# I How to keep the components of a metric tensor constant?

#### snoopies622

I've noticed that a very easy way to generate the Lorentz transformation is to draw Cartesian coordinate axes in a plane, label then ix and ct, rotate them clockwise some angle $\theta$ producing axes ix' and ct', use the simple rotation transformation to produce ix' and ct', then just divide out the i and c accordingly. I assume this works because rotation keeps the components of the metric tensor constant, and applying a kind of pseudo-Euclidean metric $ds^2 = d(ix)^2 + d(ct)^2 = d(ix')^2 + d(ct')^2$ is consistent with the premise of special relativity.

My question is, what other linear transformations in a plane maintain the
[ 1 0
0 1 ]
metric tensor form?
Thanks.

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#### jambaugh

Science Advisor
Gold Member
Your use of Euclidean rotations will not properly reflect a Lorentz rotation because, whether you incorporate an $i$ factor for the x-coordinate or not, the quadratic form conserved by your simple rotations will be the Euclidean distance: $r = \sqrt{ |ix|^2 + |ct|^2}$. Note you can rotate over 45degrees and convert a time like displacement to a space-like displacement.

You should rather fully generalize to pseudo-Euclidean pseudo-rotations by utilizing hyperbolic trigonometric functions.
$$\left[\begin{array}{c} x' \\ ct' \end{array}\right] = \left( \begin{array}{cc} \cosh(\beta) & \sinh(\beta) \\ \sinh(\beta) & \cosh(\beta) \end{array}\right) \left[ \begin{array}{c} x \\ ct \end{array}\right]$$
[edit: Note the absence of a minus sign on the off-diagional sinh.]

The pseudo-angle $\beta$ is the boost parameter which we can relate to the relative frame velocity by $v/c = \tanh(\beta)$. This is a nice way to do things because the whole complicated addition of boost velocities problems has an elegant solution in that it is the boost parameters that add. $\beta= \beta_1 + \beta_2$.

Now there is a situation where it is valid to complexify the coordinates and perform a Wick rotation but that has to do with path integration which become path independent in the complex extension provide one accounts properly for poles. This is a matter of analytically extending the domain of the formal path integral including any pseudo-metric dependency, to the complex extension of space-time. Then using the mathematical result about path independence of the path integral (provided no poles are crossed) one can move the integral to a value-equivalent path on a real-Euclidean subspace of the complexification of the prior real pseudo-Euclidean space.

#### snoopies622

Thanks Jambaugh, good stuff.

Perhaps I should have left out special relativity altogether and simply posed my question this way: In a plane using the Cartesian coordinate system, the metric is $ds^2 = dx^2 + dy^2$. If I create a new coordinate system (x',y') by rotating the (x,y) axes through any angle, the metric using the new coordinates is $ds^2 = (dx')^2 + (dy')^2$. That is, the components of the metric tensor are exactly the same.
$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$
This is not at all true in general, so I'm wondering if there are any other linear transformations that have this property. Thanks!

#### jambaugh

Science Advisor
Gold Member
Ah, I see. Well there's a well established answer to your question. Those linear transformations which leave the metric unchanged are elements of the orthogonal group (or psuedo-orthogonal group if you're in SR and dealing with an indefinite metric).

So for example, in the plane, O(2) is the group of rotations about the origin composed with a reflection about some line through the origin.

The Lorentz group, O(3,1) is the group preserving the space-time metric of special relativity.

And we generalize these to general indefinite metrics. The orthogonal group O(p,n) preserves the metric with square norm equal to the sum of p squared terms minus the sum of another n squared terms.

And, finally, if you remove those orthogonal transformations which invert the space you have the special orthogonal group: SO(p,n).

#### snoopies622

Great, this is just what I was looking for - thanks!

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