Lorentz Transform Deduction: Question Explained

[thaiqi]

TL;DR

I read a deduction in one book, but I got a question in it.

I read in one book about the deduction of Lorentz transform. It writes:
'
$$
\begin{aligned}
t^\prime & = \xi t + \zeta x (1) \\
x^\prime & = \gamma x + \delta t (2) \\
y^\prime & = y (3) \\
z^\prime & = z (4)
\end{aligned}
$$
from (2), it gives:
$$
\begin{aligned}
{dx \over dt} = -{ \delta \over \gamma} = v (5)
\end{aligned}
$$
from (1) and (2), they give:
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\delta \over \xi} = - v (6)
\end{aligned}
$$
so,
$$
\begin{aligned}
\xi = \gamma
\end{aligned}
$$
and,
$$
\begin{aligned}
\delta = - \gamma v
\end{aligned}
$$
let
$$
\begin{aligned}
\zeta = - \eta \gamma
\end{aligned}
$$
we have:
$$
\begin{aligned}
t^\prime & = \gamma ( t - \eta x ) (7) \\
x^\prime & = \gamma ( x - v t) (8) \\
y^\prime & = y (9) \\
z^\prime & = z (10)
\end{aligned}
$$

...
we can get
$$
\begin{aligned}
\eta = v / c^2
\end{aligned}
$$
...
'
My question lies in formula (6), it can give
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\delta \over \xi} = - v
\end{aligned}
$$
as the above, but it can also give
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\gamma \over \zeta}
\end{aligned}
$$
such that:
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\gamma \over \zeta} = - {{1} \over {\eta}} = - {c^2 \over v} \neq -v (??)
\end{aligned}
$$
So why??

 

[anuttarasammyak]

Let x'=const, you calculate $\frac{dx}{dt}$.
Let x=const, you calculate $\frac{dx'}{dt'}$. Do not let t=const.

 

[Ibix]

You don't provide full details, but I presume what you've done is inverted (1) and (2) to get $x(x',t')$ and $t(x',t')$. Then when you take the differentials in these two equations you are considering $\left.\frac{dx'}{dt'}\right|_{x=\mathrm{const}}$ and $\left.\frac{dx'}{dt'}\right|_{t=\mathrm{const}}$. So yes, they are different things.

You don't want to consider holding a time coordinate constant when you are studying velocity. You are considering the velocity of an object at rest in one frame or the other, so you want to vary the time and space coordinate in one frame while you hold the spatial coordinate in the other frame constant, as @anuttarasammyak says.

 

[PeroK]

Summary:: I read a deduction in one book, but I got a question in it.

I read in one book about the deduction of Lorentz transform. It writes:
'
$$
\begin{aligned}
t^\prime & = \xi t + \zeta x (1) \\
x^\prime & = \gamma x + \delta t (2) \\
y^\prime & = y (3) \\
z^\prime & = z (4)
\end{aligned}
$$
from (2), it gives:
$$
\begin{aligned}
{dx \over dt} = -{ \delta \over \gamma} = v (5)
\end{aligned}
$$
from (1) and (2), they give:
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\delta \over \xi} = - v (6)
\end{aligned}
$$
so,
$$
\begin{aligned}
\xi = \gamma
\end{aligned}
$$
and,
$$
\begin{aligned}
\delta = - \gamma v
\end{aligned}
$$
let
$$
\begin{aligned}
\zeta = - \eta \gamma
\end{aligned}
$$
we have:
$$
\begin{aligned}
t^\prime & = \gamma ( t - \eta x ) (7) \\
x^\prime & = \gamma ( x - v t) (8) \\
y^\prime & = y (9) \\
z^\prime & = z (10)
\end{aligned}
$$

...
we can get
$$
\begin{aligned}
\eta = v / c^2
\end{aligned}
$$
...
'
My question lies in formula (6), it can give
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\delta \over \xi} = - v
\end{aligned}
$$
as the above, but it can also give
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\gamma \over \zeta}
\end{aligned}
$$
such that:
$$
\begin{aligned}
{dx^\prime \over dt^\prime} = {\gamma \over \zeta} = - {{1} \over {\eta}} = - {c^2 \over v} \neq -v (??)
\end{aligned}
$$
So why??

You need to ask some questions about what is happening here. First, $x, t$ and $x', t'$ are two coordinate systems, related by a transformation rule.

You cannot, in general, then derive an expression for $\frac{dx}{dt}$. That makes no sense.

Instead, what you can do is consider the coordinates of the origin of the primed frame over time. I.e. you consider the path $(t', x' = 0)$ in the primed frame and look at the coordinates of that path in the unprimed frame. This leads to:
$$\frac{dx}{dt} = - \frac{\delta}{\gamma}$$
Where $(t, x)$ is the specific coordinates of the origin of the primed frame.

 

[thaiqi]

You don't provide full details, but I presume what you've done is inverted (1) and (2) to get $x(x',t')$ and $t(x',t')$. Then when you take the differentials in these two equations you are considering $\left.\frac{dx'}{dt'}\right|_{x=\mathrm{const}}$ and $\left.\frac{dx'}{dt'}\right|_{t=\mathrm{const}}$. So yes, they are different things.

You don't want to consider holding a time coordinate constant when you are studying velocity. You are considering the velocity of an object at rest in one frame or the other, so you want to vary the time and space coordinate in one frame while you hold the spatial coordinate in the other frame constant, as @anuttarasammyak says.

I think you stated it very clearly. Thanks for you all.

 

[vanhees71]

What's behind the derivation of (5) is that you define the velocity of the origin $x'=0$ of the system $\Sigma'$ to be $v$, i.e., you set $x'=0$ in (2) and then get (5).

What's behind the derivation of (6) is that the origin $x=0$ of the system $\Sigma$ should move with velocity $-v$ in $\Sigma'$. For $x=0$ in (1) and (2) get uniquely (!)
$$\frac{\mathrm{d} x'}{\mathrm{d} t'}=\frac{\mathrm{d} x'}{\mathrm{d}t} \frac{\mathrm{d} t}{\mathrm{d} t'}=\frac{\delta}{\xi}=-v.$$
This is thus uniquely (!) (6). Your conclusions from (5) and (6) are correct, i.e., you indeed can reparametrize the transformation equations as you wrote to
$$t'=\gamma(x-\eta x), \quad x'=\gamma(x-v t).$$
Now you have to use that the light cone is invariant in the sense that from $x=c t$ you must get $x'=c t'$. So setting $x=c t$ you get
$$t'=\gamma t (1-c\eta), \quad x'=\gamma t (c-v).$$
Now it follows
$$\frac{x'}{t'} \stackrel{!}{=}c=\frac{c-v}{(1-c\eta)}=c \frac{1-\beta}{1-c\eta},$$
where $\beta=v/c$. From this you get $\eta=\beta/c=v/c^2$.

Now the transformations read
$$t'=\gamma \left (t-\frac{v}{c^2} x \right), \quad x'=\gamma (x-v t).$$
It's also clear that $\gamma=\gamma(v)$, i.e., the relative velocity of the reference frames must be a unique function of $v$, and you need an additional assumption to derive this function.

One plausible assumption is that the transformations build a group, i.e., if I "boost" from one inertial frame $\Sigma$ to an inertial frame $\Sigma'$ with relative velocity $v_1$ wrt. $\Sigma$ and then from $\Sigma'$ to another inertial frame $\Sigma''$ with relative velocity $v_2$ wrt. $\Sigma'$, one should get again such a boost transformation which directly transforms $\Sigma$ to $\Sigma''$ with some relative velocity $V$ wrt. $\Sigma$. Doing this (most conveniently with matrices) leads to
$$\gamma(V)=\gamma(v_1) \gamma(v_2) (1+\beta_1 \beta_2), \quad V \gamma(V)=(v_1+v_2) \gamma_1 \gamma_2.$$
One important result is the addition theorem for velocities:
$$V=\frac{v_1+v_2}{1+\beta_1 \beta_2}.$$
Another point is that the transformation must be the identity transformation for $v=0$. Further from the already assumed reciprocity theorem, i.e., that the inverse transformation of the boost with $v$ is the transformation with $-v$. This leads to
$$\gamma(-v) \gamma(v) (1-\beta^2)=1. \qquad (*)$$
Further the two directions along the $x$ axis are symmetric (or if you consider the entire 3D space, and you also want to have a Euclidean geometry for any inertial observer the space is isotropic, and especially this shows again that the $x$ and the $-x$ direction are symmetric).

Now take a clock at rest in $\Sigma=\Sigma(v=0)$, e.g., at $x=0$. Then the observer in $\Sigma'=\Sigma(v)$ sees an interval $\Delta t$ wrt. $\Sigma$ as a time interval $\Delta t'=\gamma(v) \Delta t$. An observer in the frame $\tilde{\Sigma}=\Sigma(-v)$ concludes $\Delta \tilde{t}=\gamma(-v) \Delta t$. Now since the situation for the boosts with $v$ and $-v$ are symmetric, we conclude that $\Delta t'=\Delta \tilde{t}$ and thus $\gamma(v)=\gamma(-v)$, and from that using (*)
$$\gamma^2(v)=\frac{1}{1-\beta^2},$$
which first implies that $\beta^2<1$, i.e., $|v|/c<1$, i.e., no inertial frame can have a relative velocity against any other inertial frame that's greater than $c$, i.e., the speed of light in the vacuum is the ultimate speed limit for any relative velocities of inertial reference frames.

Finally we have to decide about the sign of $\gamma(v)$. Here we want to consider orthochronous transformations, i.e., the forward lightcone $x=c t$ as measured in $\Sigma$ should be mapped to the forward lightcone in $\Sigma'$, i.e., $x'=+c t'. This implies$\gamma(v) >0$for all$v## and thus finally
$$\gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

 

[Sagittarius A-Star]

Now the transformations read
$$t'=\gamma \left (t-\frac{v}{c^2} x \right), \quad x'=\gamma (x-v t).$$

In both equations you can set ...
x := v₁ * t and v := -v₂, and then get easier
V = x'/t' by deviding the second equation by the first one. The result:

One important result is the addition theorem for velocities:
$$V=\frac{v_1+v_2}{1+\beta_1 \beta_2}.$$