# Special relativity, question about the transformation of EM field

We have defined the electromagnetic tensor field as always as :

$$F^{u \nu} = \partial^u A^{\nu} - \partial^{\nu} A^u$$

Then to see how the fields transform we have applied:

$$F'^{u \nu} = \Lambda^{u}_{\hspace{0,2 cm} \alpha} \Lambda^{\nu}_{\hspace{0,2 cm} \beta} F^{\alpha \beta}$$

where $$\Lambda$$ is the Lorentz matrix and from this we have derived how $$E$$ and $$B$$ transform

But what does it tell me that the EM tensor fields transform in that way? More precisely we have created a tensor but i think that we could create infinite tensor that behaves in different ways when we applied Lorentz transformation.

• Since $F^{\mu \nu}$ is a finite dimensional tensor and $\Lambda^\mu_{\alpha}$ and $\Lambda^{\nu}_{\beta}$ are finite dimensional matrices, you would need to take an infinite dimensional tensor product of the finite dimensional tensor $F^{\mu \nu}$ with the corresponding infinite number of finite Lorentz transforms - one transform for each index on the infinite dimensional tensor. So mathematically, it's probably possible - but physically it would be garbage. – Cinaed Simson May 7 '19 at 22:42

i think that we could create infinite tensor that behaves in different ways when we applied Lorentz transformation.

No. The fact that $$F'^{\ \mu \nu} = \Lambda^\mu_{ \ \ \alpha} \Lambda^{\nu}_{\ \ \beta} F^{\alpha\beta}$$

under Lorentz transformations is more or less the defining characteristic of a tensor. If you have some other quantity $$G$$ that does not transform in precisely this way, then it isn't a tensor.