# Lorentz transformations and coordinate transformations - some doubts

We have the Lorentz transformation $$\Lambda$$ that acts on the vectors and tensors and the meaning of the $$\Lambda$$ is that it modifies the entries of a vector or tensors ( $$V^{\mu}$$ or $$T^{\mu \nu ...}$$ ). These transformations are special in special relativity (SR) because they leave the $$ds^2$$ invariant (are they therefore called isomorphisms?).

The Lorentz transformation is in itself also a coordinate transformation, because if you have a boost of the system from S to S', for example, you can use the coordinate adapter to S' and the relation between the coordinate in S' and S is a Lorentz transformation that it can be also thought as coordinate transformation. So you have for the coordinates in S' $$x' ^{\mu} = \Lambda^{\mu} {}_{\nu} x^{\nu}$$ In SR $$\Lambda$$ is a matrix with constant elements so the infinitesimal Lorentz transformation is just $$dx'^{\mu} = \Lambda^{\mu} {}_{\nu} dx^{\nu}$$ so here $$\Lambda^{\mu} {}_{\nu}$$ is also the Jacobian of the change of coordinates. Is it correct?

When taking a generic coordinate transformation this does not leave the line element invariant in general. This transformation acts on the coordinates with a matrix again, we call it $$M^{\mu} {}_{\nu}$$ in the following way $$x' ^{\mu} = M^{\mu} {}_{\nu} x^{\nu}$$ and now $$M$$ could depend on $$x$$.

And to calculate the differential of a coordinate transformation you need the Jacobian so the infinitesimal coordinate transformation is $$dx' ^{\mu} = J^{\mu} {}_{\nu} dx^{\nu}$$  where $$J^{\mu} {}_{\nu} = \partial ( x' ^{\mu} ) / ( \partial x^{\nu} )$$

Are the components of a vector in these new coordinates are obtained using $$V^{\mu} = M^{\mu} {}_{\nu} V^{\nu}$$ ?

No, they are obtained using: $$V'^\nu=V^\mu\left(M^\nu_\mu+x^\alpha\frac{\partial M^\nu_\alpha}{\partial x^\mu}\right)$$

I do not understand the point of separating the coordinate transformation with the matrix $$M$$, if the transformation is not linear though. Are you assuming some special form of $$M$$ matrix? If not, just write the 4 transformation functions: $$x'^\nu=x'^\nu(x^\mu)$$

I think you should first and foremost look at what a vector is and try to derive the transformation properties straight from the definition. There are more possible definition of vectors and I do not know which one are you familiar with.

The most common definition for vector in physics is n-tuple of numbers that transform under coordinate transformation with this rule $$V'^\mu=\frac{\partial x'^\mu}{\partial x^\nu}V^\nu,$$ which leads to the transformation of components I have written at the beginning.

Another definition of vectors, that is used quite often when dealing with curvilinear coordinates or with curved spaces (spacetimes) is that vector is linear operator on functions on the space (spacetime), i.e.

$$V=V^\mu\frac{\partial}{\partial x^\mu}.$$ (now, you can apply your 4-vector on some function on spacetime $$f(x)$$, to get the speed of change of the function in the direction of the vector $$V$$: $$Vf=V^\mu \frac{\partial f}{\partial x^\mu}$$) You can easily see this transforms according to our rule: $$V^\mu\frac{\partial}{\partial x^\mu}=V^\mu\frac{\partial x'^\nu}{\partial x^\mu}\frac{\partial}{\partial x'^\nu}=V'^\nu \frac{\partial}{\partial x'^\nu}\Rightarrow V'^\nu= V^\mu\frac{\partial x'^\nu}{\partial x^\mu}.$$ The isomorphism between these two is given by $$(1,0,0,0)\leftrightarrow \frac{\partial}{\partial x^0},$$ and analogically for other base vectors.

These transformations are special in special relativity (SR) because they leave the ds2 invariant (are they therefore called isomorphisms?)

They are called isometries.

• Thanks a lot, this helped to clarify a bit my ideas! I indeed was (for no good reason) thinking about only linear coordinate transformations. Just one last question: when I said that the $\Lambda^{\mu}_{\nu} matrix is also the Jacobian of the coordinate transformation is that correct? Does it make sense to think about the Lorentz transformations as coordinate transformations? Commented Nov 16, 2020 at 8:37 • @Phrancesco The Lorentz transformations are coordinate transformations. Just google, them they are most often written as such. For example on wikipedia they are written in form$t'=t'(t,x)\$ and so on. Even from physics view, they compare two frames of references, i.e. they compare two sets of spatial and time coordinates. How did you arrive at the idea they are not? Commented Nov 16, 2020 at 10:03
• Yap, I think I agree. I was trying to understand if the Lorentz transformations are more than just special coordinates transformations that leave the line element invariant. If they are a different mathematical object. Commented Nov 16, 2020 at 10:07