Having trouble writing down a metric in terms of metric tensor in matrix form?

[zeromodz]

Can someone please explain to me how exactly you write down a metric, say the FLRW metric in matrix form. Say we have the given metric here.

ds^2 = dt^2 - R(t)^2 * [dw^2 + s^2 * (dθ^2 + sin^2(θ)dΦ^2)]

Thank you.

 

[atyy]

Let the coordinates be q₁,q₂,q₃,q₄.

The line element will be a sum of terms like C₁₂dq₁dq₂.

In matrix form, C₁₂ will be in row 1 column 2 of the matrix.

Edit: See Rasalhague's post for the correct version. I forgot the 1/2 for the off-diagonal terms.

 

[Rasalhague]

In your example, if q⁰ = t, q¹ = w, q² = θ, q³ = Φ (where superscripts are indices), then the matrix of coefficients will be as follows, with superscript 2 denoting an exponent:

\begin{pmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; R(t)^2 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; (R(t) \cdot s)^2 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; (R(t)\cdot s \cdot \sin(\theta))^2<br /> \end{pmatrix}

= diag(1,0,0,0) - R(t)²[diag(0,1,0,0) + s²(diag(0,0,1,0)+diag(0,0,0,sin(θ)²)].

Here diag(a,b,c,d) denotes a diagonal 4x4 matrix with diagonal entries as indicated, from top left to bottom right.

In general, given an expression of the form

ds^2 = ...,

where the values of the indices are not equal, the scalar coefficients in each term of the form

A \, dx^\mu dx^\nu \enspace (\text{no summation} )

(EDIT: Ignore the words "no summation" - a relic of previous version which I forgot to remove before posting. Sorry.) correspond to matrix entries

g_{\mu\nu} = \frac{1}{2} A.

And where the values of the indices are equal, the scalar coefficients in each term of the form

B \, (dx^\mu)^2

correspond to matrix entries

g_{\mu\mu} = B.

 

[DrGreg]

Rasalhague accidentally missed some minus signs:

\begin{pmatrix}<br /> 1 &amp; 0 &amp; 0 &amp; 0\\ <br /> 0 &amp; -R(t)^2 &amp; 0 &amp; 0\\ <br /> 0 &amp; 0 &amp; -R(t)^2 \, s^2 &amp; 0\\ <br /> 0 &amp; 0 &amp; 0 &amp; -R(t)^2 \, s^2 \, \sin^2\theta<br /> \end{pmatrix}​

 

[Rasalhague]

Oopsh... Thanks for the correction, Dr Greg!