Deriving Lorentz Transformations: Hyperbolic Functions

[Samama Fahim]

Huh? What do you mean by this? How is it connected to what I said which was a statement regarding the relation between the invariance of the spacetime interval amd the invariance of the speed of light?

It's not.

 

[vanhees71]

That's true. In the usual way of deriving the LT, going back to Einstein's famous paper of 1905, is to use his "two postulates", which is (a) the special principle of relativity, i.e., the existence and indistinguishability of inertial reference frames and (b) the independence of the speed of light from the relative velocity between the light source and any inertial observer. In addition one tacitly assumes that for any inertial observer space is a 3D Euclidean affine space (implying its symmetries, i.e., translation and rotation symmetry) and the homogeneity of time (translation invariance in time).

First from the special principle of relativity we find that a free particle is moving with constant velocity with respect to any inertial frame of reference, and this implies that the transformation between two reference frames must be linear.

Now we consider the special case that we keep the directions of the spatial Cartesian bases the same for both frames of reference and consider only boosts in $x$ direction, i.e., then
$$c t'=A c t + B x, \quad x'=C ct + D x, \quad y'=y, \quad z'=z. \qquad (1)$$
We have assumed without loss of generality that the origins of space and time in both frames are chosen to be the same (if not, you can just redefine the coordinates by a time or space translation, which doesn't change anything because of the assumed homogeneity of time and space).

Now the 2nd postulate tells us that the wave front of a spherical em. wave switched on at $t=0$ from a source located at $\vec{x}=0$ obeys
$$c^2 t^2-\vec{x}^2=0, \qquad (2)$$
and the same must hold in $\Sigma'$, i.e., from (2) it necessarily follows also
$$c^2 t^{\prime 2}-\vec{x}^{\prime 2}=0,$$
i.e., there must be some factor $\alpha$ such that
$$c^2 t^2 - \vec{x}^2=\alpha (c^2t^{\prime 2}-\vec{x}^{\prime 2}).$$
Plugging in (1) you find
$$c^2 t^2 - \vec{x}^2=\alpha [(A c t + B x)^2 - (C ct + D x)^2-y^2-z^2]. \qquad (3)$$
Since this must hold for all $\vec{x}$ comparing the coefficients of $y^2$ and $z^2$ on both sides of the equation, it follows $\alpha=1$. So (3) reads with $\alpha=1$
$$c^2 t^2 -x^2 = (A^2-C^2) c^2 t^2 + 2 (AB-CD) ct x + (B^2-D^2) x^2.$$
Since this must be true for all $(ct,x)$ you find
$$A^2-C^2=1, \quad AB-CD=0, \quad B^2 - D^2=-1.$$
The rest then follows as shown by @Orodruin in #25.

 

[throw]

On p. 18, Robinson writes:
" corresponds to This corresponds to a vector with only a temporal component and no spatial component."

Why is that?

He explains why on the page, referencing equation (1.74). If you apply (1.75) to a four-vector and use $\beta = 0$ (what this implies for $\sinh$ and $\cosh$ is on p.18 explicitly too) you can see this explicitly.

 

[Samama Fahim]

He explains why on the page, referencing equation (1.74). If you apply (1.75) to a four-vector and use $\beta = 0$ (what this implies for $\sinh$ and $\cosh$ is on p.18 explicitly too) you can see this explicitly.

When $\beta = 0$, the transformation matrix we have, using 1.78, is the identity matrix. Applying this matrix to a four vector should leave this vector unchanged. Shouldn't it? And if this 4-vector has non-zero spatial components, they should be left untouched as well.