# Why can't a ladder sliding down a wall go faster than light?

I know the title sounds ridiculous, but when I took calculus in college we had a problem like this. The result seemed unreal to me, but I can't figure out the reason why it doesn't work.

This was the gist of the question:

Suppose there is a ladder leaning against a wall, as shown in the figure below. The ladder is $$2$$ meters long. The ladder is on a conveyor belt moving away from the wall at a constant speed of $$1$$ meter per second.

How fast will the end of the ladder against the wall be moving downward once it hits the ground? So here's how we solved this: Let $$h$$ be the distance from the ground to the top of the ladder, and $$w$$ be the distance from the wall to the other end of the ladder, and let $$l$$ be the length of the ladder.

By the Pythagorean theorem,

$$h^2+w^2=l^2$$

Take the derivative of both sides of the equation with respect to time, and note that the length of the ladder is constant:

$$2h\frac{dh}{dt}+2w\frac{dw}{dt}=2l\frac{dl}{dt}=0$$

Solving for $$\frac{dh}{dt}$$ gives us:

$$\frac{dh}{dt}=-\frac{w}{h}\frac{dw}{dt}$$

Suppose at time $$t=0$$ the ladder is vertical. Then $$w=t$$ and $$h=\sqrt{l^2-w^2}=\sqrt{4-t^2}$$. Take the limit of $$\frac{dh}{dt}$$ as $$t\to 2$$:

$$\lim_{t\to2^-}\frac{dh}{dt}=\lim_{t\to2^-}-\frac{w}{h}\frac{dw}{dt}=\lim_{t\to2^-}-\frac{t}{\sqrt{4-t^2}}\left(1\frac{\text{m}}{\text{s}}\right)=\infty$$

The equations are saying that the top of ladder is approaching infinite speed as it falls towards the ground!

This doesn't seem like a physically possible result, but what part of this setup is not phyiscally possible?

• In classical physics there is no limit of $c$. Though I'm not sure that's the whole story here. – zhutchens1 Jul 4 '18 at 20:01
• You assume that the end-point of the ladder is fixed at $x=0$ by some kind of pulley mechanism, otherwise, the ladder will disconnect from the wall as soon as the gravitational acceleration can't keep accelerating it by the rate required for the connection to be maintained (and then it will only be accelerated by the weight forces). – Sebastian Riese Jul 4 '18 at 20:13
• @SebastianRiese Ok, then suppose that the ladder is fixed to the wall by some mechanism. – Peter Olson Jul 4 '18 at 20:30

The question should have a different title. The question you pose at the end does not require relativity to answer.

There is a perennial question about ladders sliding, namely, when does it leave the floor. You have a variation on this.

Ladders have finite mass, and so finite moment of inertia.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

Your solution assumes the ladder remains in contact with the wall and the belt. It won't. It isn't clear to me if the ladder is fastened to the belt or just resting on it.

But you need to work out when the force on the wall, or the belt, falls to zero. At that point the ladder will no longer contact at one end, and your formula for vertical speed will no longer be valid.

• So what if you use some mechanism to force the ladder to stay in contact with the wall and the belt? For example a pulley or a wheel track that the ladder is affixed to? Is there still something that would prevent the ladder from increasing its speed asymptotically? – Peter Olson Jul 4 '18 at 21:16
• The limit to the vertical ladder speed on the wall would be a function of the stiffness and strength of the ladder, wheel track, wall, and conveyor. Only if you assume that all parts are infinitely stiff and that there is no speed limit (speed of light) would you get an infinite speed. In other words, all the parts would start to stretch/deform/fracture before the ladder reaches horizontal. – James Jul 5 '18 at 12:35

In special relativity, the length is no longer $x^2 + y^2 = l^2$, but $x^2 + y^2 - (ct)^2 = s^2$ where $s$ is no longer the length of the ladder, but the spacetime interval. This means that the Pythagorean theorem no longer holds and that the length of the ladder becomes observer-dependent.

To elaborate @dainelunderwood's correct answer, classically, the ladder's a rigid body. And in this case it's a "straight line" as far as your geometric argument's concerned. But when special relativity's involved ($\frac vc\not\ll1$), rigid bodies no longer necessarily exhibit rigid-body geometry.

The "classic" textbook example (though google's not immediately coughing up a reference) is a car driving over a pothole at speed $v\sim c$. And suppose the diameter of the pothole is just about the same as the length of the car. So consider that situation from the point of view of a passenger inside the car. To him, the pothole seems foreshortened, whereby the car doesn't fall through the pothole (albeit, "it's going to be a bumpy ride"). But from the point of view of a bystander on the ground, it's the car that's foreshortened, whereby it does fall through.

So which is it? Ultimately, the car does fall through the pothole. If (like Daniel said) you calculate the special-relativistic geometry carefully, then you find the car "bends" through the hole, assuming a kind of dog-leg shape during the fall. What happens to your ladder is analogous. Obviously, there's nothing wrong (nothing non-physical) about the one-meter-per-second conveyor-belt end of your ladder. But then the wall-end will necessarily curve upward as its downward $v$ approaches $c$. (But I'll leave it for "extra credit" to figure out the exact analytic geometry shape of that curve:)