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The Lorentz factor is ubiquitous in Special Relativity and is used to express "how much the measurements of time, length, and other physical properties change for an object while that object is moving."

However, in my fairly introductory study of General Relativity, mention of the Lorentz factor disappears. Does the Lorentz factor have any relevance in General Relativity and if so, what is its intuitive meaning?

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$γ$ is a geometrical factor that would appear in Euclidean geometry also if it were taught in a slightly different way. A transformation between two Cartesian coordinate systems with a common origin can be written $\begin{pmatrix} \cos θ & \sin θ \\ -\sin θ & \cos θ \end{pmatrix}$ where $θ$ is the angle between them. If you parametrize it by a slope $m=\tan θ$ instead, the rotation matrix can be written $\displaystyle \frac{1}{\sqrt{1+m^2}} \begin{pmatrix} 1 & m \\ -m & 1 \end{pmatrix}$. Likewise, a transformation between Lorentzian coordinate systems can be written either $\begin{pmatrix} \cosh α & \sinh α \\ \sinh α & \cosh α \end{pmatrix}$ or $\displaystyle \frac{1}{\sqrt{1-v^2}} \begin{pmatrix} 1 & v \\ v & 1 \end{pmatrix}$, where $α$ is a hyperbolic angle (rapidity) and $v=\tanh α$.

Einstein used $v$ in his original 1905 paper, and the teaching of special relativity is still largely stuck in 1905. For general relativity, he switched to an abstract four-tensor notation, where velocities are typically represented not by $v$ or $α$ but by a vector tangent to the worldline (four-velocity), and where it's uncommon to use Lorentzian coordinate systems at all, much less need to transform between two of them. $γ$ has the same meaning as in special relativity, and you will see it occasionally, but it's less common to need it.

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Abstracting the answer by @benrg, the time-dilation factor $\gamma$ is essentially the dot-product between future timelike unit-vectors---think of it as "cosine" (the hyperbolic-cosine of the rapidity between two 4-velocities), which is used to project down components with respect to a 4-velocity. Tensorially, it would appear as $\gamma=g_{ab}u^a v^b$.

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  • $\begingroup$ And can one can prove this by assuming a frame in which $u^{a}=(1, 0, 0, 0)$ and they would measure $v^b=(\gamma, \gamma \vec{v})$? $\endgroup$
    – user246795
    Commented Jan 27, 2023 at 9:18
  • $\begingroup$ Related physics.stackexchange.com/q/494953/246795 (as proof of $\gamma$ result) $\endgroup$
    – user246795
    Commented Jan 27, 2023 at 9:31
  • $\begingroup$ @user246795 I think that would be fine, presumably $(1,0,0,0)$ is in rectangular $(t,x,y,z)$-coordinates in the vector space at an event. $\endgroup$
    – robphy
    Commented Jan 27, 2023 at 14:50
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Briefly speaking, the Lorentz factor $\gamma$ is associated with a Lorentz transformation $\Lambda^{\mu}{}_{\nu}$ in SR, more specifically a Lorentz boost. In GR Lorentz transformations $\Lambda^a{}_b$ transform the vielbein $e^a{}_{\mu}$.

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The Lorentz factor turns up in GR when physical observables are involved, particularly under transformations between different (local) frames. Sadly there is usually little attention given to such observables in textbooks and courses, which is why you haven't seen the Lorentz factor there. The formula $\gamma = -\langle\mathbf u,\mathbf v\rangle := -g_{\mu\nu}u^\mu v^\nu$ between two 4-velocities at the same point (as mentioned in robphy's answer), does appear in Carroll §2.5 and Hartle §5.6, but not many others. Their "relative 3-velocity" $\gamma^{-1}\mathbf v - \mathbf u$, as one might term it, does appear in Misner, Thorne & Wheeler §2.8 eqn 2.35. You will find the Lorentz factor a lot in the "spacetime splitting" literature by Bini, Jantzen, and others in this field, but it gets fairly technical.

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