Lorentz Transformation: Coefficients a_{nm}(u) & Inverse Relations

[Petar Mali]

$$x'=a_{11}x+a_{12}y+a_{13}z+a_{14}t$$

$$y'=a_{21}x+a_{22}y+a_{23}z+a_{24}t$$

$$z'=a_{31}x+a_{32}y+a_{33}z+a_{34}t$$

$$t'=a_{41}x+a_{42}y+a_{43}z+a_{44}t$$

$$\vec{u}=u\vec{e}_x$$

Coefficients $$a_{nm}=a_{nm}(u)$$

Why I suppose that coefficients are function only of velocity $$u$$?

Inverse relations

$$x=a_{11}'x'+a_{12}'y'+a_{13}'z'+a_{14}'t'$$

$$y=a_{21}'x'+a_{22}'y'+a_{23}'z'+a_{24}'t'$$

$$z=a_{31}'x'+a_{32}'y'+a_{33}'z'+a_{34}'t'$$

$$t=a_{41}'x'+a_{42}'y'+a_{43}'z'+a_{44}'t'$$

$$a_{nm}'(u)=a_{nm}(-u)$$

Equations of transformations are linear (time and space are homogeneous).

That means from linearity of transformations $$\Rightarrow$$ time and space are homogeneous?

Why now I can say

$$y'=a_{22}y$$

$$z'=a_{33}z$$

$$t'=a_{41}x+a_{44}t$$?

Thanks for your answer!

 

[starthaus]

$$x'=a_{11}x+a_{12}y+a_{13}z+a_{14}t$$

$$y'=a_{21}x+a_{22}y+a_{23}z+a_{24}t$$

$$z'=a_{31}x+a_{32}y+a_{33}z+a_{34}t$$

$$t'=a_{41}x+a_{42}y+a_{43}z+a_{44}t$$

$$\vec{u}=u\vec{e}_x$$

Coefficients $$a_{nm}=a_{nm}(u)$$

Why I suppose that coefficients are function only of velocity $$u$$?

Because the presence of a relative speed is the only thing that differentiates frames F and F'.

Inverse relations

$$x=a_{11}'x'+a_{12}'y'+a_{13}'z'+a_{14}'t'$$

$$y=a_{21}'x'+a_{22}'y'+a_{23}'z'+a_{24}'t'$$

$$z=a_{31}'x'+a_{32}'y'+a_{33}'z'+a_{34}'t'$$

$$t=a_{41}'x'+a_{42}'y'+a_{43}'z'+a_{44}'t'$$

$$a_{nm}'(u)=a_{nm}(-u)$$

Equations of transformations are linear (time and space are homogeneous).

That means from linearity of transformations $$\Rightarrow$$ time and space are homogeneous?

Reverse the arrow direction and you will get the correct implication.

Why now I can say

$$y'=a_{22}y$$

$$z'=a_{33}z$$

$$t'=a_{41}x+a_{44}t$$?

Thanks for your answer!

You can do this only in the specific case when the velocity between frames F and F' is perpendicular on the axes y and z. In other words, the coefficients affecting the formulas for y' and z' do not depend on the relative speed v.

 

[Petar Mali]

Thanks a lot!

@starthaus

You can do this only in the specific case when the velocity between frames F and F' is perpendicular on the axes y and z. In other words, the coefficients affecting the formulas for y' and z' do not depend on the relative speed v.

Yes, but why is that? Is that consequence of isotropy of space? Can you say some more words about this?

 

[starthaus]

Thanks a lot!

@starthaus

You can do this only in the specific case when the velocity between frames F and F' is perpendicular on the axes y and z. In other words, the coefficients affecting the formulas for y' and z' do not depend on the relative speed v.

Yes, but why is that? Is that consequence of isotropy of space? Can you say some more words about this?

No, it is a consequence of the fact that there is no reason to assume that speed affects the dimensions orthogonal to it.
Now, the formulas for the arbitrary motion, you can find the general Lorentz transformation https://www.physicsforums.com/blog.php?b=1959 . As you can see, in this case the transformation matrix is not sparse anymore.