X and x' in two coordinate relativistic systems

[C_Ovidiu]

let's say we have two coordinate systems ( S-OXYZ and S'-O'X'Y'Z') , S' moving with velocity v away from S . At t=0 O=O'.

According to Lorentz' transformations we have :

At t=0 x in as a function of x' :
$$x=\frac{x'}{\sqrt{1-\beta^2}}=x'\gamma$$

Now, say I know x(I just found it above) , and I want to find x'
$$x'=x\gamma=x'\gamma^2=>\gamma^2=1$$

Where does this come from and what have I missed ?
Thank you !

 

[JesseM]

let's say we have two coordinate systems ( S-OXYZ and S'-O'X'Y'Z') , S' moving with velocity v away from S . At t=0 O=O'.

According to Lorentz' transformations we have :

At t=0 x in as a function of x' :
$$x=\frac{x'}{\sqrt{1-\beta^2}}=x'\gamma$$

Why are you just looking at cases where t=0? The general transformation would be $$x' = \gamma (x - vt)$$, so if t=0 then you do have $$x' = \gamma x$$ (you seem to have reversed x and x' though, unless you meant to write t'=0).

Now, say I know x(I just found it above) , and I want to find x'
$$x'=x\gamma=x'\gamma^2=>\gamma^2=1$$

I agree with $$x' = x \gamma$$ in the case of t=0, but where are you getting $$x \gamma = x' \gamma^2$$? Are you taking the reverse Lorentz transform $$x = \gamma (x' + vt')$$ and setting t'=0 as well, thus getting $$x = \gamma x'$$? The problem is that if you set both t=0 and t'=0, the only coordinates where both time coordinates are 0 are (x=0, t=0) and (x'=0, t'=0), so obviously if x'=0 then you can have $$x' = x' \gamma^2$$ even when $$\gamma$$ is not 0.

 

[C_Ovidiu]

Tnx.