- #1
kent davidge
- 933
- 56
If we have a plane wave, usually in Relativity notation it is written as ##A^\alpha = a^\alpha \exp(i x_\alpha k^\alpha)##. (I know we need to take the real part in the end). In cartesian coordinates, and two dimensions say, that ##x_\alpha k^\alpha## would be ##x^\alpha k_\alpha = x k_x + y k_y##. How does it reads in polar coordinates? ##x^\alpha' k_\alpha' = r k_r + \theta k_\theta = r [(\partial x/ \partial r)k_x + (\partial y/ \partial r)k_y] + \theta [(\partial x/\partial \theta )k_x + (\partial y/\partial \theta )k_y] = r (\cos\theta k_x + \sin\theta k_y) + \theta[(-r\sin\theta ) k_x + (r\cos\theta ) k_y]##. The first term is just equal to its version in Cartesian coordinates. But then there is that second term, which makes ##x^\alpha' k_\alpha' \neq x^\alpha k_\alpha##. I already expected this result from the fact that ##x^\alpha## do not transform like, and thus are not, four vectors. So a product like ##x^\alpha k_\alpha## will not be a scalar.
But then what does this result mean? A field like ##A^\alpha = a^\alpha \exp(i x_\alpha k^\alpha)## will not transform as a vector if the exponent is not a scalar. But it should transform as a vector. So what we do?
But then what does this result mean? A field like ##A^\alpha = a^\alpha \exp(i x_\alpha k^\alpha)## will not transform as a vector if the exponent is not a scalar. But it should transform as a vector. So what we do?