What is the Unique Metric Tensor in this Line Element?

[Breo]

Hello,

this is the metric I am talking about:

$$ ds^2= (dt - A_idx^i)^2 - a^2(t)\delta_{ij}dx^idx^j $$

I never see one like this. How the metric tensor matrix would be?

 

[Matterwave]

How do you think you can obtain the metric tensor from the space-time interval that you gave?

 

[Breo]

How do you think you can obtain the metric tensor from the space-time interval that you gave?

I think the matrix would be:
$$ g_{\mu\nu} =
\left( \begin{array}{ccc}
1 & 2A_1 & 2A_2 & 2A_3 \\
0 & A_1^2 - a^2 & 0 & 0 \\
0 & 0 & A_2^2 - a^2 & 0 \\
0 & 0 & 0 & A_3^2 - a^2 \end{array} \right) $$

 

[Matterwave]

I think you are missing some more off-diagonal terms. The first parenthesis looks like it will also give terms that look like $dx^1 dx^2$ etc. Make sure to foil correctly. :)

Also note that the metric must be a symmetric matrix, and your matrix is definitely not symmetric.

 

[Breo]

Do you mean:

$$ (A_i dx^i)^2 = A_i^2 dx^idx^j $$ or $$ 2 . A_i dt . dx^i \longrightarrow 2A_i dtdx^i + 2A_i dx^i dt $$

So with the latter I get:

$$ g_{\mu\nu} =
\left( \begin{array}{ccc}
1 & 2A_1 & 2A_2 & 2A_3 \\
2A_1 & A_1^2 - a^2 & 0 & 0 \\
2A_2 & 0 & A_2^2 - a^2 & 0 \\
2A_3 & 0 & 0 & A_3^2 - a^2 \end{array} \right) $$

 

[Matterwave]

Notice that $A_i dx^i=A_1 dx^1+A_2 dx^2+A_3 dx^3$, then you want to take $(dt-A_1 dx^1-A_2 dx^2-A_3 dx^3)^2$ what do you get? You can see that there will be not only terms like $dt dx^1$ etc in there, there will be terms like $dx^1 dx^2$, which are all 0 in your matrix.

 

[Mentz114]

... there will be terms like $dx^1 dx^2$, which are all 0 in your matrix.

Are you sure ? I think $a^2(t)\delta_{ij}dx^idx^j$ means diagonal spatial elements only.

The metric an FLRW type expanding cosmology with anisotropic matter flow.

 

[Matterwave]

Are you sure ? I think $a^2(t)\delta_{ij}dx^idx^j$ means diagonal spatial elements only.

The metric an FLRW type expanding cosmology with anisotropic matter flow.

I don't see how you can avoid the off diagonal terms when clearly the squaring of the first term will give you terms like $dx^1 dx^2$. There's a sum inside the square. I'm talking about the term $(dt-A_i dx^i)^2$

 

[Breo]

Mmm! nice! I just only have a doubt, as I know the metric tensor is symmetric, when I obtain, for example, -2A₂A₃dx²dx³ should I write also in the ds² formula the symmetric term -2A₃A₂dx³dx² ? i think it is not necesary, right?

Well the matrix I obtained now is:

$$
\left( \begin{array}{ccc}
1 & -2A_1 & -2A_2 & -2A_3 \\
-2A_1 & A_1^2 - a^2 & 2A_1 A_2 & 2A_1 A_3 \\
-2A_2 & 2A_1 A_2 & A_2^2 - a^2 & 2A_2 A_3 \\
-2A_3 & 2A_1 A_3 & 2A_2 A_3 & A_3^2 - a^2 \end{array} \right)
$$

 

[Mentz114]

I don't see how you can avoid the off diagonal terms when clearly the squaring of the first term will give you terms like $dx^1 dx^2$. There's a sum inside the square. I'm talking about the term $(dt-A_i dx^i)^2$

Yes, I missed that sum. Very weird. Otherwise the metric is not unusual.

 

[Matterwave]

Mmm! nice! I just only have a doubt, as I know the metric tensor is symmetric, when I obtain, for example, -2A₂A₃dx²dx³ should I write also in the ds² formula the symmetric term -2A₃A₂dx³dx² ? i think it is not necesary, right?

Well the matrix I obtained now is:

$$
\left( \begin{array}{ccc}
1 & -2A_1 & -2A_2 & -2A_3 \\
-2A_1 & A_1^2 - a^2 & 2A_1 A_2 & 2A_1 A_3 \\
-2A_2 & 2A_1 A_2 & A_2^2 - a^2 & 2A_2 A_3 \\
-2A_3 & 2A_1 A_3 & 2A_2 A_3 & A_3^2 a^2 \end{array} \right)
$$

To check your answer, try to obtain your original $ds^2=(dt-A_i dx^i)^2-a^2 \delta_{ij} dx^i dx^j$ by using $ds^2=g_{\mu\nu}dx^\mu dx^\nu$. Match the two sides to see if they give you the same expression. :)

 

[Breo]

So I must write explicitly the 16 terms in the ds² expression and the matrix seems right :)

 

[Matterwave]

To check to make sure. But notice that since $g_{\mu\nu}=g_{\nu\mu}$ when you take the sum $g_{\mu\nu}dx^\mu dx^\nu$ you will get terms like $dx^1 dx^2$ and then another term like $dx^2 dx^1$ repeated, with the same factor in front.

What I'm getting at is I think you maybe have a factor of 2 off on some of your off diagonal terms, so you might want to double check.

 

[Breo]

To check to make sure. But notice that since $g_{\mu\nu}=g_{\nu\mu}$ when you take the sum $g_{\mu\nu}dx^\mu dx^\nu$ you will get terms like $dx^1 dx^2$ and then another term like $dx^2 dx^1$ repeated, with the same factor in front.

What I'm getting at is I think you maybe have a factor of 2 off on some of your off diagonal terms, so you might want to double check.

Oh, I did not notice. If my intuition does not fail, the off-diagonal terms in the ds² equation when you obtain something like: -2A2A3dx²dx³ must be splitted in two terms dividing by two? so you would have: -A2A3dx²dx³ - A3A2dx³dx² ?

 

[Matterwave]

Oh, I did not notice. If my intuition does not fail, the off-diagonal terms in the ds² equation when you obtain something like: -2A2A3dx²dx³ must be splitted in two terms dividing by two? so you would have: -A2A3dx²dx³ - A3A2dx³dx² ?

Yeah, basically.

 

[Breo]

:D

$$
\left( \begin{array}{ccc}
1 & -A_1 & -A_2 & -A_3 \\
-A_1 & A_1^2 - a^2 & A_1 A_2 & A_1 A_3 \\
-A_2 & A_1 A_2 & A_2^2 - a^2 & A_2 A_3 \\
-A_3 & A_1 A_3 & A_2 A_3 & A_3^2 - a^2 \end{array} \right)
$$

 

[Breo]

So now in order to define a natural vierbein I must diagonalize this matrix:

$$ g_{\mu\nu} = e^{\alpha}_{\mu}\eta_{\alpha\beta}e^{\beta}_{\nu} $$

right?

 

[Matterwave]

You must diagonalize the metric into the form diag(-1,1,1,1) or diag(1,-1,-1,-1) depending on the signature.

 

[DrGreg]

So now in order to define a natural vierbein I must diagonalize this matrix:

$$ g_{\mu\nu} = e^{\alpha}_{\mu}\eta_{\alpha\beta}e^{\beta}_{\nu} $$

right?

Yes. (In the sense that Matterwave just said.)

But here is a hint. You should find it easier to work with the original form of the metric in post #1, rather than the matrix you have just calculated.

 

[Breo]

Mmm interesting.

I must find an analytical tranform to obtain something like $(\alpha dt^2 +\beta_i (dx^i)²)$ from $(dt - A_i dx^i)^2$... I am wondering how. Maybe second grade equations... roots... ?

 

[DrGreg]

Mmm interesting.

I must find an analytical tranform to obtain something like $(\alpha dt^2 +\beta_i (dx^i)²)$ from $(dt - A_i dx^i)^2$... I am wondering how. Maybe second grade equations... roots... ?

You are aiming to get something that is a sum and difference of squares. But the formula in post #1 already is a sum and difference of squares...

 

[Breo]

And no squares, aswell. That doesn't matter? I thought I should find the terms of a diagonal matrix from which I would get the vierbeins.

 

[DrGreg]

$$e^0_\mu dx^\mu = dt - A_i dx^i\\ e^i_\mu dx^\mu = a \, dx^i$$

 

[Breo]

$$e^0_\mu dx^\mu = dt - A_i dx^i\\ e^i_\mu dx^\mu = a \, dx^i$$

Sorry for my "blindness",this is the first time I have to deal with a non-diagonal metric. I see more than 4 different terms in your equations.

 

[DrGreg]

Use post #24 to define all 16 components of $e^\alpha_\mu$ and then put into post #17. You ought to get post #16 for $g_{\mu\nu}$.

If you succeed, then think why it worked.

 

[Breo]

I am not...

I had just calculated 4 terms: $$ g_{00} = 1 \\ g_{11} = A_1^2-3a^2 \\ g_{01}=-A_1 \\ g_{12} = A_1 A_2 -3a^2 $$

What I did was fix the next equation:

$$ g_{\mu\nu} = e^0_{\mu}(1)e^0_{\nu} + e^1_{\mu}(-1)e^1_{\nu} + e^2_{\mu}(-1)e^2_{\nu} + e^3_{\mu}(-1)e^3_{\nu} $$

 

[DrGreg]

$$ g_{\mu\nu} = e^0_{\mu}(1)e^0_{\nu} + e^1_{\mu}(-1)e^1_{\nu} + e^2_{\mu}(-1)e^2_{\nu} + e^3_{\mu}(-1)e^3_{\nu} $$

That's correct, so I think you must be getting some of the values for $e^i_{\mu}$ wrong ($i=1,2,3$).

You need to solve for $e^i_{\mu}$$$e^1_0 dt + e^1_1 dx^1 + e^1_2 dx^2 + e^1_3 dx^3 = a \, dx^1$$etc.

 

[Breo]

I obtain this:

$$ e^0_0 = 1; \space e^0_1=-A_1; \space e^0_2=-A_2; \space e^0_3=-A_3 \\
e^1_1= a \\
e^2_2= a \\
e^3_3= a $$

The rest, zeros.

Now I am thinking how to set this up in order to obtain an adequate vierbein.

Could be this? (I am not sure as I had never seen a vierbein for a non-diagonal metric):

$$ e^1 = dt - A_1dx^1 - A_2 dx^2 - A_3 dx^3 \\
e^2=adx^1 \\
e^3=adx^2 \\
e^4=adx^3$$

(I fixed the upper indices to my usual notation)

By the way, could you explain me what's the mathematical reasoning of the #14 post?

 

[pervect]

Maybe this is too much help, but consider the transformation

$dp = dt - A_1 dx^1 - A_2 dx^2 - A_3 dx^3$

1) Is the transformation linear?
2) If you write the metric in terms of dp and $dx^i$, instead of dt and $dx^i$, what do you get?

 

[Breo]

The same metric with different coords?

$$ ds^2 = dp^2 - \delta_{ij} a^2dx^idx^j $$

Which is diagonal. This seems too much easy :-/

I wrote this on mobile. Hope it worked xD

 

[Matterwave]

The same metric with different coords?

$$ ds^2 = dp^2 - \delta_{ij} a^2dx^idx^j $$

Which is diagonal. This seems too much easy :-/

I wrote this on mobile. Hope it worked xD

It seems too easy because Pervect basically gave it to you haha.

You have diagonalized the metric, but you still need to put it in terms of diag(1,-1,-1,-1) so you have to do one more transformation.

 

[Breo]

Mmm as it seems to work with the signs I bet you are talking about this:

$$ ds^2 = dp^2 - a^2 [(dx^1)^2 + (dx^2)^2 + (dx^3)^2] $$

removing the delta expression?

 

[Matterwave]

You still have the $a^2$ there. What you wrote is just explicitly what you had in post #30, you just wrote out every term. It is identical with the one in #30 though. You need to go 1 more step.

 

[Breo]

I am still wondering :-/

 

[Matterwave]

I am still wondering :-/

I'm not sure what other hints I can give other than just telling you the answer haha. Recall what you did for the spherical coordinates? What did the metric look like there, and how did you find a triad?