# Dark energy? Are we just making things harder than they need to be Lorentz?

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#### Messenger

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##### Dark energy? Are we just making things harder than they need to be Lorentz?
« on: 15/03/2012 21:38:57 »
If you are a mathematical physicist, you might be horrified at my lack of formalism, so instead allow me to pose some questions to the readership at large (but feel free to pipe in of course!).  It might be helpful if you are familiar with some simple calculus and vector analysis.  You can find my derivation at http://vixra.org/abs/1203.0025 (arxiv rejected it).

I have posted my thoughts into another forum, http://www.bautforum.com/showthread.php/129309-Negative-Mass-Interpretation-of-General-Relativity, hoping for some robust criticism of my logic or maths.  It doesn't seem to be forthcoming there, but this may be just due to registration problems.

One of the main downsides to general relativity is the difficulty in grasping what the equations fundamentally mean and their complexity.  In spite of their accuracy, and the vast amount of analysis that has gone into understanding them (the majority of which I am not familiar with), the accelerating expansion in 1998 was completely unpredicted.  If you would, allow for just a moment a simpler (and perhaps naive) take on the Einstein field equation.  While you may not agree, your criticism will help me sort out my thoughts and perhaps we will both learn something new.  Perhaps you can read through http://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy/  so as to familiarize yourself with the issues.

So, forgetting what anybody else has told you this equation means, this is my take on:

$$R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R=G_{\mu \nu}+g_{\mu \nu}\Lambda$$

The left hand side of this equation is pretty well known, and you don't have to be familiar with what it means or does.  The important part here is the right hand side.  The term with the G in it is referred to as Einstein's curvature tensor and the term with $$\Lambda$$ is the mysterious cosmological constant.  Firstly, lets dispense with any mystery of what this right hand side means.  G is nothing more than what is called a tensor and $$\Lambda$$ is what we can refer to as a constant of integration.  An easy way to think of a tensor is as nothing more than a point with a bunch of arrows pointing out of it.  Being able to include in $$\Lambda$$ just means that, taken together, we can make those arrows however we want as long as the come out pointing the right direction and the right length.  Mathematical physicists have known this for a long long time.  But there is a problem.  You can make the exact same arrows with a big $$\Lambda$$ or none at all.  This has a fancy name called "gauge invariance" but it can also mean what I like to think of as "Am I measuring what I think I am?"  When Einstein used this equation, he found exactly what he was looking for: An equation where if one throws in the mass-energy of a body it will accurately predict the precession of Mercury's perehilion, magnitudes of gravitational lensing and simplify down to Newton's equation for gravity (among other things).  It didn't exactly make physical sense, and there were some grave conceptual problems, but it sure is accurate.  Everything after this was trivial, until 1998.  After reading that NASA page, you should know there is a problem with our view of the universe.  Whether anyone has told you or not, something is wrong big time with our physics.

Now that we are all comfortable with the quantum vacuum, dark energy , etc. what if we were to instead take that right hand side, assign the cosmological constant as the total value of mass-energy of the vacuum, and subtract off a tensor of remaining mass-energy of the quantum vacuum.  Preposterous you might say!  What a crazy universe that would be!  Really? Hmmm, there is more than one way to make a gradient.  Take a look at the top and bottom of this picture.

The magnitudes are only to help you visualize what I mean, but the individual values aren't important.  The only thing important is the rate of change (derivative).

The top part represents what we think of as the regular Newtonian attractive potential gradient and the bottom as an illustration of how to make an equivalent repulsive gradient.  This gives you an idea of what $$\vec{g}=-\nabla\Phi=-\frac{GM}{r^{2}}\hat{\vec{r}}=-\frac{\Lambda_{\mathrm{vac}} c^{2}r}{6}\hat{\vec{r}}+\frac{G\rho_{\mathrm{res}}V}{r^{2}}\hat{\vec{r}}$$ means.

But that can't be, gravity comes from the presence of mass! Like us!  Read carefully through that NASA page again.  All may not be as it seems.

If you look through my paper, you will see that what we end up with is a repulsive gravity that should look just like Newtonian gravity, but at a certain point (I came up with $$r=(\frac{6G\rho_{\mathrm{res}}V}{\Lambda_{\mathrm{vac}} c^{2}})^{1/3}$$) a small 1kg mass would no longer be attracted but would be repulsed away instead.  Therefore, for small distances gravity should look just like we experience it.  For distances past a certain point though, everything would just keep getting pushed away faster and faster.  Kind of like that NASA image.

So, does it match up with the data from the expansion?  Don't know yet, but am looking for any interested scientists or people who just want to know about this crazy universe we live in.

I didn't want to just copy and paste what I had in the other forum into here, but stay tuned and ask questions!  If you have any good constructive criticism, that is always welcome!
« Last Edit: 15/03/2012 22:41:20 by Messenger »

#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be?
« Reply #1 on: 15/03/2012 23:12:52 »
You do understand that I, as well as most others, will have to reread this several times

Let me see. Einstein once created the cosmological constant to explain why 'gravity', or the mass coupled to 'gravity' didn't fall into itself, right? So what he had in his equation was something, of what magnitude? That if he didn't counteract it, it would make all matter 'meet'.  So he suggested a equivalent counterforce called the cosmological constant to counterbalance it as he believed in a static universe.

As we found an accelerating expansion of the universe instead 1998 theorists "By bracketing the expansion history of the universe between today and when the universe was only approximately 380,000 years old, the astronomers were able to place limits on the nature of the dark energy that is causing the expansion to speed up. (The measurement for the far, early universe is derived from fluctuations in the cosmic microwave background, as resolved by NASA's Wilkinson Microwave Anisotropy Probe, WMAP, in 2003.)

Their result is consistent with the simplest interpretation of dark energy: that it is mathematically equivalent to Albert Einstein's hypothesized cosmological constant, introduced a century ago to push on the fabric of space and prevent the universe from collapsing under the pull of gravity. (Einstein, however, removed the constant once the expansion of the universe was discovered by Edwin Hubble.)"

So from a undefined 'equivalence' to a defined 'magnitude', if i get it right, but with the same name 'The cosmological constant'? and it is this defined amount you use? or is it a undefined 'equivalence'?

"assign the cosmological constant as the total value of mass-energy of the vacuum, and subtract off a tensor of remaining mass-energy of the quantum vacuum. "

Why do you subtract the 'remaining mass-energy of the quantum vacuum.' Isn't the quantum vacuum and the mass-energy of a vacuum the same thing? I am most certainly missing something here?
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#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #2 on: 15/03/2012 23:23:47 »
I will have to read that pdf you wrote for sure, and get some sleep first.
==

Don't get me wrong here, I like the idea, it would be a nice solution, although rather mind boggling as you introduce a gravity that is both 'attractive' as well as repulsive at long distances, but that is good Keeps ones mind working.

But I'm still trying to see how you came up with it.
« Last Edit: 15/03/2012 23:38:04 by yor_on »
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#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #3 on: 15/03/2012 23:24:16 »
This is good, as it sometimes what is in my head doesn't quite make it to the page.  What I mean is that if you were to take the total quantized energy of the vacuum (flat space-time for a certain volume) with no matter or energy in it and assign that to $$\Lambda$$, then subtract off the mass-energy of a certain body, say the Sun, then assign that smaller value to a tensor (still extremely HUGE, I call $$\Pi$$), it is still formulaically equivalent to Einstein's energy-momentum tensor BUT if we calculate how gravity works with this alternate tensor, we see something I think spectacular.  It appears not that there is negative pressure, but that WE are a reduction in mass-energy compared to the vacuum.

« Last Edit: 15/03/2012 23:28:46 by Messenger »

#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #4 on: 15/03/2012 23:41:35 »
Heh, sweet, that one I think I can follow
And welcome to TNS btw..
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#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #5 on: 15/03/2012 23:48:18 »
And that was mind boggling Messenger. A reduction?
Now that makes my night

It's a 'good hypothesis' as you get so many 'answers' from it.
Some of them should be testable, hopefully.
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#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #6 on: 16/03/2012 03:52:53 »
Here are some aids in visualizing this (and remember, as hard as this is all to believe, it is all derived from the fundamental theorem of calculus, general relativity and the accelerating expansion):

All gravity is repulsive, but when two bodies come within a certain range near each other, the repulsion between them decreases.  Looking at the repulsive Newtonian equation,

$$\vec{g}=-\frac{\Lambda_{\mathrm{vac}} c^{2}r}{6}\hat{\vec{r}}+\frac{G\rho_{\mathrm{res}}V}{r^{2}}\hat{\vec{r}}$$

the second term on the right hand side is dominant when r is small.  This would be how we experience Newtonian gravity.  As a test body goes away from the main mass, r gets bigger , and the last term decreases.  This means that the repulsion between the two bodies increases, but the repulsion on all other sides is constant.  At some value for r, the two terms on the right hand side are equivalent, and the repulsion on that test body is equivalent on all sides. As that test body goes even farther though, the repulsion gets even larger between the two bodies.  From our point of view it would appear that the attractive gravity has decreased to zero and started becoming repulsive.

Another way to think of it is if you were looking at a large mass some distance away from you, say a sun, and for whatever reason the mass of that sun grows, the second term decreases.  The repulsion between you and the sun also decreases.

A body alone in the universe must require a certain amount of vacuum mass-energy to exist, and lower the mass-energy of the vacuum around them. Provided that two bodies have their required amount, the space between them is repulsive.  When they come within the mass-energy range of each other, though, that energy reduction is additive and is greatest in the line between them.

A good way to think of it is that we experience the world as a vivid color image, but our physics works in the negative of the picture.

The easiest way to derive this back to the Einstein field equation is to look at those gradients and understand them well.  Then go back through Einstein's derivation that I have listed in the references.  What you end up finding is that if the EFE holds true, then this equation should very accurately predict the expansion of the universe.  If you goto http://en.wikipedia.org/wiki/Spacetime_symmetries, you will see that it accomplishes the three relativistic preserving properties.

For the expansion of the universe, imagine that you cast a handful of flour onto a smooth tile floor.  Every flour particle repulses each in the vicinity, but that repulsion is reduced if it is in a certain range.  It would appear as attractive gravity and they would start randomly moving towards each other.  After awhile there would be clusters and an average distance between them.  When that average distance goes past that tipping point calculated from each cluster then there should be a switch over to all the clusters moving apart faster and faster.

Although I don't agree with negative mass particles, a good visualization can be found by Hyoyoung Choi at http://www.youtube.com/watch?v=SRUqQM2FfNU  Think of the pink particles as us (pits in the energy of vacuum mass-energy), and the negative particles as the Lorentzian aether.

Anybody know how to increase tex size?
« Last Edit: 16/03/2012 04:36:26 by Messenger »

#### MikeS

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #7 on: 16/03/2012 08:36:31 »
I'm no mathematicial but I did learn Calculus 50 odd years ago.  50 years with no use of calculus and it is all but forgotten.  So I have no idea as to the mathematics behind Messengers idea but it does have a certain appeal.

We normally consider gravity to be an attractive 'force'.  Please forgive the term force.  Perhaps it would be equally valid to think of gravity as being due to the effect of space-time being squeezed.  Where there is considerable gravity as in a super-cluster then mass is squeezed into a smaller volume at the expense of time which dilates.

In the intergalactic voids between super-clusters there is little gravity to squeeze time so space (measured distance) increases as time contracts.

Looked at this way the effect of gravity appears to be attractive at short distance as matter is being pushed together but repulsive at very great distance as space is stretched (being pushed apart) or being created.

If we think of gravity as squeezing space time and all that it contains then it can be both attractive at short distance and repulsive at great distance.

Energy creates matter, creates gravity causing entropy to increase.  This is what happens at short distance.  At great distance where relatively free from the restrictions of gravity, energy dissipates creating space whilst increasing entropy.
« Last Edit: 16/03/2012 08:54:38 by MikeS »

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #8 on: 16/03/2012 12:41:06 »
Here is what is bothering me in the back of my mind.  General relativity works great for systems such as our solar system.  If it doesn't work well at large scales though, but this reversal works well for both, why should we think any calculations for very small scales, such as the Large Hadron Collider, are in any way representative of reality?  If a black hole were possible, it doesn't suck anything in, there is pressure in towards a very large decrease in mass-energy.  I have no way to know if particle collision experiments are safe, but I see no reason to trust the judgement of CERN.  Am I over reacting?

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #9 on: 16/03/2012 12:52:00 »
Interesting analysis Mike.

#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #10 on: 16/03/2012 13:04:00 »
Yes Messenger, you put your thumb on a thing that is bothering me too.
I'm getting more and more interested in what 'scales' do with 'reality'.

Scales and probability goes together in my mind.
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#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #11 on: 16/03/2012 13:17:54 »
As for Cern?

Depends on if you're right, right
And even if you are, you can't 'know' for sure.
What about the hard radiation they use as a proof for black holes being unable to form?

I'm still trying to see this one
You've had some time working it through, I need it too )

And thinking of it, if you are right we should find black holes rather alone?
Or put it as this, if they gets created in a system of a lot of stars the 'force' is attractive.
But if we find one alone, then maybe?

They should be very 'small' if so as they will repulse before attract?
But I'm still trying to wrap my mind around the idea, and it would most probably not constitute a 'proof'
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#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #12 on: 16/03/2012 13:33:11 »
I have read the arguments that far worse collisions happen all the time, but if you have any links you would like to post that would be great too.  You are correct though, I have no way to know.  It is actually far more likely that I have gotten something wrong here
As I read somewhere, proofs exist in math but not in physics.  Only a consensus.

As for Cern?

Depends on if you're right, right
And even if you are, you can't 'know' for sure.
What about the hard radiation they use as a proof for black holes being unable to form?

I'm still trying to see this one
You've had some time working it through, I need it too )

And thinking of it, if you are right we should find black holes rather alone?
Or put it as this, if they gets created in a system of a lot of stars the 'force' is attractive.
But if we find one alone, then maybe?

They should be very 'small' if so as they will repulse before attract?
But I'm still trying to wrap my mind around the idea, and it would most probably not constitute a 'proof'

#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #13 on: 16/03/2012 13:55:00 »
Yep, it would be cool to get your math examined, I'm definitely not certified for that though
But your ideas and conclusions are still very interesting, as well as your descriptions of how you see the math.
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#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #14 on: 16/03/2012 16:36:32 »
One more analogy, Lorentzian invariance of space-time. Note that these are only 3 dimensional analogies, 4 dimensional motion is harder to imagine but I am working on that.

You may have read that the metric is a property of space-time and it has negative pressure.  If you view space-time as a jello, and we are decreases in density in it, then you get a natural Lorentz invariance where a negative energy density is proportional to a positive pressure.  This would be where $$g_{\mu\nu} =diag(-1,1,1,1)$$ comes from.

#### Æthelwulf

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #15 on: 17/03/2012 02:32:43 »
So many scientists today appreciate dark energy as a must. They are wrong. In fact, the root of the problem is the Big Bang. There are in fact atleast 15 major problems with the BB. In fact, the static universe can answer for more than what the 15 major problems of BB can account for. I will find you a link.

#### Æthelwulf

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #16 on: 17/03/2012 02:34:22 »
In fact after a quick read, there are at least 30 problems with the BB. Here, you will find the top ten http://metaresearch.org/cosmology/top10BBproblems.asp

enjoy.

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #17 on: 17/03/2012 02:38:10 »
Heh. The list is from 1997, so the first one seems to be out
Static universe models fit the data better than expanding universe models.

but I think you would find the simulations listed by Icarus2 on bautforum.com in the ATM threads very interesting.

#### Æthelwulf

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #18 on: 17/03/2012 02:40:38 »
All I need to know is that the BB is faulty premise. Even the reminent BB hypothesis of an leftover of radiation is in fact the average temperature radiated by star systems. The BB is wrong.

So wrong in fact, it does not apply to a correct model. The fact we have not unified it yet is an evidence this is the wrong approach.

#### Æthelwulf

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #19 on: 17/03/2012 02:42:36 »
You must read these implications. Scientists alike have been deluded by this ''phenomenon''. Even by someone like me, who supported such a method for years to come... till now. I've realized myself the implications of these mistakes. They are evident. Some scientists would rather bury their heads in the quantum sand.

#### Æthelwulf

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #20 on: 17/03/2012 02:44:34 »
''Icarus2 on bautforum.com in the ATM threads very interesting.''
I know him, and have seen his speculations. No harm to the man, but he is wrong, atleast in representing his work in some scientific format, like a rigourous math. So no, I would not be interested, no real disrespect to him though.

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #21 on: 17/03/2012 03:04:21 »
I see from one of your posts that you believe in a static universe.   Does that mean you don't trust the interpretation that distances are increasing between galaxies?

#### MikeS

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #22 on: 17/03/2012 05:53:52 »
I see from one of your posts that you believe in a static universe.   Does that mean you don't trust the interpretation that distances are increasing between galaxies?

Messenger I don't  know what posts you have read on this forum but I suspect that the Hubble red shift as an interpretation of distance and age of the universe may be wrong.  I believe you can get the same effect from time contraction (speeding up) over cosmological time.  There have been heated arguments over this but as far as I am aware there is very little evidence to confirm the interpretation of the Hubble red shift as being correct.  The cosmological red shift may be due to a mixture of expansion and time contraction.

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #23 on: 17/03/2012 06:13:20 »
Well, I am open to any explanation as long as it helps explain certain phenomena and I can logically/mathematically follow it, even with some effort.  Have any links I can read up on it?

#### MikeS

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #24 on: 17/03/2012 06:35:55 »
Messenger

Hypothesis # 1
According to GR gravity is warped space time due to mass and energy.
What if the 'energy' component of that is negative? (Energy dissipates, entropy)
Gravity would be attractive where there is mass and repulsive where there is no mass (energy being everywhere but it's repulsive gravity being overwhelmed by the attractive gravity of mass).
This equates to gravity being attractive at short range and repulsive at long range which is what we see.

Hypothesis # 2
Even if 'energy' has attractive (or no) gravity it's effect may be overwhelmed by energies tendency to disperse.  Possibly when dispersing free from the influence of gravity it 'creates' space.
If this is correct then it's the same outcome as above.
Gravity would (seem to) be attractive where there is mass and repulsive where there is no mass (energy being everywhere).
This again equates to gravity being attractive at short range and repulsive at long range which is what we see.

[In answer to you request for links, a search for Hubble red-shift, cosmological red-shift, cosmological time should produce many links.]
« Last Edit: 17/03/2012 06:37:28 by MikeS »

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #25 on: 17/03/2012 06:46:38 »
Mike, it might interest you to read my paper very carefully when you get time.  What it shows you is that if GR is inverted, it naturally gives you attractive gravity up close, and a predictable point where it becomes repulsive, as well as how to quantize space-time via quantum harmonic oscillation modes.

#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #26 on: 17/03/2012 22:41:43 »
Ive been thinking a little about your idea about matter being a reduction. Maybe you can relate it to indeterminacy. Imagine that first BB, without a point, just that 'energy'. And that energy is a pressure, and a pressure should be a temperature sooner or later. A temperature though needs a 'space' to exist in as it is consisting of the 'energy' of rest mass 'jiggling' and interacting. That would place temperature as a secondary phenomena, with pressure (as we get a 'space' from it) as the first principle.

So pressure, does it need a space? Can you imagine a pressure without? Or does that follow directly from the emergence of a pressure. But 'energy' then, that 'energy' that we expect to exist before that 'space' came to be? And there we can look at QM:s interpretation of the 'vacuum energy' or 'zero point energy'. You can expect it to exist in every 'empty' point of space. A space that is observer defined, not defined as an absolute. And in we have indeterminacy, that something that we extrapolate into outcomes, gaining a arrow in those rare circumstances it express itself. So is 'space' of a earlier order than matter, containing what made matter? It seems so to me?
==

Take 'earlier' with a big pinch of salt here, you can imagine a 'empty space' at least two ways. Assume a Black hole cleaning out a 'space' of all mass. As the BH is a singularity it does not fit the physics we use, our physics end at the event horizon, the rest pure theory, Or you can assume a 'space' with 'gravity' as a result of that 'energy' pressuring on some boundaries. Or, you might assume that 'space' doesn't need a metric (gravity) to exist, even though it to me then becomes a dimension less point. What I mean by 'dimension less point' here is that I expect it to be unmeasurable practically. and what I can't measure isn't 'there' classically, although you theoretically can give it any 'dimensions' you want, excluding the arrow. Also that has to do with how I expect a existing point inside SpaceTime to exist, it should as soon as it is in a arrow be 'four dimensional'.

There are probably more ways.
=

And we have a problem with the arrow here, if I assume a 'empty space' is that enough for a arrow?
Maybe if the space is ? one dimensional, nah, at least two dimensions is needed. One for the arrow, the other for ?? Whatever you want to define it as. Strings, loops etc. Or you include 'gravity' in which case you need a 'four dimensional space', as I do. Maybe you can assume the dimensions to fluctuate at 'different points' though,  remember that distance is a meaningless concept at the origin, meaning that there are different combinations with the one we see being a 'semi stable' four dimensional continuum, including the arrow. But all of them should need the equivalent of a arrow to 'grow', as a guess.

Without the arrow we're back to indeterminacy.
« Last Edit: 18/03/2012 01:43:36 by yor_on »
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#### yor_on

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #27 on: 18/03/2012 00:38:21 »
Hmm, maybe you need to expand on "but that WE are a reduction in mass-energy compared to the vacuum."

Are you questioning E=MC^2, or is it a statement of that, if we could measure on the vacuum properly, that we would find it of a higher 'energy density' than expected. I presumed the second, but thinking again I'm unsure?

"The greatest energy source by far consists of mass itself. This energy, E = mc2, where m = ρV, ρ is the mass per unit volume, V is the volume of the mass itself and c is the speed of light. This energy, however, can be released only by the processes of nuclear fission, nuclear fusion, or the annihilation of some or all of the matter in the volume V by matter-antimatter collisions. Nuclear reactions cannot be realized by chemical reactions such as combustion. Although greater matter densities can be achieved, the density of a neutron star would approximate the most dense system capable of matter-antimatter annihilation possible. A black hole, although denser than a neutron star, doesn't have an equivalent anti-particle form."

Is there any way to test that one? What we know is matter, and there the experiments supports Einsteins definition.

What I can say from my definitions is that indeterminacy, which I prefer, doesn't define any specific 'energy' for any 'point of space', at least not singularly. Using 'virtual particles' as your definition you can state it as there is no limit to their 'energy'. And then the question becomes if there is no limit and I made a observation of all points in a space, would that still be true?

You better define your thoughts there further
Also it would be worthwhile to see what definition of 'cosmological constant' you refer too?
Einsteins undefined 'equivalence' or ""By bracketing the expansion history of the universe between today and when the universe was only approximately 380,000 years old, the astronomers were able to place limits on the nature of the dark energy that is causing the expansion to speed up."
==

Rereading you "the total quantized energy of the vacuum" But there is no definition of that one, or is there?

"Canonically, if the field at each point in space is a simple harmonic oscillator, its quantization places a quantum harmonic oscillator at each point. Excitations of the field correspond to the elementary particles of particle physics. Thus, according to the theory, even the vacuum has a vastly complex structure and all calculations of quantum field theory must be made in relation to this model of the vacuum."

But then you have Lorentz contractions, redefining that 'space', and I would assume time dilations to have an importance for the harmonic oscillators 'oscillations' too. There is no way you can define that universally in Einsteins relativity?  All definitions using 'a clock and ruler' measuring will only have a local importance, as I see it?
==

To clarify my thinking there. all measurements/experiments are done relative the local observer, using his local clock and his local ruler. That invalidates any definition of absolute universe, as long as you don't assume that the 'true measure' of the universe is strictly conceptual, as in expressed through Lorentz transformations. And if using that you will find a 'whole' universe impossibly complex to define, as you will need to Lorentz transform from each 'point' of it to each other point, to finally get some conceptual statistical average. I don't think that is possible. (And that is assuming a static 'still picture', not including the arrow, relative motion and accelerations.)
« Last Edit: 18/03/2012 01:27:14 by yor_on »
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#### Messenger

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• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #28 on: 18/03/2012 02:02:04 »
Great questions!

I haven't explored this part quite as in depth, but maybe you can help point out any flaws.  Not questioning $$E=mc^2$$ at all.

Lets start with the definition of mass-energy in this other sense.  If we perceive mass as $$m=\rho V$$, but in the paper for the source of gravitation our perception of mass is $$m=(\Lambda\frac{c^{2}}{8\pi G}-\rho_{\mathrm{res}})V$$
so our perception of energy would have to be
$$E=mc^2=(\Lambda\frac{c^{4}}{8\pi G}-\rho_{\mathrm{res}}c^2) V$$
so what I take from this equation is that the entire perceptible energy to us includes not only the mass of the object but the gravitational field it exhibits.

"What I can say from my definitions is that indeterminacy, which I prefer, doesn't define any specific 'energy' for any 'point of space', at least not singularly. Using 'virtual particles' as your definition you can state it as there is no limit to their 'energy'. And then the question becomes if there is no limit and I made a observation of all points in a space, would that still be true?"

This seems to me to get at the heart of renormalization.  I don't see how there could not be a limitation.  If by indeterminacy, you are referring to concepts based on Schrodinger's equation, I would have to bow out of that for now.

As far as I can tell,  the cosmological constant in the original EFE with the curvature tensor G is nothing but an abstract equation.  By itself it has no physical significance, and the cosmological constant is nothing more than the constant in the equation.  However, when the curvature tensor is lined up with the stress-energy tensor, so that $$G_{\mu\nu}=\kappa T_{\mu\nu}$$ then it becomes a very big question of what the cosmological constant represents.  As I see it, it was fine when it originally represented the reference point of empty vacuum, but as Zeldovich realized, empty space isn't exactly empty but his calculations of the energy within it (and which should have affected gravitation) had no relationship to reality (and it is even worse now).  Whether or not there are zero-point energy fluctuations at all points in space-time seems to still be an open question.  If there were, in my view this would just slow the accelerating expansion in the same manner as during the early universe when matter was spread out.

I view the cosmological constant as nothing more than a reference point, whereby we can measure curvature of space-time based on energy differences.

"By bracketing the expansion history of the universe between today and when the universe was only approximately 380,000 years old, the astronomers were able to place limits on the nature of the dark energy that is causing the expansion to speed up."

This is a reference to the accelerating expansion phase we are in.  This would be the logical next option to pursue but I think this needs some more in depth thought on energy conservation in regards to the repulsion.

I am not sure what you mean when you say Einstein's "equivalence".  Do you mean the equivalence principle?  I think we might be able to make some headway into why inertial mass is equivalent to gravitational mass.

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #29 on: 18/03/2012 02:22:57 »
Sorry, just read this again and see what you mean by indeterminacy. Still not quite sure I follow, but let me put something to you about your energy and pressure analogy.  If the metric $$g_\mu\nu$$ is a property of space-time, and the positive energy of space-time has a positive pressure, but in order for any structures to exist within that energy they must be a decrease.  Therefore, those structures would see themselves as having positive energy and positive pressure, but from the point of view of the space-time they are a decrease in energy within that positive pressure.  This would be where the opposing signs come from in the metric.  When velocities increase, energies contents increase, length contractions occur and from the point of view of the universe the energy decrease is getting deeper (as well as my analogy heh).  I am trying to boil some things down to simplicity but could easily run into some logical errors in my thinking.

Ive been thinking a little about your idea about matter being a reduction. Maybe you can relate it to indeterminacy. Imagine that first BB, without a point, just that 'energy'. And that energy is a pressure, and a pressure should be a temperature sooner or later. A temperature though needs a 'space' to exist in as it is consisting of the 'energy' of rest mass 'jiggling' and interacting. That would place temperature as a secondary phenomena, with pressure (as we get a 'space' from it) as the first principle.

So pressure, does it need a space? Can you imagine a pressure without? Or does that follow directly from the emergence of a pressure. But 'energy' then, that 'energy' that we expect to exist before that 'space' came to be? And there we can look at QM:s interpretation of the 'vacuum energy' or 'zero point energy'. You can expect it to exist in every 'empty' point of space. A space that is observer defined, not defined as an absolute. And in we have indeterminacy, that something that we extrapolate into outcomes, gaining a arrow in those rare circumstances it express itself. So is 'space' of a earlier order than matter, containing what made matter? It seems so to me?
==

Take 'earlier' with a big pinch of salt here, you can imagine a 'empty space' at least two ways. Assume a Black hole cleaning out a 'space' of all mass. As the BH is a singularity it does not fit the physics we use, our physics end at the event horizon, the rest pure theory, Or you can assume a 'space' with 'gravity' as a result of that 'energy' pressuring on some boundaries. Or, you might assume that 'space' doesn't need a metric (gravity) to exist, even though it to me then becomes a dimension less point. What I mean by 'dimension less point' here is that I expect it to be unmeasurable practically. and what I can't measure isn't 'there' classically, although you theoretically can give it any 'dimensions' you want, excluding the arrow. Also that has to do with how I expect a existing point inside SpaceTime to exist, it should as soon as it is in a arrow be 'four dimensional'.

There are probably more ways.
=

And we have a problem with the arrow here, if I assume a 'empty space' is that enough for a arrow?
Maybe if the space is ? one dimensional, nah, at least two dimensions is needed. One for the arrow, the other for ?? Whatever you want to define it as. Strings, loops etc. Or you include 'gravity' in which case you need a 'four dimensional space', as I do. Maybe you can assume the dimensions to fluctuate at 'different points' though,  remember that distance is a meaningless concept at the origin, meaning that there are different combinations with the one we see being a 'semi stable' four dimensional continuum, including the arrow. But all of them should need the equivalent of a arrow to 'grow', as a guess.

Without the arrow we're back to indeterminacy.

#### Messenger

• Jr. Member
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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #30 on: 18/03/2012 02:58:12 »
This might help to, take a look at around the 20:55 point.  I view those bright virtual particle colors as decreases in energy with respect to the vacuum.

#### Messenger

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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #31 on: 18/03/2012 03:56:17 »
Here is a thought that might help.  In the equations of general relativity, keep all of the formalism and accuracy but mirror the magnitudes of everything. Then with the cosmological constant rewrite the Newtonian equation.  In any way you view this, the cosmological constant and the Newtonian gradient end up with opposing vectors.  At some radius, the two end up being equal, and in our current view some mysterious force is attributed to the cosmological constant that causes an acceleration in the expansion in addition to no path of how to match this up with quantum field theory.
The vectors in the mirror image though, just show that at the point of when they become equal, the repulsion is no longer reduced enough to keep them from moving apart.  The combined curvature lessens until it becomes equal to the back ground curvature, and nothing to keep them together.

#### Messenger

• Jr. Member
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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #32 on: 18/03/2012 04:36:23 »
Another note: I have no idea if the flyby anomaly is related or not, but it is interesting to see the similarities between the precession of the perihelion in the picture here
http://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity

and in this video I made with a negative mass ring (represents the repulsive gravity)

I don't have the background to do the calculations that would be needed for the flyby anomaly or the pioneer anomaly (as someone earlier asked, sorry for the late reply).

#### MikeS

• Neilep Level Member
• 1044
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #33 on: 18/03/2012 05:29:42 »
clip
"A black hole, although denser than a neutron star, doesn't have an equivalent anti-particle form."

Perhaps it does.  A white hole.  Do white holes exist in our Universe?
No we would not expect them to as the Universe is made essentially from matter.
Maybe they did at the start of the Universe.  That would account for quasars.

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #34 on: 18/03/2012 15:57:53 »
If you are a mathematical physicist, you might be horrified at my lack of formalism, so instead allow me to pose some questions to the readership at large (but feel free to pipe in of course!).  It might be helpful if you are familiar with some simple calculus and vector analysis.  You can find my derivation at http://vixra.org/abs/1203.0025 (arxiv rejected it).

I've only just started reading your paper. At first I glanced the OP quickly and gave you my answer.

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #35 on: 18/03/2012 16:20:27 »
Ok, I've read some of it. One equation at least so far seems to appear to be wrong.

First of all, your dimensions here are suspicious.

rho = E/V = 1/V sum 1/2 hbar  omega

Does this place use tex?

Anyway, rho is

M/V

You have equalled this to Mc^2/V. Also, what was you motive for moving V outside the summation with a 1 over it? That effectively removed the volume from your equations. For instance, imagine I had

A = partial^2 psi/partial x^2

and B = e^-ikx e^i omega

Thus

A = -sum 2m omega a*(k) B

It is customary to remove the 2m, and by doing this one does

1/2m A = -sum omega a*(k)

Just a quibble so far. I haven't seen that equation before in the context you have given it. the rest of the oscillator equation seems familiar.

#### yor_on

• Naked Science Forum GOD!
• 12188
• (Ah, yes:) *a table is always good to hide under*
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #36 on: 18/03/2012 19:56:29 »
Heh, we both seems to 'get of' on thinking And when I do it keeps growing and one finds it a he* of a problem to stop before it grows too unruly, to those not being inside my mind seeing how it grew.

Before I forget. Take a look at this one.
Determining dynamical equations is hard. It seems to have such a relevance to my thoughts about 'static and dynamic systems' as Lorentz transforming a 'system'.

And the 'undefined' equivalence I was referring to was the original cosmological constant in where I gather he took the opposite 'force' of what he expected was necessary to stop all mass to fall in on itself. Maybe 'undefined' is wrong there? But as an assumption you don't need the exact value, you just need an 'equivalent' factor keeping the invariant mass and universe 'static', as was his first thought, before the 'expansion'?

But the equivalence principle is GR, so you're right, haven't thought about that one yet, still trying to understand where this might take me
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#### yor_on

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• 12188
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##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #37 on: 18/03/2012 20:03:01 »
"I am trying to boil some things down to simplicity but could easily run into some logical errors in my thinking." Nothing wrong in that. If you can make it follow logically in words there's a greater chance that your math has something to say too. As long as we expect the universe to follow a logic of some sort, 'emergence', non-linear or/and linear.

If it doesn't all math can be questioned.
==

Hmm, I wrote "Maybe if the space is ? one dimensional, nah, at least two dimensions is needed. One for the arrow, the other for ?? Whatever you want to define it as. Strings, loops etc."

Why I find it hard to see it as one dimensional is that it 'jiggles', and is under tension. Both tension and jiggling presumes an arrow to do in as I see it, if you don't want the arrow to come out of the very fact that 'tension/jiggling' comes to be?

But that seems to me as me lifting myself by my hair, if you see what I mean? Not logically following, if I don't presume an 'emergence' for both the arrow and the tension. But where would that 'emergence' build up?
In what?

0uch, string theory is hard to follow.
==

There is one possibility more naturally, but then we're back to 'two dimensions', assuming that 'the arrow' and 'time' are two concepts, where the arrow rests in 'time'. That would make 'time' hidden to us, but allowing what we see to be 'one-dimensional'.

( And every time I wonder over that one, I end up wondering about indeterminacy
« Last Edit: 18/03/2012 20:24:00 by yor_on »
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #38 on: 18/03/2012 20:53:57 »
Which equation number was this? I want to make sure I can check my notes to see what my motive was.  Use the the word "tex" surrounded by [] to start and "/tex" surrounded by [] to end.

Ok, I've read some of it. One equation at least so far seems to appear to be wrong.

First of all, your dimensions here are suspicious.

rho = E/V = 1/V sum 1/2 hbar  omega

Does this place use tex?

Anyway, rho is

M/V

You have equalled this to Mc^2/V. Also, what was you motive for moving V outside the summation with a 1 over it? That effectively removed the volume from your equations. For instance, imagine I had

A = partial^2 psi/partial x^2

and B = e^-ikx e^i omega

Thus

A = -sum 2m omega a*(k) B

It is customary to remove the 2m, and by doing this one does

1/2m A = -sum omega a*(k)

Just a quibble so far. I haven't seen that equation before in the context you have given it. the rest of the oscillator equation seems familiar.

#### yor_on

• Naked Science Forum GOD!
• 12188
• (Ah, yes:) *a table is always good to hide under*
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #39 on: 18/03/2012 20:58:09 »
I expect this thread to grow organically

And yes Mike, I wondered about that too, but??
Lets pick Messengers ideas apart (ahem:) first, and see where it goes
==

" If the metric is a property of space-time, and the positive energy of space-time has a positive pressure, but in order for any structures to exist within that energy they must be a decrease.  Therefore, those structures would see themselves as having positive energy and positive pressure, but from the point of view of the space-time they are a decrease in energy within that positive pressure. "

What you are referring to here is the notion in QM that 'space' isn't the lowest state of 'oscillation', as an assumption, Space becoming something of a 'Mexican hat' according to Brian Greene (if I remember right). And you turn it around suggesting that we, to 'space' would contain a lower state of 'energy'? Or maybe I could see it as that this concept would be wrong too, possibly?

But "researchers in quantum optics have created special states of fields in which destructive quantum interference suppresses the vacuum fluctuations. These so-called squeezed vacuum states involve negative energy. More precisely, they are associated with regions of alternating positive and negative energy."

And Negative Energy: From Theory to Lab.  Hmm, rereading this I find some rather far-fetched conclusions, that I won't agree too. Sorry about that, was looking for a online source, describing squeezed states in more detail, but got sloppy.

This one seems more to the point.. But I will see if I can find something more descriptive later.

« Last Edit: 18/03/2012 21:18:38 by yor_on »
"BOMB DISPOSAL EXPERT. If you see me running, try to keep up."

#### Messenger

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• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #40 on: 18/03/2012 21:02:15 »
"I am trying to boil some things down to simplicity but could easily run into some logical errors in my thinking." Nothing wrong in that. If you can make it follow logically in words there's a greater chance that your math has something to say too. As long as we expect the universe to follow a logic of some sort, 'emergence', non-linear or/and linear.

If it doesn't all math can be questioned.

Good point.
==

Hmm, I wrote "Maybe if the space is ? one dimensional, nah, at least two dimensions is needed. One for the arrow, the other for ?? Whatever you want to define it as. Strings, loops etc."

Why I find it hard to see it as one dimensional is that it 'jiggles', and is under tension. Both tension and jiggling presumes an arrow to do in as I see it, if you don't want the arrow to come out of the very fact that 'tension/jiggling' comes to be?

But that seems to me as me lifting myself by my hair, if you see what I mean? Not logically following, if I don't presume an 'emergence' for both the arrow and the tension. But where would that 'emergence' build up?
In what?

0uch, string theory is hard to follow.

My eyes glaze over, are you able to follow some of it?
==

There is one possibility more naturally, but then we're back to 'two dimensions', assuming that 'the arrow' and 'time' are two concepts, where the arrow rests in 'time'. That would make 'time' hidden to us, but allowing what we see to be 'one-dimensional'.

( And every time I wonder over that one, I end up wondering about indeterminacy

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #41 on: 18/03/2012 21:04:46 »
I expect this thread to grow organically

And yes Mike, I wondered about that too, but??
Lets pick Messengers ideas apart (ahem:) first, and see where it goes
==

" If the metric is a property of space-time, and the positive energy of space-time has a positive pressure, but in order for any structures to exist within that energy they must be a decrease.  Therefore, those structures would see themselves as having positive energy and positive pressure, but from the point of view of the space-time they are a decrease in energy within that positive pressure. "

What you are referring to here is the notion in QM that 'space' isn't the lowest state of 'oscillation', as an assumption, Space becoming something of a 'Mexican hat' according to Brian Greene (if I remember right).

And you turn it around suggesting that we, to 'space' would contain a lower state of 'energy'?
----
Ah! Nice visual, yes!
---
Or maybe I could see it as that this concept would be wrong too, possibly?

But "researchers in quantum optics have created special states of fields in which destructive quantum interference suppresses the vacuum fluctuations. These so-called squeezed vacuum states involve negative energy. More precisely, they are associated with regions of alternating positive and negative energy."

And Negative Energy: From Theory to Lab.

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #42 on: 18/03/2012 21:15:20 »
Ah, I see, with the quantum harmonic oscillator.  That was an equation from reference [12]
S. E. Rugh and H. Zinkernagel, Studies In History and
Philosophy of Science Part B: Studies In History and
Philosophy of Modern Physics 33, 663 (2002)
http://philsci-archive.pitt.edu/398/1/cosconstant.pdf
Eq. 6

I was using it as an example of how we might be able to match up the energy density from the harmonic oscillator to the energy density from the metric equation.  I will have to examine it more in depth to make sure I didn't transcribe it incorrectly.  Note that in the reference, I think that there is a typo on the order of magnitude, as I think it showed $$10^144$$ and I assumed it was actually supposed to read 114, as that was more in line with what I had read elsewhere (looks like it was fixed in this version, think I also looked at the arxiv version).  I think the number is more of a demonstration of how incorrect something must be rather than a precise estimate.

Ok, I've read some of it. One equation at least so far seems to appear to be wrong.

First of all, your dimensions here are suspicious.

rho = E/V = 1/V sum 1/2 hbar  omega

Does this place use tex?

Anyway, rho is

M/V

You have equalled this to Mc^2/V. Also, what was you motive for moving V outside the summation with a 1 over it? That effectively removed the volume from your equations. For instance, imagine I had

A = partial^2 psi/partial x^2

and B = e^-ikx e^i omega

Thus

A = -sum 2m omega a*(k) B

It is customary to remove the 2m, and by doing this one does

1/2m A = -sum omega a*(k)

Just a quibble so far. I haven't seen that equation before in the context you have given it. the rest of the oscillator equation seems familiar.
« Last Edit: 18/03/2012 21:21:18 by Messenger »

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #43 on: 18/03/2012 21:49:43 »
And now that I think of it, I need to fix that reference placement so it is more evident where the equation came from. Woops.

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #44 on: 19/03/2012 02:00:02 »
Right, I checked the link. It appears to be the sixth equation in your reference. I am still a bit confused, never -- not once have I seen a density be described as

$$\frac{E}{V}$$ [1]

rather you always see it as

$$\frac{M}{V}$$ [2]

Which would mean this is not a usual density? Which still doesn't add up... for me at least.  If equation 2 has dimensions of density, then equation 1 cannot? No, is there something I am missing? Am I being stupid here?

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #45 on: 19/03/2012 02:15:28 »
Ok I have a new criticism. You say that on page 5 that the equation of state has $$\frac{p}{3}$$ for a single dimension. What you really meant is that $$3$$ accounts for the three dimensions of space? Fair, or did I mix you up. Is this really what you meant?

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #46 on: 19/03/2012 03:28:42 »
Right, I checked the link. It appears to be the sixth equation in your reference. I am still a bit confused, never -- not once have I seen a density be described as

$$\frac{E}{V}$$ [1]

rather you always see it as

$$\frac{M}{V}$$ [2]

Which would mean this is not a usual density? Which still doesn't add up... for me at least.  If equation 2 has dimensions of density, then equation 1 cannot? No, is there something I am missing? Am I being stupid here?

Hm, I hadn't looked at that in depth.  Might not be on here for a few days but I will look at that when I get back.

#### Messenger

• Jr. Member
• 27
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #47 on: 19/03/2012 03:30:34 »
Ok I have a new criticism. You say that on page 5 that the equation of state has $$\frac{p}{3}$$ for a single dimension. What you really meant is that $$3$$ accounts for the three dimensions of space? Fair, or did I mix you up. Is this really what you meant?

Yes, I modeled that after the same reason for the 3 in the Friedmann equation for a single dimension, but now looking at it again that should have been $$\frac{\rho}{3}$$, no?

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #48 on: 19/03/2012 03:58:56 »
Yea, that would read the density expressed in three dimensions.

#### Æthelwulf

• Sr. Member
• 358
##### Re: Dark energy? Are we just making things harder than they need to be Lorentz?
« Reply #49 on: 19/03/2012 04:01:30 »
Yea, that would read the density expressed in three dimensions.

Maybe I should add that the equation of state in the Friedmann model has a constant $$\omega$$ in it. So the full equation is really

$$p = \rho c^2 \omega$$

then

$$\frac{p}{\omega} = \rho c^2$$