Edwards, Tangherlini, Selleri transformations and their inverse

[bernhard.rothenstein]

Edwards, Tangherlini, Selleri propose synchrony parameter dependent transformation equations we have discussed here. Call them direct transformations. They also their inverse version. As I see they are not used. Is there a special reason for that. Are they of interest?
Thanks

 

[DrGreg]

There's no need to quote an inverse transformation unless the argument you are making requires such a transformation. I think I've probably seen inverse Selleri/Tangherlini transforms somewhere, though I can't remember where. It's quite easy to calculate.

The Edwards transform is between two arbitrarily synced systems and so is "its own inverse" in the sense that the Lorentz transform is "its own inverse", you just need to change the values of the parameters. (E.g. v to -v in the Lorentz case.)

 

[bernhard.rothenstein]

There's no need to quote an inverse transformation unless the argument you are making requires such a transformation. I think I've probably seen inverse Selleri/Tangherlini transforms somewhere, though I can't remember where. It's quite easy to calculate.

The Edwards transform is between two arbitrarily synced systems and so is "its own inverse" in the sense that the Lorentz transform is "its own inverse", you just need to change the values of the parameters. (E.g. v to -v in the Lorentz case.)

My problem is with the Selleri transformation There in I the clocks are standard synchronized whereas in I using the so called external synchronization. If we know the direct transformations the inverse ones are not obtainable by the rule which works in the case when in both frames standard clock synchronization takes place i.e. change the sign of V and interchange the primed with unprimed same physical quantities.
As allways respect and thanks.

 

[DrGreg]

My problem is with the Selleri transformation There in I the clocks are standard synchronized whereas in I using the so called external synchronization. If we know the direct transformations the inverse ones are not obtainable by the rule which works in the case when in both frames standard clock synchronization takes place i.e. change the sign of V and interchange the primed with unprimed same physical quantities.
As allways respect and thanks.

If you know the direct transformation, finding the inverse is just mathematical algebra, in this case, solving two simultaneous equations, or, equivalently, inverting a 2x2 matrix.

Changing the sign of V etc won't work because whereas the forward transform is from isotropic to anistropic coordinates, the reverse (inverse) transform is from anisotropic to istropic coordinates, so you would not expect the "same" formula to apply. If should also be pointed out that V is measured within the isotropic coordinates (as dx/dt for the "moving" observer relative to the "stationary" observer). The velocity V' of the "stationary" observer relative to the "moving" observer as measured in the "moving" observer's anisotropic coordinates (dx'/dt') will not be -V.

[tex]x' = \gamma(x - Vt)[/tex]
[tex]t' = t/\gamma[/tex]​

has inverse

[tex]x = (x' + \gamma^2 Vt')/\gamma[/tex]
[tex]t = \gamma t'[/tex]​

from which it follows that

[tex]V' = -\gamma^2 V[/tex]​

So the rule in this case is to replace V by [itex]-\gamma^2 V[/itex] and [itex]\gamma[/itex] by [itex]1/\gamma[/itex].

 

[bernhard.rothenstein]

If you know the direct transformation, finding the inverse is just mathematical algebra, in this case, solving two simultaneous equations, or, equivalently, inverting a 2x2 matrix.

Changing the sign of V etc won't work because whereas the forward transform is from isotropic to anistropic coordinates, the reverse (inverse) transform is from anisotropic to istropic coordinates, so you would not expect the "same" formula to apply. If should also be pointed out that V is measured within the isotropic coordinates (as dx/dt for the "moving" observer relative to the "stationary" observer). The velocity V' of the "stationary" observer relative to the "moving" observer as measured in the "moving" observer's anisotropic coordinates (dx'/dt') will not be -V.

[tex]x' = \gamma(x - Vt)[/tex]
[tex]t' = t/\gamma[/tex]​

has inverse

[tex]x = (x' + \gamma^2 Vt')/\gamma[/tex]
[tex]t = \gamma t'[/tex]​

from which it follows that

[tex]V' = -\gamma^2 V[/tex]​

So the rule in this case is to replace V by [itex]-\gamma^2 V[/itex] and [itex]\gamma[/itex] by [itex]1/\gamma[/itex].

Thanks

I know that considering the relative positions of the I and I' reference frame from I at a given time t I can derive the say direct transformation for the space coordinates of an event
taking into account length contraction. If I consider the same situation from I at a time t' and taking into account length contraction I can derive the inverse transformation of the space coordinate. Combining the two equations I can derive the direct and the inverse transformations for the time coordinates of the same event.
Do you know a way to do the same thing but starting with the time dilation?