Metric of 4-D space time

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  • Thread starter Apashanka
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Main Question or Discussion Point

Our 4-D space is ##x^1,x^2,x^3 ,t##.
Our sub-manifold is defined by ##(x^1,x^2,x^3)##
Therefore for this sub-manifold to be maximally symmetric and for which the tangent vector ##\frac{∂}{∂t}(\hat t)## orthogonal to this sub-manifold
The metric becomes,
##ds^2=g(t)dt^2+f(t)(dr^2+r^2d\Omega^2)##
From the known metric for 4-D space and comparing this with above ##g(t)=-1## and for static ##f(t)=1## and for expanding ##f(t)=a(t)^2## ,is it the case ??
And can we say that the hypersurface to any t is orthogonal to the increament direction of t??
 

Answers and Replies

  • #2
PeterDonis
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the known metric for 4-D space
First, spacetime, not space. You got this right in the title of the thread.

Second, what metric for 4-D spacetime? There is no single "known" metric; it depends on what spacetime you want to look at.

comparing this with above ##g(t)=-1## and for static ##f(t)=1## and for expanding ##f(t)=a(t)^2## ,is it the case ??
Is what the case? What are you asking?

can we say that the hypersurface to any t is orthogonal to the increament direction of t??
If you mean, is every hypersurface of constant ##t## orthogonal to the vector ##\partial / \partial t##, you should be able to answer that from the last thread we had on that subject.
 

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