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The Lorentz transformation does not transform, since it is not an object living on the spacetime manifold in the way that vectors and tensors do. In general, the objects that we think of as "vectors" or "tensors" are elements of the tensorial powers of the (co)tangent spaces at every point of the spacetime manifold. Under any coordinate transformation ...


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Indeed, to add to orion's answer, the definition of a vector (and / or spinor) in physics (as opposed to the mathematical definition as an element of a linear space over a field) is most often stated in terms of how the object concerned transforms (e.g. see my answer here) when "co-ordinate transformations" are made: more precisely, when one switches between ...


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Since, to take a specific example; for two $4$-vectors $a^{\mu}, b^{\mu}$ it's not too difficult to show that the scalar product \begin{eqnarray} a \cdot b &=& a_{\mu}b^{\mu} \\ &=& \eta_{\mu \nu}a^{\mu}b^{\nu} \end{eqnarray} is invariant under boosts. Now, what are the transformations that leave this scalar product (and by ...


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It is a transform, not a tensor. Tensors describe a physical quantity at a selected point in time-space and have to transform accordingly. But the Lorentz matrix doesn't describe any physical quantity in a single frame, it's just a change of variables between two coordinate frames. You can understand this by analogy with 3D rotations. Transforms are composed ...



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