Why is there a wiwj at the end in the kinetic energy expression?

[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 w_iw_j at the end, and not just w_i². 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?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 w_iw_j at the end, and not just w_i²?

 

[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 E_xE_y?

 

[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

$$<br /> P_i^{(2)} = \sum_{jk}{\chi_{ijk}E_jE_k}<br />$$

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 w_iw_j at the end, and not just w_i²?

Sure it does. The i and j are different and terms like w_xw_y are not part of the sum over w_i².