# Why is the 1-dimensional wave equation called like that when it seems to be 2-dimensional?

The wave equation in one dimension traveling along a string is: $$\frac{∂^2y}{∂x^2} = \frac1{v^2} \frac{∂^2y}{∂t^2}$$ but this equation has 3 variables $$x, y,$$ and $$t$$, shouldn't it be in 2 dimensions according to the 2D plane $$(x,y)$$ the string is free to move up and down (the $$y$$ component) and right to left (the $$x$$ component), so why is that the 1 dimension equation of the wave?

I can't think of a wave in 1 dimension, it has an amplitude which is in the $$y$$-axis and a wavelength in the $$x$$-axis so how this equation describe waves in 1D?

• it is called 1d since it propagates in one direction , 3d waves travel in 3 directions and still have an amplitude. – trula Sep 25 '19 at 10:32
• It's like in the big picture the amplitude isn't considered a new dimension, but what if you zoom in, the wave in a pond, for example, moving in a 2 dimensional plane, would have another dimension which is the height of the wave known as the amplitude, that's my point about why 1d waves are more like 2d – user144435 Sep 25 '19 at 10:41
• Well, the question is, do you consider a straight line $y(x) = mx + b$ a two dimensional object? It definitely has $y$ and $x$ components in the 2D plane. Is a straight line two-dimensional then? – Ezze Sep 25 '19 at 13:40
• I have edited the title, I hope you don't mind. The previous one was too confrontational, and reducing your chances of getting answers. – Javier Sep 25 '19 at 13:42
• @Ezze This is a very good point, and one that often confuses physicists studying formal geometry. For example, the $n$-sphere is the sphere that is parametrized by $n$ parameters, not the sphere that lives in $n$ spatial dimensions. So the 1-sphere is the circle, despite us usually thinking of circles as 2-dimensional. Likewise, the 2-sphere is the conventional "3D" sphere, and so on. The distinction is important, but is rarely one we have to think about in physics. – probably_someone Sep 25 '19 at 13:45

More generally, if the wave equation $$\Box y=0$$ is satisfied for a scalar field $$\mathbb{R}^{n+1}~\ni~ (\vec{x},t)\quad \stackrel{y}{\mapsto} \quad y(\vec{x},t) ~\in~ \mathbb{R},$$ with spacetime $$\mathbb{R}^{n+1}$$ as domain, and with a 1-dimensional target space $$\mathbb{R}$$, we speak of the wave equation in $$n+1$$-dimensional spacetime, or equivalently, the wave equation in $$n$$-dimensional space.

The 1D wave equation is called like that because it has only one independent space variable, $$x$$. That's it. The 2D equation has two variables, etc.

You are correct that an oscillating rope sweeps out a 2-dimensional plane; in fact, by moving one end in a circle instead of up and down, you can make it occupy a 3D region. But the equation doesn't care about that: it's just an equation for a function that depends on $$x$$ and $$t$$. If you consider the 3D behavior of the rope then you will have more functions, but they will still only depend on $$x$$ and $$t$$, so the equation (or equations) will still be 1-dimensional.

To summarize: the dimensionality of the wave equation refers to the number of independent space variables, and not the dimensions of the movement. There are, as far as I know, two good reasons for this:

• Not every wave is a moving rope. For example, we could imagine something like, say, temperature obeying a wave equation in 3D space$$^1$$. The temperature is not moving in any direction like a rope is, so it doesn't make much sense to assign a physical dimension to it, but it does depend on the three variables $$(x,y,z)$$, so we'd talk about the 3D wave equation.

• The mathematical properties of the solutions depend mostly on the number of independent variables, and not so much on the number of functions, so it makes more sense to classify the equations by the number of variables and call that the dimension.

$$^1$$I'm not actually sure if there is a situation in which the temperature obeys the wave equation, but this is just an example.

• Electromagnetic waves are an example of a non-matter wave. – Bill N Sep 25 '19 at 13:51
• @BillN: Yes, but they are 3D vectors, so my point about ignoring the dimension of the functions would not be as strong. Also they are more abstract, and frequently hard to understand for this very reason. – Javier Sep 25 '19 at 13:52
• And to add another consideration, if the wave is longitudinal, the $y$ variable describes motion parallel to the propagation, $x$, and not a 2nd spatial coordinate. – Bill N Sep 25 '19 at 13:56