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I am reading Kleppner.(Lorentz transformations) He said,we take the most general transformation relating the coordinates of a given event in the two systems to be of the form $$x'=Ax +Bt, y'=y, z'=z, t'=Cx +Dt,$$ and then he found out the constants considering four cases in which we know a priori how an event appears in the two systems.

but why the transformations are linear? He said ,a nonlinear transformation would predict acceleration in one system even if the velocity were constant in the other. But i think thats what happen when we consider Lorentz force(without electric field) if $v$(velocity of a charged particle) is zero in one inertial frame then there is no Lorentz force on the particle(hence the particle has no acceleration in that frame). but this may not be the case in other inertial frames(where the velocity of the charged particle is not zero)? whats wrong here?

Qmechanic
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Paul
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    A non-linear transformation leads to pseudo-forces if the coordinates are being interpreted as Euclidean coordinates. There is nothing wrong with analyzing physics in such coordinate systems, indeed generalized coordinates in Lagrangian and Hamiltonian mechanics are doing just that with enormous success. As for the second part of the question: a boost will not just change the velocity of the particle, but also transform the fields, so that a particle that is not accelerated in one system will also not be accelerated in another. – CuriousOne Dec 28 '14 at 06:28
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    Related: http://physics.stackexchange.com/q/12664/2451 and links therein. – Qmechanic Dec 28 '14 at 06:38

1 Answers1

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Linearity follows from:

  1. Translation invariance (where we mean translation in space and time): the image of a vector $\vec{AB}$ joining events $A$ and $B$ under a Lorentz transformation $\mathscr{L}:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$ is unaffected by the addition of any offset added to both ends $A$ and $B$;

  2. Continuity: The Lorentz transformation $\mathscr{L}:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$ is a continuous map.

To see how this plays out, our translation invariance axiom is encoded:

$$\mathscr{L}(X+Y)-\mathscr{L}(Y) = \mathscr{L}(X)-\mathscr{L}(0);\quad\forall\,X,\,Y\in\mathbb{R}^{1+3}\tag{1}$$

If we define $h:\mathbb{R}^{1+3}\to\mathbb{R}^{1+3}$ by $h(Z)=\mathscr{L}(Z)-\mathscr{L}(0)$ then it follows from (1) alone that:

$$h(X+Y)=h(X)+h(Y);\quad\forall\,X,\,Y\in\mathbb{R}^{1+3}\tag{2}$$

This is the famous Cauchy functional equation generalized to $3+1$ dimensions. For one, real dimension, the only continuous solution is $h(X)\propto X$; there are other solutions, but they are everywhere discontinuous, as shown in Section 1.5 of the Hewitt and Stromberg reference I give at the end. It is easy to broaden the Hewitt-Stromberg argument to any number of dimensions, so that, given our continuity postulate, we must have:

$$\mathscr{L}(X) = \Lambda\,X + \Delta\tag{3}$$

where $\Lambda$ is a linear operator - a $4\times4$ matrix and $\Delta$ a spacetime offset. Given again our translation invariance postulate, we can translate the image (3) to cancel the offset, whence we see that we can, without loss of generalness, take the Lorentz transformation to be linear and homogeneous:

$$\mathscr{L}(X) = \Lambda\,X\tag{4}$$

Reference

E. Hewitt & K. R. Stromberg, "Real and Abstract Analysis" (Graduate Texts in Mathematics), Springer-Verlag, Berlin, 1965. Chapter 1, section 5 constructs all solutions to the Cauchy equation $f:\mathbb{R}\to\mathbb{R};\,f(x+y)=f(x)+f(y)$. These are worth looking at, the discontinuous ones are truly weird and wonderful.

Selene Routley
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