This recent news article (here is the original, in German) says that

Shouryya Ray, who moved to Germany from India with his family at the age of 12, has baffled scientists and mathematicians by solving two fundamental particle dynamics problems posed by Sir Isaac Newton over 350 years ago, Die Welt newspaper reported on Monday.

Ray’s solutions make it possible to now calculate not only the flight path of a ball, but also predict how it will hit and bounce off a wall. Previously it had only been possible to estimate this using a computer, wrote the paper.

What are the problems from this description? What is their precise formulation? Also, is there anywhere I can read the details of this person's proposed solutions?


3 Answers 3


This thread(physicsforums.com) contains a link to Shouryya Ray's poster, in which he presents his results.

So the problem is to find the trajectory of a particle under influence of gravity and quadratic air resistance. The governing equations, as they appear on the poster:

$$ \dot u(t) + \alpha u(t) \sqrt{u(t)^2+v(t)^2} = 0 \\ \dot v(t) + \alpha v(t) \sqrt{u(t)^2 + v(t)^2} = -g\text, $$

subject to initial conditions $v(0) = v_0 > 0$ and $u(0) = u_0 \neq 0$.

Thus (it is easily inferred), in his notation, $u(t)$ is the horizontal velocity, $v(t)$ is the vertical velocity, $g$ is the gravitational acceleration, and $\alpha$ is a drag coefficient.

He then writes down the solutions

$$ u(t) = \frac{u_0}{1 + \alpha V_0 t - \tfrac{1}{2!}\alpha gt^2 \sin \theta + \tfrac{1}{3!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right) t^3 + \cdots} \\ v(t) = \frac{v_0 - g\left[t + \tfrac{1}{2!} \alpha V_0 t^2 - \tfrac{1}{3!} \alpha gt^3 \sin \theta + \tfrac{1}{4!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right)t^4 + \cdots\right]}{1 + \alpha V_0 t - \tfrac{1}{2!}\alpha gt^2 \sin \theta + \tfrac{1}{3!}\left(\alpha g^2 V_0 \cos^2 \theta - \alpha^2 g V_0 \sin \theta\right) t^3 + \cdots}\text.$$

From the diagram below the photo of Newton, one sees that $V_0$ is the inital speed, and $\theta$ is the initial elevation angle.

The poster (or at least the part that is visible) does not give details on the derivation of the solution. But some things can be seen:

  • He uses, right in the beginning, the substitution $\psi(t) = u(t)/v(t)$.

  • There is a section called "...öße der Bewegung". The first word is obscured, but a qualified guess would be "Erhaltungsgröße der Bewegung", which would translate as "conserved quantity of the motion". Here, the conserved quantity described by David Zaslavsky appears, modulo some sign issues.

  • However, this section seems to be a subsection to the bigger section "Aus der Lösung ablesbare Eigenschaften", or "Properties that can seen from the solution". That seems to imply that the solution implies the conservation law, rather than the solution being derived from the conservation law. The text in that section probably provides some clue, but it's only partly visible, and, well, my German is rusty. I welcome someone else to try to make sense of it.

  • Also part of the bigger section are subsections where he derives from his solution (a) the trajectory for classical, drag-free projectiles, (b) some "Lamb-Näherung", or "Lamb approximation".

  • The next section is called "Verallgemeneirungen", or "Generalizations". Here, he seems to consider two other problems, with drag of the form $\alpha V^2 + \beta$, in the presence of altitude-dependent horizontal wind. I'm not sure what the results here are.

  • The diagrams to the left seem to demonstrate the accuracy and convergence of his series solution by comparing them to Runge-Kutta. Though the text is kind of blurry, and, again, my German is rusty, so I'm not too sure.

  • Here's a rough translation of the first part of the "Zusammanfassung und Ausblick" (Summary and outlook), with suitable disclaimers as to the accuracy:

  • For the first time, a fully analytical solution of a long unsolved problem
  • Various excellent properties; in particular, conserved quantity $\Rightarrow$ fundamental [...] extraction of deep new insights using the complete analytical solutions (above all [...] perspectives and approximations are to be gained)
  • Convergence of the solution numerically demonstrated
  • Solution sketch for two generalizations

EDIT: Two professors at TU Dresden, who have seen Mr Ray's work, have written some comments:

Comments on some recent work by Shouryya Ray

There, the questions he solved are unambiguously stated, so that should answer any outstanding questions.

EDIT2: I should add: I do not doubt that Shouryya Ray is a very intelligent young man. The solution he gave can, perhaps, be obtained using standard methods. I believe, however, that he discovered the solution without being aware that the methods were standard, a very remarkable achievement indeed. I hope that this event has not discouraged him; no doubt, he'll be a successful physicist or mathematician one day, should he choose that path.

  • 2
    $\begingroup$ Link to image of Shouryya Ray's poster is now dead. $\endgroup$
    – Qmechanic
    Commented Jul 9, 2012 at 20:00
  • $\begingroup$ There's a picture of S Ray with the poster here (Wayback Machine / stack.imgur), though it's probably not the high-res image referenced here. $\endgroup$ Commented Aug 25, 2019 at 16:25

It is indeed quite difficult to find information on why exactly this project has attracted so much attention. What I've pieced together from comments on various websites and some images (mainly this one) is that Shouryya Ray discovered the following constant of motion for projectile motion with quadratic drag:

$$\frac{g^2}{2v_x^2} + \frac{\alpha g}{2}\left(\frac{v_y\sqrt{v_x^2 + v_y^2}}{v_x^2} + \sinh^{-1}\biggl|\frac{v_y}{v_x}\biggr|\right) = \text{const.}$$

This applies to a particle which is subject to a quadratic drag force,

$$\vec{F}_d = -m\alpha v\vec{v}$$

It's easily verified that the constant is constant by taking the time derivative and plugging in the equations of motion

$$\begin{align}\frac{\mathrm{d}v_x}{\mathrm{d}t} &= -\alpha v_x\sqrt{v_x^2 + v_y^2} \\ \frac{\mathrm{d}v_y}{\mathrm{d}t} &= -\alpha v_y\sqrt{v_x^2 + v_y^2} - g\end{align}$$

The prevailing opinion is that this has not been known before, although some people are claiming to have seen it in old textbooks (never with a reference, though, so take it for what you will).

I haven't heard anything concrete about how this could be put to practical use, although perhaps that is part of the technical details of the project. It's already possible to calculate ballistic trajectories with drag to very high precision using numerical methods, and the presence of this constant doesn't directly lead to a new method of calculating trajectories as far as I can tell.

  • $\begingroup$ There is a discussion on Reddit about this subject, which describes the problem and a verification of the solution. See reddit.com/r/worldnews/comments/u7551/… $\endgroup$
    – jbatista
    Commented May 28, 2012 at 22:02
  • $\begingroup$ @jbatista yeah, that's one of the sources I was getting my information from. $\endgroup$
    – David Z
    Commented May 28, 2012 at 22:05
  • 8
    $\begingroup$ So it sounds like a very neat result, and definitely impressive for a high school student; but not exactly worth a "Kid out-thinks Newton" headline. Junky science journalism, as always. $\endgroup$
    – Colin K
    Commented May 28, 2012 at 22:19
  • 6
    $\begingroup$ The MathExchange cross-post math.stackexchange.com/q/150242 also copy+pastes the Reddit discussion; particularly salient is that it cites a result by G. W. Parker published in Am.J.Phys. 45 (1977) 606-610 discussing the same problem. This makes it even more interesting to find out about how Ray obtained his result. $\endgroup$
    – jbatista
    Commented May 28, 2012 at 22:19
  • $\begingroup$ Wayback Machine capture of the broken image link. $\endgroup$ Commented Aug 25, 2019 at 16:28

I) Here we would like to give a Hamiltonian formulation of a point particle in a constant gravitational field with quadratic air resistance

$$\tag{1} \dot{u}~=~ -\alpha u \sqrt{u^2+v^2}, \qquad \dot{v}~=~ -\alpha v \sqrt{u^2+v^2} -g. $$

The $u$ and $v$ are the horizontal and vertical velocity, respectively. A dot on top denotes differentiation with respect to time $t$. The two positive constants $\alpha>0$ and $g>0$ can be put to one by scaling the three variables

$$\tag{2} t'~=~\sqrt{\alpha g}t, \qquad u'~=~\sqrt{\frac{\alpha}{g}}u, \qquad v'~=~\sqrt{\frac{\alpha}{g}}v. $$

See e.g. Ref. [1] for a general introduction to Hamiltonian and Lagrangian formulations.

II) Define two canonical variables (generalized position and momentum) as

$$\tag{3} q~:=~ -\frac{v}{|u|}, \qquad p~:=~ \frac{1}{|u|}~>~0.$$

(The position $q$ is (up to signs) Shouryya Ray's $\psi$ variable, and the momentum $p$ is (up to a multiplicative factor) Shouryya Ray's $\dot{\Psi}$ variable. We assume$^\dagger$ for simplicity that $u\neq 0$.) Then the equations of motion (1) become

$$\tag{4a} \dot{q}~=~ gp, $$ $$\tag{4b} \dot{p}~=~ \alpha \sqrt{1+q^2}. $$

III) Equation (4a) suggests that we should identify $\frac{1}{g}$ with a mass

$$\tag{5} m~:=~ \frac{1}{g}, $$

so that we have the standard expression

$$\tag{6} p~=~m\dot{q}$$

for the momentum of a non-relativistic point particle. Let us furthermore define kinetic energy

$$\tag{7} T~:=~\frac{p^2}{2m}~=~ \frac{gp^2}{2}. $$

IV) Equation (4b) and Newton's second law suggest that we should define a modified Hooke's force

$$\tag{8} F(q)~:=~ \alpha \sqrt{1+q^2}~=~-V^{\prime}(q), $$

with potential given by (minus) the antiderivative

$$ V(q)~:=~ - \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$ $$\tag{9} ~=~ - \frac{\alpha}{2} \left(q \sqrt{1+q^2} + \ln(q+\sqrt{1+q^2})\right). $$

Note that this corresponds to an unstable situation because the force $F(-q)~=~F(q)$ is an even function, while the potential $V(-q) = - V(q)$ is a monotonic odd function of the position $q$.

It is tempting to define an angle variable $\theta$ as

$$\tag{10} q~=~\tan\theta, $$

so that the corresponding force and potential read

$$\tag{11} F~=~\frac{\alpha}{\cos\theta} , \qquad V~=~- \frac{\alpha}{2} \left(\frac{\sin\theta}{\cos^2\theta} + \ln\frac{1+\sin\theta}{\cos\theta}\right). $$

V) The Hamiltonian is the total mechanical energy

$$ H(q,p)~:=~T+V(q)~=~\frac{gp^2}{2}- \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$ $$\tag{12}~=~\frac{g}{2u^2} +\frac{\alpha}{2} \left( \frac{v\sqrt{u^2+v^2}}{u^2} + {\rm arsinh}\frac{v}{|u|}\right). $$

Since the Hamiltonian $H$ contains no explicit time dependence, the mechanical energy (12) is conserved in time, which is Shouryya Ray's first integral of motion.

$$\tag{13} \frac{dH}{dt}~=~ \frac{\partial H}{\partial t}~=~0. $$

VI) The Hamiltonian equations of motion are eqs. (4). Suppose that we know $q(t_i)$ and $p(t_i)$ at some initial instant $t_i$, and we would like to find $q(t_f)$ and $p(t_f)$ at some final instant $t_f$.

The Hamiltonian $H$ is the generator of time evolution. If we introduce the canonical equal-time Poisson bracket

$$\tag{14} \{q(t_i),p(t_i)\}~=~1,$$

then (minus) the Hamiltonian vector field reads

$$\tag{15} -X_H~:=~-\{H(q(t_i),p(t_i)), \cdot\} ~=~ gp(t_i)\frac{\partial}{\partial q(t_i)} + F(q(t_i))\frac{\partial}{\partial p(t_i)}. $$

For completeness, let us mention that in terms of the original velocity variables, the Poisson bracket reads

$$\tag{16} \{v(t_i),u(t_i)\}~=~u(t_i)^3.$$

We can write a formal solution to position, momentum, and force, as

$$ q(t_f) ~=~ e^{-\tau X_H}q(t_i) ~=~ q(t_i) - \tau X_H[q(t_i)] + \frac{\tau^2}{2}X_H[X_H[q(t_i)]]+\ldots \qquad $$ $$\tag{17a} ~=~ q(t_i) + \tau g p(t_i) + \frac{\tau^2}{2}g F(q(t_i)) +\frac{\tau^3}{6}g \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))} +\ldots ,\qquad $$ $$ p(t_f) ~=~ e^{-\tau X_H}p(t_i) ~=~ p(t_i) - \tau X_H[p(t_i)] + \frac{\tau^2}{2}X_H[X_H[p(t_i)]]+\ldots\qquad $$ $$ ~=~p(t_i) + \tau F(q(t_i)) +\frac{\tau^2}{2}\frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))}$$ $$\tag{17b} + \frac{g\alpha^2\tau^3}{6} \left(q(t_i) + \frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right) +\ldots ,\qquad $$ $$ F(q(t_f)) ~=~ e^{-\tau X_H}F(q(t_i)) $$ $$~=~ F(q(t_i)) - \tau X_H[F(q(t_i))] + \frac{\tau^2}{2}X_H[X_H[F(q(t_i))]] + \ldots\qquad $$ $$ \tag{17c}~=~ F(q(t_i)) + \tau \frac{g\alpha^2p(t_i)q(t_i)}{F(q(t_i))} +\frac{g(\alpha\tau)^2}{2}\left(q(t_i) +\frac{g\alpha^2 p(t_i)^2}{F(q(t_i))^3}\right) +\ldots ,\qquad $$

and calculate to any order in time $\tau:=t_f-t_i$, we would like. (As a check, note that if one differentiates (17a) with respect to time $\tau$, one gets (17b) multiplied by $g$, and if one differentiates (17b) with respect to time $\tau$, one gets (17c), cf. eq. (4).) In this way we can obtain a Taylor expansion in time $\tau$ of the form

$$ \tag{18} F(q(t_f)) ~=~\alpha\sum_{n,k,\ell\in \mathbb{N}_0} \frac{c_{n,k,\ell}}{n!}\left(\tau\sqrt{\alpha g}\right)^n \left(p(t_i)\sqrt{\frac{g}{\alpha}}\right)^k \frac{q(t_i)^{\ell}}{(F(q(t_i))/\alpha)^{k+\ell-1}}. $$

The dimensionless universal constants $c_{n,k,\ell}=0$ are zero if either $n+k$ or $\frac{n+k}{2}+\ell$ are not an even integer. We have a closed expression

$$ F(q(t_f)) ~\approx~ \exp\left[\tau gp(t_i)\frac{\partial}{\partial q(t_i)}\right]F(q(t_i)) ~=~ F(q(t_i)+\tau g p(t_i)) $$ $$ \tag{19} \qquad \text{for} \qquad ~ p(t_i)~\gg~\frac{ F(q(t_i))}{\sqrt{\alpha g}}, $$

i.e., when we can ignore the second term in the Hamiltonian vector field (15).

VII) The corresponding Lagrangian is

$$\tag{20} L(q,\dot{q})~=~T-V(q)~=~\frac{\dot{q}^2}{2g}+ \frac{\alpha}{2} \left(q \sqrt{1+q^2} + {\rm arsinh}(q)\right) $$

with Lagrangian equation of motion

$$\tag{21} \ddot{q}~=~ \alpha g \sqrt{1+q^2}. $$

This is essentially Shouryya Ray's $\psi$ equation.


  1. Herbert Goldstein, Classical Mechanics.

$^\dagger$ Note that if $u$ becomes zero at some point, it stays zero in the future, cf. eq.(1). If $u\equiv 0$ identically, then eq.(1) becomes

$$\tag{22} -\dot{v} ~=~ \alpha v |v| + g. $$

The solution to eq. (22) for negative $v\leq 0$ is

$$\tag{23} v(t) ~=~ -\sqrt{\frac{g}{\alpha}} \tanh(\sqrt{\alpha g}(t-t_0)) , \qquad t~\geq~ t_0, $$

where $t_0$ is an integration constant. In general,

$$\tag{24} (u(t),v(t)) ~\to~ (0, -\sqrt{\frac{g}{\alpha}}) \qquad \text{for} \qquad t ~\to~ \infty ,$$


$$\tag{25} (q(t),p(t)) ~\to~ (\infty,\infty) \qquad \text{for} \qquad t~\to~ \infty.$$

  • $\begingroup$ Interestingly, one may generalize to arbitrary power laws $\dot{u}= -\alpha u (u^2+v^2)^{\frac{r}{2}}$ and $\dot{v}= -\alpha v (u^2+v^2)^{\frac{r}{2}} -g$. Here $r$ is a real power with $r\neq 0,-1$. The canonical coordinates are then $q:=-\frac{v}{|u|}$ and $p:=|u|^{-r}$ with eoms $\dot{q}=\frac{g}{|u|}=gp^{\frac{1}{r}}$ and $\dot{p}=r\alpha(1+q^2)^{\frac{r}{2}}=:F(q)=-V^{\prime}(q)$. The potential $V(q)$ is (essentially) a hypergeometric function. The Hamiltonian reads $H(q,p):=\frac{g}{1+r^{-1}}p^{1+r^{-1}}+V(q)$. $\endgroup$
    – Qmechanic
    Commented Feb 23, 2014 at 19:02

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