# Has the time "dimension" no spatial extension?

It's just think that if we measure 1 second with a clock we should be able to "see" a 300'000km long piece of something in space or not ?

Or does the time extension only has to be understood as a set of numbers indicating timelaps, so that there is no "geometry" of time ?

• PeroK

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jbriggs444
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None of the above. It is just a four dimensional space time geometry. There is no physical fabric on which a clock could leave a 300,000 km brown streak.

• Dale and PeroK
So the fourth dimension were in some sense hidden to the eye ?

Rephrasing the question were like this, maybe more precise :

Is it possible to have a metric coefficient for time depending only on r but in external geometry the manifold coordinate for time depends also on let say the polar angle ?

I know that GR works only with the metric and hence shall be independent of the embedding. I think this is based on Gauß theorem of intrinsic curvature, but if I remember well this says the intrinsic curvature does not depend on reparametring the manifold, but somewhere the shape of the manifold has to be given.

Like we give a manifold via ##f:\mathbb{R}^2\rightarrow\mathbb{R}^3## then the curvature does not change if we compose f with a diffeomorphism ##g:\mathbb{R}^2\rightarrow\mathbb{R}^2##.

But it seems to me that the intrinsic curvature is but depending on f of course, but not g ?

Since for example symmetry arguments would depend on f too, one needs to find f.

I was asking myself which work should be done, like :

Find ##M1\in M_{5\times 5} s.t. \exists P\in M_{4\times 5} s.t. P^T M1 P=(g_{\mu\nu})## where the latter is the GR given metric and then find ##f:\mathbb{R}^4\rightarrow\mathbb{R}^5## s.t ##M1=\langle e_i|e_j\rangle## with as usual ##e_i=D_i f##

But then since this 5th dimension is unknown shall it be considered only mathematically a euclidean background metric hence ##e_i^Te_j## or an asymptotic vanishing dimension with ##\lim_{\epsilon\rightarrow 0}e_i^T\cdot diag(1,1,1,-1,\epsilon)e_j##

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