Lorentz Transformation Derivation Question

[jk22]

I wanted to make a derivation of the Lorentz transformation :

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

The conservation of the quadratic form $c^2t'^2-x'^2$ yields the equations:

$$A^2-B^2/c^2=1\\D^2-E^2/c^2=-1/c^2\\AD=BE/c^2$$

Hence $B=c\sqrt{E^2-1}$,$D=\sqrt{E^2-1}/c$,$A=\pm E$.

The speed of the origin seen from primed frame shall be $-v$ hence :

$$\frac{dx'}{dt'}|_{x=0}=-v$$

That's where my question comes : how should the differential be computed ? Is it not that now B is a function of v and t with $B'(v,t)=\frac{\partial B}{\partial v}$ and $\dot{B}=\frac{\partial B}{\partial t}$ leading to

$$dx'=d(Bt)=B'tdv+\dot{B}tdt+Bdt$$

??

 

[strangerep]

That's where my question comes : how should the differential be computed ? Is it not that now B is a function of v and t [...]

If B were a function of t, your original transformations would be nonlinear, whereas you're assuming linear transformations upfront in this "derivation".

Also, since you're (presumably) transforming from one inertial observer's frame to another (i.e., where the original and new observer frames experience 0 acceleration) you can treat v as a constant when computing the differentials.

[Aside/Rant: deriving the LT by assuming the invariant interval is just plain cheating.]

HTH.

 

[jk22]

Well at the end I found :

$$E(v,t)=\frac{-Lv/t+\sqrt{L^2v^2/t^2+(c^2+L^2/t^2)(c^2-v^2)}}{c^2-v^2}$$

L is a constant of integration, a length.

We recover the usual Lorentz factor in the limit $t\rightarrow\infty$ and in other cases the limit $v\rightarrow c$ seems to exist.

But what does t mean ?

 

[vanhees71]

There's some misunderstanding though your ansatz is correct. Due to the special principle of relativity your coefficients $A$, $B$, $D$, and $A$ must not depend on $t$ and/or $x$, because the transformation must be linear. So they can only be functions of the velocity $v$ between the frames.

The final step is indeed to get $v$ into the game. You have to express that frame $\Sigma'$ moves with velocity $v$ (in $x$-direction) wrt. frame $\Sigma$, i.e., the origin of $\Sigma'$, which is described by $x'=0$ should lead to the trajectory in terms of the coordinates in $\Sigma$, $x=v t$, which in your coordinates reads
$$x'=0 \; \Rightarrow \; A x + B t=0 \; \Rightarrow \; x=-\frac{B}{A} t \; \Rightarrow \; v=-\frac{B}{A}.$$
That should lead to the correct result for the Lorentz boost.

 

[jk22]

So the coefficients can depend only on $v$ but not $t$.

Btw how do we know that they do not depend on the acceleration ? Are there experiments about this ?

 

[PeterDonis]

how do we know that they do not depend on the acceleration ?

What acceleration?

 

[Gaussian97]

[Aside/Rant: deriving the LT by assuming the invariant interval is just plain cheating.]

Not really, you can prove the invariance of the interval only by symmetry arguments, without any Lorentz transformation.

 

[vanhees71]

Or you can simply say the form invariance of the spacetime interval expressed in vector components wrt. different inertial (Minkowski) coordinates defines the corresponding transformations, leading to the Lorentz transformations.

It's a very similar argument that leads to the definition of rotations in Euclidean space as those transformations which leave the Euclidean distance invariant, when expressed in terms of Cartesian components. Another way to put it is that rotations map a right-handed Cartesian basis to another right-handed Cartesian basis.

 

[Ibix]

Btw how do we know that they do not depend on the acceleratio

Acceleration of what? The Lorentz transformations relate the behaviour of one set of inertial (i.e. non-accelerating) clocks and rulers to another such set. So nothing is accelerating.

You certainly can derive relations for accelerating clocks and rulers, but these are not the Lorentz transforms.

 

[Nugatory]

Btw how do we know that they do not depend on the acceleration ? Are there experiments about this ?

Don’t lose sight of the problem statement... you started with “derive the transformations between the coordinates used in two inertial frames” so you’re assuming no acceleration.