocity υ\upsilon, changing only its direction and not the norm Tυ2T\propto \upsilon^2, and the velocity is always perpendicular to the radius, such that it doesn’t change either, conserving Φr2\Phi\propto r^2 in the process.

In particular, the centrifugal force does not perform work at fixed rr: That’s a reflection of d’Alembert’s principle that the virtual work done by the constraint force is zero. Here, the constraint force is the radial centripetal force and the virtual displacement takes place in the tangential direction as the natural direction of motion of the system consistent with the constraints. Both are perpendicular to each other, making the virtual work vanish.

Hamilton’s perspective

Now it’s time to choose polar coordinates (r,φ)(r,\varphi) that are adapted to the system, in the sense that they respect the boundary condition: The last term of the Lagrange-function (Eq. 1) vanishes automatically, leaving only the kinetic term:

L=12r2φ˙2(7)\mathcal{L} = \frac{1}{2}r^2\dot{\varphi}^2 \tag{7}

There can not be a term r˙2/2\dot{r}^2/2 associated with the kinetic energy of radial motion. This would imply a changing radius and a violated boundary condition. Hamilton would immediately identify φ\varphi as a cyclic coordinate ,

Lφ=0,(8)\frac{\partial\mathcal{L}}{\partial\varphi} = 0, \tag{8}

and associate an conserved canonical momentum with it,

pφ=Lφ˙=r2φ˙=r2ωwithdpφdt=ddtLφ˙=Lφ=0(9)p_\varphi = \frac{\partial\mathcal{L}}{\partial\dot{\varphi}} = r^2\dot{\varphi} = r^2\omega \quad\text{with}\quad \frac{\mathrm{d}p_\varphi}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial\mathcal{L}}{\partial\dot{\varphi}} = \frac{\partial\mathcal{L}}{\partial\varphi} = 0 \tag{9}

otherwise known as (specific) angular momentum, pφ=r2ωLp_\varphi = r^2\omega \equiv L. Carrying out a Legendre-transform to arrive at the (specific) Hamilton-function H\mathcal{H} leads to

H=pφ×φ˙(pφ)L(φ˙(pφ))=L22r2(10)\mathcal{H} = p_\varphi\times \dot{\varphi}(p_\varphi) - \mathcal{L}(\dot{\varphi}(p_\varphi)) = \frac{L^2}{2r^2} \tag{10}

Not only is the total energy H\mathcal{H} conserved as L\mathcal{L} does not depend on tt, but in addition we get that p˙φ=L˙=H/φ=0\dot{p}_\varphi = \dot{L} = -\partial\mathcal{H}/\partial\varphi = 0 in conservation of angular momentum, and φ˙=+H/pφ=+H/L=L/r2=ω\dot{\varphi} = +\partial\mathcal{H}/\partial p_\varphi = +\partial\mathcal{H}/\partial L = L/r^2 = \omega, showing that φ=ωt\varphi = \omega t increases linearly with tt. I find it awesome how a proper choice of generalised coordinates adapted to the system gets rid of all boundary conditions and constraint forces, but introduces a potential L2/(2r2)\propto L^2/(2r^2), keeping any particle with nonzero angular momentum from reaching r=0r=0, ultimately making sure that planetary systems or hydrogen atoms are stable. The derivative H/r\partial\mathcal{H}/\partial r closes the circle (pun intended), as it recovers

Hr=L2r3=rω2=υ2r,(11)\frac{\partial\mathcal{H}}{\partial r} = -\frac{L^2}{r^3} = -r\omega^2 = -\frac{\upsilon^2}{r}, \tag{11}

for the acceleration needed to set up the system with identical angular momentum at different rr. To me, this result is a bit remarkable, because the centrifugal acceleration is a gradient of the energy, but not a gradient of a potential: The Lagrange-function has only a kinetic term in our coordinate choice.

Lie’s perspective

At last, I’d like to try a clearly overengineered derivation! Instead of tracking the motion of a particle on a circular path, we can rotate the coordinate system. Any vector r\mathbf{r} changes in time according to

drdt=ω×r(12)\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \boldsymbol{\omega}\times\mathbf{r} \tag{12}

if rotated in a plane perpendicular to ω\boldsymbol{\omega} (which is parallel to the angular momentum L\mathbf{L}). From this point of view, ω×\boldsymbol{\omega}\times is the rotation operator for any vector. Clearly, this leads to ω×(ω×r)\boldsymbol{\omega}\times\left(\boldsymbol{\omega}\times\mathbf{r}\right) as the acceleration d2r/dt2\mathrm{d}^2\mathbf{r}/\mathrm{d}t^2 for circular motion and for constant angular velocities – you’ll find this in every mechanics textbook.

But let’s try to rotate the vector r\mathbf{r} explicitly, by means of a rotation operator properly generated as a Lie-group element:

R=exp(σωt)withσ=(0+110)(13)R = \exp\left( \sigma\omega t \right) \quad\text{with}\quad \sigma = \left( \begin{array}{cc} 0 & +1\\ -1 & 0 \end{array} \right) \tag{13}

Here, σ\sigma is one of the Pauli-matrices, and the antisymmetry of σ\sigma makes RR orthogonal. Any vector r\mathbf{r} is rotated by application of RR,

R=exp(σωt)=n(σωt)nn!=(cos(ωt)+sin(ωt)sin(ωt)cos(ωt))(14)R = \exp\left(\sigma\omega t\right) = \sum_n\frac{(\sigma\omega t)^n}{n!} = \left( \begin{array}{cc} \cos(\omega t) & +\sin(\omega t)\\ -\sin(\omega t) & \cos(\omega t) \end{array} \right) \tag{14}

as shown by explicit computation, where orthogonality conserves the norm. The velocity υ\boldsymbol{\upsilon} is proportional to ωσR\omega\sigma R and consequently, the acceleration a\mathbf{a} to ω2σ2R\omega^2\sigma^2 R. Then, the norm υ2\upsilon^2 comes out as ω2r2\omega^2 r^2 and the acceleration a2a^2 as ω4r2\omega^4r^2 (using σtσ=id\sigma^t\sigma = \mathrm{id} as well as RtR=idR^t R = \mathrm{id}, which follows as a consequence). Combining both leads yet again to a=ω2r=υ2/ra = \omega^2 r = \upsilon^2/r as well as to a vanishing scalar product between υ\boldsymbol{\upsilon} and a\mathbf{a}, as an expression of orthogonality.

Summary

Centrifugal forces are difficult to grasp: We moved from explicit accelerations in circular motion to constraints, which get soon replaced by potentials; only to arrive at a centrifugal potential that is no potential at all! But the fundamental issue is not touched on in classical mechanics: its property of being related to accelerated frames of reference and the equivalence between inertial and gravitational forces. It’d be amazing to see how centrifugal forces (and Coriolis and Euler-forces) follow consistently from a gravitational theory – perhaps in a new article for Phoebe!


Editor’s note: The background image was made by Ethan Hoover and published on Unsplash (from: https://unsplash.com/de/fotos/menschen-die-im-vergnugungspark-fahren-zyU2gQ9mWLM).

Comments
0
comment
No comments here
Why not start the discussion?