# Cosmological Constant, Einstein equation Quick Question

## Main Question or Discussion Point

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?

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Matterwave
Gold Member
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.

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".
Thanks for your reply. So $\nabla_a g_{bc}=0$ implies $\nabla^a g_{ab}=0$ ?
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?

Matterwave
Gold Member
Thanks for your reply. So $\nabla_a g_{bc}=0$ implies $\nabla^a g_{ab}=0$ ?
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.

DrGreg
Gold Member
Thanks for your reply. So $\nabla_a g_{bc}=0$ implies $\nabla^a g_{ab}=0$ ?
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?
$$\nabla^a g_{ab}=\nabla^d g_{db}=g^{da}\nabla_a g_{db}=\sum_{a,d} \left( g^{da} \times 0 \right) = 0$$

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?

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

DrGreg
Gold Member
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
$$\nabla^a g_{ab}=\nabla^d g_{db}=g^{da}\nabla_a g_{db}=g^{da}0_{adb} = 0$$
where $0_{adb} = 0$

Matterwave
Gold Member
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 $A_a=g_{ab}A^b$.

ChrisVer
Gold Member
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.

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?

Matterwave
Gold Member
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.

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?

Matterwave
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?