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Cycloid is a type of trajectory which is traced by a point on the circumference of a planar circle rolling without slipping on a surface. It turns out that this is the solution to the Brachistochrone problem, that is the curve along which a material particle may slide without friction in the least time, under the acceleration due to gravity or any uniform force field (such as a charged particle in an electric field). Besides, the path that light takes through an optically non-homogeneous medium is also a cycloid. This intriguing connection of the resulting trajectory being the same for these phenomena has been explored and is explained beautifully by Steven Strogatz in this video.

However, the motion of a charged particle in a uniform crossed electro-magnetic (EM) field is also a cycloid. Is there an intuitive explanation (perhaps from a Hamiltonian perspective) why this correspondence exists? It is tempting to guess the Hamiltonian perspective to explain this equivalence of the resulting dynamics since all these phenomena must follow from minimizing an action. Additionally, please feel free to contribute any ideas or cite any references addressing any relationship with the Fermat's principle of least action of light (cycloid in non-homogeneous medium) which in effect is also a crossed EM field and the trajectory of a charged particle in a uniform crossed EM field?

kbakshi314
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    The principle of least action applies to all action in physics. The example you give is not about light but about a particle moving under a constrained EM field. So, the equations of motion would come from that same place as for gravity or any other problem, Newton's Law (for non-relativistic motion). –  Feb 28 '20 at 13:17
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    Also, I am confused about your use of the term cycloid. I believe this is a specific curve traced out by a point on a moving wheel and NOT the solution of a particle moving under the influence of gravity. Can you clarify this in your question? What specifically are you asking about? Thanks. –  Feb 28 '20 at 13:18
  • I have edited the question to improve on the lack of clarity in the original question by providing links to relevant sources and background material. Please let me know in the comments section if the edited version is unclear. – kbakshi314 Feb 28 '20 at 20:14
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    Thank you, that is much better. In physics all phenomenon is derived from a principle of least action, from a Lagrangian. In all these cases I'd suspect that the functional form of the Lagrangian is identical hence the similar solutions. But that is a mathematical explanation and perhaps not very physically intuitive. –  Feb 28 '20 at 20:30
  • Although I agree with your comment that the connection between various phenomena exhibiting identical equations of motion which lead to the cycloidal trajectory, I was expecting a more direct answer. Note, that the corresponding mechanics-optics connection was shown using pure geometry. Please let me know if you have any concrete suggestions about the commonality of the Lagrangian for these phenomena or any technical literature. A professor once mentioned to me to explore Maxwell's laws for this question, but I do not possess the background to get my hands dirty based on just that lead. – kbakshi314 Mar 01 '20 at 06:49
  • Related to Brachistochrone-Cycloid: (1)What is the position as a function of time for a mass falling down a cycloid curve?.(2)How to determine radius of curvature of cycloid using centripetal acceleration?...and not only. – Frobenius Mar 01 '20 at 07:45
  • @ggcg The Brachistochrone curve is a Cycloid curve. – Frobenius Mar 01 '20 at 07:50
  • Thanks for the comments. I understand that the Brachistochrone is a cycloid. I encourage watching this video which explains the intuition (as opposed to the mathematical deductive reasoning) behind the fact that the Brachistochrone is a cycloid. At face value, these appear to be disparate problems, which on closer examination via geometry reveals that they (Brachistochrone-rolling without slipping) in fact enjoy a special connection. My question is on this exact connection except for the Brachistochrone-crossed EM case. – kbakshi314 Mar 01 '20 at 08:14
  • @kb314 I don't yet know any way of using minimization to prove this, but as you asked for ideas :- The way I derive this result is by changing the frame of reference from the ground frame to a frame with velocity $E/B$ in the direction of $\mathbf{E} × \mathbf{B}$. In this frame, after applying the Joules-Bernoulli Equations, we get the new value of fields. Due to the new values, the $\mathbf{E}=0$, and $\mathbf{B}$ stays the same..... –  Mar 05 '20 at 07:48
  • ....This scenario corresponds to uniform circular motion. Now if we go back to the ground frame, we obtain a cycloid. Of course, in this method I have considered that $E/B$ is negligible compared to the speed of light. –  Mar 05 '20 at 07:48
  • Thanks a lot @FakeMod! I was not aware of the Joules-Bernoulli equations. Although the fact that this field leads to a cycloidal charged particle trajectory does not require any assumptions on $E/B$, this might be a great lead. If you can add this comment as the answer (hopefully showing more work), I will accept it. – kbakshi314 Mar 05 '20 at 10:32
  • @kb314 I would love to, but I am a bit busy. If I ever get some time, I will come back to write the answer. I don't want to write the answer in any hurry, because the solution is elegant, and as you like 3B1B, so I want to write a 3Blue1Brown-ish answer which will definitely require a lot of time. :) –  Mar 05 '20 at 11:32
  • @FakeMod, take your time. I have known this problem for 12 years now, so there is no hurry! And I will read the solution at leisure if it is elegant! – kbakshi314 Mar 05 '20 at 12:01
  • @kb314 Till then, you can see the solution here. (However it's not exactly the way I do it, but it's 90% the same) –  Mar 05 '20 at 12:42

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I’m sorry to disappoint you, but in general the motion of the charged particle is not a cycloid. Dare I say it? In the textbook page you linked, Feynman was wrong! The path in Fig. 29-20 does not have cusps, so it’s not a cycloid. Of course, Feynman wasn’t wrong about the physics, just the definition of a cycloid.

We can follow the derivation provided in one of the comments above to see the details of the motion (except that I'll follow the motion of an electron instead of a positively charged particle). In the coordinates where the electric field is zero, the electron goes in a circle with a radius $\rho$ (the Larmor radius) and angular velocity $\Omega$ (the cyclotron frequency). If there is no initial motion in the $z$ direction, then \begin{align} x’ &= \rho\cos\Omega t\\ y' &= -\rho\sin\Omega t. \end{align} Back in the original coordinates, we need to add the drift velocity, which for $\mathbf{B}$ in the $z$ direction is $(E/B) \hat{\mathbf{y}}$, and we have \begin{align} x &= \rho\cos\Omega t\\ y &= \frac{E}{B}t - \rho\sin\Omega t. \end{align}

The path described by these equations is a trochoid, which is a generalization of a cycloid. The image (from Wolfram MathWorld) shows that troichoids can be traced out by an arm added to a rolling circle.

Illustrations of trocoid

If the relationship between the linear motion and the rotation is just right (as in the middle image), you will get a cycloid, but that's a special case.

As for light following a cycloid path in a non-homogeneous medium, that is also not true in general. If you look closely at the lecture you cited, you'll see that Fermat's principle is specifically applied to a velocity changing as $v = \sqrt{2 g y}$. It would be remarkable if an optical medium happened to have this property, but that's not the intent of the author. Fermat's principle is actually being applied to the heavy particle rolling down a track.

In summary, there is no profound principle at work here. Just geometry.

A. Newell
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  • (1/2) Thanks for the answer. I will accept it unless there is a better addition. I am aware that the trajectory of the particle we're discussing could be a prolate or curtate cycloid, depending on the ratio $\frac{E}{B}$. Additionally, thanks for clarifying the related Bernoulli's argument about the path of light in a non-homogeneous medium. Note that the minimum principle results in the cycloid $y(1+{y'}^2)=const.$ on minimizing the action (time of motion) as $\underset{y}{\min} \int_{x_0}^{x_1} \sqrt{\frac{1+{y'}^2}{2gy}} dx$. – kbakshi314 Mar 08 '20 at 06:11
  • (2/2) The question then pertains to whether we can write an action functional from which the cycloidal (whether curtate or prolate) path can be derived for the charged particle on application of the minimum principle, and if so, it's similarity with the action in the Brachistochrone problem (for the special case of $\frac{E}{B}$) which results in an exact cycloidal trajectory). Finally, it can indeed turn out that the two trajectories are identical purely due to aspects of geometry, but a variational understanding, if it exists, could lend some insight into this interesting co-incidence. – kbakshi314 Mar 08 '20 at 06:13
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    It seems that Feynman was right after all and I should have looked harder for extended definitions of cycloid. As far as I am concerned, though, the big surprise is that the solution to the Brachistochrone problem always gives you a cycloid, regardless of your starting and ending points (as long as you end lower than you started). Perhaps there's some deeper significance to that. By contrast, the E x B drift is a very specific geometry, and the construction method for trochoids is an obvious imitation of the motion. – A. Newell Mar 08 '20 at 20:16
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    @A. Newell why is the drift velocity equal to $\frac EB$? – Shub Aug 11 '22 at 16:56