Kinetic energy and tensors

[Niles]

Hi

If i want to express the kinetic energy for some angular momentum, I can write


T=\frac{1}{2}\sum_{ij}{I_{ij}\omega_i\omega_j}


I cannot quite see why we have wiwj at the end, and not just wi2. It is not that obvious to me. I have read several examples regarding polarization and electric fields, and they make perfectly good sense. But in the case with the KE, I'm a little confused. Can somebody perhaps shed some light on this?

Best,
Niles.

[quZz]

In general bodies are not isotropic and coordinate axis are arbitrary, so I_{ij} =/= 0, i =/= j.
But since I_{ij} = I_{ji}, you can always choose axis in such a way that I_{ij} = 0, i =/= j.

[Niles]

In general bodies are not isotropic and coordinate axis are arbitrary, so I_{ij} =/= 0, i =/= j.
But since I_{ij} = I_{ji}, you can always choose axis in such a way that I_{ij} = 0, i =/= j.

Thanks. But does that also explain why wiwj at the end, and not just wi2?

[Niles]

Or e.g. another example: Why is it that we have two different factors of E in the expression for the second-order polarization?


P_i^{(2)} = \sum_{jk}{\chi_{ijk}E_jE_k}


What I cannot understand in this case is that the electric field comes in with a certain direction, so why do we even consider elements such as ExEy?

[quZz]

In general, the expression quadratic in \omega will have the form I_{ij}\omega_{i}\omega_{j}.

Since I_{ij} is symmetric you can choose a special coordinate system where it is diagonal. In this special case you will have I_{11}\omega_1^2 + I_{22}\omega_2^2 + I_{33}\omega_3^2.

[Niles]

In the case with



P_i^{(2)} = \sum_{jk}{\chi_{ijk}E_jE_k}



can we also always represent the choose a coordinate system where x it is diagonal?

[quZz]

don't know =) but what if not?

[K^2]

Thanks. But does that also explain why wiwj at the end, and not just wi2?
Sure it does. The i and j are different and terms like wxwy are not part of the sum over wi².