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 <br /> ds^2 = d(ix)^2 + d(ct)^2 = d(ix&#039;)^2 + d(ct&#039;)^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.

 

[jambaugh]

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&#039; \\ ct&#039; \end{array}\right] = \left( \begin{array}{cc} \cosh(\beta) &amp; \sinh(\beta) \\ \sinh(\beta) &amp; \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&#039;)^2 + (dy&#039;)^2. That is, the components of the metric tensor are exactly the same.
<br /> \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{pmatrix} <br />
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]

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!