Invited (peer) review of article(s) by Otto Rössler

Background
Yesterday, I addressed Dr. Otto Rössler on the merits of his general position.

(Emphasis added)
He replied in part:

As soon as he describes me as his "enemy," I am concerned that in his adversarial casting of our roles, he has abandoned the quest for scientific truth in favor of a rhetorical exercise in lawyering. I replied in part:

This was not taken in the manner intended, and the following request was made:

And I accepted the request. My reply in part:

Obviously, here we have trained specialists in physics and math. I don't claim to exceed their expertise in many areas, but I think I know a failure to learn general relativity when I see it.

Ground Rules and Appeal to Moderation

So setting aside my 6 numbered requests for clarification on Rössler's blog for his later reply on his chosen forum, I prefer this site to format a peer review for its ability to typeset mathematics and attribute quotes. If this is inappropriate cross-site discussion or an inappropriate sub-forum, I apologize and request that you work with me to find a better solution.

The primary paper is
[Rössler2012] Otto E. Rossler "Einstein's equivalence principle has three further implications besides affecting time: T-L-M-Ch theorem (“Telemach”)" African Journal of Mathematics and Computer Science Research 5( 3), pp. 44-47, (9 February, 2012) http://www.academicjournals.org/AJMCSR/PDF/pdf2012/Feb/9 Feb/Rossler.pdf

Also cited:
[Rössler1997] OE Rossler, C. Giannetti, "Cession, twin of action (La cesión: hermana gemela de la acción)" In: Arte en la era electronica: Perspectivas de una nueva estética (ed. by C. Giannetti), (Barcelona: Associación de Cultura Temporánia L’Angelot, and Goethe-Institut Barcelona 1997), p.124.
[Rössler2007] OE Rossler "Abraham- like return to constant c in general relativity: gothic-R theorem demonstrated in Schwarzschild metric." Paper allegedly accepted for publication in Chaos, Solitons and Fractals.
[Rössler2009] OE Rossler "Abraham- like return to constant c in general relativity: gothic-R theorem demonstrated in Schwarzschild metric." revised draft at http://www.wissensnavigator.com/documents/Chaos.pdf
[Rössler2010] OE Rossler, D Fröhlich "The weight of the Ur-Kilogram" http://www.achtphasen.net/index.php/plasmaether/2010/12/11/p1890

This effort may be doomed, but I appeal to posters to keep the comments civil and the arguments directed at the text, not the person.

 

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I would like to contribute to this thread if I have time. Unfortunately time is something I don't have too much of at the moment.

I have no problem with this thread being here, and in many ways it's a more important discussion to have than the usual haranguing that goes on here. To avoid the thread being derailed I will be rather more strict with off topic and general chit chat type posts than I normally would be.

 

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Issue #1. This article in published in a math journal carried by few libraries, and is labeled as "Review" despite claiming to introduce a new theorem which is described in the abstract as a "new finding".

Issue #2. Umlauts don't appear where expected, such as for Rössler or Tübingen.

Issue #3. The @yahoo.com email address is not actually associated with the listed academic association. Indeed, Prof. Dr. Otto Rössler (pensioniert) is listed only as a "Former professor and faculty member" http://www.mnf.uni-tuebingen.de/fac...che-chemie/institut/institutsangehoerige.html

Issue #4. The abstract does not summarize the physics or math content. Indeed, it calls into question if the author even knows what a theorem is. "The TeLeMaCh theorem ... is bound to transform metrology if correct." Why add "if correct" if the theorem was proved?

Skimming, a different paper, we find a sketch of the "gothic-R theorem."

 

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As I understand it, if a space-time has zero Weyl curvature tensor, then it has a Weyl symmetry, and it admits arbitrary conformal mapping of coordinates, even when the trace of the Riemann tensor is non-zero. So if I can demonstrate that the only space times Rössler (and Cook) looked at have Weyl symmetry then I will have gone a long way towards showing that Rössler has not been studying GR so much as hastily generalizing from a particular geometry.

Also, in a new post, Rössler has doubled-down on the Rössler2012 paper http://lifeboat.com/blog/2012/05/i-...ng-colleagues-who-in-2008-defamed-my-gothic-r

For him, this is personal. Let's try and be professional.

 

This doesn't seem right, the second expression doesn't follow from differentiating the first one.

 

Please elaborate. I admit, that I left u in there without effect, from the related outgoing Eddington–Finkelstein coordinates. But for the last expression I just needed v, r, and the angles.
\(\frac{d}{dx} \left( x + a \ln ( \frac{x}{a} -1 ) \right) =\frac{d}{dx} \left( x + a \ln ( x - a ) + a \ln \frac{1}{a} \right) = \frac{d}{dx} x + a \frac{d}{dx} \ln ( x - a ) + a \frac{d}{dx} \frac{1}{a} = 1 + \frac{a}{x - a} + 0 = \frac{x}{x - a} = \frac{1}{1 - \frac{a}{x}}\)

 

Rössler himself points out a source of further information about Rössler2007/Rössler2009

CERN links to two reports that claim R's paper is no way to learn General Relativity.

Nicolai and Giulini write:

Which I think means the Weyl curvature of the Schwarzschild geometry is not zero and my previous supposition was too generous to Rössler. Time to calculate this. Wald's textbook seems easier to handle than M-T-W, but both are at home.

 

Mod note: 4 not very edifying posts deleted.

 

Rpenner, if you don't feel I can offer any assistance to you, please just say the word and I will politely withdraw. But for now can I say that this raised a flag:

Take care that you don't construct an argument that can be demolished in an instant via "rpenner has merely employed a consensus view to justify a consensus view". Also take a close look at the coordinate speed of light and at \( \frac{d \tau}{dt} = \sqrt{1 - \frac{2 G M}{c^2 r}}\). Imagine your clock is a light clock at 2GM = c²r, and read up on frozen stars. Kevin Brown's mathspages has a nice little article called The Formation and Growth of Black Holes. I'd say he sides with the consensus view, but gives the frozen star a fair hearing. It might be worth your while to then look at the gravastar which features"a void in the fabric of space and time".

 

Farsight, my clock can't stay at \(r = \frac{2 GM}{c^2}\) which is why there is a coordinate singularity there. There is no geometric singularity. That was the purpose of exploring alternate coordinate systems.

The consensus view that the coordinates are imaginary creations is factual. The consensus view that coordinates that display features not in the geometry are bad coordinates dates back to map-making of the Earth. Greenland is not larger in area than South America. The mathematics of manifolds only require that the manifold is locally well-described as flat. The more curved the geometry, the less distance you need to go before your local coordinate system seems distorted. France and England point in different directions when they point up this only seems slightly odd. Australia and England point in radically different directions. In the geometry of black holes, the future light cone interior to the hole points in a different direction than the exterior, and while coordinates can obscure this, they cannot change it.

The defining difference between space-time exterior to the black hole and the event horizon is that one has 3 degrees of spatial freedom exterior to a black hole, while at the event horizon one has no option to return to the exterior. Constrained to the interior of a light cone that nowhere points outward from the black hole, the geometry of space time constrains ones future to successively smaller r coordinates. Eventually, general relativity loses predictive power and our ability to do empirically justified physics ends -- well inside the black hole.

I believe Kevin Brown puts the nail in frozen stars when he considers accretion. The event horizon doesn't grow if the infalling matter doesn't wind up inside the black hole. The Schwarzschild solution is a description of a frozen space-time but it was a specific solution to general relativity when there was no matter in the rest of the universe. Knowing one solution to a physical theory is no substitute for knowing the physical theory.

Discussing the hypothetical idea of gravastars and its idiosyncratic requirements on quantum gravity theory has no merit on this thread. In http://arxiv.org/abs/1111.4187, authors Chan, da Silva, Senna, Villas da Rocha admit that http://arxiv.org/abs/gr-qc/0701154 may place "very strict observational constraints on the existence of such stars."

 

rpenner: I have to dash but for now, I said make it a light clock for a reason. Mentally strip away the actual apparatus leaving only the light. And note that hailstones grow.

 

It's honestly hard to tell if you're being serious or if you've descended into self parody.

Farsight, given every single journal you sent your work to rejected it, as has every single physics researcher who has ever seen it, you can't provide a single working model from your work and you regularly show a lack of understanding for well understood phenomena why do you think you're in a position to be telling anyone how things work, particularly with closing sentences like "And note that hailstones grow". You openly admit you don't read actual physics textbooks and you demonstrably lack the necessary mathematical physics to understand papers on ArXiv in any more than the most superficial manner (and you view everything through your own rose tinted glasses of bias and self deception), so you're hardly in a position to tell people what current physics says either, particularly someone like rpenner who has demonstrable understanding in many areas and has had to explain things to you in the past.

By every rational yardstick you have provided no reason for anything to think you're in a position of knowledge. In fact you've been demonstrably dishonest on occasions, pushing your own work by being a hypocrite.

 

Well, isn’t Cook’s paper wrong since it is in conflict with Birkhoff's theorem?

You’ve had to have seen this blog before, right?

http://elnaschiewatch.blogspot.com/2011/04/new-venue-for-otto-rosslers-lhc-ravings.html

They even mention you.

http://elnaschiewatch.blogspot.com/...howComment=1337853447108#c1058184113173416614

 

I haven't found the 2007 version yet, but I found a 2008 version.

[Rössler2007b] OE Rössler "Abraham-Solution to Schwarzschild Metric Implies That CERN Miniblack Holes Pose a Planetary Risk" http://www.wissensnavigator.ch/documents/OTTOROESSLERMINIBLACKHOLE.pdf
[Rössler2008] OE Rossler "Abraham- like return to constant c in general relativity: gothic-R theorem demonstrated in Schwarzschild metric." http://lhc-concern.info/wp-content/uploads/2009/01/fullpreprint.pdf

Sorry this is taking so long, but I'm trying to introduce tensors and specifically the Riemann tensor in a way that is legible to people who only feel comfortable with multivariable calculus.

 

Much of Rössler's problem seems to be that he wants to talk about physics but he spends all of his time with coordinates. As a result [Rössler2012] tries to claim that in the Rindler coordinates corresponding to a world seen by an observer in constant proper acceleration that the spatial extent of objects in a certain direction is really scaled by certain amounts, despite that these Rindler coordinates describe completely flat space-time and therefore nothing should be "really" scaled to suit Rössler's aesthetics.

The important thing to know about tensors is that tensors are linear geometric objects. While they have representations in various coordinate basis, the important thing is that they are linear in all their slots. Once a tensor is known in one coordinate basis, we have the option of changing coordinate basis and keeping the geometry the same.

An important tensor is a vector, or tensor of (1,0) type, written \(v^a\) where a, b, c, d, e, etc are placeholders for the basis vectors. A dual vector, a tensor of type (0,1) is written as \(\omega_a\). The tensor product is indicated by placing two or more tensors next to each other, e.g. \(v^a v^b = \left(v \otimes v\right)^{ab}\). A tensor contraction reduces the tensor in rank and generalizes the trace of a matrix. Tensor contraction is indicated by having the same letter in an upper and lower slot, e.g. \(m_{abc}^{db} = \sum_e m_{aec}^{de} = \tilde{m}_{ac}^d\). To keep the geometric content of the linearity of the upper and lower slots we are only allowed to contract an upper slot with a lower slot.

When it is important to talk about an element or slice of a tensor in a specific coordinate system, we can use the letters that identify the coordinate basis. But when talking about elements and slices, we need to remember that we may not be talking about geometry. Also, some symbols (\(\partial, \nabla, \Gamma\)) are used that look like tensors but are not actually tensors.

The Physics of Rindler coordinates

For a potentially curved manifold, we use an infinitesimal coordinate basis and a metric which generalizes the concept of Euclidean distance. A metric is symmetric and non-degenerate.

Assuming k is an acceleration > 0, and x > 0 we define the Rindler coordinates for constant proper acceleration k in the basis of infinitesimal motions in orthogonal directions t, x, y, z as:

\((ds)^2 = - \frac{k^2 x^2}{c^2} (dt)^2 + (dx)^2 + (dy)^2 + (dz)^2 = g_{ab} v^a v^b\)

Since the metric (0,2) tensor is diagonal in this basis, it's trivial to find its (2,0) inverse. In fact, a lot of the math in this section simplifies specifically because g is, in this case, diagonal. Written as matrices, we have:

\(g_{ab} = \begin{pmatrix} -\frac{k^2 x^2}{c^2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \; , \quad \quad \quad g^{ab} \equiv (g^{-1})^{ab} = \begin{pmatrix} -\frac{c^2}{k^2 x^2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\)

We use the metric tensor and its inverse to raise an lower indices on other tensors. If \(v^a\) is a vector, we define the corresponding dual vector as \(\omega_b = g_{ab} v^a\). Since the metric and its inverse are diagonal in this basis, when we contract over it, many sums simplify. We can write \(\color{blue} g_{aa}\) or \(\color{blue} g^{aa}\) to talk about this diagonal as a (0,1) or (1,0) vector. Hopefully this won't be confusing.

See [Wald1984], pp. 17-49.

Ordinary derivatives we write as if they were tensors, for example \(\partial_a v^b \equiv \frac{\partial v^b}{\partial e^a}\) and \(\partial_t v^b \equiv \frac{\partial v^b}{\partial t}\).

So we have the following ordinary derivatives:
\(\partial_t g_{ab} = \partial_y g_{ab} \partial_z g_{ab} = 0 \; , \quad \quad \quad \partial_x g_{ab} = \begin{pmatrix} -\frac{2 k^2 x}{c^2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}\) or \(\partial_a g_{bc} = -\frac{2 k^2 x}{c^2} \delta_a^x \delta_b^t \delta_c^t\).

Ordinary derivatives obviously reflect the properties of the coordinate system chosen to evaluate them. To get away from coordinate artifacts and back to the geometry of the space-time manifold, we need a covariant derivative -- a generalization of a derivative that respects the geometry of the metric.

We introduce the Christoffel symbol \(\Gamma_{ab}^c = \Gamma_{ba}^c\) which will tell us how to relate a covariant derivative to the ordinary coordinate-based derivative.
\(\nabla_a f = \partial_a f \\ \nabla_a \nabla_b f = \partial_a \partial_b f - \Gamma_{ab}^c \nabla_c f \\ \nabla_a v^b = \partial_a v^b + \Gamma_{ac}^b v^c \\ \nabla_a \omega_b = \partial_a \omega_b - \Gamma_{ab}^c \omega_c\)

So we compute the Christoffel symbol as \(\Gamma_{ab}^c = \frac{1}{2} g^{cd} \left( \partial_a g_{bd} + \partial_b g_{ad} - \partial_d g_{ab} \right)\) Since the metric is diagonal in this coordinate basis, this simplifies for us to: \(\Gamma_{ab}^c = \frac{1}{2} g^{cc} \left( \partial_a g_{bc} + \partial_b g_{ac} - \partial_c g_{ab} \right)\)

The only possible non-zero terms are \(\Gamma_{tt}^x = \frac{1}{2} g^{xx} ( - \partial_x g_{tt} ) = \frac{k^2 x}{c^2}\) and \(\Gamma_{tx}^t = \Gamma_{xt}^t = \frac{1}{2} g^{xx} (\partial_x g_{tt} ) = \frac{1}{x}\)

Now we should compute: \(\nabla_a g_{bc} = \partial_a g_{bc} - \Gamma_{ab}^d g_{cd} - \Gamma_{ac}^d g_{bd}\). Because our metric is diagonal, this simplifies to \(\nabla_a g_{bc} = \partial_a g_{bc} - \Gamma_{ab}^c g_{cc} - \Gamma_{ac}^b g_{bb}\) or for what we know about this specific metric: \(\nabla_a g_{bc} = -\frac{2 k^2 x}{c^2} \delta_a^x \delta_b^t \delta_c^t + \frac{k^2 x}{c^2} \delta_a^t \delta_b^x \delta_c^t + \frac{k^2 x}{c^2} \delta_a^x \delta_b^t \delta_c^t - \frac{k^2 x}{c^2} \delta_a^t \delta_b^t \delta_c^x + \frac{k^2 x}{c^2} \delta_a^t \delta_b^t \delta_c^x + \frac{k^2 x}{c^2} \delta_a^x \delta_b^t \delta_c^t - \frac{k^2 x}{c^2} \delta_a^t \delta_b^x \delta_c^t = 0\). This covariant derivative is, in a rigorous sense, the only derivative-like operator that works like this with respect to the metric.

Now that we have the geometrically natural covariant derivative, \(\nabla_a\), we can talk about curvature in terms of the Riemann tensor which is defined as the (1,3) tensor with the following property for all smooth dual-vector fields: \(\nabla_a \nabla_b \omega_c - \nabla_b \nabla_a \omega_c = R_{abc}^d \omega_d\) and therefore, \(R_{abc}^d = \partial_b \Gamma_{ac}^d - \partial_a \Gamma_{bc}^d + \Gamma_{ac}^e \Gamma_{eb}^d - \Gamma_{bc}^e \Gamma_{ea}^d\). In a coordinate basis, the Riemann tensor has 256 components. However, even without exploiting necessary symmetries, we only have to consider 7, as the rest are trivially zero.

\(R_{ttt}^t = \Gamma_{tt}^x \Gamma_{xt}^t - \Gamma_{xx}^x \Gamma_{xt}^t = 0 \\ R_{ttx}^x = \Gamma_{tx}^t \Gamma_{tt}^x - \Gamma_{tx}^t \Gamma_{tt}^x = 0 \\ R_{txt}^x = \partial_x \Gamma_{tt}^x - \Gamma_{xt}^t \Gamma_{tt}^x = \frac{k^2}{c^2} - \frac{k^2}{c^2} = 0 \\ R_{txx}^t = \partial_x \Gamma_{tx}^t + \Gamma_{tx}^t \Gamma_{tx}^t = - \frac{1}{x^2} + \frac{1}{x^2} = 0 \\ R_{xtt}^x = - \partial_x \Gamma_{tt}^x + \Gamma_{xt}^t \Gamma_{tt}^x = - \frac{k^2}{c^2} + \frac{k^2}{c^2} = 0 \\ R_{xtx}^t = - \partial_x \Gamma_{tx}^t - \Gamma_{tx}^t \Gamma_{tx}^t = \frac{1}{x^2} - \frac{1}{x^2} = 0 \\ R_{xxt}^t = \partial_x \Gamma_{xt}^t - \partial_x \Gamma_{xt}^t + \Gamma_{xt}^t \Gamma_{tx}^t - \Gamma_{xt}^t \Gamma_{tx}^t = 0\)

Thus, the Riemann tensor, \(R_{abc}^d\), the Ricci tensor, \(R_{ac} \equiv R_{abc}^b\), the scalar curvature \(R \equiv R_a^a\), and the Weyl tensor (which is in 4 dimensions, \(C_{abcd} = R_{abc}^e g_{ed} - \frac{1}{4} \left( g_{ac} R_{db} + g_{bd} R_{ca} - g_{ad} R_{cb} - g_{bc} R_{da} \right) + \frac{1}{6} R \left( g_{ac}g_{db} - g_{ad}g_{cb} \right)\) are all zero. This is flat space-time geometry.

Geodesics of flat space-time in Rindler coordinates

See [Wald1984], p. 41.

In the coordinate basis, we have
\( \frac{d^2 t}{d \lambda^2} + \frac{2}{x} \frac{dt}{d \lambda} \, \frac{dx}{d \lambda} = 0 \\ \frac{d^2 x}{d \lambda^2} + \frac{k^2 x}{c^2} \left( \frac{dx}{d \lambda} \right)^2 = 0 \\ \frac{d^2 y}{d \lambda^2} = \frac{d^2 z}{d \lambda^2} = 0 \\ ds^2 = - \frac{k^2 x^2}{c^2} (dt)^2 + (dx)^2 + (dy)^2 + (dz)^2\)

Thus \(y(\lambda) = K_1 + \lambda K_2, \quad z(\lambda) = K_3 + \lambda K_4, \quad \frac{dt(\lambda)}{d \lambda} = \frac{K_5}{x^2(\lambda)}\)

\(K_6 = ds^2/d\lambda^2 = - \frac{k^2 (K_5)^2}{c^2 x^2(\lambda)} + \left( \frac{dx(\lambda)}{d \lambda} \right)^2 + (K_2)^2 + (K_4)^2\)

.... to be continued?

 

Geodesics of flat space-time in Rindler coordinates (continued)

Sorry -- to be clearer -- the geometry of space time gives us the ability to slide a vector \(v^a\) along a smooth curve where the local tangent is \(t^a = \begin{pmatrix} \frac{d t}{d \lambda} \\ \frac{d x}{d \lambda} \\ \frac{d y}{d \lambda} \\ \frac{d z}{d \lambda} \end{pmatrix}\). And where the notion of sliding is natural to the geometry, we call that parallel transport of the vector v, satisfying \(t^a \nabla_a v^b = t^a \left( \partial_a v^b + \Gamma_{ac}^b v^c \right) = 0\) .

Geodesics are the straightest possible lines in the geometry -- so if you parallel transport the tangent vector of a geodesic along itself, you get another tangent vector. Thus they satisfy: \(t^a \partial_a t^b + t^a \Gamma_{ac}^b t^c \) which gives us the coordinates of the curve (parametrized by λ) as \(t^a \frac{d t^c}{d t^a} + \Gamma_{ab}^c t^a t^b = 0\) which gives us with the chain rule: \(\ddot{t} + \Gamma_{xt}^t \dot{x} \dot{t} + \Gamma_{tx}^t \dot{t} \dot{x} = \ddot{t} + 2 x^{\tiny -1} \dot{x} \dot{t} = 0\), \(\ddot{x} + \Gamma_{tt}^x \dot{t}^2 = \ddot{x} + \frac{k^2}{c^2} x \dot{t}^2 = 0\), \(\ddot{y} = \ddot{z} = 0\). (Here \(\dot{f} =\frac{df}{d\lambda}\) and \(\ddot{f} = \frac{d^2 f}{d\lambda ^2}\).

Solving:
\(\ddot{t} + 2 x^{\tiny -1} \dot{x} \dot{t} = 0 \\ \frac{\ddot{t}}{\dot{t}} + 2 \frac{\dot{x}}{x} = 0 \\ \ln \dot{t} + 2 \ln x + C = 0 \\ \dot{t} x^2 = K_5 = \frac{c^2 a}{k}\)
\(\ddot{x} + \frac{k^2}{c^2} x \dot{t}^2 = \ddot{x} + \frac{k^2}{c^2} x \frac{c^4 a^2}{k^2 x^4} = \ddot{x} + \frac{c^2 a^2}{x^3} = 0 \\ \ddot{x} x^3 = - c^2 a^2 \)
Try \(x = \sqrt{f}\)
\(f^{\tiny \frac{3}{2}} \frac{d^2 \quad}{d \lambda^2} \sqrt{f} = \frac{1}{2} f \ddot{f} - \frac{1}{4} \dot{f}^2 = -c^2 a^2 \\ 2 f \ddot{f} - \dot{f}^2 = - 4 c^2 a^2 \\ 1 = \frac{\dot{f}^2 - 4 c^2 a^2}{2 f \ddot{f}} \\ \frac{ 2 \dot{f} \ddot{f}}{\dot{f}^2 - 4 c^2 a^2} = \frac{\dot{f}}{f} \\ \ln \left( \dot{f}^2 - 4 c^2 a^2 \right) = \ln f + K_7 \\ \dot{f} = \pm \sqrt{ e^{K_7} f + 4 c^2 a^2 } \\ \frac{ \dot{f} }{ \sqrt{ e^{K_7} f + 4 c^2 a^2 } } = \pm 1 \\ 2 e^{- K_7} \sqrt{ e^{K_7} f + 4 c^2 a^2 } = K_8 \pm \lambda \\ f = \frac{1}{4} e^{K_7} \left( \lambda - K_8 \right)^2 - 4 c^2 a^2 e^{- K_7} \\ x = \sqrt{ b \left( \lambda - \lambda_{\tiny 0} \right)^2 - \frac{c^2 a^2}{b} } \)
\(\dot{t} = \frac{c^2 a}{k} \frac{1}{b \left( \lambda - \lambda_{\tiny 0} \right)^2 - \frac{c^2 a^2}{b}} \\ t = t_{\tiny 0} + \frac{-c}{k} \tanh^{\tiny -1} \frac{ b \left( \lambda - \lambda_{\tiny 0} \right) }{a c}\)
\(y = y_{\tiny 0} + y_{\tiny v} \lambda\)
\(z = z_{\tiny 0} + z_{\tiny v} \lambda\)

 

Is this it?
http://people.ee.ethz.ch/~aganz/webtest/chweb2005/pub/hubble_movassagh.pdf

Ya, but is there any chance that you could include a little layman intermezzo? That lifeboat project is extremely weird and a little creepy.

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BTW, how in hell can he be a jack of all trades? In regards to his publications, I don't understand how university affiliation works? Why would they allow this? :shrug:

 

No -- that appears to be a random paper from CSF1991-2009 -- a journal of low reputation but high citation metrics due to self-publiblication of papers with multiple self citations under former Editor-in-Chief MS El Naschie.

Quirin Schiermeier "Self-publishing editor set to retire" Nature 456 p. 432 (27 November 2008)

On March 2, 2009 the publisher ended its relationship with El Naschie and continued processing the backlog of articles accepted for a year. On March 16, 2010 CSF was relaunched under a new editorial board. A lawsuit over the above Nature news article was tried in October-November 2011 and no judgement published.

 

I don’t know, RP. Crazy people do make life more interesting but maybe you shouldn’t mess around with this one.

Added in proof: reception, erratum, confirmation.pdf

I Dreamed of Dying Last Night-Time Stood Still-Then the Dream-Giving Instance-Let It Run Again

Can Anyone Explain Why CERN Fears Nothing more than a Scientific Safety Conference?

Is CERN about to trigger the worst imaginable accident with an odds of 1 to 6?.pdf

 

I know, I know, I’m just a snoopy housewife. I don’t get it, though. Didn’t Domenico Giulini and Hermann Nicolai already dispute this?

http://environmental-impact.web.cern.ch/environmental-impact/Objects/LHCSafety/NicolaiFurtherComment-en.pdf

http://environmental-impact.web.cern.ch/environmental-impact/Objects/LHCSafety/NicolaiComment-en.pdf

Alright, from here on out, I’ll stay out of your topic, but I don't think he's going to answer you. Carry on.

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