Jump to content

hyper-space has 10D ??


Widdekind

Recommended Posts

in general, GR metric matrices (can) have (up to) ten independent components…



so…



[math]ds^2 = \begin{array}{r} \left( dt dx dy dz \right) \end{array} g_{\mu \nu} \begin{bmatrix} dt \\ dx \\ dy \\ dz \end{bmatrix}[/math]



and we would want to embed the 4D manifold, into a higher dimensional hyperspace, of some unknown number of dimensions, which was flat, such that



[math]ds^2 = dx_1^2 + dx_2^2 + dx_3^2 + dx_4^2 + \cdots + dx_N^2[/math]



you'd have to imagine mapping each manifold-enscribing coordinate (t,x,y,z), into an (unknown number) of "hyper-space coordinates":



[math]\begin{bmatrix} t \\ x \\ y \\ z \end{bmatrix} \longrightarrow[/math][math] \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \cdots \\ x_N \end{bmatrix}[/math]



so that, when, on the manifold, you began at point (t,x,y,z), and sought a small step to another nearby point (t+dt,x+dx,y+dy,z+dz), you were actually taking a straight step, through a higher-D hyperspace:



[math]dx_1 = \frac{\partial x_1}{\partial t} dt + \frac{\partial x_1}{\partial x} dx + \frac{\partial x_1}{\partial y} dy + \frac{\partial x_1}{\partial z} dz[/math]


[math]dx_2 = \frac{\partial x_2}{\partial t} dt + \frac{\partial x_2}{\partial x} dx + \frac{\partial x_2}{\partial y} dy + \frac{\partial x_2}{\partial z} dz[/math]



etc.



so, how many dimensions would the higher-D hyper-space require ??



on the RHS:



[math] = \left( \frac{\partial x_1}{\partial t} dt + \frac{\partial x_1}{\partial x} dx + \frac{\partial x_1}{\partial y} dy + \frac{\partial x_1}{\partial z} dz \right)^2 + \left(\frac{\partial x_2}{\partial t} dt + \frac{\partial x_2}{\partial x} dx + \frac{\partial x_2}{\partial y} dy + \frac{\partial x_2}{\partial z} dz\right)^2 + \cdots[/math]



you would wind up, with all of the same sorts of manifold-four-coordinate-differential pairs, e.g. [math]dt^2, dt dx[/math], on the RHS, as on the LHS...



but on the RHS, those differentials would be multiplied, by partial derivatives of the unknown hyper-coordinate mapping functions...


whereas on the LHS, those same differentials would be multiplied, by the 10 independent components of the metric matrix...



so, to have as much mathematical flexibility, on the RHS, as on the LHS, you'd have to pick 10 independent coordinate mapping functions, i.e. 10 independent coordinates..


i.e. hyper-space has 10D



essentially, you're re-casting all of the curvature information, w/in the metric matrix' components, into "dis-entangled" linearly independent orthogonal coordinate components, of a higher-D hyper-space



is this correct ??




simple analogy:



a 1D line, can require 3D, to curve through... you can construct 1D linear line-land spaces, which curve more complexly, than solely simply thru 2D on a plane page of paper




extension:



if space-time itself has more dimensions, e.g. 5, then hyper-space would have to have even more dimensions, to accommodate the complex curvatures conceivable, w/in the fabric of space-time, e.g. [math]5!^+ = 15[/math].

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.