Can this type of transformation be non-linear?

[snoopies622]

I've finally worked out a derivation of the Lorentz transformation that doesn't use the now out of favor i^2=-1, but it still has one weak spot: it assumes that the transformation is linear. It seems quite reasonable to me that it would be linear since it has to graph straight lines on to straight lines (since the laws of mechanics should be the same in both reference frames) but how can I go from that fact to

x' = Ax + Bt
t' = Dx + Et

where A,B,D and E are constants without any doubt? Is it mathematically possible for a transformation that requires any straight line in one coordinate system to become a straight line in the other coordinate system to assume some other, non-linear form?

 

[fresh_42]

The transformation you wrote is linear by construction: $\begin{bmatrix}x'\\t' \end{bmatrix}=\begin{bmatrix}A&B\\ D&E \end{bmatrix}\cdot \begin{bmatrix}x\\t \end{bmatrix}$. To get something non-linear you will have to use non-linear terms.

Are you talking about one straight line or all straight lines? You could add a constant term, so it still transforms lines into lines, but isn't linear anymore, however, affine linear.

 

[snoopies622]

I start with the premise that all straight lines in (x,t) are transformed into straight lines into (x',t') and vice versa and that (0,0) in one coordinate system is (0,0) in the other. Does it then follow that the transformation must look like

x' = Ax + Bt
t' = Dx + Et

where A,B, D, E are constants, and how do I know this for certain? Thanks.

 

[fresh_42]

I start with the premise that all straight lines in (x,t) are transformed into straight lines into (x',t') and vice versa and that (0,0) in one coordinate system is (0,0) in the other. Does it then follow that the transformation must look like

x' = Ax + Bt
t' = Dx + Et

where A,B, D, E are constants, and how do I know this for certain? Thanks.

You want to show, that for a given transformation $f$ it has to hold: $f$(straight)=straight which means $f(\lambda \vec{a}+\mu \vec{b})= \lambda f(\vec{a})+\mu f(\vec{b})$ for all $\vec{a},\vec{b},\lambda,\mu\,.$