# Orthogonality of modes

I want to know what it exactly means when it is said the modes of vibration are Orthogonal.I understand what it means mathematically but what is its physical interpretation?

• It means that the different modes of motion of the system that you are studying are decoupled. In other words, they can be 'told apart', as they are physically distinct ways for the system to evolve in time. The catch is that each mode has a distinct energy, so that the total energy of a linear combination made up with different modes of vibration may be written as a sum over their energies (weighed with the weights of the linear combination). Which is nice by itself, but also because the quantum-mechanic analogue of this idea is one of the most useful tools in Quantum Mechanics. – user17581 Jan 14 '13 at 16:13
• I've got no time right now to post a complete passable answer, if I come later and no one else has written an answer, I may give a more elaborate response than this one. – user17581 Jan 14 '13 at 16:15
• See also normal modes on Wikipedia. – Qmechanic Jan 14 '13 at 19:19

Concepts rather than math? Let me give this a shot:

By orthogonal modes of vibration, they mean that you cannot construct any one mode with any linear combination of the others. Moreover, you can't even construct part of one mode through any linear combination of the others.

What does that mean?

I'm going to have to get a little mathy here for the concept, but I hope it still works: Consider the modes, $f_0 = 1$, $f_1 = cos(\omega t)$, $f_2 = sin(\omega t)$, $f_3 = cos(2 \omega t)$ and $f_4 = sin(2 \omega t)$. This corresponds to holding steady, vibrating at a fundamental, vibrating at a fundamental of a different phase, and vibrating at twice the fundamental in two phases.

There's no way you can construct any one of these from a linear combination of all or some of the others. No linear combination of, say, $A f_1 + B f_2$, where $A$ and $B$ are constants, will give you $f_3$. You'll just get something that looks like $C sin(\omega t - \phi)$ (where, again, $C$ and $\phi$ are constants whose values I'm not calculating here).

On the other hand, consider $sin^2(\omega t)$. This might look orthogonal to our functions above, but really it's not, because it's a linear combination of $f_0$ and $f_1$: $sin^2(\omega t) = \frac{1 - 2 cos(\omega t)} {2} = \frac {f_0 + f_1} {2}$. So any orthogonal system of vibrations that included $sin^2(\omega t)$ could not include either $f_0$ or $f_1$. On the other hand, it is orthogonal to $f_2$, $f_3$, and $f_4$. The system one puts together, however, must include modes that are totally orthogonal to all other modes. They usually also try for "completeness" (being able to describe any arbitrary function with a linear combination of base modes) and "normality" (each mode has, in some sense, an "amplitude" of 1), but that stuff is beyond the concept of orthogonality.

I've tried to make this as non-mathy as possible, with really only moderate success; I hope it helps!.

• @user17581,@daaxix,@thatnerd thanks a lot – user12448 Jan 16 '13 at 14:01

It means that they cannot "interfere" basically, since two orthogonal modes are not at all "similar", aka the if one mode is projected to an orthogonal one, the result is "zero".

It also means, that for any other arbitrary function in the span of the orthogonal modes, you can write that arbitrary function as a weighted sum of those orthogonal modes uniquely! So you can identify an arbitrary function with some (possibly infinite) string of numbers corresponding to the weights!

Energy (Edit)

Suppose that we have some basis functions $f({\bf{r}}), g({\bf{r}})$, and some superposition (addition) of those functions $h({\bf{r}}) = \alpha f({\bf{r}})+\beta g({\bf{r}})$. Then usually the energy is found through the norm, i.e., $$\|h({\bf{r}})\|=\int_{\mathbb{R}^n} h({\bf{r}})h^*({\bf{r}})d^n r$$ where ${\bf{r}} \in \mathbb{R}^n$, $\alpha,\beta\in \mathbb{C}$ and the integral above is $n$ integrals, each over infinity. If we expand $h({\bf{r}})$, then we get $$\|h({\bf{r}})\|=\int_{\mathbb{R}^n} [\alpha f({\bf{r}})+\beta g({\bf{r}})][\alpha^* f^*({\bf{r}})+\beta^* g^*({\bf{r}})]d^n r\\ =|\alpha|^2\int_{\mathbb{R}^n} f({\bf{r}})f^*({\bf{r}})d^n r+|\beta|^2\int_{\mathbb{R}^n} g({\bf{r}})g^*({\bf{r}})d^n r\\ =|\alpha|^2\|f({\bf{r}})\| + |\beta|^2\|g({\bf{r}})\|,$$ since $$\alpha\beta^*\int_{\mathbb{R}^n} f({\bf{r}})g^*({\bf{r}})d^n r=0\\ \alpha^*\beta\int_{\mathbb{R}^n} f^*({\bf{r}})g({\bf{r}})d^n r=0$$ by the orthogonality. So when energy is defined this way (it often is) then the energies contributed by orthogonal modes do indeed add and contribute individually to the system.

• @user17581,@daaxix If I am getting it right then it means that if two modes are orthogonal then there are independent of each others?They individually contribute to the energy of the whole system ? – user12448 Jan 15 '13 at 13:19
• see my edit above...usually this is true – daaxix Jan 15 '13 at 15:15
• It get the concept now. One more query. While doing the expansion of h(r) is it convolution that you have used ?? – user12448 Jan 15 '13 at 16:35
• @user 12448: As daaxix said, this 'separability' is usually true. The point is, the only way that such a property wouldn't be true is by changing the dynamical problem, i.e., to introduce additional terms in the system so that the differential equation changes. Normal modes, after all, are nothing more than a set of orthogonal solutions to the differential equation for a vibrating system, you may understand them as eigenfunctions of a certain differential operator which is a signature of the dynamical system. If the system's dynamical equations change, so do its eigenfunctions. (continues) – user17581 Jan 16 '13 at 9:10
• And if you stop for a while and think about it, it is the whole idea behind perturbation theory, both in the classical (which would be the present case) and quantum versions, as both versions eventually come down to the same idea, as the whole 'eigenfunction' idea is nothing more than linear algebra applied to differential equations. – user17581 Jan 16 '13 at 9:21