science_blog_fun

Can Generative AI solve hard physics problems?

Short answer: very likely yes, it would crack.

Here is why, in a physically grounded way:

1. Impact speed from a third story

A third-story height is roughly 8–12 meters. Ignoring air resistance, the egg hits the water at about: [ v \approx \sqrt{2gh} \approx 12\text{–}15 ,\text{m/s} ] That’s around 25–35 mph.

2. Water behaves “hard” at high speed

Although water feels soft at low speeds, at high impact speeds it behaves much more like a solid surface. This is because:

This leads to a large deceleration force over a very short time.

3. Eggshell strength is limited

Eggshells are:

Hitting water at speed creates:

It’s possible (but unlikely) if:

Even then, it’s a borderline case.

Further modelling

Let us take a typical chicken egg:

Ignoring air resistance, the speed on impact is

[ v=\sqrt{2gh}\approx \sqrt{2(9.8)(9\text{–}12)} \approx 13\text{–}15\ \text{m/s}. ]

Now the key issue is: how quickly does the water stop the egg?

If the egg is brought to rest over a distance (d), then the average stopping force is roughly

[ F_{\text{avg}} \approx \frac{mv^2}{2d}. ]

Using (m=0.06\ \text{kg}) and (v \approx 14\ \text{m/s}):

[ F_{\text{avg}} \approx \frac{0.06\cdot 14^2}{2(0.01)} \approx 590\ \text{N} ]

[ F_{\text{avg}} \approx 295\ \text{N} ]

[ F_{\text{avg}} \approx 118\ \text{N} ]

A second estimate comes from drag:

[ F_d \approx \tfrac12 \rho C_d A v^2 ]

with water density (\rho \approx 1000\ \text{kg/m}^3), egg cross-sectional area (A \approx 1.5\times 10^{-3}\ \text{m}^2), and (C_d \sim 0.8\text{–}1). That gives

[ F_d \approx 100\text{–}180\ \text{N} ]

right at impact.

For comparison, a chicken egg typically fractures under loads of only tens of newtons in ordinary compression, though the exact number depends a lot on orientation and shell defects.

So the comparison is:

That means the impact force is very plausibly several times larger than the shell can tolerate.

So the physical answer is: yes, it would very likely crack.

The important subtlety is that there is no single “actual force” here — the force rises and falls during impact, so what matters is the peak force and the impact duration, not just one number. The numbers above show why water does not necessarily protect the egg much at that speed.

Plot

Analysis for sand

Forthcoming

Mathematics problem(s)

Forthcoming

Irodov problem(s)

Image analysis (forthcoming)

I asked ChatGPT to solve a problem from the book Irodov (Figure 1).

Solution by ChatGPT

Little projectile + moving-wall. I will assume:

Let (t) be the time from launch to impact. The collision position satisfies [ x_{\text{ball}}(t)=u_x t,\qquad x_{\text{wall}}(t)=D - v t, ] so [ u_x t = D - v t \quad\Rightarrow\quad (u_x+v)t = D. \tag{1} ]

In the wall frame the horizontal velocity of the ball before impact is (u_x+v); an elastic bounce reverses that horizontal component, so transforming back to the lab frame the horizontal velocity immediately after impact is [ v_x’ = -\big(u_x+2v\big). ] Thus the time from impact until the ball returns to the origin (horizontal condition (x_{\text{collision}}+v_x’ t_r=0)) is [ t_r=\frac{u_x t}{u_x+2v}. \tag{2} ]

Vertical motion is unaffected by the collision. Let the collision height be [ y_c = u_y t-\tfrac12 g t^2, ] and the vertical velocity just after collision (v_{y,c}=u_y-gt). Requiring the ball to return to (y=0) after time (t_r) gives [ 0 = y_c + v_{y,c}t_r -\tfrac12 g t_r^2. ] This simplifies (collecting terms) to [ u_y(t+t_r) - \tfrac12 g (t+t_r)^2 = 0 \quad\Rightarrow\quad t+t_r=\frac{2u_y}{g}. \tag{3} ]

Substitute (t_r) from (2) into (3): [ t + \frac{u_x t}{u_x+2v} = \frac{2u_y}{g}. ] Solve for (t). Noting (1+\dfrac{u_x}{u_x+2v}=\dfrac{2(u_x+v)}{u_x+2v}) we get [ t\frac{2(u_x+v)}{u_x+2v}=\frac{2u_y}{g} \quad\Rightarrow\quad t=\frac{u_y}{g},\frac{u_x+2v}{u_x+v}. ]

Finally substitute (u_x=v_0\cos\alpha,;u_y=v_0\sin\alpha): [ \boxed{ ; t ;=; \frac{v_0\sin\alpha}{g},\frac{v_0\cos\alpha+2v}{,v_0\cos\alpha+v,};. } ]

(Checks: for (v=0) this gives (t=v_0\sin\alpha/g) — the half flight time as expected; the formula requires (v_0\cos\alpha+v\neq0).)

Proof by Lean

Unverified

Another problem in Krotov

image

This problem is a classic application of constrained motion in kinematics. The key is to recognize that the total length of the inextensible thread remains constant, which relates the velocities of the two rings.

1. Define the Geometry and Coordinates

Let’s set up a coordinate system where:

From the problem description, the thread starts at fixed point $A’$, passes through ring $O’$, and ends at ring $O$. Thus, the total length $L$ is composed of two segments:

  1. Segment $A’O’$: Its length is simply the vertical distance from the ceiling to ring $O’$, which is $y_1$.
  2. Segment $O’O$: Its length, let’s call it $s$, is the distance between the two rings. Using the Pythagorean theorem: \(s = \sqrt{d^2 + (y_2 - y_1)^2}\)

The total length is: \(L = y_1 + \sqrt{d^2 + (y_2 - y_1)^2}\)

2. Relate the Geometry to Angle $\alpha$

According to the diagram, $\alpha$ is the angle $\angle AOO’$. Since rod $AB$ is vertical, the vertical distance between the rings is the adjacent side to this angle in the right triangle formed by the horizontal distance $d$. \(\cos \alpha = \frac{y_2 - y_1}{s}\)

3. Differentiate with Respect to Time

Since the thread is inextensible, its total length $L$ is constant, so $\frac{dL}{dt} = 0$. Differentiating the length equation: \(0 = \frac{dy_1}{dt} + \frac{d}{dt} \left( \sqrt{d^2 + (y_2 - y_1)^2} \right)\) \(0 = v_1 + \frac{1}{2\sqrt{d^2 + (y_2 - y_1)^2}} \cdot 2(y_2 - y_1) \cdot \left( \frac{dy_2}{dt} - \frac{dy_1}{dt} \right)\)

We know that $\frac{dy_1}{dt} = v_1$ (velocity of $O’$) and $\frac{dy_2}{dt} = v_2$ (velocity of $O$). Substituting these and the expression for $\cos \alpha$: \(0 = v_1 + \cos \alpha \cdot (v_2 - v_1)\)

4. Solve for $v_2$

Now, we rearrange the equation to find the velocity of ring $O$: \(v_1 + v_2 \cos \alpha - v_1 \cos \alpha = 0\) \(v_2 \cos \alpha = v_1 \cos \alpha - v_1\) \(v_2 = \frac{v_1(\cos \alpha - 1)}{\cos \alpha}\) \(v_2 = v_1 \left(1 - \frac{1}{\cos \alpha}\right)\)

Final Answer

The velocity $v_2$ of ring $O$ is: \(v_2 = v_1 (1 - \sec \alpha)\)

Note on Direction: Since $\alpha$ is an acute angle, $\sec \alpha > 1$, making $v_2$ a negative value. This indicates that while ring $O’$ moves downwards, ring $O$ must move upwards to maintain the thread length. The magnitude of the velocity is: \(|v_2| = v_1 (\sec \alpha - 1)\)

Kinematics of Constrained Motion: Two Rings on Rods

1. The Setup

We define the vertical positions of the rings from the ceiling (the top line AA’):

2. The Geometry

The thread is composed of two parts:

  1. The vertical segment from the ceiling to ring O’: y1
  2. The diagonal segment between ring O’ and ring O: s

Using the Pythagorean theorem for the diagonal segment:

s = sqrt( d² + (y2 - y1)² )

Total length L is constant:

L = y1 + sqrt( d² + (y2 - y1)² )

3. Relating the Angle (alpha)

From the diagram, the angle alpha is at ring O. Based on the right triangle formed between the rings:

4. Step-by-Step Derivation

Since the thread is inextensible, the rate of change of its total length is zero (dL/dt = 0).

Differentiating the length equation with respect to time:

  1. 0 = v1 + [ (y2 - y1) / s ] * (v2 - v1)
  2. Substitute the cosine relationship: 0 = v1 + cos(alpha) * (v2 - v1)
  3. Expand and isolate v2: 0 = v1 + v2 * cos(alpha) - v1 * cos(alpha) v1 * (cos(alpha) - 1) = -v2 * cos(alpha)

Final Result

The velocity of ring O (v2) is:

v2 = v1 * (1 - 1/cos(alpha))

Or, using the secant function:

v2 = v1 * (1 - sec(alpha))

“Gadgets” for proving computability class from textbook by Chris Moore

Fusing a scientific AI agent with the Ramanujan Machine poster here (an algorithmic framework that discovers conjectures for fundamental constants) creates a powerful pipeline. Here is how that architecture would likely look and function.

1. The Architecture of a “Ramanujan-Agent” Fusion

In this setup, the Ramanujan Machine acts as the “Intuition Engine,” and the AI Agent acts as the “Research Lead.”

2. The Discovery Loop: Human + AI + Machine

The synergy you are pursuing would likely follow this cycle:

Phase Actor Task
Generation Ramanujan Machine Scours numerical space to find a new, unproven identity.
Contextualization AI Agent Searches existing literature to see if this is a known identity or a transformation of one.
Formalization AI Agent Writes the conjecture in Lean code.
Strategic Guidance Human Decides if the conjecture is “beautiful” or mathematically significant enough to pursue.
Proof Search AI + Prover Uses AlphaProof-style reinforcement learning to find a formal proof.

3. Current AI Tools for Solving Math Problems

If you are interested in solving problems today using these tools, here is the “stack” I recommend exploring:

A. Formal Methods (The Gold Standard)

B. Neural-Symbolic Tools

C. Computational Discovery

A Note on the “Human” Element: > While the Ramanujan Machine is brilliant at finding patterns, it lacks mathematical taste. It might find a billion identities that are technically true but utterly boring. Your role in this synergy is to be the “Curator of Significance”—directing the AI’s immense processing power toward problems that actually move the needle in physics or number theory.

Would you like me to help you draft a Python script using SymPy to test a specific mathematical conjecture, or would you prefer to dive deeper into how Lean 4 works?

Key Takeaways 🧩🚀 From the paper above:

Narratives emphasizing autonomous “AI scientists,” the underrecognition of data and infrastructure work, misaligned incentives, and gaps between domain experts and machine-learning researchers all limit the impact of AI on scientific discovery

We call for reframing AI for science as a collective social project, where sustainable collaboration and equitable participation are treated as prerequisites for achieving technical progress.

... and infrastructure inequities, which concentrate power within privileged institutions [25].

Knuth

The system could search for patterns in known theorems and generate plausible new statements.

Examples:

“If theorem X holds under assumptions A, B, C, what is the weakest assumption set that still works?” “What analogous lemma should hold if I replace this algebraic structure with that one?” “Can the system detect useful intermediate lemmas that humans usually discover informally?”

Prompt template for long AI tasks with multiple agents

AI and Students and education and pedagogy

AI and medicine

Physics and Lean