Are Lorentz-tensors tensors? It might sound like a stubid question, but I really have some problems in unterstanding it.
When defining tensor fields on a given manifold $\mathcal{M}$, one find the following transformation rule:
$${\displaystyle {\hat {T}}_{j'_{1}\dots j'_{q}}^{i'_{1}\dots i'_{p}}\left({\bar {x}}^{1},\ldots ,{\bar {x}}^{n}\right)={\frac {\partial {\bar {x}}^{i'_{1}}}{\partial x^{i_{1}}}}\cdots {\frac {\partial {\bar {x}}^{i'_{p}}}{\partial x^{i_{p}}}}{\frac {\partial x^{j_{1}}}{\partial {\bar {x}}^{j'_{1}}}}\cdots {\frac {\partial x^{j_{q}}}{\partial {\bar {x}}^{j'_{q}}}}T_{j_{1}\dots j_{q}}^{i_{1}\dots i_{p}}\left(x^{1},\ldots ,x^{n}\right).}$$
where $(x^{1},\dots,x^{n})$ and $(\bar{x}^{1},\dots,\bar{x}^{n})$ denotes local charts of the manifold. In physics this is often directly used as the definition of tensors.
Now to my question: A Lorentz tensor is defined to be a object with some indices, which transforms like a tensor under Lorentz transformations: So e.g.
$$F^{\mu^{\prime}\nu^{\prime}}(x^{\prime})={\Lambda^{\mu^{\prime}}}_{\mu}{\Lambda^{\nu^{\prime}}}_{\nu}F^{\mu\nu}(x)$$
where $\Lambda$ satisfies $\Lambda^{T}\eta\Lambda=\eta$ with $\eta=\operatorname{diag}(1,-1,-1,-1)$ and where the underlying manifold is the Minkowski-space.
According to the definition of tensors, $F^{\mu\nu}$ has to have this transformation law for every coordinate transformation, while in the definition of Lorentz tensors, $F^{\mu\nu}$ has to have this transformation law only for these special group of $\Lambda$'s....
 A: This is a great question. It strikes at the heart of what manifold we are actually considering. For physics, the actual manifold is rarely made clear and we often just see statements about allowed coordinate changes. Something rather special happens in the Lorentz transformation case:
$$ x^{\mu'} = \Lambda^{\mu'}_{\mu}x^{\mu} $$
so as $\Lambda$ is a constant matrix which solves $\Lambda^T \eta \Lambda  = \eta$ when we differentiate there is no coordinate-dependence for $\Lambda^{\mu'}_{\mu}$ and
$$ \frac{\partial x^{\mu'}}{\partial x^{\nu}} = \Lambda^{\mu'}_{\mu} \frac{\partial x^{\mu}}{\partial x^{\nu}} = \Lambda^{\mu'}_{\mu} \delta^{\mu}_{\nu} =  \Lambda^{\mu'}_{\nu} $$
If the coordinate change resembled the given rule for Lorentz transformations $\Lambda$ was allowed a coordinate dependence (I'll replace $\Lambda$ with $A$ for clarity of comparison) then:
$$ \frac{\partial x^{\mu'}}{\partial x^{\nu}} = \frac{ \partial A^{\mu'}_{\mu}}{\partial x^{\nu}}x^{\nu} + A^{\mu'}_{\nu} $$
so there is no nice relation between the rule for coordinate change and the corresponding rule for transformation of partial derivatives. Suppose that $F$ is a Lorentz tensor, I'll stick with rank one since it suffices, we're given pointwise that:
$$ F^{\mu'} = A^{\mu'}_{\mu}F^{\mu}. $$
I'll be
$$ F^{\mu'} = \left( \frac{\partial x^{\mu'}}{\partial x^{\mu}} - \frac{ \partial A^{\mu'}_{\mu}}{\partial x^{\nu}}x^{\nu} \right)F^{\mu}. $$
There is no nice way to remove the derivative term to get the general tensor law  you want. This is not a proof, and I think to answer your question better I'd need to blather on about domains of coordinate charts and the general idea that a given point set can be given inequivalent differentiable structures. The concept of a manifold can be specialized to just take the coordinate changes whose derivatives make for nice coordinate changes. Probably the better way to think about this is with the frame bundle and the concept of associated subbundles which limit the allowed transformations in the ways we are interested. There is also the larger question of Riemannian vs. semi-Riemannian. That is something else you'd need to be aware of to properly understand the question you begin to ask. Still, it's is a great question. It's the first question I asked my advisor: "what is a tensor". His response: "what kind". That's it, there are many kinds of tensors depending on what structure you wish to preserve. I'll leave it at that for now. Supper is ready.
