Let L be a Lorentz transformation in one dimension of space and one of time. Let V and W be the two world lines of light traveling in the positive and negative directions. Then V and W are eigenspaces of L (constancy of speed of light). If F is the linear map taking V into W and vice versa, with FF=1, then FLF is the inverse of L and det(F) = 1, which implies det(L) = 1. If (t1,t2) are coordinates with respect to V and W (i.e. for a point p, light travels t1 along V and t2 along W to get to p.), then L is represented by the 2x2 matrix diag(u,1/u), where u and 1/u are the eigenvalues of V and W.
3
-
1$\begingroup$ As I wrote in another comment, u is the Doppler factor, which Bondi calls the k-factor and uses it with radar method to develop special relativity using what he calls the k-calculus. $\endgroup$– robphyCommented Dec 18, 2020 at 23:07
-
$\begingroup$ I have to excuse for self-advertising, but: physics.stackexchange.com/a/227128/102232 $\endgroup$– Gyro GearlooseCommented Dec 18, 2020 at 23:41
-
$\begingroup$ @GyroGearloose Here is a visualization associated with that approach desmos.com/calculator/iswdppnuxs , which shows the analogies between Euclidean geometry and the geometries of special and Galilean relativity (by tuning E).The "clock diamonds" for the Minkowskian case use the u eigenvalue (Bondi's k-factor) to be reshaped [while preserving its edge-directions (eigenvectors) and area (unit determinant)] so that its timelike diagonal is along the observer 4-velocity. (The underlying rotated grid [not shown] is associated with the light-cone coordinates, with axes along V and W.) $\endgroup$– robphyCommented Dec 19, 2020 at 14:01
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
No not really. The Lorentz transformation can be written in the form of a hyperbolic rotation \begin{align} \left( \begin{array}{cc} \cosh (w) & \sinh (w) \\ \sinh (w) & \cosh (w) \\ \end{array} \right) \end{align} where $\cosh(w)=\gamma$, $\sinh(w)=-(v/c)\gamma= -\beta/\gamma$, with $\gamma=1/\sqrt{1-\beta^2}$. Your derivation is just one where $u=e^{w}$. Moreover, unless you can connect the eigenvalues to the parameter $\beta=v/c$, it's not terribly useful to have everything in terms of $u$.
answered Dec 18, 2020 at 22:01
-
2$\begingroup$ The parameter u is the Doppler factor, sometimes called the Bondi-k factor. The Bondi k-calculus goes quite far using just k and the radar method. My “relativity on rotated graph paper” method relies heavily on it. Since it is an eigenvalue of the boost, it is a mathematically natural variable to work with. $\endgroup$– robphyCommented Dec 18, 2020 at 23:04
-
1$\begingroup$ @robphy sure but you just did provide an interpretation for this eigenvalue, which the OP did not do. Moreover you have just ipso facto answered the OP’s question in the negative (and made me learn about Bondi k-calculus). $\endgroup$ Commented Dec 18, 2020 at 23:09
-
$\begingroup$ It should be $\sinh w=+(v/c)\gamma=\beta\gamma$, where $\beta=\tanh w=v/c$. Note that $u= \exp w=\cosh w +\sinh w = (\cosh w)(1+\tanh w) =\frac{1+\tanh w)}{\sqrt{1-\tanh^2 w}} =\sqrt{\frac{ 1+\tanh w}{1-\tanh w}}=k$. $\endgroup$– robphyCommented Dec 19, 2020 at 14:33
$\begingroup$
$\endgroup$
2
Sorry for reading too quickly. That derivation looks fine, but it's not what I would really call particularly new. Your derivation centers on several assumptions:
- Lorentz transformations are linear, so they can be represented by matrices
- The speed of light is the same in both directions, and the inverse of a boost is a boost with the same speed in the opposite direction (space is isotropic and boosts compose with one another)
- The speed of light is the same in all reference frames
These are the fundamental assumptions baked into all derivations of the Lorentz transformations, in one way or another. Your derivation is different from most that I see because you're comfortable jumping straight into a new coordinate system in which the Lorentz transformation matrix is diagonal. In that sense, it is new to me, but it is ultimately just the typical one viewed in rotated coordinates.
answered Dec 18, 2020 at 21:46
-
$\begingroup$ In the new coordinates I defined, they are diagonal! Remember, any symmetric matrix can be diagonalized. So the standard form, which is symmetric can be diagonalized by a change of basis. That is essentially what I've done. $\endgroup$ Commented Dec 18, 2020 at 21:53
-
$\begingroup$ Many thanks for all the comments. I am interested in axiomatic (and coordinate free) Special Relativity What are the basic assumptions? Speed of light constant in all frames is basic. Isotropy of space too. What if we say that boosts preserve non-acceleration of motion (i.e. maps lines to lines)? That will give linearity. Won't it also give composition of boosts? Is "the inverse of a boost is a boost with opposite velocity" a basic assumption. Or does it follow from the others? $\endgroup$ Commented Dec 20, 2020 at 17:06