Can We Explain the Cosmological Constant as a Tensorial Concept?

[jk22]

Can we explain the cosmological constant in this way ?

The Einstein tensor is derived from the Ricci tensor and one property is that, like the stress-energy tensor, its covariant derivative shall vanish.

Since the covariant derivative of the metric vanishes it can be added to the EFE as $+\Lambda g_{\mu\nu}$

But then we could add any power of the metric like $+\Lambda_1 g_{\mu\rho}g^\rho_\nu$ ?

 

[Ibix]

But then we could add any power of the metric like $+\Lambda_1 g_{\mu\rho}g^\rho_\nu$ ?

Contract over $\rho$ and this becomes $\Lambda_1 g_{\mu\nu}$. This is true for any product of repetitions of the metric that only has two free indices.

 

[PeterDonis]

Since the covariant derivative of the metric vanishes it can be added to the EFE

Yes, this is actually the more or less standard argument among physicists for why the cosmological constant term should be expected to be present.

 

[jk22]

Does $(g_{\mu\nu})^2$ also give the metric ?

 

[PeterDonis]

Does $(g_{\mu\nu})^2$ also give the metric ?

What do you mean by $(g_{\mu\nu})^2$?

 

[Ibix]

Does $(g_{\mu\nu})^2$ also give the metric ?

Interpreting $\left(g_{\mu\nu}\right)^2$ literally as $g_{\mu\nu}g_{\mu\nu}$ it's an illegal expression. Interpreting it as $g_{\mu\nu}g^{\mu\nu}$ it's $\delta_\mu^\mu=4$.

 

[haushofer]

Interpreting $\left(g_{\mu\nu}\right)^2$ literally as $g_{\mu\nu}g_{\mu\nu}$ it's an illegal expression. Interpreting it as $g_{\mu\nu}g^{\mu\nu}$ it's $\delta_\mu^\mu=4$.

A similar problem arises when you naively would construct a mass term for the graviton in spin-2 field theories, as compared to spin-0 and spin-1.

 

[Ibix]

A similar problem arises when you naively would construct a mass term for the graviton in spin-2 field theories, as compared to spin-0 and spin-1.

Learning QFT is still on my to-do list. One of these days...

 

[jk22]

I mean a special case of $\lambda_{\mu\nu}^{a_1b_1a_2b_2...a_nb_n}g_{a_1b_1}g_{a_2b_2}...g_{a_nb_n}$

 

[DrGreg]

Does $(g_{\mu\nu})^2$ also give the metric ?

What do you mean by $(g_{\mu\nu})^2$?

I mean a special case of $\lambda_{\mu\nu}^{a_1b_1a_2b_2...a_nb_n}g_{a_1b_1}g_{a_2b_2}...g_{a_nb_n}$

So, do you mean $\lambda_{\mu\nu}^{ab}g_{ab}$? What is $\lambda$?

 

[jk22]

$\lambda$ were a 2+2n constant tensor. It lacks a sum over n to be complete.

 

[PeterDonis]

$\lambda$ were a 2+2n constant tensor.

What 2 + 2n constant tensor? What are you talking about?

 

[Ibix]

I think OP is proposing $\lambda$ as a tensorial cosmological constant (or at least, asking if such a thing is possible). As I think I've said to you before, @jk22, you must always define your terms. I think I've guessed your meaning, but it is a guess and guesswork is no basis for scientific communication.

Assuming my guess is correct, I think the answer is no. The construction proposed reduces to $\lambda_{\mu\nu}{}^{a_1}{}_{a_1}{}^{a_2}{}_{a_2}...{}^{a_n}{}_{a_n}=\lambda_{\mu\nu}$, a rank two tensor, which you are only free to add if its covariant derivative is zero. So this is either an over-complicated way of writing $\Lambda g_{\mu\nu}$ or it's not allowed.