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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?

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

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  • $\begingroup$ 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 ... $\endgroup$ – amilton moreira Apr 15 at 19:17
  • $\begingroup$ 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 .... $\endgroup$ – amilton moreira Apr 15 at 19:22
  • $\begingroup$ 3. Is a rotation of the observer (the apparatus) in this case the spinor get a fase $\endgroup$ – amilton moreira Apr 15 at 19:23
  • $\begingroup$ There are no spinors here, and you have chosen Minkowski coordinates given by $\eta$. You already specified them, they cannot be polar. $\endgroup$ – Charles Francis Apr 15 at 19:46
  • $\begingroup$ 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 $\endgroup$ – amilton moreira Apr 15 at 20:05
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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

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

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

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