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space time curvature


alejandrito20

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hello

i understand that in a flat space the metric is [math]\eta_{uv}dx^udx^v[/math]...i know that this means that the light follows straight geodesic in this space time...

 

but ¿what would means that metric is [math]f(t)\eta_{uv}dx^udx^v[/math] where f(t)=infinite in t=0 and f(t)=0 in t=infinite.....obvious i understand the matematics, but physically ¿what means?.....for example..¿what means that in bing bang in t=0 f(t)= infinite????

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Assuming something like [math]t =x^{0} [/math] the metric you present looks conformally equivalent to the Minkowski metric. (Not sure if the [math]f(t)= 0[/math] or the [math]f(t)= \infty[/math] course trouble, so maybe mod that and the statement that the function is positive).

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Assuming something like [math]t =x^{0} [/math] the metric you present looks conformally equivalent to the Minkowski metric. (Not sure if the [math]f(t)= 0[/math] or the [math]f(t)= \infty[/math] course trouble, so maybe mod that and the statement that the function is positive).

 

yes [math]t =x^{0} [/math] , [math]f(t=0)=\infty[/math],[math]f(t=\infty)=0[/math], f(t) is positive.

 

the metric in t= 0 is [math]\infty \eta_{uv}dx^u dx^v[/math], in [math]t=\infty[/math] is [math]0\eta_{uv}dx^u dx^v[/math].....

 

physically....¿what would mean this???

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