# Wave equation - v speed or velocity?

## Main Question or Discussion Point

Consider the classical wave equation.

Does the v in the equation stand for the wave speed or the wave velocity? The image says v is the wave velocity but I am not sure about that as other sources like Wikipedia say that v is the wave speed. So, what is v really - speed or velocity?

I think it's wave velocity since v is squared, so that allows for the possibility for v being positive or negative. And the direction of travel is the direction of the wavevector.

What do you think?

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Ken G
Gold Member
I think that when v is squared, there is no distinction. You can use vector notation where the "wave vector" k is a vector in the direction of propagation, but the wave equation results for any direction for k, the actual direction appears in the boundary conditions to the equation, not in the equation itself. So you can interpret v as having a specified direction that the equation doesn't care about, or you can imagine that v has no specified direction until it emerges from the boundary conditions-- the equation is the same either way.

vanhees71
Gold Member
2019 Award
Here, it's of course the speed, since it enters only as $v^2$, which is a scalar (under rotations).

I think that when v is squared, there is no distinction. You can use vector notation where the "wave vector" k is a vector in the direction of propagation, but the wave equation results for any direction for k, the actual direction appears in the boundary conditions to the equation, not in the equation itself. So you can interpret v as having a specified direction that the equation doesn't care about, or you can imagine that v has no specified direction until it emerges from the boundary conditions-- the equation is the same either way.
I see. I did not solve wave equations before, so I had no idea that boundary conditions are needed to specify the complete problem. So, thanks for the insight.

Here, it's of course the speed, since it enters only as $v^2$, which is a scalar (under rotations).
But that contradicts Ken G, doesn't it?

I have no idea how $v^2$ is a scalar under rotations. Do you mean that if you imagine a coordinate system with three degrees of freedom for velocity, then if you rotate $v^2$ in the coordinate system, $v^2$ remains the same?

Matterwave
Gold Member
$v^2$ is just a number. It doesn't matter if you got it by taking $|\vec{v}|^2$ or by $\vec{v}\cdot\vec{v}$. The result is the same.

$v^2$ is just a number. It doesn't matter if you got it by taking $|\vec{v}|^2$ or by $\vec{v}\cdot\vec{v}$. The result is the same.
I see! So, you agree with Ken G that we have the freedom to decide if we should take the quantity as either speed or velocity?

So, vanhees71 is wrong, I suppose?

vanhees71
Gold Member
2019 Award
Of course $v^2=\vec{v}^2$ whereever you get a vector quantity from. The wave equation reads
$\frac{1}{v^2} \frac{\partial^2 \phi}{\partial t^2}-\Delta \phi=0.$
Any solution can be expressed via plain waves
$$\phi(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^3} \exp[-\mathrm{i} \omega(\vec{k}) t] \left [A(\vec{k}) \exp[+\mathrm{i} \vec{k} \cdot \vec{x}] + B(\vec{k}) \exp[-\mathrm{i} \vec{k} \cdot \vec{x}] \right ].$$
$$\omega(\vec{k})=v |\vec{k}|.$$
As you see again from the very beginning of the derivation, $v$ is a scalar quantity! It's usually called the phase velocity of the wave although it's a scalar quantity.