JustinLevy
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If we start with the Lorentz transformation
<br /> \begin{align*}<br /> ct' &= \gamma (ct - \beta x) \\<br /> x' &= \gamma (x - \beta ct) \\<br /> y' &= y \\<br /> z' &= z<br /> \end{align*}<br />
with the usual \beta = v/c, \gamma = 1/\sqrt{1-\beta^2}
and take the limit c \rightarrow \infty, then we get:
<br /> \begin{align*}<br /> t' &= t \\<br /> x' &= x - v t \\<br /> y' &= y \\<br /> z' &= z<br /> \end{align*}<br />
ie. we get the Galilean transformations.
But what of spacetime? Can we still discuss a spacetime manifold? If I look at the line element in an inertial frame:
ds^2 = g_{ab} dx^a dx^b = c^2 dt^2 - dx^2 - dy^2 - dz^2
taking c \rightarrow \infty seems to render it as nonsense (or at least I am unsure how to interpret it).
Is there a way to have a Galilean spacetime?
And if we instead make Galilean symmetry just a local symmetry, can we obtain something like a "Galilean version" of GR?
<br /> \begin{align*}<br /> ct' &= \gamma (ct - \beta x) \\<br /> x' &= \gamma (x - \beta ct) \\<br /> y' &= y \\<br /> z' &= z<br /> \end{align*}<br />
with the usual \beta = v/c, \gamma = 1/\sqrt{1-\beta^2}
and take the limit c \rightarrow \infty, then we get:
<br /> \begin{align*}<br /> t' &= t \\<br /> x' &= x - v t \\<br /> y' &= y \\<br /> z' &= z<br /> \end{align*}<br />
ie. we get the Galilean transformations.
But what of spacetime? Can we still discuss a spacetime manifold? If I look at the line element in an inertial frame:
ds^2 = g_{ab} dx^a dx^b = c^2 dt^2 - dx^2 - dy^2 - dz^2
taking c \rightarrow \infty seems to render it as nonsense (or at least I am unsure how to interpret it).
Is there a way to have a Galilean spacetime?
And if we instead make Galilean symmetry just a local symmetry, can we obtain something like a "Galilean version" of GR?
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