View Full Version : Does the cosmological constant push ?

tommac

2008-Apr-23, 04:09 PM

Sorry if this is a stupid question but does the cosmological constant push?

The thought is that gravity due to a contracting of space-time seems to pull stuff towards an object.

Does the cosmological constant push things away from a vacuum?

I think this effect would be small as it would be equal to the gravitational pull at the point where space-contracts 2x the cosmological constant. Which I believe would we the gravitational pull well outside of our galaxy ( but not smart enough to do the math here)

However I am not sure how to determine this ... because mass can deduced to a singularity of center of gravity. However can a vacuum be deduced to such a singularity? Can a gradient of gravity be created? Is gravity scalar?

If gravity is scaler and a gradiant can be created. Then I would proposed that light will tend to travel in a straight line not in a perfect vacuum but rather where the gravitational pull exactly offsets the cosmological constant.

publius

2008-Apr-23, 06:54 PM

Tommac,

There are a lot of misconceptions about things there in your post. These things are not trivial, and it's very easy to "get ideas in your head" based on words being used to describe what can only be rigorously described mathematically.

First gravity is not a "contraction of space-time". Gravity is due to the *curvature* of space-time. And here, "gravity" is not the same thing as the Newtonian notion of a g field, but is better thought of as gradients in that g field (the tidal forces). The Newtonian 'g' is more of a coordinate thing than anything invariant.

Space-time is described in General Relativity by a 4 dimensional geometric construct, a manifold. There are are many ways a 4D manifold can curve and it requires a rank-4 tensor to describe. While constraints and symmetries greatly reduce the number of indepedent components from 256 (4x4x4x4), there are still a lot of components.

And while I'm not sure exactly what you're driving at, gravity is not a "scalar". Newtonian gravity is a vector field. GR gravity is far more than that, a *tensor* field. In terms of potentials, Newtonian gravity can be completely described as the gradient of a scalar potential function. Now, EM requires a vector potential function (4-vector or 1 scalar plus 1 three-vector function). So thus we can say that Newtonian gravity derives from a scalar field, while EM requires a vector field potential.

However, the equivalent of the potential in GR gravity is the space-time metric, which is a rank-2 tensor. There are 10 independent components there. IOW, GR gravity is a far more complex mathematical entity.

-Richard

publius

2008-Apr-23, 07:49 PM

Now, Lambda, the cosmological constant. GR gravity is described by what is called the Einstein Field Equation (EFE). The EFE is a very complex non-linear tensor partial differential equation (which could be written out as a complicated mess of coupled non-linear partial differential equations for the 10 independent components of the metric mentioned above). Solving that mess is no trivial to say the least.

But what it does is relate the curvature of space-time to the mass-energy-momentum content of that space-time. IOW, the solution of the EFE, the metric, tells us the "shape" of space-time to the mass-energy-momentum content of that space-time.

That's the plain EFE. When Einstein derived GR, the universe was thought to be a static, unchanging thing. However, simple solutions of the EFE predicted a dynamic space-time. The universe should be flying apart or contracting due to its own gravity. It could be doing a lot of things depending on the initial conditions and the mass-energy content, but it would not be just sitting there.

So Einstein added the cosmological constant, which is a term that simply "cancels out" the gravity of the mass-energy and would allow a static universe. Then came Hubble, and so Einstein called this his "greatest blunder", as he could've predicted the Hubble expansion.

And that's the way it was until recently when the "acceleration" of expansion was discovered. Lambda can cancel out a contraction, but increase it and can go the other way and drive the expansion faster and faster, depending on the mass-energy content.

The simplest solutions to the EFE are the "empty" space-time solutions, zero mass-energy content. The solution of the plain EFE with no cosmological constant and zero source terms is, with appropriate boundary conditions (these are all important), simply flat Minkowski space-time, the playground of SR.

Now, the empty solution to the EFE with a positive cosmological constant is a "hyperbolic" manifold of constant, positive curvature, known as deSitter space-time. In "normal" geometries with a positive definite metric signature, constant curvature is simply spherical. That is, a 2D surface of constant positive curvature is the surface of a sphere. Well, space-time has the time part acting opposite of the space part in the "norm" function, and it turns out constant positive curvature there is a hyperbolic type of "thing".

DeSitter space-time is the simplest "expanding space" space-time there is, and can serve as a toy model for our universe (out to a few billion light years or so before it deviates too much from what we actually have).

With the appropriate choice of static coordinates, which can be thought as plopping a certain coordinate system around a test particle embedded in that space-time and calling that the origin, r = 0 of a spherical spatial coordinate system, we can consider the Newtonian limit (small r).

Here, in Newtonian terms it looks like space is filled with a constant *negative* mass density (this density is proportional to lambda) which gives a radially repulsive gravitational field.

The result is that world lines of test particles fly apart from each other in all directions. A little spherical cloud of test particles would expand with time like a balloon. From the viewpoint of any of the particles, everything else is flying away radially.

That can be described in various ways. If you adopt the standard cosmological co-moving coordinate system, you say space is "expanding" between all the particles.

For the currently accepted value of lambda in our universe, this repulsive acceleration is very small locally, going as (tiny number)*r. I forget just how tiny it is, but basically it would take billions of years two particles initially one meter apart to get two meters. It is thus insignificant on local galactic scales. However, separate them on the order of a billion light years, and things fly apart pretty good.

Anyway, that's roughly how you can think of Lambda behavior.

-Richard

tommac

2008-Apr-23, 08:01 PM

Gotcha!!! thanks.

Tommac,

There are a lot of misconceptions about things there in your post. These things are not trivial, and it's very easy to "get ideas in your head" based on words being used to describe what can only be rigorously described mathematically.

First gravity is not a "contraction of space-time". Gravity is due to the *curvature* of space-time. And here, "gravity" is not the same thing as the Newtonian notion of a g field, but is better thought of as gradients in that g field (the tidal forces). The Newtonian 'g' is more of a coordinate thing than anything invariant.

Space-time is described in General Relativity by a 4 dimensional geometric construct, a manifold. There are are many ways a 4D manifold can curve and it requires a rank-4 tensor to describe. While constraints and symmetries greatly reduce the number of indepedent components from 256 (4x4x4x4), there are still a lot of components.

And while I'm not sure exactly what you're driving at, gravity is not a "scalar". Newtonian gravity is a vector field. GR gravity is far more than that, a *tensor* field. In terms of potentials, Newtonian gravity can be completely described as the gradient of a scalar potential function. Now, EM requires a vector potential function (4-vector or 1 scalar plus 1 three-vector function). So thus we can say that Newtonian gravity derives from a scalar field, while EM requires a vector field potential.

However, the equivalent of the potential in GR gravity is the space-time metric, which is a rank-2 tensor. There are 10 independent components there. IOW, GR gravity is a far more complex mathematical entity.

-Richard

tommac

2008-Apr-23, 08:06 PM

Believe it or not I agree 100% with everything you wrote here.

Now, Lambda, the cosmological constant. GR gravity is described by what is called the Einstein Field Equation (EFE). The EFE is a very complex non-linear tensor partial differential equation (which could be written out as a complicated mess of coupled non-linear partial differential equations for the 10 independent components of the metric mentioned above). Solving that mess is no trivial to say the least.

But what it does is relate the curvature of space-time to the mass-energy-momentum content of that space-time. IOW, the solution of the EFE, the metric, tells us the "shape" of space-time to the mass-energy-momentum content of that space-time.

That's the plain EFE. When Einstein derived GR, the universe was thought to be a static, unchanging thing. However, simple solutions of the EFE predicted a dynamic space-time. The universe should be flying apart or contracting due to its own gravity. It could be doing a lot of things depending on the initial conditions and the mass-energy content, but it would not be just sitting there.

So Einstein added the cosmological constant, which is a term that simply "cancels out" the gravity of the mass-energy and would allow a static universe. Then came Hubble, and so Einstein called this his "greatest blunder", as he could've predicted the Hubble expansion.

And that's the way it was until recently when the "acceleration" of expansion was discovered. Lambda can cancel out a contraction, but increase it and can go the other way and drive the expansion faster and faster, depending on the mass-energy content.

The simplest solutions to the EFE are the "empty" space-time solutions, zero mass-energy content. The solution of the plain EFE with no cosmological constant and zero source terms is, with appropriate boundary conditions (these are all important), simply flat Minkowski space-time, the playground of SR.

Now, the empty solution to the EFE with a positive cosmological constant is a "hyperbolic" manifold of constant, positive curvature, known as deSitter space-time. In "normal" geometries with a positive definite metric signature, constant curvature is simply spherical. That is, a 2D surface of constant positive curvature is the surface of a sphere. Well, space-time has the time part acting opposite of the space part in the "norm" function, and it turns out constant positive curvature there is a hyperbolic type of "thing".

DeSitter space-time is the simplest "expanding space" space-time there is, and can serve as a toy model for our universe (out to a few billion light years or so before it deviates too much from what we actually have).

With the appropriate choice of static coordinates, which can be thought as plopping a certain coordinate system around a test particle embedded in that space-time and calling that the origin, r = 0 of a spherical spatial coordinate system, we can consider the Newtonian limit (small r).

Here, in Newtonian terms it looks like space is filled with a constant *negative* mass density (this density is proportional to lambda) which gives a radially repulsive gravitational field.

The result is that world lines of test particles fly apart from each other in all directions. A little spherical cloud of test particles would expand with time like a balloon. From the viewpoint of any of the particles, everything else is flying away radially.

That can be described in various ways. If you adopt the standard cosmological co-moving coordinate system, you say space is "expanding" between all the particles.

For the currently accepted value of lambda in our universe, this repulsive acceleration is very small locally, going as (tiny number)*r. I forget just how tiny it is, but basically it would take billions of years two particles initially one meter apart to get two meters. It is thus insignificant on local galactic scales. However, separate them on the order of a billion light years, and things fly apart pretty good.

Anyway, that's roughly how you can think of Lambda behavior.

-Richard

mugaliens

2008-Apr-26, 04:28 PM

For the currently accepted value of lambda in our universe, this repulsive acceleration is very small locally, going as (tiny number)*r. I forget just how tiny it is, but basically it would take billions of years two particles initially one meter apart to get two meters. It is thus insignificant on local galactic scales. However, separate them on the order of a billion light years, and things fly apart pretty good.

Anyway, that's roughly how you can think of Lambda behavior.

-Richard

According to NASA (http://apod.nasa.gov/apod/ap960513.html), it's about 80 km per second between two points located 300,000 light years apart.

Using a more understandable scale, that means that the nearest star, which is four light years distant, is moving away from us at a rate of around 1m, or 3 feet, per second.

Put another way, the distance to our nearest star is 2.34628E+13 feet. So the rate expressed in terms of a percentage becomes 1.27862E-11%, or, 0.00000000001279%. That's can also be expressed at a rate of 1 part per 78,209,280,000 (78.2 billion).

Tiny?

Yep!

speedfreek

2008-Apr-26, 04:43 PM

But of course the nearest star to our sun is not actually moving away from us due to cosmic expansion, as we are both part of the same gravity-bound system, the Milky Way.

alainprice

2008-Apr-28, 06:41 PM

It actually is moving away from us, just not in a measurable fashion.

You're not allowed to say something is happening everywhere, except here. Unless you are talking about crime in your small town.

phunk

2008-Apr-28, 08:22 PM

It actually is moving away from us, just not in a measurable fashion.

You're not allowed to say something is happening everywhere, except here. Unless you are talking about crime in your small town.

I don't know the details of the relationship between us and the centauri system, but "it is moving away" is not necessarily true. While expansion is effectively a 'force' pushing everything apart, if there are also forces pushing things back together, you have to look at the sum of the forces to determine the final outcome.

For example, the earth is not moving away from the sun due to expansion. It may be orbiting a hair farther out than it would in a universe without expansion, but it is not getting farther. The force of the sun's gravity and the force of expansion add together to determine the size of the orbit, which is constant (if you ignore the other gravitational influences).

Neverfly

2008-Apr-28, 08:26 PM

For example, the earth is not moving away from the sun due to expansion. It may be orbiting a hair farther out than it would in a universe without expansion, but it is not getting farther. The force of the sun's gravity and the force of expansion add together to determine the size of the orbit, which is constant (if you ignore the other gravitational influences).

This isn't how I understood it at all:doh:

phunk

2008-Apr-28, 08:38 PM

I could be wrong, how do you understand it?

Neverfly

2008-Apr-28, 08:43 PM

As I understood it, expansion effect in the solar system is negligible.

You had said a "hair further out" which I don't know if I should take that literally or not...

You said after that : "The force of the sun's gravity and the force of expansion add together to determine the size of the orbit" and from what i have been reading about Celestial mechanics, expansion doesn't play a role in Solar Orbits.

I could be wrong too- that's why I only said that is now how I understand it- A more learned one may come along and clear it up;)

phunk

2008-Apr-28, 09:09 PM

I 'a hair' almost literally. :) The difference is probably immeasurable at the scale of 1 AU.

By my quick calculation, using a rough estimate of 70km/s/Mpc for the hubble constant, we get an acceleration of 3.3*10-7m/s2 at a distance of 1 AU due to expansion. The acceleration due to the sun's gravity at 1 AU is roughly 5.9*10-3m/s2. 4 orders of magnitude difference between them means it's probably immeasurable. We'd need to know the mass and orbit of everything in the solar system to a rediculous accuracy to see if the earth's orbit is any bigger than it would be via gravity alone.

phunk

2008-Apr-28, 09:15 PM

My point to alainprice is simply that while expansion is everywhere, it scales with distance such that you can't say that everything is moving apart due to expansion, because local effects are many orders of magnitude stronger. You have to look at the sum of the forces for each case.

Neverfly

2008-Apr-28, 09:24 PM

My point to alainprice is simply that while expansion is everywhere, it scales with distance such that you can't say that everything is moving apart due to expansion, because local effects are many orders of magnitude stronger. You have to look at the sum of the forces for each case.

Ah, ok. Now I'm following you again.

You can see plainly though, how that one line (Expansion and gravity together determines orbit) would throw me for a loop.;)

phunk

2008-Apr-28, 09:29 PM

Yeah you're right, Earth's orbit is too small of an example.

publius

2008-Apr-28, 10:30 PM

I 'a hair' almost literally. :) The difference is probably immeasurable at the scale of 1 AU.

By my quick calculation, using a rough estimate of 70km/s/Mpc for the hubble constant, we get an acceleration of 3.3*10-7m/s2 at a distance of 1 AU due to expansion. The acceleration due to the sun's gravity at 1 AU is roughly 5.9*10-3m/s2. 4 orders of magnitude difference between them means it's probably immeasurable. We'd need to know the mass and orbit of everything in the solar system to a rediculous accuracy to see if the earth's orbit is any bigger than it would be via gravity alone.

Phunk,

I think that's way too high -- I calculated that back in some thread here a while back, and IIRC, the repulsive acceleration is much smaller than that at 1AU, on the 10^(-double digits), I want to say 10^-20 , but I'm not sure. It is well below any threshold of measurement locally.

When I get back home, somebody remind me and I'll crunch it again.

10^-7 m/s^2 would be quite detectable. The famous Pioneer anamoly is around 10^9 m/s^2.

-Richard

publius

2008-Apr-28, 10:35 PM

Here it is -- wasn't hard to find at all:

http://www.bautforum.com/questions-answers/68167-what-local-effect-lambda-de-orbits.html

The repulsive g(r) at 1AU is ~10^-25 m/s^2, about 1 part in 10^20 of the solar g!

The local g due to Lamdba goes as (c^2*Lambda/3) r.

-Richard

phunk

2008-Apr-28, 10:38 PM

I thought it looked a too high, I should have double checked my math.

Speaking of too high, I think you missed a - in here. :)

The famous Pioneer anamoly is around 10^9 m/s^2.

publius

2008-Apr-28, 11:26 PM

Right, that's 10^-9 m/s^2. :)

And, at any rate, note that c^2*L/3 is a tidal looking expression, acceleration per length, call it a_r. While the expression g(r) = a_r * r is a Newtonian limit (at large r, just like with Schwarzschild, things would go very different from Newton), it is nonetheless the exact GR tidal force a co mover feels.

-Richard

trinitree88

2008-May-01, 01:03 AM

As I understood it, expansion effect in the solar system is negligible.

You had said a "hair further out" which I don't know if I should take that literally or not...

You said after that : "The force of the sun's gravity and the force of expansion add together to determine the size of the orbit" and from what i have been reading about Celestial mechanics, expansion doesn't play a role in Solar Orbits.

I could be wrong too- that's why I only said that is now how I understand it- A more learned one may come along and clear it up;)

Neverfly. You're on the money....and if the axial orientation of the type 1-a supernovae "standard candles" follows the seen tendency of Cepheid variables, to be aligned in common relative to their host galactic planes....then it may disappear altogether, as they would then constitute non-representative viewing angles when brightest. pete

Neverfly

2008-May-01, 01:54 AM

Neverfly. You're on the money....and if the axial orientation of the type 1-a supernovae "standard candles" follows the seen tendency of Cepheid variables, to be aligned in common relative to their host galactic planes....then it may disappear altogether, as they would then constitute non-representative viewing angles when brightest. pete

Now, trinitree, I think you're makin' fun of me:p ;)

alainprice

2008-May-01, 04:52 AM

My point to alainprice is simply that while expansion is everywhere, it scales with distance such that you can't say that everything is moving apart due to expansion, because local effects are many orders of magnitude stronger. You have to look at the sum of the forces for each case.

Exactly, the words I was looking for.

While expansion is universal, individual objects obviously have other influences(the sum of all forces).

tommac

2008-May-01, 03:00 PM

Exactly, the words I was looking for.

While expansion is universal, individual objects obviously have other influences(the sum of all forces).

I can agree with this also. However there is another question. Sorry guys ...

If the expansion of the universe is the magical dropping of space in between things ... but we all seem to be agreeing that it gets canceled out at least in part by gravity ( sum of all forces ) ( I thought the expansion of the universe was not a force ) , then does that mean that the space is magically dropped in but then somehow removed ?

For the way I was looking at things ( although I have been told I was wrong )

I have seen the expansion of space as the expansion of space-time #2 in another thread. I can then see the sum of all forces argument working as we have an expansion of space-time ( - curvature ) then a contraction of space time ( + curvature ) if you sum up the two contactions ( sum up the two curves ) you will end up with a contraction of space-time since that effect is much stronger than the effect of the expansion.

I have been told that this is without a doubt flat out wrong. And I am sure there will be at least three posts after this from people telling me that they have answered this question in other posts and that i need to read and all of that ... please try not to post these ...

but how can one say that the effect is cancelled out unless the forces are similar? If the forces are similar ( gravitational curvature + expansion of the universe ) then logically they need to be related :

If 3 meteres are magically appearing in the expansion of space time then 7 meters must disappear in the contraction due to gravity

OR

If 1 meter-second expands to 3 meters-seconds from the expansion but then is reduced to .2 meters-seconds from the contraction due to gravity.

Does anyone understand my logic here ... I am not saying that I am right ... I am just confused and humbly asking for clarification.

And please I beg of your dont off topic post her with repremands telling me that you have answered this in another post because i have been reading EVERYTHING and although I may be stubborn I am not sure if this question has been answered clearly.

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