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If you take the homogenous wave equation:

$$-\Delta_x u(x,t) + \frac{1}{c^2} \frac{\partial^2 u}{\partial^2 t} (x,t) \ = \ 0 \ \ \mathrm{in} \ \Omega \times (0, \infty),$$

with some proper initial- and boundary conditions and make the ansatz:

$$u(x,t):= e^{-i \omega t} v(x),$$

i.e. we seek a time-harmonic wave to the angular frequency $\omega \geq 0 $.
Inserting this ansatz into the equation, we get the eigenvalue problem:

$$-\Delta v(x) \ = \ \underbrace{\frac{1}{c^2} \omega^2}_{:= \ \lambda} v(x) \ \ \mathrm{in} \ \Omega.$$

Now, if i solve this eigenvalue problem and got a eigenvalue $\lambda$, i can retrieve the angular frequency by $\omega = \sqrt{\lambda \ c^2}$.
Here, i'm trying to model the 2D case of acoustic waves ($\Omega \subset \mathbb{R}^2$), but the 3D-case could be also interesting. Doing some research, i found out (for example here), that $c$ is in my case the speed of sound, i.e. for example $c$ = 343 m/s in air at 20°C. I assume, since $c$ is dealing with meters, my domain $\Omega$ has also the dimension m$^2$ (or m$^3$ in 3D-case).
But, by the upper formula, $\omega = \sqrt{\lambda \ c^2}$, the angular frequency would have dimension "m/s" (meters per second) - that doesn't sound right, shouldn't this be of dimension s$^{-1}$?

I would be glad if anyone, who's more experienced in physics, could explain this. Thanks in advance!

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What I am missing in your question, is the dimension of $\lambda$. You can see from your second formula, that the dimension of $\lambda$ is $\text{m}^{-2}$ (because of the Laplacian).

Further, to add a bit more information,

Normally we would write your Eigenvalue problem as

$$\Delta v = - k^2 v.$$

We use $k^2$ here mostly because it will make the answer look nicer (no square roots). We then call $k$ the wavenumber

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  • $\begingroup$ Thanks! I always thought of eigenvalues as dimensionless numbers. But what exactly do you mean with "because of the laplacian"? I know that $u(x,t)$ is the pressure and should be of dimension $\frac{kg}{m s^2}$, but how exactly do you know what dimension this $v(x)$ has? I mean, the time dimension (1/s) could be in that $e^{-i \omega t}$-part after doing that ansatz, but what about the remaining part? $\endgroup$ – Rfield Aug 24 '14 at 9:31
  • $\begingroup$ @Rfield The dimension of $v$ does not matter, because it is on both sides. $\Delta=d^2/dx^2$ and $dx$ has dimension of meter, hence $[\lambda]=\text{m}^{-2}$. $\endgroup$ – Bernhard Aug 24 '14 at 10:00
  • $\begingroup$ @Rfield Let's just look at your definition $\lambda:=\omega^2/c^2$. It is easily seen from it that dimension of $\lambda$ is $\mathrm{m}^{-2}$. $\endgroup$ – firtree Aug 24 '14 at 11:05

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