Friday, April 25, 2008

Constraining Theories of the Speed of Light

A successful Theory satisfies far more constraints than speculations about repulsive energy. While the latter violates the First Law of Thermodynamics, Theory conserves energy even in the case of Black Hole entropy. This was a response to a "Brief Communication" in NATURE 2002 by Tamara Davis, Paul Davies and Doug Linaweaver. No conspiracy suspected, but unfortunately by the time this was prepared NATURE was no longer publishing brief communications. Presently this little paper is seeking another home. Enjoy!

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20 Comments:

Anonymous Paul Neilson said...

Louise

Check out Unzicker's paper on the physics.gen-ph arXiv 0708.3518v4 "A Look at The Abandoned Contributions to Cosmology of Dirac, Sciama and Dicke".

It looks at various consequences of VSL theories. I liked the part about Gravitation arising out of electromagnetism as a requirement of varying c.

Paul Neilson

3:17 PM  
Blogger CarlBrannen said...

Here's a link to Unzicker's physics/0708.3518 (2007), yes a very interesting article. He quotes Dirac: "The laws may be changing, and in particular, quantities which are considered to be constants of nature may be varying with cosmological time. Such variations would completely upset the model makers."

My favorite line is "Dicke started from Einstein’s idea of light deflection caused by a lower c in the vicinity of masses."

This can in itself be turned into a cosmology. Let the universe be initially populated by matter without light or gravitational influence. As time goes on, matter will feel the influence of other matter farther and farther away. This is equuivalent to a change in the presence of nearby matter and causes light to slow down.

The nearby matter is defined by "R". The speed of light varies due to this increasing with time. In addition, the clocks of matter depends on the amount of nearby matter. The result is Riofrio's cosmology.

3:16 PM  
Anonymous Anonymous said...

I keep tripping over the light! Its fantastic!

12:16 PM  
Blogger Robin said...

Hi Louise

Some interesting ideas here. I had quite a bit of exposure to VSL theories a few years back as Joao Magueijo was my MSc supervisor at Imperial College. I was particularly interested in your conclusions about the Planck quantities – as I think you are making an observation similar to one I made in an earlier paper of mine: http://arxiv.org/abs/gr-qc/0106007
I have only just come across your blog as I was doing a search for Machian GR and your name came up. I have not had the chance to read all your older posts yet, but from what I have seen already there is some fascinating stuff here – particularly because your research has many parallels with what I have been doing. For example, your paper arxiv: 0610034 explores many of the same themes as my paper referenced above. (Not suggesting plagiarism here. More a case of great minds thinking alike!) One topic that I had not come across before is that of unexplained sun luminosity data. I will definitely follow up on this. By the way, why was this paper removed from Arxiv?
You may also be interested in linking to my own site: http://www.insularinstitute.org , where you can find a few more of my publications. I am thinking of opening this up to include a discussion forum, blog, contributions from other like minded independent researchers, so let me know if you are interested in publishing any of your papers on this suite.
Finally, have you heard of the Alternative Cosmology Group? If not, head for their site at: http://www.cosmology.info
More importantly, you need to be at their conference near Seattle from 7-11 September this year, and present one or more of your papers. See details at:
http://www.cosmology.info/2008conference

Robin

1:36 AM  
Blogger L. Riofrio said...

Carl and Paul: Thanks for the link to the paper. Varying c is definitely allowed by GR, as Einstein himself wrote. Like David Wiltshire's idea of time slowing, the effect of light bending around a mass is mathematically equivalent to a changing speed of light.

Robin: Thank you very much, I am glad that this little blog has connected forward-thinking researchers around the world. I am planning on the September meeting, and look forward to reading all your work.

4:57 AM  
Anonymous Anonymous said...

Louise

I have some comments on your article from a few posts ago entitled "Unified spacetime indicated from supernova data". This post has LaTeX in it so it's easier to read in the full version.

I've tried to follow the working in your article, with dimensional constants inserted to get the correspondence with your statements. In equations (3) to (9) you step through the derivation of the Friedmann equations (10) and (11),
where I surmise that $z$ is the curvature of 3-space (either +1, -1 or 0). You don't state the form of the metric from which you derive equations (4) and (5), but since you set $z = 0$ to zero anyway we can start from the line element $ds^2$ with flat spatial component,

$\begin{eqnarray}
ds^2 = -c^2 dt^2 + r(t)(dx^2 + dy^2 + dz^2)
\end{eqnarray}}$

where $r(t)$ is a scaling factor. Although $r(t)$ is sometimes informally referred to as the "radius of the universe" it is actually dimensionless. You can see heuristically from above that $r(t)$ must be dimensionless in order for the dimensions in the line element to be consistent. This immediately shows that there is a problem with setting $r(t) = ct$ if you insist on interpreting the $c$ term as the speed of light.

Using the stress tensor for pressure-free dust and assuming constant $c$ the Einstein equations boil down to the equations you presented for $r$:

$\begin{eqnarray}
(10)\mbox{ } \frac{r''}{r} = -\frac{4\pi G\rho}{3} \\
(11)\mbox{ } \frac{(r')^2}{r^2} = \frac{8\pi G\rho}{3}
\end{eqnarray}$

(I want to emphasise here that since the Einstein tensor is made up of derivatives of the metric coefficients, if $c$ really were varying with $t$ you would end up with derivatives of $c$ on the left-hand side of these equations). The exact solution of these equations is well-known - adapted from Wald's book (with units inserted) it is

$\begin{eqnarray}
(12)\mbox{ } r(t) = (6\pi G \rho {r}^3)^{1/3} t^{2/3}
\end{eqnarray}$

Although $r$ appears on both sides, this is fine because the quantity $\rho r^3$ is a constant, interpreted as demonstrating conservation of mass. The expression $\rho = (6\pi G t^2)^{-1}$ that you stated for the density can be found by plugging $r(t)$ into either of the differential equations above.

An important point is that there is no freedom in the constant in front of $t^{2/3}$ in $r(t)$ above. If you assume that $r = \lambda t^{2/3}$ for some constant $\lambda$, then the two differential equations (10) and (11) still hold and $\rho = (6\pi G t^2)^{-1}$ whatever $\lambda$ is, since the $\lambda$'s cancel out on the left-hand side of both equations. However, taking the cube of $r$ shows that

$\begin{eqnarray}
r^3 = (\lambda t^{2/3})^3 \\
= \lambda^3 t^2 \\
= \frac{\lambda^3}{6 \pi G \rho}
\end{eqnarray}}$

so $\lambda^3$ can only be $6 \pi G \rho r^3$. This is what reveals the inconsistency in choosing $r = ct$. You claim that $r = ct$ with $GM = tc^3$ is a solution of the Friedman equations. You don't give a formula for the mass $M$ but I'm guessing from your equation (14) that you must mean $M = 2\pi^2 G \rho r^3$. Using $GM = tc^3$ to eliminate $c$ from $r = ct$ gives

$\begin{eqnarray}
r(t) = (2\pi^2 G \rho r^3 /t)^{1/3} t \\
= (2\pi^2 G \rho r^3)^{1/3} t^{2/3}
\end{eqnarray}$

This is just your equation (13) with $M$ spelled out. Comparing this to (12) above, your equation cannot be right since it implies that $2\pi^2 = 6\pi$. The key point here is that your expression for $r$ satisfies the differential equation (as does any function proportional to $t^{2/3}$) but doesn't satisfy the additional consistency condition that the scaling constant must be $(6\pi G \rho {r}^3)^{1/3}$.

5:39 AM  
Blogger Yoyo said...

Well, the latex above didn't work as well as I'd hoped. I don't blog myself but I've created one so you can see my comment in its intended form.

6:07 AM  
Blogger CarlBrannen said...

Thanks for the links, Robin.

Paul, the Unzicker article points out a couple of things. The first is that in modeling the Big Bang, we have a choice of letting space, time, and the speed of light vary.

The usual solution is to fix c and t, and to let space vary, hence the Big Bang. Louise's method is to t and let c and s vary. The calculation I made kept s and t constant and let c vary. But I think the right way to do it is to keep c and s fixed and let t vary.

The other thing about the paper is that it makes it clear that my calculation uses the wrong scaling for c as a function of the observable universe. However, if you look at the powers, the only decent way to do the CMB calculation is to make c vary as 1/t (if you let space be fixed), as this is what leads to a logarithm term, which is what is needed to scale ratios like that of 380,000 years to 14 billion years.

However, while I used the wrong scaling law, I think it can also be argued that the usual scaling law is wrong. The MOND theory lets the force of gravity drop off as 1/r instead of 1/r^2, and when you take this into account, it may be enough to get the scaling right so the CMB calculation works.

The other thing to mention is that the scaling laws in physics are strange in other places. The "quantum zeno effect" implies a modification of a very weak exponential decay (as would be perhaps appropriate for a very weak quantum gravitational field).

9:02 AM  
Blogger L. Riofrio said...

Robin's links are very interesting! He and carl will both be on the links list soon.

For yy, who has been kind enough to create a blog account to make his comments: The difference between 3 and π leads to a very interesting result. Theory predicts a stable density of:

ρf = (6πGt^2)^{-1}

If a mass M is spread among a sphere, we have an initial density:

ρi = M/V = (2 π^2 G t^2 )^{-1}

Difference between this initial density ρi and stable density ρf is:

4.507034%

Precisely what our WMAP spacecraft has measured within half a standard deviation. Below that density, quantum theory predicts that matter will be created via pair production.

The Albrecht/Maguiejo paper can be a help here. Refer to their equation (10). Here e is the deviation from critical density.

edot = (1 + e)e(adot/a)(1 + 3w) + 2(cdot/c)e

Now we have w = 0,
(adot/a) = (2/3t) and (cdot/c) = (-1/3t)

edot = e^2 (2/3t)

Is it not reassuring that the other terms cancel? When t is low e is large and a large edot drives density toward a critical value. Today, when t is billions of years and e is very nearly zero, little mass is being created.

4:57 AM  
Anonymous Paul Neilson said...

Robin

I just read your Machian GR paper. I will be studying it for a while longer.

I had Dr. Wolfgang Yourgrau for a course way back in 74. Prof Yourgrau was a student of Schrodinger's when the wave equation was created. He had Einstein as a Prof for a course in GR (I think it was GR, my memory of this is getting dim). Anyway, Dr Yourgrau published several papers on Mach's principal. He influenced me enough that I wrote my term paper for that class on Mach's principle.

So you can see that I really enjoyed your paper. Thanks for the post.

Carl - I am not sure that your scaling is wrong. But I am not fully understanding your arguments.

The observable part of the Universe is according to Mach the only part we care about as the unobservable part is disconnected from the observable part by c. But entanglement throws in a tangle so to speak and what is now unobservable could still be connected by a primordial entanglement. I have also seen arguments where gravity moves at 4/3 c. So I am not sure what role observability by light takes.

When I get some more time I will go over the scaling issue again to see if I can come to grips with it.

Louise - This is all getting very interesting. Thanks to all the contributers.

Paul Neilson

8:47 AM  
Blogger Kea said...

This comment has been removed by the author.

7:05 PM  
Blogger Kea said...

Robin, your Mach paper is FANTASTIC! You explain the Scale Time very nicely.

8:32 PM  
Blogger Yoyo said...

Louise (sorry for the long post)

I think the result you are seeing is due to a mathematical error, so looking for observational confirmation may be premature. As far as I can tell, you're using a totally standard derivation to get the Einstein-Friedmann equations which you have numbered as (10) and (11) in your article. As you note, this does indeed predict a density

$\rho = (6\pi G t^2)^{-1}$.

No argument there. However, this is the expression for $\rho$ for all time $t > 0$. There's no way for the formula for $\rho$ to change from the above to your formula for initial density $\rho_i = (2\pi^2 G t^2)^{-1}$ without inserting some additional mechanism. You still have the variable $t$ in there anyway so in what sense do you mean this to be an "initial" density? Perhaps what you are trying to say is that the spacetime starts off at some initial time $t_i$ as a 3-sphere with volume $V$ given by your formula and initial density $\rho_i$. Assuming things evolve smoothly from there obeying the E-F equations I don't see how $\rho$ can evolve from $\rho_i$ - clearly in the limit as $t \longrightarrow t_i$, $\rho$ tends to $(6\pi G {t_i}^2)^{-1}$, not $(2\pi^2 G {t_i}^2)^{-1}$. Am I missing something here?

Anyway, after thinking about this for a while I looked up your reference to Einstein (1916) and I think I see the problem: the $R$ that appears in $V$ is not the same $r$ that appears in the E-F equations. This is obvious on dimensional grounds since $R$ has units of length while $r$ is dimensionless, but here's a more detailed description.

In Ch 31, "On the possibility of a finite yet unbounded universe", Einstein introduces the "three dimensional space which was discovered by Riemann", which has volume $V = 2\pi^2 R^3$. I assume this is where you get your $V$ from - if not, could you give me a page number please? This is not referring to any exact solution in GR, it is referring to the 3-sphere embedded in $\mathbf{\mathrm{R}}^4$. In coordinates $(x, y, z, w)$ it is the set of points satisfying

$x^2 + y^2 + z^2 + w^2 = R^2$

where $R$ represents a fixed distance from the origin $(0, 0, 0, 0)$ - clearly if we interpret the coordinates as lengths then $R$ also has units of length. The 3-sphere consists of all points at distance $R$ from the origin as measured by the usual Euclidean metric, and has volume $V = 2\pi^2 R^3$. Note that it is a compact manifold embedded in flat space $\mathbf{\mathrm{R}}^4$ , and has non-vanishing curvature $1/R^2$. These points are important - the compactness means it's possible to integrate over the manifold and get a finite volume, and the non-zero curvature means it cannot be isomorphic to a flat space.

What Einstein was speculating on was that in some spacetimes it might be possible that the spacelike hypersurfaces $t = \mbox{const.}$ might have the topology of Riemann's 3-sphere. This is reasonable for a closed universe ie. the positive curvature case ($z = +1$) but it is not possible for the flat case ($z = 0$) that you use, or the open case ($z = -1$), because the curvature of the 3-sphere is strictly positive. So that rules out using this $V$ in your scenario, unless there's a topology change somewhere.

The real problem though is that the $R$ in the volume formula $V = 2\pi^2 R^3$ is not the same thing as the $r$ that appears in the Einstein-Friedmann equations. They are completely unrelated. Most standard texts on GR (Wald, MTW etc) will tell you that the $r$ that appears in the Einstein-Friedmann equations originates from a line element of the form

$ds^2 = -c^2 dt^2 + r^2(t)(dx^2 + dy^2 + dz^2)$

where $r(t)$ which is dimensionless - clearly it has to be dimensionsless for the units of the line element to make sense. It does not represent a radial distance, it is simply a scaling factor and has nothing to do with the $R$ in $V = 2\pi^2 R^3$. You can see yourself that if you plug "$r = ct$" into the line element you end up with different dimensions in the spatial components than in the temporal components.

Moreover, as far as I'm aware it's not even possible to define a "volume" or a "total mass" for the spacelike hypersurfaces of constant $t$ in this cosmology, because the topology is unknown. GR doesn't describe the topology of the spacetime, only the metric. In order to get a finite answer for the volume or mass of the universe you would need to specify a compact topology for the spacelike hypersurfaces - obviously $S^3$ won't work because of the non-vanishing curvature, but maybe there is a compact 3-manifold with vanishing curvature that will do the job.

I would be interested in seeing your response to this. In particular, how would you reconcile the fact that your $r = ct$ has dimensions of length while the $r$ in the Einstein-Friedmann equations is dimensionless?

10:39 PM  
Blogger L. Riofrio said...

For Robin, Kea, Carl and Paul: Minds do think alike, and a quick reading shows many parallels between the Unzicker papers and Robin's work. Though separated in distance (scale factor?), we are converging toward a similiar result. I am glad that you are all with the program.

For yy: This could easily decay into an argument over units, but there is time for one more reply. The mechanism to evolve from initial density ρi to stable density ρf is pair production of matter. The Einstein-Friedmann equations with k = 0 are solved by the predicted R(t) which has dimensions of length. Try it yourself.

5:28 PM  
Blogger Yoyo said...

Louise

In the course of reading your paper I did try this. I completely agree with you that any function of the form $R(t) = \lambda t^{2/3}$ for constant $\lambda$, including yours, is a solution of the the E-F equations, regardless of the units, since any units of $R$ on the LHS of the E-F equations will cancel out. I stated as much above.

My point is that if you look at where $r(t)$ originally comes from, it is clearly a dimensionless term in the spacetime metric. It doesn't make sense for $r(t)$ to suddenly acquire units of length in going from the metric to the solution of the E-F equations. It also doesn't make sense on dimensional grounds to plug $r = ct$ into the metric. I have no problem with you solving the E-F equations in isolation, but if you ignore the fact the units in your expression are different than the ones you started with then there is no connection with the Einstein field equations and the metric.

There are a couple of things that would be helpful to me here. Firstly, so I can see where your $r$ and $z$ come from, could you please state the form of the metric that you used to derive the right-hand side of equation (4) in your article.

Secondly, could you write down the components of the metric for your spacetime, that is, the components $g_{ij}$. Then it's easy to plug those into the Einstein equations and check that it's a solution.

6:37 PM  
Blogger L. Riofrio said...

The moderator looks at watch and declares question time over. Check out next post.

2:32 AM  
Blogger Yoyo said...

This comment has been removed by the author.

4:18 AM  
Blogger Yoyo said...

This comment has been removed by the author.

4:21 AM  
Blogger Yoyo said...

Thanks for the link to that post. It doesn't seem to have anything to do with the metric you're using now. I don't see how you could derive the Einstein-Friedmann equations from it. Many of the terms like $a$ and $r$ are undefined, and $a$ and the various $x$'s don't appear in the E-F equations. So it's not especially helpful in understanding your article.

I noticed that you specifically chose your equation $GM = tc^3$ so that the signature of your metric would turn out to be $(-1, -1, -1, -1)$. In that case, you're not doing relativity, which as you know has Lorentzian signature. It certainly cannot be the metric of the Einstein-de Sitter space.

I gather you're declining to answer further questions on this. It's disappointing that you didn't answer my follow-up about reconciling the units of $r = ct$ with the dimensionless $r$ in the metric. Since there are other physics types who read this blog, I hope you won't mind if I throw this open to them.

The solution $r(t)$ to the E-F equations arises from the line element of the form

$ds^2 = -c^2 dt^2 + r^2(t)(dx^2 + dy^2 + dz^2)$

Plugging in Louise's $r = ct$ to that gives

$ds^2 = -c^2 dt^2 + c^2t^2(dx^2 + dy^2 + dz^2)$

It's clear (to me at least) that this has inconsistent units. So to the physicists out there, how is it possible to reconcile the dimensionless $r$ of the metric with $r = ct$?

I am genuinely interested in getting a response from Louise or anyone else who can address this.

8:59 PM  
Blogger Yoyo said...

A few days have passed and there's been no comments from Louise or the other physicists who read this blog about my question on the units of $r = ct$, or on what the metric of Louise's spacetime is based on (that is, the form of the metric that leads to the RHS of eq (4) on this page).

Without knowing what the original $g_{ij}$ was, having $r(t)$ is not very useful. How can I know what role $r(t)$ is supposed to play in the metric? It's not even possible do simple things like evaluate it on a pair of vectors $X$ and $Y$ ie. $g_{ij}X^iY^j$, or raise and lower indices, or calculate a geodesic without knowing what $g_{ij}$ is and where $r(t)$ appears in it.

Louise, I've made a concerted effort to read and try to follow your article below. I spent several hours on this, checking calculations and looking up your references - at least as much time as I would spend refereeing a journal article. Regardless of whether your article is correct or not, it's very difficult for other readers to follow in it's current form because of the various steps that are left out and the terms that are undefined. So, to help me with this could you please write down the form of the metric that is used to get the RHS of equation (4). What I'm after is something of the form

$ds^2 = g_{ij} dx^i dx^j$

where the terms $r$, $z$ and $A$ from eq (4) appear.

5:29 AM  

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