# Reflecting on the physics of Gravitational metric

Discussion in 'Physics & Math' started by Trapped, Nov 12, 2013.

1. ### TrappedBannedBanned

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The metric given as the Schwarzschild expression is

$(1 - \frac{2GM}{c^2 r})$

in theory describes the strength of the gravitational field. This expression itself is dimensionless, you can check the units.

The redshift is the deviation of radiation from the gravitational field. In fact, it is radiation from the source which is in the gravitational field. To describe the deviation, you simply restate the metric as a ratio of the metric associated to the received signal to the source given as

$\frac{\sqrt{(1 - \frac{2GM}{c^2 r (rec)})}}{\sqrt{(1 - \frac{2GM}{c^2 r (sou)})}}$

This is a direct result of the equivalence principle.

Now, charges in a [gravitational field] are also rooted from the equivalence principle. The synchroton expression which describes this is

$\frac{2}{3}\frac{e^2}{m^2c^3}$

Keeping this as straight forward as possible, the idea is that both the redshift and the charge in the gravitational field are in fact part of the same system. Looking at the paper

http://gravityresearchfoundation.org/pdf/awarded/1968/motz.pdf

Motz approached his work by putting the original field strength of the field in the denominator of the radiation emitted by the charge. There shouldn't be in theory any reason why the denominator cannot be replaced with the correct ratio of the metrics which describes the redshift of the radiation which is emitted by the charge. Since the factor which describes this is rooted from first principles of equivalence and that a charge in a gravitational field emitting radiation is also rooted from the same principles, then it should be no surprise you can restate this approach as

$\frac{2}{3}\frac{e^2}{c^3} \frac{a^2}{\frac{\sqrt{(1 - \frac{2GM}{c^2 r (rec)})}}{\sqrt{(1 - \frac{2GM}{c^2 r (sou)})}}}$

The denominator is now describing the relativistic dynamics of the redshift which should be present in the equation which describes the charge in our gravitational field.

Is any of this unclear? I spoke to someone last night on the theory of this approach and they helped me with suggestions on things I could say. If any of this is unclear, what parts are? I have done a bit of research into this and I honestly don't see any problems in the approach.

3. ### StryderKeeper of "good" ideas.Valued Senior Member

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What would likely remain unclear is your motives for posting? If you are after some sort of peer review method, then this isn't it, discussion... possible if you are open for that and don't shoot anyone down that is critical about it, otherwise is it just about propagating a bunch of mathematics to fill the internet with bogus non-peer reviewed documentation so any documentation becomes supported by source information which is the same stuff just pasted elsewhere. (Which is where the realms of Pseudoscience rears it's ugly head, and No it wasn't me on the Grassy Knoll that placed your "debate" thread in Pseudoscience.)

That's what the moderators of the many forums, blogs and paper archives currently consider of your endeavours to "publish" this. (And in that light, they won't necessarily look so deeply at the contents)

5. ### TrappedBannedBanned

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When I first came here, I asked about the physics behind the idea; which was actually going to lead to another question I had. This was before all the nonsense which happened.

Assuming these are first principles, wouldn't it also mean that not all observers would actually agree with radiation rates? The radiation rate is determined by the Einstein shift and is intimately tied to ticking relativistic clocks. That was going to be my question.

There were some other things I wanted to talk about this approach - I really liked it and took a lot of interest in it.

7. ### TrappedBannedBanned

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As for the question of whether the content is pseudo, I think I have demonstrated quite well above that this isn't about butchering equations or making faulty premises. I thought my OP was very straight forward.

8. ### originHeading towards oblivionValued Senior Member

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It is nice to see that you have dropped the 'mysterious friend' and now are posting the pseudo-science stuff as your own. All you have to do now is admitt that you are reiku and we will have completed the journey.

9. ### TrappedBannedBanned

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I don't know what you are talking about. I posted the work above, yes. But it's not my work. I made that clear.

I don't know how you have read what I have written and still assume that it is my work. If it was, I would have just copied and pasted the original work and posted it. So go figure.

I'd prefer if you stayed out of this thread because you don't seem to have anything constructive to say.

10. ### originHeading towards oblivionValued Senior Member

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I don't object to the parts of the paper you copied (not so much anyway), I do object to your additions.

11. ### TrappedBannedBanned

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Again they are not my assertions, but I have also told you to show me, like some working idea why this is not allowed. You haven't showed me anything ... just vacuous accusations on the quality of the work.

The work in the OP, is perfectly solid. I am yet to hear a real objection?

12. ### brucepValued Senior Member

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The objection would be g_shell is a local invariant. I don't give a crap what you think it is measured from remote coordinates. What you wrote down is complete irrelevant nonsense. g_shell isn't frame dependent natural phenomena. rpenner explained it to you mathematically. Your denominator cancels to 1. Everything associated with physics. Terms and units don't exist when it cancels to 1. It's a bigger load of nonsense than expecting a remote result. You misquoted me [crankstyle in the title of your challenge for what I said... I'd have to be brain dead to say something like that] when you didn't have the balls to claim the equation as yours. You're anonymous already dude. If you think it's right then submit it for publication. This is the relativistic dynamics for g_shell [g_shell IS local phenomena]

g_shell = [M_meters/r^2_shell] / (1-2M_meter/r_shell)^1/2

I tried to show you how it works when I wrote down the derivation for the Hawking temperature from the Unruh temperature. The measurement/or calculated prediction for g_shell is frame invariant. Not frame dependent.

Last edited: Nov 12, 2013
13. ### rpennerFully WiredValued Senior Member

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Wrong.

The Schwarzschild vacuum solution to the Einstein equations at the heart of General Relativity describes a metric, a rank-2 tensor field which is written in one particular coordinate choice $(t, r, \theta, \phi)$ and choice of sign convention (+---) as:

$g_{\alpha\beta} = \begin{pmatrix} c^2 - \frac{2 G M}{r} & 0 & 0 & 0 \\ 0 & - \frac{1}{1 - \frac{2 G M}{c^2 r}} & 0 & 0 \\ 0 & 0 & - r^2 & 0 \\ 0 & 0 & 0 & - r^2 \, \sin^2 \theta \end{pmatrix}$

Thus the contraction with two infinitesimal vectors is:
$g_{\alpha\beta} \, dx^{\alpha} \, dy^{\beta} = \left( c^2 - \frac{2 G M}{r} \right) \, dt_x \, dt_y \; - \; \frac{1}{1 - \frac{2 G M}{c^2 r}} \, dr_x \, dr_y \; - \; r^2 \, d\theta_x \, d\theta_y \; - \; r^2 \, \sin^2 \theta \, d\phi_x \, d\phi_y$

The metric can be separated into its flat space-time part:
$\eta_{\alpha\beta} = { \tiny \begin{pmatrix} c^2 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & - r^2 & 0 \\ 0 & 0 & 0 & - r^2 \, \sin^2 \theta \end{pmatrix} }$
and the part that makes it different that flat space-time:
$h_{\alpha\beta} = - 2 \frac{GM}{r} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{r}{c^2 r - 2 GM} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
Then $g_{\alpha\beta} = \eta_{\alpha\beta} + h_{\alpha\beta}$.

Disputed.

Indeed, the gravitational field looks like it should be defined in terms of derivatives of the metric (or connection coefficients), not the metric itself or h which is roughly proportional to the Newtonian gravitational potential far from the singularity.

14. ### brucepValued Senior Member

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You're amazing. In my simplistic way I'd say the gravitational field is a derivative of the curvature component of the metric. A first derivative of a fundamental geometry we call spacetime. Sounds good to me. LOL. Hope I don't get slapped around to much.

15. ### TrappedBannedBanned

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I edited this thinking that I picked up Rpenner wrong.

Rpenner, the Schwarzschild expression of this form has been used in literature by the infamous Motz. It doesn't involve the length c^2t^2 which is usually attached to the terms. You must read this paper to understand the use of this dimensionless form

http://gravityresearchfoundation.org/pdf/awarded/1968/motz.pdf

16. ### TrappedBannedBanned

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Let's dispute it then.

the gravitational field I argue has two main descriptions, either given by the connection $\Gamma$ describing curvature - the metric does however contain the appropriate variables which describes the gravitational field strength.

17. ### TrappedBannedBanned

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You can also prove very simply how it describes the field strength from the original work, and I quote ''

The strength of the gravitational field, is again provided by the metric which we have rewritten in terms of energy. To evaluate how this is featured in the metric, you can show that the radius of curvature is

$r_c = \frac{c^2}{g}$

Strength of gravitational ﬁeld at the surface of a gravitating object is given as a ratio of the actual radius to the light curve radius.

$\frac{R}{r_c} = \frac{GM}{Rc^2}$

where we have used

$g = \frac{GM}{R^2}$

If

$\frac{GM}{Rc^2} << 1$

then the field is said to be weak. Notice,

$\frac{GM}{Rc^2}$

is an integral expression of the Schwarzschild metric, therefore it describes the field strength of your metric. ''

18. ### TrappedBannedBanned

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Brucep, that argument is obviously inconsistent since Motz has used the same approach. What is local invariant, the acceleration? What kind of argument is that supposed to be? The equivalence principle says there should be radiation detected from an accelerating object in some gravitational field.

I don't know why you keep bringing up this.

19. ### brucepValued Senior Member

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I realize you don't know. You're equation doesn't do what you think it does. g_shell is an invariant and the radiation bath is predicted by Unruh and Hawking. William Unruh describes radiation bath associated with a linear accelerated charge. Stephen Hawking describes the same phenomena for an accelerated observer in a gravitational field. Equivalent. Your equation says.

...=... (a^2_g/1) = ... (a^2_g)

The following is g_shell

g_shell =[M_meter / r^2_shell](1-2M_meter/r_shell)^-1/2 [gamma ~ 1 when r_shell is very large]

(1-2M_meter/r_shell)^1/2 / (1-2M_meter/r_shell)^1/2 = 1 [always = 1]

Relativistic strong field effects on measured/ or calculated g_shell are when gamma > 1. gravitational redshift / gravitational redshift = 1. Always. It's nonsense mathematics which doesn't represent real natural phenomena [physics]. You might substitute gamma into the denominator and see what your equation predicts.

Last edited: Nov 14, 2013
20. ### rpennerFully WiredValued Senior Member

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When Motz writes towards the bottom of page 3
To those that can pass examinations on the topic under discussion, he is clearly limiting himself to the case where $dr = d\theta = d\phi = 0$ such that his particle is being suspended above a black hole which is equivalent to accelerated motion in flat space-time.

Thus $1 - \frac{2 GM}{c^2 r}$ is not the metric, but a frame-dependent quantity which happens to be the $g_{tt}$ component of the metric for empty space in the $\left(t,r,\theta,\phi)$ coordinate system where the black hole is stationary, uncharged and non-rotating at the spatial origin. So your source does not support your claim that the expression is "[t]he metric given as (sic) the Schwarzschild expression (sic)".

21. ### TrappedBannedBanned

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Then there isn't any dispute since the physical parameters where set for a charge around a massively dense spacetime in the original work or even when particles are boosted in the gravitational field surrounding a black hole.

22. ### TrappedBannedBanned

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Brucep, the equation for redshift is

$1 + z = \sqrt{\frac{g_{tt}(r)}{g_{tt}(s)}}$

$z$ is your redshift. So for $z = 0$ (no gravitational redshift) then yes it comes to unity. It does not, counter to your claim is always 1.

23. ### brucepValued Senior Member

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1 + z is the cosmological redshift not the gravitational redshift. Explain what the cosmological redshift has to do with a radiation bath at g_shell. For that matter explain what the gravitational redshift, or any other redshift, has to do with a radiation bath at g_shell? No don't bother. Have a good one. Maybe Tach can explain it for you?