Recent content by Vermax

Thank you! :)

I see, however I have met with such thing (A'=P^{-1} AP) before only once(in finding min or max of some function while having some other functions as conditions) and that was only connected with examine a quadratic form as positive define or not. Could you then explain me why is such matrix...

Oh ok, was not that difficult as it looked like :) Thanks.

Ok, thank you, so I think I need more theoretical knowledge about tensors. I am also curious when we need to change refrence frame of EM field? In standard homework excercises it is rather not used? And I also would be very gratefull if you could help me with my second question presented in my...

Sorry, I feel like I am too stupid to understand it :( I would say I need only one matrix (transform operator) to transform 4x4 matrix(tensor) to another 4x4 one.

Thanks for your answer but I do not fully understand it. Lets say we have four-vector w^a and we want to have vector w^a'. We can write: w^{\alpha'} = \Lambda^{\alpha'}_{\alpha} w^{\alpha} So here lambda is a transformation (matrix - tensor), right? So why we mulitply that twice?

I understand that after writing down this: F^{ \mu v} = \partial^{\mu} A^v- \partial^v A^{\mu} We can get a nice matrix connecting E and B vectors. But I just wonder what we need this matrix for? I am a little bit confused about all this relativity in electromagnetism... And another...

Hm yes I think so, and I can trot out that I tried to prove equations from Wiki: http://en.wikipedia.org/wiki/Vector_calculus_identities And I can say that I was succesfull in almost every one :) I am still working on some more complicated but it is a matter of time I think.

Oh I see... But that was I think the most difficult part. I am glad, however, that I understad it thanks to you of course :)

Hehe quite logical! One last thing I am afraid... I understand why: \vec{k}_i E_j\nabla_j E_i=\left(\vec{E}\nabla\right)\vec{E} But how: \vec{k}_i E_j\nabla_i E_j={{1}\over{2}}\nabla E^2 Why there is 0.5 and where goes the versor k?

I stucked in the second line. =\vec{k}_i\delta_{il}\delta_{jm}E_j\nabla_l E_m-\vec{k}_i\delta_{im}\delta_{jl}E_j\nabla_l E_m=\vec{k}_i jE_j\nabla_i E_j-\vec{k}_i E_j\nabla_j E_i={{1}\over{2}}\nabla E^2-\left(\vec{E}\nabla\right)\vec{E} I think: \delta_{il}\delta_{jm} =...

I did as you told me to, and this thing: (\vec{a} \times \vec{b})_k=\epsilon_{klm} a_lb_m Looks for me (almost ofc) like a Laplace determinant expansion with the first versor. It makes more sense now for me. And rot->grad thing is not so diffcult as it looked like :) I will try to...

Hmm I just copied it from some Wiki-book. If you want to check it by yourself here is the link but it is not in english I am afraid http://pl.wikibooks.org/wiki/Fizyka_matematyczna/Zasady_zachowania_a_twierdzenia_o_właściwościach_pola_elektromagnetycznego . Yes, I must play more with this...

Just I am not too familiar with this Einstein's notation. Starting from the very first part of this equation: \vec{E}\times\left(\nabla\times\vec{E}\right)= \vec{k}_{i}\epsilon_{ijk}E_j\left(\nabla\times\vec {E}\right)_k So cross product can be written as: \vec{u} \times \vec{v}= u^j...

There is a transformation: \vec{E}\times\left(\nabla\times\vec{E}\right)= \vec{k}_{i}\epsilon_{ijk}E_j\left(\nabla\times\vec{E}\right)_k= \vec{k}_{i}\epsilon_{jk}E_j\epsilon_{klm}\nabla_l E_m= \vec{k}_i\epsilon_{ijk}\epsilon_{klm}E_j\nabla_l E_m...