# What is global Lorentz transformation and what is local Lorentz transformation?

I will consider $$\textbf{spacetime}$$ as $$(M,\eta)$$ where $$M$$ is a four dimensional $$\textbf{manifold}$$ and $$\eta$$ the metric which in this coordinates \begin{align*} x \colon M &\longrightarrow \mathbb{R}^4\\ p &\mapsto x(p)=(x_0,x_1,x_2,x_3). \end{align*} is given by $$\eta=dx^0\otimes dx^0-dx^1\otimes dx^2-dx^2\otimes dx^1-dx^3\otimes dx^3 \tag1$$

An $$\textbf{observer}$$ is a worldline $$\gamma$$ with together with a choice of basis $$O=v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda)$$ of each $$T_{\gamma(\lambda)}M$$ where the observer worldline passes, if $$\eta(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} \tag2$$

$$v_{\gamma,\gamma(\lambda)}$$ is the tangent vector of the the curve $$\gamma$$ at the point $$\gamma(\lambda)$$

In text books i've found three definition of $$\textbf{Lorentz transformation} \quad \Lambda$$

1. $$\Lambda \colon \mathbb{R}^4 \longrightarrow \mathbb{R}^4$$ is a group of coordinate transformations that leave eq.1 in the same form ,that is $$\Lambda \cdot x(p)=y(p)=(y_0,y_1,y_2,y_3)$$ such that in this coordinate $$\eta=dy^0\otimes dy^0-dy^1\otimes dy^2-dy^2\otimes dy^1-dy^3\otimes dy^3$$
2. $$\Lambda \colon M \longrightarrow M$$ a Spacetime diffeomorphism such that $$\Lambda_* \eta=\eta$$ where $$\Lambda_* \eta$$ is the pullback of the metric $$\eta$$
3. $$\Lambda \colon T_pM \longrightarrow T_pM$$ such that $$\Lambda O=O'=e'_0(\lambda) , e'_1(\lambda), e'_2(\lambda), e'_3(\lambda)$$ satisfy the eq.2 that is $$\eta(e'_a(\lambda), e'_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab}$$

My question is which is of these transformation is global Lorentz transformation and which is local?

• Minor question: What's with the off-diagonal terms $dx^1\otimes dx^2$, $dy^1\otimes dy^2$, etc? – Qmechanic Apr 15 '20 at 20:13

The three definitions are the same. They are ways of saying the same thing. Since you have a manifold $$(M,\eta)$$ this is a flat, Minkowski, spacetime. The Lorentz transformation is global on Minkowski spacetime.

In a curved spacetime the metric is usually denoted $$g$$, rather than $$\eta$$. $$g$$ is, in general, a function of time and position. At each point there is a Minkowski tangent space, meaning that the manifold is locally Minkowski (to the accuracy of measurement) and that local Lorentz transformations can be applied within a neighbourhood of each point.

• They are not the same. Suppose we are measuring the spin of a particle.1 Means just change of coordinates and since coordinates does no matter in physics we could choose say polar coordinates and the physics would be the same in other words 1 is imagination of our mind ... – amilton moreira Apr 15 '20 at 19:17
• 2. Is something that is happen in the real world that is we are rotating the observer and the spinor. I does not have a consequence in physics because it is a symmetry of the metric .... – amilton moreira Apr 15 '20 at 19:22
• 3. Is a rotation of the observer (the apparatus) in this case the spinor get a fase – amilton moreira Apr 15 '20 at 19:23
• There are no spinors here, and you have chosen Minkowski coordinates given by $\eta$. You already specified them, they cannot be polar. – Charles Francis Apr 15 '20 at 19:46
• I choosed the coordinates just to specify the metric and to define lorentz transformation as most books do . I did not say that the problem would be in this coordinates – amilton moreira Apr 15 '20 at 20:05

The basic definition of the Lorentz transformation is, given a vector space $$V$$ equipped with a Minkowski metric $$\eta$$, it is the group that leaves the norm invariant (in other words, definition 1).

In a spacetime, every tangent space can be considered as a copy of Minkowski space, ie for every $$p \in M$$, $$T_p M \cong V$$, since by Sylvester's law, we can always put the metric tensor $$g_p$$ at that point in the appropriate form by a change of basis (that basis being an orthonormal basis $$\{ e_\mu \}$$). Then at each point, you can perform a Lorentz transform of that basis.

Diffeomorphisms on a manifold are a rather large class, but there exists a subset of diffeomorphisms such that, if $$\phi \in \mathrm{Diff}(M)$$, the pushforward on a vector at $$p$$, $$\phi^* v$$, corresponds to a Lorentz transform. Just by having say

$$$$\frac{\partial x^\mu(p)}{\partial y^\nu(p)} \in \mathrm{SO}(3,1)$$$$

Locally, in the Riemann/Fermi coordinates, this is roughly equivalent to a Lorentz transform, since the normal neighbourhood is diffeomorphic to $$T_pM$$.