寻找可将给定度规简化为最大对称度规的坐标变换

Nice question! Let’s walk through simplifying this metric step by step—we’ll tackle the cross terms one at a time and end up with a well-known maximally symmetric spacetime.

Step 1: Eliminate the $dtdx$ cross term #

First, focus on the coupled $t$ and $x$ components of the original metric:

$-dt^2 -4\cosh^2(x/2)dtdx + \left(1 -4\cosh^{2(x/2)\right)dx}2$

Note that $2\cosh^2(x/2) = 1 + \cosh x$, so the coefficient of $dtdx$ simplifies to $-2(1+\cosh x)$. We can eliminate this cross term with a new time coordinate:
$$u = t + x + \sinh x$$
Taking the differential gives $du = dt + (1 + \cosh x)dx$, so we can rewrite $dt$ as $du - (1+\cosh x)dx$.

Substitute this into the $t$-$x$ portion of the metric. After expanding and canceling the cross terms (you can verify the algebra yourself), this reduces to:
$$-du^2 + \cosh^2 x , dx^2$$

Step 2: Eliminate the $dxdy$ cross term #

With the first transform applied, our metric now looks like:
$$ds^2 = -du^2 + \cosh^2 x , dx^2 + dy^2 + 2\sinh x , dxdy + dz^2$$

The $dxdy$ term has a coefficient of $2\sinh x$. We’ll define a new $y$-coordinate to get rid of this cross term:
$$v = y + \cosh x$$
Its differential is $dv = dy + \sinh x , dx$, so $dy = dv - \sinh x , dx$.

Substitute this into the $y$-$x$ terms. Expanding and canceling cross terms leaves us with:
$$dv^2 - \sinh^2 x , dx^2$$

Step 3: Combine terms to reveal the maximally symmetric metric #

Now substitute both transformations back into the full metric:
$$ds^2 = -du^2 + \cosh^2 x , dx^2 + dv^2 - \sinh^2 x , dx^2 + dz^2$$

Using the hyperbolic identity $\cosh^2 x - \sinh^2 x = 1$, the $dx^2$ terms combine to a simple $dx^2$. We’re left with:
$$ds^2 = -du^2 + dx^2 + dv^2 + dz^2$$

This is the Minkowski metric—a flat, maximally symmetric spacetime (with a zero curvature tensor, which is trivially maximally symmetric).

It’s understandable that de Sitter/AdS transforms didn’t work first; the original metric’s hyperbolic functions made it look like a curved spacetime, but it’s actually just a non-standard coordinate patch of flat spacetime. The key was targeting each cross term with coordinate shifts that use integrals of the hyperbolic coefficients from the cross terms.

内容的提问来源于stack exchange，提问作者APORIL