Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

180 submissions , 140 unreviewed
4,479 questions , 1,782 unanswered
5,140 answers , 21,866 comments
1,470 users with positive rep
695 active unimported users
More ...

  The continuous tensor calculus

+ 1 like - 0 dislike
63 views

It is usual to have tensors in general relativity with discret index in the integer numbers. I propose to have index in the real numbers so that we can have for example:

$$\tilde A^t =\int_{-\infty}^{+\infty} A^{t'} (\frac{\partial \tilde x^t}{\partial x^{t'}}) dt'$$

The differential of a function is:

$$df=\int_{-\infty}^{+\infty} \frac{\partial f}{\partial x^t} dx^t dt  $$

We have also:

$$\int_{-\infty}^{+\infty}(\frac{\partial \tilde x^t}{\partial x^{t'}})(\frac{\partial x^{t'}}{\partial \tilde x^{t''}})dt'=\delta (t-t'')$$

The sums are replaced by integrals. The points of the manifold are replaced by smooth functions. The coordinates are:

$$x^t(f)=f(t)$$

Can we make Einstein general relativity with continuous tensor calculus?

asked Jun 20 in Mathematics by Antoine Balan (255 points) [ no revision ]

You are proposing an "index set" \(I\subset{\Bbb R}\) with coordinates as mappings \(x:I\rightarrow{\Bbb R}, x\mapsto x(t)\equiv x^t\). In particular you chose \(I=(-\infty,+\infty)\). A question that occurs to me w.r.t. your first equation (coordinate transformation) is: What is the meaning of \(\partial {\tilde x}^t / \partial x^{t'}\)? Is this a well-defined object? By definition it should be the derivative of the value of the function \(\tilde x\) at point \(t\) w.r.t. the value of another function, \(x\), at point \(t'\). Once this is clarified, the next question is: Does the integral exist?

Similar questions can be asked w.r.t. \({\rm d}x^t\) and \(\partial f / \partial x^t\).

Regarding your final equation:

If you have \(x^t(f)=f(t)\), what then is \({\tilde x}^t(f)\)?

Assuming such issues can be clarified, one could also consider different measures in the integral or fractal index sets. A further possibility is arbitrary index sets (countable and non-countable) with a definition of summation as known for families in a normed vector space.

This approach might also stimulate further thought on different concepts of dimension.

Given that spacetime has (or appears to have) four dimensions (Why? I don't know. Quantum decoherence?), the question is of the relevance of this approach to general relativity. For some procedures treating the number of dimensions as a continuous quantity might be useful or even a prerequisite.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...