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\\<br /> e^i_\mu dx^\mu = a \, dx^i

 

[Breo]

e^0_\mu dx^\mu = dt - A_i dx^i\\<br /> 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}<br /> e^1_0 dt + e^1_1 dx^1 + e^1_2 dx^2 + e^1_3 dx^3 = a \, dx^1 <br />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