- #1

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ds^2 = d(ix)^2 + d(ct)^2 = d(ix')^2 + d(ct')^2 [/itex] 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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- Thread starter snoopies622
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- #1

- 775

- 17

ds^2 = d(ix)^2 + d(ct)^2 = d(ix')^2 + d(ct')^2 [/itex] 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.

- #2

- 2,311

- 299

You should rather fully generalize to pseudo-Euclidean pseudo-rotations by utilizing hyperbolic trigonometric functions.

[tex] \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][/tex]

[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.

- #3

- 775

- 17

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 [itex]ds^2 = dx^2 + dy^2 [/itex]. 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 [itex]ds^2 = (dx')^2 + (dy')^2 [/itex]. That is, the components of the metric tensor are exactly the same.

[tex]

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

[/tex]

This is not at all true in general, so I'm wondering if there are any other linear transformations that have this property. Thanks!

- #4

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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).

- #5

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- 17

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

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