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For a 3-D manifold ($r,\theta,\phi$) and sub-manifold (2-sphere)($\theta,\phi$) , $\hat r$ is orthogonal to both the 2-sphere and tangent space associated with each point of the 2-sphere. Similarly for 4-D manifold($r,\theta,\phi,t$) ,and sub-manifold(hypersurface)($r,\theta,\phi$) if $\hat t$ is orthogonal to the hypersurface(space like) for which the metric becomes $$ds^2=g(t)dt^2+f(t)\delta_{ij}\gamma_{ij}dx^idx^j$$(no cross terms between $t$ and $x^{is}$)where $x^i$ are the spatial coordinates ,is it necessary that $\hat t$ is also orthogonal to the tangent space associated with each point of the hypersurface??

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The components of the metric tensor are defined as $g_{\mu \nu} = \partial_\mu \cdot \partial_\nu$, where $\cdot$ is the scalar product and $\partial_\mu$ is a coordinate basis vector.

If a metric has no off-diagonal terms, the coordinate basis vectors are othogonal to each other.

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  • $\begingroup$ Yes it is but my question is if $\hat t$ is orthogonal to spatial hypersurface ,then whether $\hat t$ will also be orthogonal to tangent space associated with each point of the spatial hypersurface?? $\endgroup$ – Apashanka Das Mar 4 at 18:25
  • $\begingroup$ The tangent space to each point of the spacelike hypersurface is made up of the coordinate bases associated to the coordinates $(r, \theta, \phi)$. A vector, in this case $\hat t$, is orthogonal to a hypersurface, if it is orthogonal to the tangent space of the hypersurface. It is a definition. $\endgroup$ – Michele Grosso Mar 5 at 8:38

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