# How does the Lorentz transformation $\Lambda^{\mu}{}_{\nu}$ transform?

For example the Four-velocity transforms as $$U^{a'}=\Lambda^{a'}{}_{\nu}U^{\nu},$$ the Faradaytensor as $$F^{a'b'}=\Lambda_{\,\,\mu}^{a'}\Lambda_{\,\,\nu}^{b'}F^{\mu\nu}$$ or in Matrixnotation: $$F'=\Lambda F\Lambda^{T},$$ where $\Lambda^{T}$ is the Transpose of the Matrix.

But the Lorentz matrix $\Lambda^{\mu}{}_{\nu}$ is not a tensor. Does $\Lambda$ transform anyway like a second rank tensor in the same way like the Faradaytensor?

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 (multiplied) together: if you go from coordinates $a$ to $a'$ and then from $a'$ to $a''$, you multiply the transforms to get a direct transform from $a$ to $a''$.

• Hi orion, Thanks for your comment, it clarified in a precise way, why $\Lambda$ is not a tensor. – v217 Nov 13 '14 at 11:39

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 two overlapping charts of a manifold. Vectors transform like the members of tangent spaces to manifolds. "Covectors" or one-forms are linear functionals of vectors. Then tensors are simply general multilinear functionals of vectors and / or one-forms. I really like the language and teaching style of Misner Thorne and Wheeler "Gravitation" here, also replicated in the early part of Kip Thorne's lectures here.

The Lorentz transformation, on the other hand is a kind of co-ordinate transformation, and, as such, a vector / oneform / tensor must, by definition, transform in the prescribed way by it.

So tensors, vectors and n-forms are defined by how their components behave in response to co-ordinate transformations. If you like, tensors, vectors and n-forms are a kind of "software" (or "machine" to use Kip Thorne's and Misner/Thorne/Wheeler wording) that is built to a specification of how that software/machine must react to various inputs. In this analogy, the Lorentz transformation is one of the inputs for the machine addressed by the specification.

• Thank you but my question was motivated by this article (pp.3-4) arxiv.org/abs/1305.5210. In this article the author describes how the Faraday-tensor can be viewed as the generator of the general lorentz transformation. Knowing that the Faraday-2nd-rank-tensor transforms as such, I wondered that it should be interesting to find out in a general way how the Lorentz transform 'Lorentz transforms' in a general way. – v217 Nov 13 '14 at 12:44
• @user22207 Please put that reference and statement into your question, as we're all reading your question, and level, completely wrongly without this information. – WetSavannaAnimal Nov 13 '14 at 21:06
• @user22207 This question was active today, so I saw your comment again.Recently I was thinking about this kind of thing from a different perspective.The comments about the Lorentz transformation still stand, but, for the Faraday tensor,there is a simple reason it "generates" the Lorentz group and that is because it acts on a charged particle's four-velocity to map the latter to its rate of change. But four velocities don't change their norm under this (or any) action (unlike, e.g. the momentum four-vector) - it is always equal to $c$. See here – WetSavannaAnimal Sep 12 '15 at 4:41

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 $x^\mu \mapsto y^\mu(x)$, an element of the tangent space will transform as

$$v^\mu \mapsto \frac{\partial y^\mu}{\partial x^\nu}v^\nu$$

while an element of the cotangent space transforms as

$$v_\mu \mapsto \frac{\partial{x^\nu}}{\partial y^\mu}v_\nu$$

These extend by linearity to arbitrary tensor products of the (co)tangent spaces. Now, a Lorentz transformation is just a coordinate transformation such that $x \mapsto \Lambda x$, so that the derivatives acting on the (co)tangent spaces are $\Lambda$ and $\Lambda^{-1}$ at every point, respectively.

The $\Lambda$ is not a member of any (co)tangent space, it does not belong to a particular point, but is the derivative of the general coordinate transformation applied.

• Have a look at the OP's comment to my answer. We're all answering the wrong question, but then the OP hasn't really really asked that question yet! – WetSavannaAnimal Nov 13 '14 at 21:08
• @WetSavanna: Ah. That is a weird article. They first say $F$ "generates" Lorentz trafos, which holds for constant $F$ since the Lie algebra of the Lorentz group is the antisymmetric matrices, but then essentially complexify to get a rep of $SU(2) \times SU(2)$ out of the rep of the Lorentz group, and make a big deal out of the complexified EM tensor. The use of the prepotential over the usual gauge potential is wholly unclear to me. I'm not sure what question the OP precisely has, but this paper would benefit immensely from a bit more group theory and a little less blind calculation. – ACuriousMind Nov 13 '14 at 21:47
• I'm glad for your comments - I found the paper pretty incomprehensible, at least without a great deal of reading work. – WetSavannaAnimal Nov 13 '14 at 22:44

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 implication the metric) invariant? Examine this scalar product $$a' \cdot b' = (\Lambda^{\mu}_{\lambda}a^{\lambda})\eta_{\mu \nu}(\Lambda^{\nu}_{\rho}b^{\rho})$$ Clearly this works for all matrices $\Lambda$ such that $$\Lambda^{\mu}_{\lambda}\Lambda^{\nu}_{\rho}\eta_{\mu \nu} = \eta_{\lambda \rho}$$ Celarly this is just $\Lambda^{T} \eta \Lambda$ in matric notation.

• I am sure orion is right $\Lambda$ is not a Tensor. See any book on special relativity, e.g. a really good one "Relativity made relatively easy" by Andrew Steane. – v217 Nov 13 '14 at 11:37
• @user22207 qute right, had a brain melt there; updated to reflect your point. Many thanks, A. – Autolatry Nov 13 '14 at 13:29