I always feel amazed about how 2D physics can often be fascinating due to theorems in complex analysis. This article is about one among such cases.

Theorem. The conformal map w ⁣(z)\fc wz transforms the trajectory with energy B-B in potential U ⁣(z)Adw/dz2\fc Uz\ceq A\v{\d w/\d z}^2 into the trajectory with energy A-A in potential V ⁣(w)Bdz/dw2\fc Vw\ceq B\v{\d z/\d w}^2.

This result is pretty amazing in that it reveals a quite implicit duality between the two potentials, and it looks very symmetric as written.

This theorem, as I know of, was first introduced in the appendix of V. I. Arnold’s book Huygens and Barrow, Newton and Hooke. Part of this article is already covered in the relevant part of the book.

Power-law central-force potentials

Before I show the proof of it, let me first introduce it by a much more well-known example.

As we all know, Bertrand’s theorem states that the only two types of central-force potentials where all bound orbits are closed are Ur1U\propto r^{-1} (the Kepler problem) and Ur2U\propto r^2 (the harmonic oscillator). How the two potentials are special among all sorts of different central-force potentials makes people wonder if there is any connection between them. Fortunately, there is one, and it is obvious once we notice that the complex squaring transforms any center-at-origin ellipses into focus-at-origin ellipses. Inspired by this, it is easy to see that trajectories in the Kepler problem can be transformed into trajectories of harmonic oscillators under complex squaring.

You may ask, how can we notice complex squaring does the said transformation on ellipses? The observation is noticing the simple algebra


which means that the Joukowski transform zz+1/zz\mapsto z+1/z of a unit circle simply translates under complex squaring. We can then try to generalize this to circles of other radii, whose Joukowski transformations are just ellipses! (If you remember, this is the second time Joukowski transformation appears in my blog. The first time was here.)

Then, are the Kepler problem and the harmonic oscillator the only two central-force potentials whose trajectories can be transformed into each other by a complex function? The answer is no. In fact, for any trajectory in almost any power-law central-force potential, we can take some power of it to get a trajectory in another power-law central-force potential.

This result can be summarized as follows. Taking the (α/2+1)\p{\alp/2+1}th power of a trajectory with energy EE in the potential U=arαU=ar^\alp (α2\alp\ne-2) gives a trajectory with energy FF in the potential V=brβV=br^\beta, where

(α+2)(β+2)=4,b=14(α+2)2E,F=14(α+2)2a.\p{\alp+2}\p{\beta+2}=4,\quad b=-\fr14\p{\alp+2}^2E,\quad F=-\fr14\p{\alp+2}^2a.

To prove this, we just need to reparameterize the transformed trajectory in a new time coordinate τ\tau defined as dτ=zαdt\d\tau=\v z^\alp\,\d t, where zz is the complex position of the original trajectory. Then, by some calculation and utilizing the energy conservation, we can show that the parameter equation in terms of the new time coordinate satisfy the equation of motion we expect. I will not show the details here because they would be redundant once I prove the more general case using the same methods.

Corollaries and applications

There is an interesting special case, which is α=2\alp=-2. There is no potential that is dual to Ur2U\propto r^{-2}. Another interesting case is α=4\alp=-4, which is dual to itself (β=4\beta=-4). It kind of means that the coefficient in the potential is “interchangeable” with the energy, and the trajectories can be derived from each other by taking the complex reciprocal.

We can get some interesting results with a=0a=0, which is just the case of a free particle, whose trajectories are all straight lines. Since in this case we necessary have F=0F=0, we can say that the zero-energy trajectory in any power-law potential is related to a straight line by a power. From this result, we can derive some interesting corollaries. For example, the zero-energy trajectory in the Kepler problem is a parabola (square of a straight line), which is well-known. The zero-energy trajectory in Ur4U\propto-r^{-4} is a circle passing through the origin (reciprocal of a straight line), which is a pretty interesting not-so-well-known result.

Another interesting result is that, the deflection angle of an incident zero-energy particle scattered by the potential UrαU\propto-r^\alp is θ\tht under paraxial limit, if

α=2φπφ,φ=±θ2kπ,kN.\alp=\fr{2\vphi}{\pi-\vphi},\quad\vphi=\pm\tht-2k\pi,\quad k\in\bN.

This result can be easily derived by using the conformal transform of the real line (actually, a straight line that approaches the real line). The crucial part here is that kk cannot take negative integers because we need α>2\alp>-2. The reason is that, when α2\alp\le-2, paraxial zero-energy particles are bound to sink into the origin, and thus no scattering actually happens. This small pitfall indicates that the trajectory in the dual potential is not a two-side infinite straight line, either, in that limit, in contrast to being seemingly a free particle.

Some straightforward proofs

Let’s go back to the theorem I stated at the beginning of this article.

Proof. Consider a new time coordinate τ\tau defined as dτ=dw/dz2dt\d\tau=\v{\d w/\d z}^2\,\d t. Then, the motion of ww satisfies

md2wdτ2=mdtdτddt(dtdτdwdt)=mdzdw2ddt(dzdw2dwdzdzdt)=mdzdw(dzdw)((d2zdw2dwdzdzdt)dzdt+(dzdw)d2zdt2).\begin{align*} m\fr{\d^2w}{\d\tau^2} &=m\fr{\d t}{\d\tau}\fr{\d}{\d t}\p{\fr{\d t}{\d\tau}\fr{\d w}{\d t}}\\ &=m\v{\fr{\d z}{\d w}}^2\fr{\d}{\d t}\p{\v{\fr{\d z}{\d w}}^2\fr{\d w}{\d z}\fr{\d z}{\d t}}\\ &=m\fr{\d z}{\d w}\p{\fr{\d z}{\d w}}^*\p{\p{\fr{\d^2z}{\d w^2}\fr{\d w}{\d z}\fr{\d z}{\d t}}^*\fr{\d z}{\d t} +\p{\fr{\d z}{\d w}}^*\fr{\d^2 z}{\d t^2}}. \end{align*}

Here we need to substitute d2z/dt2\d^2 z/\d t^2 by the equation of motion for zz. By computing the real and imaginary parts separately, we can derive that for any holomorphic function ff, the gradient of f2\v f^2 expressed as a complex number is f2=2(df/dz)f\nabla\v f^2=2\p{\d f/\d z}^*f. Therefore, the equation of motion for zz is

md2zdt2=2Adwdz(d2wdz2).m\fr{\d^2z}{\d t^2}=-2A\fr{\d w}{\d z}\p{\fr{\d^2w}{\d z^2}}^*.

According to series reversion, we have d2w/dz2=(dw/dz)3d2z/dw2\d^2 w/\d z^2=-\p{\d w/\d z}^3\d^2 z/\d w^2. Therefore, the equation of motion for zz can also be written as

md2zdt2=2Adwdz2(dwdz)2(d2zdw2).m\fr{\d^2z}{\d t^2}=2A\v{\fr{\d w}{\d z}}^2\p{\fr{\d w}{\d z}}^{*2}\p{\fr{\d^2 z}{\d w^2}}^*.

Substitute this, and we have

md2wdτ2=dzdw(d2zdw2)(mdzdt2+2Adwdz2).m\fr{\d^2w}{\d\tau^2}=\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^* \p{m\v{\fr{\d z}{\d t}}^2+2A\v{\fr{\d w}{\d z}}^2}.

Substitute the energy conservation of the motion of zz:

12mdzdt2+Adwdz2=B,\fr12m\v{\fr{\d z}{\d t}}^2+A\v{\fr{\d w}{\d z}}^2=-B,

and we have

md2wdτ2=2Bdzdw(d2zdw2),m\fr{\d^2w}{\d\tau^2}=-2B\fr{\d z}{\d w}\p{\fr{\d^2z}{\d w^2}}^*,

which is the equation of motion for ww that we expect.

To get the energy of the motion of ww, we calculate

12mdwdτ2+Bdzdw2=12mdwdzdzdtdtdτ2+Bdzdw2=dwdz2(BAdwdz2)dzdw4+Bdzdw2=A,\begin{align*} \fr12m\v{\fr{\d w}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2 &=\fr12m\v{\fr{\d w}{\d z}\fr{\d z}{\d t}\fr{\d t}{\d\tau}}^2+B\v{\fr{\d z}{\d w}}^2\\ &=\v{\fr{\d w}{\d z}}^2\p{-B-A\v{\fr{\d w}{\d z}}^2}\v{\fr{\d z}{\d w}}^4+B\v{\fr{\d z}{\d w}}^2\\ &=-A, \end{align*}

which is the energy conservation of the motion of ww in the potential VV that we expect. \square

Noticing that we are only interested in the trajectory, we can just use Maupertuis’ principle to get a simpler proof.


S0=dz2m(BAdwdz2)=dw2m(ABdzdw2).\mcal S_0=\int\v{\d z}\sqrt{2m\p{-B-A\v{\fr{\d w}{\d z}}^2}}=\int\v{\d w}\sqrt{2m\p{-A-B\v{\fr{\d z}{\d w}}^2}}.

The abbreviated action is then exactly the same for the motion of zz and the motion of ww. Therefore, by Maupertuis’ principle, for any physical trajectory of zz, the trajectory of ww is also physical. \square

Details worth noting

Invertibility of the conformal map

There are two different definitions of a conformal transformation in two dimensions. One is that a function defined on an open subset of C\bC is conformal iff it is holomorphic and its derivative is nowhere zero. The other is that a function is conformal iff it is biholomorphic (is bijective and has a holomorphic inverse).

You may think here I have adopted the second definition because when I say V ⁣(w)Bdz/dw2\fc Vw\ceq B\v{\d z/\d w}^2, I am implicitly assuming that I can take the inverse of w ⁣(z)\fc wz to get the function z ⁣(w)\fc zw and then take the derivative of it. However, if that is the case, an immediate problem is that then the duality between the Kepler problem and the harmonic oscillator, from which I introduced the more general result in the first place, would not be actually covered by the “more general” result. This is because zz2z\mapsto z^2 is not biholomorphic (because it is not injective).

Then, why did this never become a problem when we were studying the duality between the Kepler problem and the harmonic oscillator? All we have talked about is how we can derive a trajectory in the Kepler problem by squaring the trajectory of a harmonic oscillator, but we have not discussed about how we can reverse this process, as an essential part of the duality. You may think the reverse of the process would be totally natural given how symmetric our theorem is regarding the two potentials. However, the reverse is not actually well-defined since the inverse of squaring, i.e., taking the square root, is not a single-valued function. Nevertheless, it is still well-defined in some sense: starting with whichever branch we like, tracing one point on the trajectory of the Kepler problem, and moving it along this trajectory for two cycles, we will end up with a trajectory of the harmonic oscillator if we take the square root of the position and ensure we always choose the branch so that the mapping is continuously done.

What about other power-law central potentials? In those cases, we have non-closed trajectories, so we cannot just move along the trajectory for two cycles. For example, if we take w=z3w=z^3, then the potential would be U=9Az4U=9A\v z^4. For any non-closed trajectory, we can uniquely map it to a trajectory of the potential V=Bw4/3/9V=B\v w^{-4/3}/9. However, we cannot uniquely do the reverse mapping. There would be three different trajectories in the potential UU that can be mapped to the same trajectory in VV, and we can in turn map the trajectory in VV to any of the three trajectories in UU depending on which branch we choose.

Therefore, to generalize this for more general potentials, we can use similar arguments. Because zwz\mapsto w has non-zero derivative everywhere in our considered region, it is everywhere locally invertible by the Lagrange inversion theorem. We can then bijectively map the trajectories in the two dual potentials locally for every small (and finite) segment and then patch them together to get the global correspondence between the two trajectories. This mapping may not be well-defined globally, but the trajectories can still be considered dual to each other. If the potential also becomes multi-valued due to the mapping wzw\mapsto z being multi-valued, then we should imagine this situation like this: at some point, the potential may be different when the particle visit here for the second time. This case does not happen if we only look at power-law potentials, but it does happen for more general cases.

What makes this sense of duality weaker is that one trajectory can be dual to multiple different trajectories. A case worth noting is that sometimes one trajectory can be mapped to infinitely many different trajectories. This happens when the trajectory runs around a logarithmic branch point. However, we can gain the sense of duality back if we can also consider the case where zwz\mapsto w is multi-valued. The notion of conformal transformation is now too limited to cover this case, a better notion is a global analytic function, which generalizes the notion of analytic function to allow for multiple branches.

Requirements for the potential

Not any potential can be expressed as Adw/dz2A\v{\d w/\d z}^2. How can we determine whether a potential can be expressed in this form?

Theorem. A continuous potential UU can be expressed in the form of Adw/dz2A\v{\d w/\d z}^2 (where w ⁣(z)\fc wz is a conformal transformation) iff one of the following conditions is met:

  • UU is zero everywhere, or
  • lnU\ln\v U is a harmonic function on the domain of UU.

Proof. First, prove the necessity.

An obvious requirement is that the potential must be positive everywhere or negative everywhere (or zero everywhere, but that is trivial). The sign is determined by the sign of AA. Therefore, without loss of generality, we can assume A=1A=1 because we can always absorb a factor of A\sqrt{\v A} into ww and adjust the overall sign of UU accordingly.

We can decompose (dw/dz)2\p{\d w/\d z}^2 in the polar form

(dw/dz)2=dw/dz2eiφ=Ueiφ,\p{\d w/\d z}^2=\v{\d w/\d z}^2\e^{\i\vphi}=U\e^{\i\vphi},

where φ\vphi is a real function of zz. Applying the Cauchy–Riemann equations to (dw/dz)2\p{\d w/\d z}^2 gives

ix(dwdz)2=y(dwdz)2    i(eiφxU+iUeiφxφ)=eiφyU+iUeiφyφ.\i\partial_x\p{\fr{\d w}{\d z}}^2=\partial_y\p{\fr{\d w}{\d z}}^2 \implies\i\p{\e^{\i\vphi}\partial_xU+\i U\e^{\i\vphi}\partial_x\vphi} =\e^{\i\vphi}\partial_yU+\i U\e^{\i\vphi}\partial_y\vphi.

Equate the real and imaginary parts, and we have


Use the symmetry of second derivatives on φ\vphi, and we have

xyφyxφ=0    xxUU+yyUU=0.\partial_x\partial_y\vphi-\partial_y\partial_x\vphi=0 \implies\partial_x\fr{\partial_xU}U+\partial_y\fr{\partial_yU}U=0.

In the language of vector analysis, this is just 2lnU=0\nabla^2\ln U=0.

Considering the case where UU is negative everywhere, we have that lnU\ln\v U is a harmonic function.

Then, prove the sufficiency.

The case where UU is zero everywhere is trivial. Otherwise, because lnU\ln\v U is defined everywhere on the domain of UU, we must have UU is non-zero everywhere. Because UU is continuous, we have UU is either positive everywhere or negative everywhere.

Without loss of generality, assume UU is positive everywhere. Let φ\vphi be the harmonic conjugate of lnU\ln U. Then, lnU+iφ\ln U+\i\vphi is a holomorphic function. We can then define

dwdz=Ueiφ/2,\fr{\d w}{\d z}=\sqrt U\e^{\i\vphi/2},

which is also a holomorphic function. \square

From now on, we will call this requirement on UU as being log-harmonic for obvious resons.

We should notice that whether UU is log-harmonic does not respect that any potential can have an additive constant and still be essentially the same potential. An immediate example is that a function that is positive everywhere may be negative somewhere if we add a constant to it. We may then want to ask whether UU can be log-harmonic if we allow it to be added an additive constant. It is easy to do this: we can just apply the same test to U+CU+C, and see if there is some CC that makes it work. To illustrate, solve the equation 2lnU+C=0\nabla^2\ln\v{U+C}=0 for CC, and then see whether it is a constant over the whole complex plane.

A property of log-harmonic functions is that the product of two log-harmonic functions is also log-harmonic.

Trajectories that run out of the domain

Trajectories often run out of the domain of the potential. For example, in the discussions about power-law potentials before, though not emphasized, the origin is outside the domain of the potential because it is either a pole or a zero of dw/dz\d w/\d z (except the trivial case where ww is simply proportional to zz). Another example that is rather overlooked is that unbound trajectories go to infinity while infinity is often not in the domain of the potential, either.

What need to take care of is that, when the trajectories run out of the domain, the trajectory is cut off there, and the rest of the trajectory is never considered (even if it may come back to the domain again later). Take the Kepler problem ane the harmonic oscillator as an example. If a trajectory of the harmonic oscillator passes through the origin, which is outside the domain, the trajectory degrades from a closed ellipse to a segment. If you take the square of a segment passing through the origin, you will get a broken line folded into itself, which looks like a particle in the Coulomb field may sink into the origin and then goes back along the exact path it came along. This would confusing if it were physical.

Arbitrariness in the construction of the conformal map

The construction of zwz\mapsto w is not unique for a given UU.

Rotation and translation

First, we can observe that the substitution wwweiθ+w0w\to w'\ceq w\e^{\i\tht}+w_0 does not change dw/dz\v{\d w/\d z} (nor thus UU). The real number θ\tht is a function of zz in principle, but if we want ww to be holomorphic on a connected region, then θ\tht must be a constant (except the trivial case where w=0w=0).

The dual trajectory does change, though, but the dual potential VV is also changed, too. Because dz/dw=dz/dw\v{\d z/\d w'}=\v{\d z/\d w}, we have

V ⁣(w)=V ⁣(w)=V ⁣((ww0)eiθ).\fc{V'}{w'}=\fc Vw=\fc V{\p{w'-w_0}\e^{-\i\tht}}.

Therefore, the dual trajectory and the dual potential are also rotated and translated by the same amount.


Before introducing scaling, I need to add some words about the unit systems. In the above discussions, I have never mentioned what units or dimensions do z,w,A,Bz,w,A,B have. The natural way of thinking is to let z,wz,w have the dimension of length and let A,BA,B have the dimension of energy. However, this is not the only way of thinking. We will later see that the zz-space and the ww-space can have totally different dimensions.

The dimensions or units of variables in a physical formula can be totally different from what they were originally intended to be. For example, when a particle is rotating, its motion needs to satisfy r˙=ω×r\dot{\mbf r}=\bs\omg\times\mbf r, where ω\bs\omg is the angular velocity. However, although r\mbf r has the dimension of length when it is first introduced, this formula is satisfied by any rotating vectors. A typical example is that the angular momentum changes according to this formula when a rigid body is doing precession. For another example, in classical mechanics and general relativity, the coordinates used to describe the motion of a particle are often not in the dimension of length, but have all sorts of dimensions. For another example that is less well-known, just because the Berry connection has the same gauge transformation as the electromagnetic potential, a bunch of formulas that are useful in electromagnetic theory can be applied to the Berry connection to define all sorts of interesting quantities with rich physical implications. The units of Berry connection are, however, very unimportant because they are literally arbitrary.

Therefore, what does a unit system actually bring us in a physical theory? The only thing it brings us is the ability to conveniently see in what aspects our theories are invariant under the scaling of some quantities. For example, in classical mechanics, we can scale the mass and the potential of any system with the same factor, and then the system will still behave the same in terms of the time-dependent length-based motion. This is because the part of the dimension of energy that is independent of length and time is to the first power of the dimension of mass. For similar reasons, we can derive another two scaling invariances, one about length-scaling and the other about time-scaling. In quantum mechanics, we suffer one less such scaling invariances because of the existence of \hbar; in special relativity, we suffer one less such scaling invariances because of the existence of cc; and in general relativity, we suffer two less such scaling invariances because of the existence of GG and cc. This is the incentive of introducing natural units in physics: they give us a more clear image of how our theory can be scaled leaving the physics invariant.

As for dimensional analysis, the essence of it is to find the required form of theory so that it satisfies some sort of scaling invariance. For example, we can use dimensional analysis to derive that the frequency of a harmonic oscillator is proportional to the square root of the ratio of the stiffness to the mass. We know this must be correct because this is the only theory that is consistent with the three scaling invariances that must be satisfied by any theories under the framework of classical mechanics.

Now, consider the scaling in ww, i.e., www/Cw\to w'\ceq w/C for some non-zero real number CC. The potential UU can be kept invariant by scaling AAC2AA\to A'\ceq C^2A. However, we cannot change BB if we want to leave the trajectory of zz unchanged because it is determined by the energy of the trajectory of zz. Therefore, the dual potential VV would be scaled to

V ⁣(w)=C2V ⁣(w)=C2V ⁣(Cw).\fc{V'}{w'}=C^2\fc Vw=C^2\fc V{Cw'}.

This means that physics is unchanged if length is scaled by CC and energy and potential are both scaled by C2C^2. This corresponds to one of the three scaling invariances in classical mechanics that we talked about before.

What is interesting here is that the length-scaling in the ww-space is done independently of that in the zz-space. This means that the length dimension in the two systems are independent of each other, so the two systems can have totally different unit systems.

Canonical transformation of time

The transformation from zz to ww seems like a coordinate transformation, which is covered by canonical transformations. However, here we have an additional requirement about the form of the Hamiltonian:

H=pz22m+U ⁣(z),K=pw22m+V ⁣(w),H=\fr{p_z^2}{2m}+\fc Uz,\quad K=\fr{p_w^2}{2m}+\fc Vw,

where KK is the transformed Hamiltonian (or called the Kamiltonian in the jargon of canonical transformations). This is not generally true because the transformation in the generalized momentum is restrictively determined when the transformation in the generalized coordinate is already given. From the proof of the original theorem, we can see that a transformation in time is a must, which is given by dτ=dw/dz2dt\d\tau=\v{\d w/\d z}^2\,\d t.

The problem is that the canonical transformations covered in most textbooks usually do not allow for a transformation in time, but only for a transformation in the canonical variables. Therefore, I need to first address the problem of integrating the transformation of time into the theory of canonical transformations. I will not do this for the most general case, but only for the case general enough for the purpose of explaining the case interesting this article.

Change in the time variable in the stationary-action principle

Before diving into the general canonical transformation, let’s first consider the case where the transformation is only in the time variable.

Consider a system with the Lagrangian L ⁣(q,q˙)\fc L{q,\dot q} (not explicitly dependent on time). Then, the action can be expressed as

S=t1t2L ⁣(q,q˙)dt.S=\int_{t_1}^{t_2}\fc L{q,\dot q}\d t.

The same integral can be expressed in terms of a new time variable τ\tau as

S=τ1τ2L ⁣(q,q˚τ˙)dττ˙,S=\int_{\tau_1}^{\tau_2}\fc L{q,\mathring q\dot\tau}\fr{\d\tau}{\dot\tau},

where q˚dq/dτ\mathring q\ceq\d q/\d\tau is the generalized velocity in the new time variable. The transformed Lagrangian, or what I want to call the Magrangian1, is then

M ⁣(q,q˚)L ⁣(q,q˚τ˙)1τ˙.\fc M{q,\mathring q}\ceq\fc L{q,\mathring q\dot\tau}\fr1{\dot\tau}. (1)(1)

For the case that we are concerning, τ˙\dot\tau is a positive real function of qq but does not (explicitly) depend on tt. The limits τ1,τ2\tau_1,\tau_2 satisfy the condition

τ2τ1=t1t2τ˙ ⁣(q)dt.\tau_2-\tau_1=\int_{t_1}^{t_2}\fc{\dot\tau}q\,\d t.

This relation is crucial. When finding the variation δS\dlt S, we are fixing t1,t2t_1,t_2. However, we cannot fix both τ1,τ2\tau_1,\tau_2 because their difference is dependent on the path q ⁣(t)\fc qt. What we can do is to fix τ1\tau_1 and to let τ2\tau_2 have a variation given by

δτ2=t1t2τ˙ ⁣(q)δqdt=τ1τ2τ˙ ⁣(q)τ˙ ⁣(q)δqdτ,\dlt\tau_2=\int_{t_1}^{t_2}\fc{\dot\tau'}q\dlt q\,\d t =\int_{\tau_1}^{\tau_2}\fr{\fc{\dot\tau'}q}{\fc{\dot\tau}q}\dlt q\,\d\tau,

where τ˙\dot\tau' is the derivative (or gradient, in higher dimensions) of τ˙\dot\tau as a function of qq. As can be seen, only if τ˙\dot\tau is a constant (i.e., τ\tau is simply an affine transform of tt) does δτ2\dlt\tau_2 vanish for any δq\dlt q.

Using the well-known variation of the action when there is variation in the time coordinate, we have

δS=τ1τ2(MqddτMq˚)δqdtK ⁣(q ⁣(τ2),q˚ ⁣(τ2))δτ2,\dlt S=\int_{\tau_1}^{\tau_2} \p{\fr{\partial M}{\partial q}-\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q}}\dlt q\,\d t -\fc{K}{\fc q{\tau_2},\fc{\mathring q}{\tau_2}}\dlt\tau_2,


K ⁣(q,q˚)q˚Mq˚M\fc K{q,\mathring q}\ceq\mathring q\fr{\partial M}{\partial\mathring q}-M

is the energy (or the Kamiltonian, but as a function of generalized coordinates and velocities) of the system.

A quick check of this variation

Because q ⁣(τ2)\fc q{\tau_2} is fixed, we have

q ⁣(τ2)=q ⁣(τ2+δτ2)+δq ⁣(τ2+δτ2)=q ⁣(τ2)+q˚ ⁣(τ2)δτ2+δq ⁣(τ2)    δq ⁣(τ2)=q˚ ⁣(τ2)δτ2.\fc q{\tau_2}=\fc q{\tau_2+\dlt\tau_2}+\fc{\dlt q}{\tau_2+\dlt\tau_2} =\fc q{\tau_2}+\fc{\mathring q}{\tau_2}\dlt\tau_2+\fc{\dlt q}{\tau_2} \implies\fc{\dlt q}{\tau_2}=-\fc{\mathring q}{\tau_2}\dlt\tau_2.

Now, calculate the variation of the action:

δS=τ1τ2(Mqδq+Mq˚δq˚)dτ+M ⁣(q ⁣(τ2),q˚ ⁣(τ2))δτ2.\dlt S=\int_{\tau_1}^{\tau_2} \p{\fr{\partial M}{\partial q}\dlt q+\fr{\partial M}{\partial\mathring q}\dlt\mathring q}\d\tau +\fc M{\fc q{\tau_2},\fc{\mathring q}{\tau_2}}\dlt\tau_2.

Recall the derivation of the Euler–Lagrange equation. For the second term in the integrand, we can integrate by parts to get

τ1τ2Mq˚δq˚dτ=Mq˚δqτ1τ2τ1τ2ddτMq˚δqdτ=Mq˚q˚τ2δτ2τ1τ2ddτMq˚δqdτ.\int_{\tau_1}^{\tau_2}\fr{\partial M}{\partial\mathring q}\dlt\mathring q\,\d\tau =\abar{\fr{\partial M}{\partial\mathring q}\dlt q}{\tau_1}^{\tau_2} -\int_{\tau_1}^{\tau_2}\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q}\dlt q\,\d\tau =\abar{-\fr{\partial M}{\partial\mathring q}\mathring q}{\tau_2}\dlt\tau_2 -\int_{\tau_1}^{\tau_2}\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q}\dlt q\,\d\tau.

Substitute this back into the expression for δS\dlt S, and we have the desired result.

If we let the first term in δS\dlt S vanish, we would get the well-known Euler–Lagrange equation:

MqddτMq˚=0.\fr{\partial M}{\partial q}-\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q}=0. (2)(2)

However, that term is not zero because there is another term in δS\dlt S. If we want the Euler–Lagrange equation to be satisfied, we need the second term to be zero. This means that either KK is zero or δτ2\dlt\tau_2 is zero. The latter case will lead us to the trivial case because we have just derived that δτ2\dlt\tau_2 is zero only if τ˙\dot\tau is a constant. The former case can be satisfied, however. If the Euler–Lagrange equation is satisfied, then KK is a conserved quantity due to the symmetry of MM in τ\tau-translation. Then, if KK happens to be zero at some point, it will be zero over the whole motion, and the stationary-action principle will be satisfied by the motion between any two points.

We can explicitly show that Equation 2 can be derived from the original Euler–Lagrange equation under the zero-energy condition.

Proof. We need to first derive the condition of zero energy in the old time variable. Take derivatives of Equation 1 with respect to q˚\mathring q, and we have

Mq˚=Lq˙τ˙1τ˙=Lq˙.\fr{\partial M}{\partial\mathring q}=\fr{\partial L}{\partial\dot q}\dot\tau\fr1{\dot\tau} =\fr{\partial L}{\partial\dot q}.

Therefore, the Kamiltonian is

K=Mq˚q˚M=Lq˙q˙τ˙Lτ˙=Hτ˙,K=\fr{\partial M}{\partial\mathring q}\mathring q-M=\fr{\partial L}{\partial\dot q}\fr{\dot q}{\dot\tau}-\fr L{\dot\tau} =\fr H{\dot\tau}, (3)(3)

where Hq˙L/q˙LH\ceq\dot q\partial L/\partial\dot q-L is the original Hamiltonian. This relation means that the condition K=0K=0 is equivalent to the condition H=0H=0.

Then, use Equation 1 to explicitly calculate the lhs of Equation 2:

MqddτMq˚=(Lq+Lq˙q˚τ˙ ⁣(q))1τ˙ ⁣(q)Lτ˙ ⁣(q)τ˙ ⁣(q)21τ˙ ⁣(q)ddtLq˙=(LqddtLq˙)1τ˙ ⁣(q)+(Lq˙q˙L)τ˙ ⁣(q)τ˙ ⁣(q)2=0.\begin{align*} \fr{\partial M}{\partial q}-\fr{\d}{\d\tau}\fr{\partial M}{\partial\mathring q} &=\p{\fr{\partial L}{\partial q}+\fr{\partial L}{\partial\dot q}\mathring q\fc{\dot\tau'}q}\fr1{\fc{\dot\tau}q} -L\fr{\fc{\dot\tau'}q}{\fc{\dot\tau}q^2}-\fr1{\fc{\dot\tau}q}\fr{\d}{\d t}\fr{\partial L}{\partial\dot q}\\ &=\p{\fr{\partial L}{\partial q}-\fr{\d}{\d t}\fr{\partial L}{\partial\dot q}}\fr1{\fc{\dot\tau}q} +\p{\fr{\partial L}{\partial\dot q}\dot q-L}\fr{\fc{\dot\tau'}q}{\fc{\dot\tau}q^2}\\ &=0. \end{align*} \square

Specifying τ\tau vs. specifying τ˙\dot\tau

We will see that specifying τ˙\dot\tau, which is what we have done in the above discussion, is pretty different from specifying τ\tau. The latter is much simpler, but the former is the one that is used for the conformal duality between potentials. Although I do not have to discuss what the transformation should look like when we specify τ\tau instead of τ˙\dot\tau, I will still do this because I need to point it out that it is quite different from the case we have discussed.

Recall that the canonical transformation is just a transformation of coordinates in the phase space that preserves the canonical one-form up to a total differential. Adding the idea of time transformation into this has a difficulty that time is not a coordinate in the phase space. Including the time coordinate, the actual one-form that needs to be preserved is

dS=pdqHdt,\d S=p\,\d q-H\,\d t,

which is exactly the total differential of the action. Therefore, we have

pdqHdt=PdQKdτ+dG,p\,\d q-H\,\d t=P\,\d Q-K\,\d\tau+\d G, (4)(4)

where P,QP,Q are the new canonical variables, KK is the transformed Hamiltonian, and GG is called the generating function of the canonical transformation. Assume τ\tau and GG are both functions of q,Q,tq,Q,t. Then, we have

pdqHdt=PdQK(τqdq+τQdQ+τtdt)+Gqdq+GQdQ+Gtdt.p\,\d q-H\,\d t=P\,\d Q -K\p{\fr{\partial\tau}{\partial q}\,\d q+\fr{\partial\tau}{\partial Q}\,\d Q+\fr{\partial\tau}{\partial t}\,\d t} +\fr{\partial G}{\partial q}\,\d q+\fr{\partial G}{\partial Q}\,\d Q+\fr{\partial G}{\partial t}\,\d t.

Compare the coefficients of dq,dQ,dt\d q,\d Q,\d t on both sides, and we have

p+KτqGq=0,PKτQ+GQ=0,HKτt+Gt=0.p+K\fr{\partial\tau}{\partial q}-\fr{\partial G}{\partial q}=0,\quad P-K\fr{\partial\tau}{\partial Q}+\fr{\partial G}{\partial Q}=0,\quad H-K\fr{\partial\tau}{\partial t}+\fr{\partial G}{\partial t}=0. (5)(5)

These equations determines Q,P,KQ,P,K. They will satisfy Hamilton’s equation:

dQdτ=KP,dPdτ=KQ.\fr{\d Q}{\d\tau}=\fr{\partial K}{\partial P},\quad \fr{\d P}{\d\tau}=-\fr{\partial K}{\partial Q}.

An example

Consider the Hamiltonian H=p+qH=p+q. The motion is

q=q0+t,p=p0t.q=q_0+t,\quad p=p_0-t.

Consider the new time variable τ=t/q\tau=t/q and the generating function G=QqG=Qq. With Equation 5 and the expression for HH and τ\tau, we have a set of five equations:

{pK1q2tQ=0,P+q=0,HK1q=0,τ=tq,H=p+q    {q=P,p=QPτ1τ,K=(PQ)P1τ,t=Pτ,H=QP1τ.\begin{dcases} p-K\fr1{q^2}t-Q=0,\\ P+q=0,\\ H-K\fr1q=0,\\ \tau=\fr tq,\\ H=p+q \end{dcases}\implies\begin{dcases} q=-P,\\ p=\fr{Q-P\tau}{1-\tau},\\ K=\fr{\p{P-Q}P}{1-\tau},\\ t=-P\tau,\\ H=\fr{Q-P}{1-\tau}. \end{dcases}

With the expression for the Kamiltonian KK, we get the motion of Q,PQ,P:

Q=(2τ)τ1τP0+(1τ)Q0,P=P01τ.Q=\fr{\p{2-\tau}\tau}{1-\tau}P_0+\p{1-\tau}Q_0,\quad P=\fr{P_0}{1-\tau}.

This is consistent with the motion of q,pq,p as can be verified with calculation.

It seems that specifying τ\tau is much easier than specifying τ˙\dot\tau. We can easily discuss the most general case and perfectly recover the equation of motion without having to impose a bizarre condition like the zero energy. This is because specifying τ˙\dot\tau is, in some sense, more general than specifying τ\tau: we can always find the total derivative of τ\tau for any form of it, but we cannot always find τ\tau given the form of τ˙\dot\tau because of limitations on the integrability.

The conformal transformation as a canonical transformation

Now, we can discuss the conformal transformation as a canonical transformation. The procedure is pretty analogous to that in the previous section, but this time the conclusion would only be valid under the zero-energy condition.

Denote the real and imaginary parts of zz as x,yx,y, and the real and imaginary parts of ww as X,YX,Y. The Cauchy–Riemann equations give

uXx=Yy,vXy=Yx.u\ceq\fr{\partial X}{\partial x}=\fr{\partial Y}{\partial y},\quad v\ceq\fr{\partial X}{\partial y}=-\fr{\partial Y}{\partial x}.

Here u,vu,v are two real functions defined for convenience. They can either be functions of x,yx,y or functions of X,YX,Y, depending on which are more convenient. With u,vu,v, we have

dX=udx+vdy,dY=vdx+udy,\d X=u\,\d x+v\,\d y,\quad\d Y=-v\,\d x+u\,\d y,

The time transformation is given by

τ˙=dwdz2=u2+v2.\dot\tau=\v{\fr{\d w}{\d z}}^2=u^2+v^2.

The original Hamiltonian is


(the last term is added because we want it to be zero during the motion). Substitute these into Equation 4, and we have (dG=0\d G=0)

pxdx+pydy(px2+py22m+A(u2+v2)+B)dt=PX(udx+vdy)+PY(vdx+udy)K(u2+v2)dt.\begin{align*} &p_x\,\d x+p_y\,\d y-\p{\fr{p_x^2+p_y^2}{2m}+A\p{u^2+v^2}+B}\d t\\ ={}&P_X\p{u\,\d x+v\,\d y}+P_Y\p{-v\,\d x+u\,\d y}-K\p{u^2+v^2}\d t. \end{align*}

Then, after some calculations, we have perfectly the expected result

px=uPXvPY,py=vPX+uPY,K=PX2+PY22m+Bu2+v2+A.p_x=uP_X-vP_Y,\quad p_y=vP_X+uP_Y,\quad K=\fr{P_X^2+P_Y^2}{2m}+\fr{B}{u^2+v^2}+A.

The condition K=0K=0 specifies the energy of the dual trajectory.

  1. For unknown reasons, the transformed Hamiltonian is called the Kamiltonian just because we often use the symbol KK to represent it. However, there is not a similar convention for the transformed Lagrangian, so I would like to use the letter MM and call it the Magrangian. The surname “Lagrange” is originated from the French phrase la grange (meaning “the barn”), and correspondingly “Magrange” may refer to the French phrase ma grange (meaning “my barn”). This pun then can make “Magrangian” kind of mean “my Lagrangian”.↩︎