Problem with signs in the Lorentz Transformation equations

[spaghetti3451]

Homework Statement

The following is Exercise 2.1 from from Ray d'Inverno's 'Introducing Einstein's Relativity.'

(a) Write down the Galilean transformation from observer $S$ to observer $S'$, where $S'$ has velocity $v_1$ relative to $S$.
(b) Find the transformation from $S'$ to $S$ and state in simple terms how the transformations are related.
(c) Write down the Galilean transformation from $S'$ to $S''$, where $S''$ has velocity $v_2$ relative to $S'$.
(d) Find the transformation from $S$ to $S''$.
(e) Prove that the Gailiean transformations form an Abelian (commutative) group.

2. Homework Equations

3. The Attempt at a Solution

(a) My answer is $$x' = x - v_1 t,\ y'=y,\ z'=z,\ t'=t,$$ which matches with the textbook answer. So, no problem here.
(b) My answer is $$x = x' + v_1 t,\ y=y',\ z=z',\ t=t',$$ which again matches with the textbook answer. So, no problem here.
(c) My answer is $$x''=x'- v_2 t,\ y''=y',\ z''=z',\ t''=t',$$ but the textbook answer is $$x''=x'+ v_2 t,\ y''=y',\ z''=z',\ t''=t'.$$
(d) My answer is $$x''=x- ( v_ 1 + v_2 ) t,\ y''=y,\ z''=z,\ t''=t,$$ but the textbook answer is $$x''=x+ ( v_ 1 + v_2 ) t,\ y''=y,\ z''=z,\ t''=t.$$

Can someone please tell me if my answers are wrong, or if the textbook answers are wrong?

 

[BvU]

They way you put it the book is wrong and you are right.

It's a small miracle that you agree in a) and b) since both you and the book assume $v_1$ is along the x-axis of $S$ and $S'$, and that both clocks and coordinate systems coincide at t = 0.

 

[spaghetti3451]

Alright, then, here's my attempt to part (e) of the question.The group elements of the Galilean transformations are the velocities $v$ of $S'$ relative to $S$.

Consider the boost from $S$ to $S'$, and then the boost from $S'$ to $S''$, so that $x''=x- ( v_ 1 + v_2 ) t,$ so that $v = v_1 + v_2$. So, the group of Galilean transformations is closed.

Consider the boost from $S$ to $S''$, and then the boost from $S''$ to $S'''$, so that $x''' = x'' - v_3 t = x - (v_1 + v_2 + v_3) t$. Also, consider the boost from $S$ to $S'$, and then the boost from $S'$ to $S'''$, so that $x''' = x' - (v_2 + v_3) t = x - (v_1 + v_2 + v_3) t$. So, the group multiplication operation is associative.

$v = 0$ is the identity element, because it transforms $(x, y, z, t)$ into $(x, y, z, t)$.

$-v$ is the inverse of $v$, because the consecutive application of $v$ and $-v$ transforms $(x, y, z, t)$ into $(x, y, z, t)$.

The above four properties prove that the Galilean transformations form a group.

Furthermore, consider a boost by velocity $v_1$, and then a boost by velocity $v_2$, so that $x''' = x'' - v_2 t = (x' - v_1 t) - v_2 t = x' - (v_1 + v_2) t$. Also, consider a boost by velocity $v_2$, and then a boost by velocity $v_1$, so that $x''' = x'' - v_1 t = (x' - v_2 t) - v_1 t = x' - (v_1 + v_2) t$. So, the group of Galilean transformations is Abelian (commutative).Do you think my proof is correct?

 

[BvU]

Does it help if I say yes ? You really want to convince yourself, and it shouldn't make a difference if I say "well, it looks good to me !" (which I am perfectly happy to do, even if it probably is against PF rules ("we look at them more as guidelines ")

 

[spaghetti3451]

Thank you!