Lorentz via rotation

706
8

Main Question or Discussion Point

To me the easiest way to arrive at the Lorentz transformation is by rotation, using one dimension for space and the other for the product of time, the speed of light and the square root of minus one. This seems justifiable to me if one starts with the premise that ds^2 =ds'^2. I can see how if one assumes that the speed of light is the same in both frames of reference then ds^2=0 implies ds'^2=0 (and vice versa). But I don't know why ds^2=ds'^2 in general. Must one find the Lorentz transformation in some other way in order to arrive at this equality?
 

Answers and Replies

1,986
5
To me the easiest way to arrive at the Lorentz transformation is by rotation, using one dimension for space and the other for the product of time, the speed of light and the square root of minus one. This seems justifiable to me if one starts with the premise that ds^2 =ds'^2. I can see how if one assumes that the speed of light is the same in both frames of reference then ds^2=0 implies ds'^2=0 (and vice versa). But I don't know why ds^2=ds'^2 in general. Must one find the Lorentz transformation in some other way in order to arrive at this equality?
Whatever works easiest for you!
But notice that this only works for flat spacetimes.
 
pam
455
1
ict was popular 100 years ago. It has lost fashion, because it can't be extended to GR, and it complicates relativistic QM, with two different i's.
 
706
8
Thanks to you both. I didn't realize that ict wouldn't work for non-flat spacetimes, but I guess it would yield a complex ds^2 (and who wants that?).
 

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