- #1

Wallace

Science Advisor

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## Main Question or Discussion Point

So there has been a few threads over the last few days where the whole issue of dark energy and/or the cosmological constant has been discussed in the context of whether it is a valid 'solution' to GR or if it violates some principle of relativity or even insults Einsteins 'legacy', whatever that is supposed to mean

I thought it would be good to consolidate the discussion to a single thread and also explain this issue in a bit more detail.

I'd like to start at the start, with the Einstein Field equations as we know and love them. The simplest description goes like this (ignoring some factors of [tex]\pi[/tex] which I will do throughout for simplicity):

[tex] G_{\mu\nu} = T_{\mu\nu} [/tex]

where the left hand side is the Einstein tensor, this is the 'geometric side' of the equation (I'll use this terminology later, so make note of it) while the right hand side is the stress-energy tensor, which describes the energy content.

We can expand the left hand side a bit by knowing that

[tex] G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R [/tex]

where [tex] R_{\mu\nu} [/tex] is the Ricci tensor, [tex] R [/tex] is the Ricci scalar and [tex] g_{\mu\nu} [/tex] is the metric tensor. Note that both the Ricci tensor and the Ricci scalar depend entirely on the metric tensor and hence the entire Einstein tensor can be described by just the metric tensor and its derivatives. Now, the conservation of energy and momentum is described by the Bianchi Identities such that

[tex] \Delta_{\mu}G^{\mu\nu}=0 [/tex] and [tex] \Delta_{\mu}T^{\mu\nu}=0 [/tex]

Now, watch carefully as I pull a cosmological constant out of my hat....

If the LHS (the 'geometric' side) also has an additional term so that is becomes:

[tex] R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R +\Lambda g_{\mu\nu} [/tex]

then the Bianchi identities, and hence the validity of the solution, are unchanged, since the covariant derivative of the metric tensor vanishes by definition.

The oft repeated adage of GR that "matter tells space how to curve, the curvature of space tell matter how to move" is as applicable whether or not [tex]\Lambda[/tex] is zero or not. The equivalence principle or any other principle of relativity is unaffected.

In fact there is an error in suggesting the we can 'add' a CC term to the Einstein tensor. The truth is that mathematically,

Now, lets review the history of this little beastie we call [tex]\Lambda[/tex]. When Einstein first proposed the field equations he thought that the inevitable result of setting [tex]\Lambda[/tex] to zero was that the universe would either have to be expanding or contracting, but that it could not be static. Since at the time people believe the Universe was static he set [tex]\Lambda[/tex] to be non-zero since this allowed a static, though unstable solution.

Only a few years later Hubble discovered the expansion of the Universe and therefore there was no longer the need to have a static solution, hence the value of [tex]\Lambda[/tex] of zero was a better fit and simpler solution for the data as it stood at that time. Einstein suggested it was a blunder on his part to make [tex]\Lambda[/tex] non-zero to get a static solution instead of using a zero value to predict the universe should be expanding or contracting.

However the key point about the above was that regardless of the value you think is appropriate for the [tex]\Lambda[/tex] term to have, it was always known to be a perfectly valid term from a theoretical point of view. The 'blunder' related to fitting to data, but the existence of the [tex]\Lambda[/tex] term is not only allowed but suggested by the form of the Einstein Field Equations.

So, having established the [tex]\Lambda[/tex] is not a fudge factor, hack or addition to GR but fundamentally part of the theory lets now look at how dark energy relates to [tex]\Lambda[/tex].

We now need to turn to the right hand side of the Einstein Field equation, the stress energy tensor [tex]T_{\mu\nu}[/tex] defined as:

[tex] T^{\mu\nu} = (P + \frac{\rho}{c^2})u^{\mu}u^{\nu} + Pg^{\mu\nu} [/tex]

where P is the pressure and [tex]\rho[/tex] is the density of the energy in the universe. We can further simplify this by defining the equation of state of the energy as

[tex] w = \frac{P}{\rho} [/tex]

This gives us

[tex] T^{\mu\nu} = \rho(w + \frac{1}{c^2})u^{\mu}u^{\nu} + w\rho g^{\mu\nu} [/tex]

For a homogenous and isotropic and infinite universe, the normalization of the four velocity for the fluid being at rest of

[tex] u^{\mu}u^{\nu} = diag[c^2,0,0,0] [/tex]

simplifies this to:

[tex] T^{\mu\nu} = \rho(w+1) + w\rho g^{\mu\nu} [/tex]

where we can just incorporate the factors of [tex] c^2 [/tex] into the value of [tex] \rho [/tex]

Now, we could now solve the field equations if we know the density and equation of state of the material in the Universe. The solution we get for this is known as the Friedmann equations. I'm not going to detail this solution as it's not necessary for this discussion, all we need to know is that the scale factor a(t) falls out of this solution and the form of this depends on the density and equation of state of the energy in the universe.

For matter, w=0, for radiation, w=1/3. But what would happen if say the equation of state was w=-1 for a component of the stress energy tensor?

The stress energy tensor for this component would become simply:

[tex] T^{\mu\nu} = -\rho g^{\mu\nu} [/tex]

Notice now that this is precisely the form of the cosmological constant if we replace the energy density by [tex]\Lambda[/tex]. If we took it over the other side (the 'geometric' side) of the field equations the solution is exactly the same.

Therefore as far as GR and cosmology is concerned, a non zero [tex]\Lambda[/tex] is identical to the presence of an energy with equation of state of -1. The interpretation of this is different, so if the term is on the right hand side we would call it 'vacuum energy' while if it was on the left it is just 'how gravity works', rather than being the effect of some energy.

Particle physicists would care about which side the term was on, since non-zero vacuum energy would need to be explained but a geometric term would leave the standard model of particle physics alone. But cosmologist don't care so much, observationally they are equivalent.

Now, the point about dark energy is that the current data suggests that there is a component of the universe with equation of state close to -1. However it does not have to be exactly -1 given current data. In fact the value may change with time. We call this component dark energy as a generic term, since we do not know the micro physical theory describing it. A specific candidate for dark energy would be something with w exactly =-1 for all time, in which case it might not be an 'energy' at all, just 'how gravity works' (geometric term) or vacuum energy (energy term).

The reason we introduce dark energy at all is because that is what the data says is there, we do not know how to fit the data we have in the framework of GR without having this energy component in the model.

So is Dark Energy a 'fudge factor'? Not at all! Ask yourself what is physics? It is the process of describing observations by mathematical models containing particles, fields and forces to explain what is observed. At one point in time we did not know about magnetism. Then odd things were observed and magnetic fields proposed to explain what was going in. We didn't know about gravity till it was realized that it was needed to explain the observations of the motions of planets. We didn't always think electron, quarks etc etc existed until the data suggested that they must.

This is what dark energy is. It is the prediction of some new physics required in order to match theory with what is observed. Not all new predictions turn out to be true since other things may be proposed that turn out to fit the observations better and provide a coherent theory. Dark energy may one day go the way of spontaneous generation or the fluid theory of heat, if that is what observations and theory tell us.

Thankyou for reading this far! One last point I would like to make is that I find it odd when people point out they are being 'open minded' by refusing to believe what the data is telling us. Of course we need to consider all possible explanations of the data, whether the explanation contains dark energy or not, no one would argue with this. However the data overwhelmingly points to the existence of dark energy. Despite this there are folks whose theoretical prejudice against its existence leads them to scoff at the very idea that it could exist and insist that models explaining observations that work far less well are more reasonable. This is not being open minded, this is in fact being closed minded to the possibility of the new physics implied by dark energy.

I thought it would be good to consolidate the discussion to a single thread and also explain this issue in a bit more detail.

I'd like to start at the start, with the Einstein Field equations as we know and love them. The simplest description goes like this (ignoring some factors of [tex]\pi[/tex] which I will do throughout for simplicity):

[tex] G_{\mu\nu} = T_{\mu\nu} [/tex]

where the left hand side is the Einstein tensor, this is the 'geometric side' of the equation (I'll use this terminology later, so make note of it) while the right hand side is the stress-energy tensor, which describes the energy content.

We can expand the left hand side a bit by knowing that

[tex] G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R [/tex]

where [tex] R_{\mu\nu} [/tex] is the Ricci tensor, [tex] R [/tex] is the Ricci scalar and [tex] g_{\mu\nu} [/tex] is the metric tensor. Note that both the Ricci tensor and the Ricci scalar depend entirely on the metric tensor and hence the entire Einstein tensor can be described by just the metric tensor and its derivatives. Now, the conservation of energy and momentum is described by the Bianchi Identities such that

[tex] \Delta_{\mu}G^{\mu\nu}=0 [/tex] and [tex] \Delta_{\mu}T^{\mu\nu}=0 [/tex]

Now, watch carefully as I pull a cosmological constant out of my hat....

If the LHS (the 'geometric' side) also has an additional term so that is becomes:

[tex] R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R +\Lambda g_{\mu\nu} [/tex]

then the Bianchi identities, and hence the validity of the solution, are unchanged, since the covariant derivative of the metric tensor vanishes by definition.

The oft repeated adage of GR that "matter tells space how to curve, the curvature of space tell matter how to move" is as applicable whether or not [tex]\Lambda[/tex] is zero or not. The equivalence principle or any other principle of relativity is unaffected.

In fact there is an error in suggesting the we can 'add' a CC term to the Einstein tensor. The truth is that mathematically,

*it is always there*we can just choose to set the value of [tex]\Lambda[/tex] to zero, if that is what experiment tells us to do.Now, lets review the history of this little beastie we call [tex]\Lambda[/tex]. When Einstein first proposed the field equations he thought that the inevitable result of setting [tex]\Lambda[/tex] to zero was that the universe would either have to be expanding or contracting, but that it could not be static. Since at the time people believe the Universe was static he set [tex]\Lambda[/tex] to be non-zero since this allowed a static, though unstable solution.

Only a few years later Hubble discovered the expansion of the Universe and therefore there was no longer the need to have a static solution, hence the value of [tex]\Lambda[/tex] of zero was a better fit and simpler solution for the data as it stood at that time. Einstein suggested it was a blunder on his part to make [tex]\Lambda[/tex] non-zero to get a static solution instead of using a zero value to predict the universe should be expanding or contracting.

However the key point about the above was that regardless of the value you think is appropriate for the [tex]\Lambda[/tex] term to have, it was always known to be a perfectly valid term from a theoretical point of view. The 'blunder' related to fitting to data, but the existence of the [tex]\Lambda[/tex] term is not only allowed but suggested by the form of the Einstein Field Equations.

So, having established the [tex]\Lambda[/tex] is not a fudge factor, hack or addition to GR but fundamentally part of the theory lets now look at how dark energy relates to [tex]\Lambda[/tex].

We now need to turn to the right hand side of the Einstein Field equation, the stress energy tensor [tex]T_{\mu\nu}[/tex] defined as:

[tex] T^{\mu\nu} = (P + \frac{\rho}{c^2})u^{\mu}u^{\nu} + Pg^{\mu\nu} [/tex]

where P is the pressure and [tex]\rho[/tex] is the density of the energy in the universe. We can further simplify this by defining the equation of state of the energy as

[tex] w = \frac{P}{\rho} [/tex]

This gives us

[tex] T^{\mu\nu} = \rho(w + \frac{1}{c^2})u^{\mu}u^{\nu} + w\rho g^{\mu\nu} [/tex]

For a homogenous and isotropic and infinite universe, the normalization of the four velocity for the fluid being at rest of

[tex] u^{\mu}u^{\nu} = diag[c^2,0,0,0] [/tex]

simplifies this to:

[tex] T^{\mu\nu} = \rho(w+1) + w\rho g^{\mu\nu} [/tex]

where we can just incorporate the factors of [tex] c^2 [/tex] into the value of [tex] \rho [/tex]

Now, we could now solve the field equations if we know the density and equation of state of the material in the Universe. The solution we get for this is known as the Friedmann equations. I'm not going to detail this solution as it's not necessary for this discussion, all we need to know is that the scale factor a(t) falls out of this solution and the form of this depends on the density and equation of state of the energy in the universe.

For matter, w=0, for radiation, w=1/3. But what would happen if say the equation of state was w=-1 for a component of the stress energy tensor?

The stress energy tensor for this component would become simply:

[tex] T^{\mu\nu} = -\rho g^{\mu\nu} [/tex]

Notice now that this is precisely the form of the cosmological constant if we replace the energy density by [tex]\Lambda[/tex]. If we took it over the other side (the 'geometric' side) of the field equations the solution is exactly the same.

Therefore as far as GR and cosmology is concerned, a non zero [tex]\Lambda[/tex] is identical to the presence of an energy with equation of state of -1. The interpretation of this is different, so if the term is on the right hand side we would call it 'vacuum energy' while if it was on the left it is just 'how gravity works', rather than being the effect of some energy.

Particle physicists would care about which side the term was on, since non-zero vacuum energy would need to be explained but a geometric term would leave the standard model of particle physics alone. But cosmologist don't care so much, observationally they are equivalent.

Now, the point about dark energy is that the current data suggests that there is a component of the universe with equation of state close to -1. However it does not have to be exactly -1 given current data. In fact the value may change with time. We call this component dark energy as a generic term, since we do not know the micro physical theory describing it. A specific candidate for dark energy would be something with w exactly =-1 for all time, in which case it might not be an 'energy' at all, just 'how gravity works' (geometric term) or vacuum energy (energy term).

The reason we introduce dark energy at all is because that is what the data says is there, we do not know how to fit the data we have in the framework of GR without having this energy component in the model.

So is Dark Energy a 'fudge factor'? Not at all! Ask yourself what is physics? It is the process of describing observations by mathematical models containing particles, fields and forces to explain what is observed. At one point in time we did not know about magnetism. Then odd things were observed and magnetic fields proposed to explain what was going in. We didn't know about gravity till it was realized that it was needed to explain the observations of the motions of planets. We didn't always think electron, quarks etc etc existed until the data suggested that they must.

This is what dark energy is. It is the prediction of some new physics required in order to match theory with what is observed. Not all new predictions turn out to be true since other things may be proposed that turn out to fit the observations better and provide a coherent theory. Dark energy may one day go the way of spontaneous generation or the fluid theory of heat, if that is what observations and theory tell us.

Thankyou for reading this far! One last point I would like to make is that I find it odd when people point out they are being 'open minded' by refusing to believe what the data is telling us. Of course we need to consider all possible explanations of the data, whether the explanation contains dark energy or not, no one would argue with this. However the data overwhelmingly points to the existence of dark energy. Despite this there are folks whose theoretical prejudice against its existence leads them to scoff at the very idea that it could exist and insist that models explaining observations that work far less well are more reasonable. This is not being open minded, this is in fact being closed minded to the possibility of the new physics implied by dark energy.

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