# Cosmological Constant, Einstein equation Quick Question

• binbagsss
In summary: Thanks. And is it referred to divergences in both cases : ##\nabla_a T^{abc...}=0## and ##\nabla^a T^{abc...}=0## i.e - whether the index on 'nabla' is up or down. Is one more conventional than the...A "covariant divergence" is a divergence in that it has a contraction on the index with which you are taking the derivative of. Saying "the divergence is 0" means ##\nabla_a T^{abc...}=0##. Saying "the covariant derivative is 0" means ##\nabla_a T^{bcd...}=0

#### binbagsss

So Einstein Equation: ##G_{uv}= 8 \pi G T_{uv} ##,

Justifying the cosmological constant can be included is done by noting that ## \bigtriangledown^{a}g_{ab} =0 ## and so including it on the LHS, conservation of energy-momentum tensor still holds.

I'm not sure why ## \bigtriangledown^{a}g_{ab} =0 ##. The source I'm using says to 'recall' this, and it is talking about the FRW tensor.

The only thing I can think of is the fundamental theorem of Riemannian geometry : ## \bigtriangledown_{a}g_{bc}= 0 ##. But this doesn't does look right as it has 3 free indicies, not 1, and a lower indice instead of a upper on the ## \bigtriangledown ##

On a side note, I think I am confused between 'divergence' and 'covariant derivative', when we say ## \bigtriangledown_{a} T^{ab} = 0 ##, conservation of energy-momentum tensor that its 'divergence' is zero, is this saying it's convariant derivative is zero?

The requirement that ##\nabla^a g_{ab}=0## arises from the more stringent requirement that ##\nabla_a g_{bc}=0## as is the case if the connection is to be metric compatible. In other words, this is one of the defining attributes of a "metric connection".

A "covariant divergence" is a divergence in that it has a contraction on the index with which you are taking the derivative of. Saying "the divergence is 0" means ##\nabla_a T^{abc...}=0##. Saying "the covariant derivative is 0" means ##\nabla_a T^{bcd...}=0##. Some times the covariant derivative is only 0 along some curve (e.g. for geodesic tangents along geodesics) and in that case, we mathematically specify ##u^a\nabla_a T^{bcd...}=0## meaning it's parallel transported along that curve.

Matterwave said:
The requirement that ##\nabla^a g_{ab}=0## arises from the more stringent requirement that ##\nabla_a g_{bc}=0## as is the case if the connection is to be metric compatible. In other words, this is one of the defining attributes of a "metric connection".

How would you show this? Would you argue that a,b,c range over 0,1,2,3 (In 4-d) and then this is the particular case of a=b? And then raise an index?

binbagsss said:
How would you show this? Would you argue that a,b,c range over 0,1,2,3 (In 4-d) and then this is the particular case of a=b? And then raise an index?

If ##\nabla_a g_{bc}## applies for all ##a,b,c##, then it should certainly apply for when ##a=b## right? And then you can raise the index like you say...but it is certainly the case tht ##g^{ab}\times 0=0## so I'm not sure why you want to carry out this step.

binbagsss said:
How would you show this? Would you argue that a,b,c range over 0,1,2,3 (In 4-d) and then this is the particular case of a=b? And then raise an index?
$$\nabla^a g_{ab}=\nabla^d g_{db}=g^{da}\nabla_a g_{db}=\sum_{a,d} \left( g^{da} \times 0 \right) = 0$$

Matterwave said:
so I'm not sure why you want to carry out this step.
But isn't ##\nabla_a ## covariant derivaitve? I'm not sure what ##\nabla^{a} ## is called, but don't they describe different things physically?

DrGreg said:
$$\nabla^a g_{ab}=\nabla^d g_{db}=g^{da}\nabla_a g_{db}=\sum_{a,d} \left( g^{da} \times 0 \right) = 0$$
thanks. although I'm not sure I understand the summation in the last equality, as the d,a do not appear twice.

binbagsss said:
thanks. although I'm not sure I understand the summation in the last equality, as the d,a do not appear twice.
I suppose I could have written it as
$$\nabla^a g_{ab}=\nabla^d g_{db}=g^{da}\nabla_a g_{db}=g^{da}0_{adb} = 0$$

binbagsss said:
But isn't ##\nabla_a ## covariant derivaitve? I'm not sure what ##\nabla^{a} ## is called, but don't they describe different things physically?

Why would they in the presence of a metric? Any vector ##A^a## has an associated co-vector ##A_a=g_{ab}A^b##.

Another way to see that is instead of taking the Einstein equations as you got them , with indices down, to write them equivalently with indices up:

$G^{\mu \nu} = 8 \pi G T^{\mu \nu}$

Then you can insert the cosmological constant by using [the more used to you convention of the covariant derivative] $\nabla_a g^{bc} =0 \Rightarrow \nabla_a g^{ac}=0$. However it's just what other people have already posted here.

Matterwave said:
A "covariant divergence" is a divergence in that it has a contraction on the index with which you are taking the derivative of. Saying "the divergence is 0" means ##\nabla_a T^{abc...}=0##. Saying "the covariant derivative is 0" means ##\nabla_a T^{bcd...}=0##. Some times the covariant derivative is only 0 along some curve (e.g. for geodesic tangents along geodesics) and in that case, we mathematically specify ##u^a\nabla_a T^{bcd...}=0## meaning it's parallel transported along that curve.
Thanks. And is it referred to divergences in both cases : ##\nabla_a T^{abc...}=0## and ##\nabla^a T^{abc...}=0## i.e - whether the index on 'nabla' is up or down. Is one more conventional than the other?

binbagsss said:
Thanks. And is it referred to divergences in both cases : ##\nabla_a T^{abc...}=0## and ##\nabla^a T^{abc...}=0## i.e - whether the index on 'nabla' is up or down. Is one more conventional than the other?

When you see two indices which are the same, it should always be one upper index and one lower index. It should never be both up like in your second equation.

Matterwave said:
When you see two indices which are the same, it should always be one upper index and one lower index. It should never be both up like in your second equation.
Apologies typo , the nabla with the second indice was suppose to read ##\nabla^{a}T_{abc} ##- are they both referred to as divergence and is one more conventional than the other?

binbagsss said:
Apologies typo , the nabla with the second indice was suppose to read ##\nabla^{a}T_{abc} ##- are they both referred to as divergence and is one more conventional than the other?

Yes, they are both "divergences".

## 1. What is the Cosmological Constant?

The Cosmological Constant, denoted by the Greek letter Λ (Lambda), is a term in Einstein's general theory of relativity that represents the energy density of the vacuum of space. It was introduced by Albert Einstein in 1917 to account for the observed expansion of the universe.

## 2. How does the Cosmological Constant affect the universe?

The Cosmological Constant has a significant impact on the expansion of the universe. It causes the expansion to accelerate, counteracting the gravitational pull of matter and making the universe expand at an increasing rate. This has been confirmed by observations of distant supernovae and the cosmic microwave background radiation.

## 3. What is the relationship between the Cosmological Constant and the Einstein equation?

The Cosmological Constant is a term in the Einstein equation, which is the mathematical framework for Einstein's general theory of relativity. It is represented by Λ and is one of the components that describe the curvature of spacetime.

## 4. Is the value of the Cosmological Constant constant?

No, the value of the Cosmological Constant is not constant. It was initially introduced by Einstein as a way to balance out the gravitational pull of matter and keep the universe static. However, later observations showed that the universe is expanding, and the value of the Cosmological Constant was adjusted to account for this expansion. The exact value of Λ is still a subject of scientific study and debate.

## 5. How does the Cosmological Constant relate to dark energy?

The Cosmological Constant is often used interchangeably with the term "dark energy." This is because the observed effects of the Cosmological Constant, such as the accelerating expansion of the universe, are similar to those of dark energy. However, the exact nature of dark energy is still unknown, and it is possible that it is not related to the Cosmological Constant at all.