The Ricci Flow For The Scalar Curvature And Ricci Tensor

[Dubbelosix]

The Ricci flow is like a covariant derivative acting on [math]R[/math],

 

[math]\nabla R =\ ^{g}\nabla R[/math]

 

where [math]^{g}\nabla[/math] is the covariant derivative for some Riemannian metric and the curvature scalar is just [math]R \equiv R_{ijkl}g^{ik}g^{jl}[/math] (using contractions) and for the Ricci curvature tensor we have 

 

[math]\nabla_{\gamma} R^{\alpha }_{\beta \mu \nu } = \frac{\partial R^{\alpha }_{\beta \mu \nu }}{\partial x^{\gamma }} + \Gamma ^{\alpha }_{\sigma \gamma }R^{\sigma }_{\beta \mu \nu } - \Gamma^{\sigma }_{\gamma \beta }R^{\alpha }_{\sigma \mu \nu } - \Gamma ^{\sigma }_{\gamma \mu }R^{\alpha }_{\beta \sigma \nu } - \Gamma ^{\sigma }_{\gamma \nu }R^{\alpha }_{\beta \mu \sigma }[/math]

 

The d'Alembertian operator is the generalized covariant derivative in relativity. It's form is simply:

 

[math]\frac{\partial R}{\partial t} = \alpha \Box^2 R = -\frac{\alpha}{c^2}\frac{\partial^2 R}{\partial t^2} + \alpha \nabla^2 R[/math]

 

Where we use the squared definition of [math]\Box[/math] to highlight the squared property of the Laplace operator. I arrived at this equation independently but later find out a similar equation has been suggested. I can only trace a non-relativistic form suggested by Sivaram and Arun. All three of these equations characterise a peculiar property that geometry flows like heat. In fact, it has even been suggested in literature that the Ricci flow [is] the heat equation for a Riemannian manifold. These equations were arrived at from an investigation into how temperature related to gravity. 

Edited December 28, 2017 by Dubbelosix

[Vmedvil]

woo, I can easily use dubblesiox's equations now with that Laplace operator "∇"and d'Alembertian operator there. 

 

being 

 

Which can fit into the Schrodinger's equation spot or be added or subtracted as long as it equals zero with the other parts. 

 

Edited December 28, 2017 by Vmedvil

[Vmedvil]

Be careful, there are two definitions for the d'Alembert. [math]\Box[/math] has only first derivatives, just like [math]\nabla[/math] but [math]\Box^2[/math] is the definition of the second derivative form, which is the form you posted above. 

 

Can't you root and then transform between them?

[Shustaire]

[math]\frac{\partial R}{\partial t} = \alpha \Box^2 R = -\frac{\alpha}{c^2}\frac{\partial^2 R}{\partial t^2} + \alpha \nabla^2 R[/math]

 

Where we use the squared definition of [math]\Box[/math] to highlight the squared property of the Laplace operator. I arrived at this equation independently but later find out a similar equation has been suggested.

 

 

so Your claiming to arrive at the D'Ambertion operator that has been around since the 17 century? Long before GR even existed?

[Shustaire]

Good, as I hate plagiarism.

Edited December 31, 2017 by Shustaire

[Shustaire]

Do you know what the d'Alembertian is?

 

Because there is lot more to what you quoted than just the operator.

Of course I do. I simply provided a quick lookup link previously

 

Any analytical Calculus textbook covers it. Can you provide the left and right wave equations involved?

Edited December 31, 2017 by Shustaire

[Shustaire]

Excuse me ?

 

Are you sure you know that the D'Alembertian is a 1 dimensional wave equation?

 

Thank you for proving to me you plaguarized the above

 

http://mathworld.wolfram.com/dAlembertsEquation.html

 

I was right about this site, its full of crackpots and no moderation.

 

Tell me did the fact that Reimann geometry deals with curve fitting escape you with the use of tangent vectors of which you can have left and right hand tangent vectors escape you.

 

You derived the above my arse.

 

Maybe you should study the connection to PDE's in terms of Reimann geometry a bit closer.

Edited December 31, 2017 by Shustaire

[Shustaire]

Yeah well you claimed to have derived an equation using the D'Alembertian operator and didn't even know its involvement in PDEs and curve fitting.

 

Which quite frankly is in most differential geometry textbooks.

Edited December 31, 2017 by Shustaire

[Shustaire]

Then why didn't you answer my question correctly then.

 

Instead you answered with a totally unrelated topic.

Edited December 31, 2017 by Shustaire

[Shustaire]

No you are what are you describing in geometry? What are you using Vectors not particles fool. Ever looked at path integrals under Reimann geometry.

 

Those are frigging wave functions

 

handedness also refers to covariant and contravariance under geometry. Your flipping coordinate basis.

 

https://www.evl.uic.edu/ralph/508S98/coordinates.html

Edited December 31, 2017 by Shustaire

[Shustaire]

Read this on PDE's

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.mathtube.org/sites/default/files/lecture-notes/Lamoureux_Michael.pdf&ved=0ahUKEwiliPzKuLPYAhULzmMKHep_DZoQFgguMAQ&usg=AOvVaw3_mYcou2fXEe8xIMkXeTuP

In answer to the question I had on D'Alamberton see section 1.3

more specifically section 2.5

Edited December 31, 2017 by Shustaire

[Shustaire]

I'm ignoring you now. I answered any question you asked. The rest is irrelevant and you are wasting my time. 

 

oh really then its amazing I'm correct isn't it

[Shustaire]

Yeah not relevant to my work. Secondly, on section 1.3 it talks about a one dimensional wave equation and the second order Dirac operator. 

 

Then your not doing Reimann geometry

Shall I prove that as well ?

Edited December 31, 2017 by Shustaire

[Shustaire]

I'm saying the left and right handed solutions are not needed in this. You're very annoying, do you tell your mom how to wash the dishes as well?

 

did you miss the coordinate basis right hand rule?

 

Maybe you should look it up sometime.

 

What you hate being shown wrong wanna go run home and cry like a little baby instead of learning the proper way. Or are you disappointed in being shown you missed basic lessons needed to derive those equations you posted.

I have no problem delivering insults right back at you.

 

 I literally do not care what happens to this account

 

Your equations you posted use PDE's not ODE's so the link I provided applies .

Edited December 31, 2017 by Shustaire

[Shustaire]

 

The heat energy density is defined in the following way in simple Cartesian coordinates,

 

[math]\mathbf{Q} = - k \nabla^2 T= -k (\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2})[/math]

 

However, the d'Alembert operator just involves an extra term and is nothing too complicated,

 

[math]\mathbf{Q} = - k \Box^2 T= - k (\frac{\partial^2 T}{\partial \tau^2} - \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial z^2}) = k( \frac{\partial^2 T}{\partial \tau^2} + \nabla^2T)[/math]

 

 

for example

[Shustaire]

Yes I know what it is, the reason why I don't need it is because you simply attach a unit vector to the system and that tells you the direction you are going. In fact, that is actually the common practice for a diffusion equation.

 

 

Now go away, stop pestering me. 

 

Nah I like buggin people that pretend to know something they don't there is no nice about me

[Shustaire]

 

. I arrived at this equation independently but later find out a similar equation has been suggested. I can only trace a non-relativistic form suggested by Sivaram and Arun. All three of these equations characterise a peculiar property that geometry flows like heat. In fact, it has even been suggested in literature that the Ricci flow [is] the heat equation for a Riemannian manifold.

 

 

especially when you made this claim but didn't even understand how PDE's work