How does a superhero get back into superhero shape? That's the problem Thor has in the latest trailer for Thor: Love and Thunder, where we see the Norse god trying to exercise with something like battle ropes. These are basically just two super-thick ropes that you shake up and down, which might seem silly, but it's a legit workout. And doing it the Thor way makes it even more difficult: Instead of using ropes, he uses very thick chains.

I love superhero movies, because situations like this just bring up some really great physics questions, such as: How much harder is it to exercise with a battle chain instead of a battle rope? Is this what it would actually look like if you shook a giant chain? And why does a wave move down a rope, anyway?

When you shake one end of a string (or rope or chain), you create a disturbance or a displacement that travels down its length. A wave on a string might look something like this:

The string is stretched out in the horizontal direction, which we’ll call the x-direction. Each part of the string will have a different x-value. The vertical direction will then be the y-direction. That means that every piece of the string has both an x-value and a y-value. With these two variables, y can be defined as a mathematical function of x to describe the shape of the string, as shown in the picture above.

The string's shape also changes with time as the wave moves along it. So to fully describe the vertical position of each part of the string, we need to show y as a function of both position (x) and time (t).

The motion of this disturbance is governed by the wave equation. This is a differential equation that gives a relationship between the way the string changes with time (t) and the shape of the string, or how it changes with its position (x).

OK, calm down. I told you it was a differential equation. That's why there are ∂ symbols in there—they are partial derivatives. All this says is that the vertical acceleration of the string (represented by ∂2y/∂t2) is proportional to the curvature of the string (represented by ∂2y/∂x2). The proportionality constant for this relationship is the square of the wave velocity. If you want a more complete (although complicated) derivation, here you go.

Here's the awesome thing: This isn’t just for strings. You can also use this equation to describe waves in water, the air (sound), and the ground (seismic waves). It even shows that the relationship between electric and magnetic fields can produce an electromagnetic wave, which is exactly how light is able to travel through empty space as a wave.

However, in the case of Thor's battle rope, we’ll stick to a wave on a “string.” In this case, the wave velocity depends on  the tension in the string (T) and its linear density—meaning its weight per unit of length (μ).

If you increase the string’s linear density from a rope to a giant chain, this will make the wave travel more slowly.

We can estimate both the tension and linear density of Thor's chain, but first we should build a model of a wave on a string. You can’t really understand something until you can model it. But you also can’t know if that model is legitimate until you compare it with something real. So, let's do just that.

I want to make a simple wave and measure three things: its velocity, the tension on the string, and the linear mass density of the string. That shouldn't be too difficult. For the string, I'm actually going to use a strand of plastic beads with a string length of 1.2 meters and a mass of 25 grams. Right there, I can calculate the linear mass density at μ = 0.0208 kg/m.

For the tension, I'm going to place the string of beads on a flat table with a pulley mounted on the edge. Then I can let the string hang over the pulley with a weight connected to it. This will produce a tension in the string due to the gravitational force.

Using a hanging mass of 20 grams creates a string tension of 0.196 Newtons. If the wave equation is legit, then a wave on this string should travel with a speed equal to 3.07 meters per second, using the square root of T/μ.

Great, but does this agree with an actual wave? Let's find out. Here's what happens when I give the beads a quick flick to produce a wave:

I can get the speed of this wave using the meterstick on the table and my favorite video analysis tool, Tracker Video Analysis. I can mark the location of the wave in each frame to get the following position-time plot:

Since the velocity is defined as the time-rate of change of position, the slope of this plot should give the velocity. That puts this wave speed at 2.85 m/s, which is pretty close to the theoretical prediction. I'm happy with that.

But what if I want to look at the speed of a wave in a giant metal chain, instead of a string of beads? I actually don't have one of these things lying around—and I probably couldn't move it anyway. So let's build a computational model.

Here's my idea: I'm going to let the chain be made of a bunch of point masses connected by springs, like this:

A spring exerts a force that is proportional to the amount of stretch (or compression). This makes them very useful. Now I can look at the positions of all the masses in this model and determine how much each connecting spring is stretched. With that, it's a fairly simple step to calculate the net force of each mass.

Of course, with the net force I can find the acceleration for each piece using Newton's second law: Fnet = ma. The problem with this spring force is that it's not constant. As the masses move, the stretch of each spring changes and so does the force. It’s not an easy problem. But there is a solution that uses a bit of magic.

Imagine that we calculate the forces on each mass of this modeled series of springs. Now suppose that we just consider a very short interval of time, like maybe 0.001 seconds. During this interval, the beads do indeed move—but not that much. It's not a huge stretch (pun intended) to assume that the spring forces don't change. The shorter the time interval, the better this assumption becomes.

If the force is constant, it's not too difficult to find the change in velocity and position of each mass. However, by making the problem simpler, we've just made more problems. In order to model the motion of the beaded string after just 1 second, I would need to calculate the motion for 1,000 of these time intervals (1/0.001 = 1,000). No one wants to do that many calculations—so we can just make a computer do it. (This is the main idea behind a numerical calculation.)

If you want to see all the details of building a mass-spring model of a string of beads, I have all of that here. (Warning, it's long.) But the real test is to see if a mass-spring model of a string of beads can produce a wave velocity just like a real string. Here is a mass-spring model with the same linear density and the same tension as the real string of beads, using 34 pieces:

If I track the horizontal position of the highest point on the string, I get the following plot:

I can fit a linear function (just like I did with the video analysis) to get a slope of 2.95 meters per second. That's the wave speed from the model—it's pretty much the same value as for the actual string of beads. That's a win.

We are going to need to make some estimates, but we can use the same wave equation to look at Thor's massive chain. Let's start off with the wave speed. Again, using video analysis I can plot the motion of one of the waves on the chain. I'm going to need some type of distance scale, so I'll just set the height of Thor at 1.9 meters, which is the height of the real human named Chris Hemsworth who plays him. With that, I get the following plot:

That puts the wave speed at 4.56 meters per second. So, what force would it take for Thor to get this kind of wave speed? The wave speed on a string depends on both the tension on the chain and its linear mass density. Let's estimate the density and use that to calculate the required tension Thor would need to pull on that chain.

I'm going to guess that, if you remove the holes, the chain has an equivalent diameter of 15 centimeters. If the chain is made of steel, it could have a volume density of about 8,000 kilograms per cubic meter. With these values, the chain would have a linear mass density of 141 kilograms per meter. In order to get the wave speed in the video, Thor would need to pull with a force of 2,940 Newtons, or 658 pounds. That doesn't seem so bad—at least not for the god of thunder.

OK, what about a normal human with a normal battle rope? Here is a rope with a length of 30 feet and a weight of 26 pounds. That gives it a linear mass density of 1.29 kilograms per meter. In order to get a wave moving at the same speed as in the Thor trailer, a person would need a pull force of 26.8 Newtons, or 6 pounds. So Thor needs to pull about 100 times harder than a human. I don't think that's too much to ask. I'm pretty sure he could do that. But I guess when getting back into shape, it's best to start light and work your way up to heavier stuff. So my advice to the Norse god is: Start with a rope until you’re ready for the steel chain.