Orthogonality of modes

Question

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?

Answer 1

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!.

Answer 2

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!

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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.