- #1

jk22

- 723

- 24

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 ?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jk22
- Start date

- #1

jk22

- 723

- 24

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 ?

- #2

jbriggs444

Science Advisor

Homework Helper

- 11,576

- 6,225

- #3

jk22

- 723

- 24

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##

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##

Last edited:

Share:

- Replies
- 23

- Views
- 477

- Last Post

- Replies
- 13

- Views
- 430

- Last Post

- Replies
- 5

- Views
- 428

- Last Post

- Replies
- 1

- Views
- 329

- Last Post

- Replies
- 1

- Views
- 209

- Replies
- 10

- Views
- 374

- Replies
- 6

- Views
- 373

- Last Post

- Replies
- 3

- Views
- 231

- Replies
- 1

- Views
- 232

- Replies
- 17

- Views
- 701