I Linearity of the Lorentz transformations

kent davidge

There are several ways to show that the Lorentz transformations must be linear. What's the best/more intuitive argument in your opinion?

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Nugatory

Mentor
How would they behave under translations if they were non-linear?

kent davidge

How would they behave under translations if they were non-linear?
I dont know what you mean by this. The Lorentz transformations are by definition when you do not perfom translations. If you include them, you have the Poincaré transformations.

Pencilvester

Now, we are particularly interested in inertial coordinate systems, that is we like coordinates where free particles go in a straight line at constant velocity, following Newton's first law. Any coordinates where all free particles have straight lines as their worldlines are inertial coordinates, so if we want to study the transformations from one inertial frame to another inertial frame then we want to study transformations that map straight lines to other straight lines.

The simplest such transformation is a linear transformation, which is the form chosen in the derivation you cited. So the reason for choosing that generalization is that it is the simplest generalization that has the necessary property of mapping straight lines to straight lines.
This seems like an intuitive explanation to me posted by @Dale. This is quoted from this thread, post 24:

dextercioby

Homework Helper
There are several ways to show that the Lorentz transformations must be linear. What's the best/more intuitive argument in your opinion?
Well, first semi-argument (weak) is: moving to the limit v/c -> 0, one should recover the space-time linear Galilei transformations. Next semi-argument (a stronger one): assume they are quadratic, that is
X' = A(v) X^2. Now study the movement of a point P of the spherical wave of light issued from the source O from two different frames in motion with speed v with respect to each other and which are linked by this quadratic transformation. How do you show the speed of light is constant in these two frames and equal to c?

strangerep

There are several ways to show that the Lorentz transformations must be linear. [...]
It's not true to say that they "must" be linear.

See my post #26 in this thread for references.

Summary:

The most general transformation of Lorentz-like velocity-boosting type is a fractional-linear transformation (since they preserve the property of zero acceleration, i.e., map inertial systems among themselves) and also satisfy the following conditions:

1) The transformations must be well-defined on (at least) an open neighbourhood of the spacetime origin.

2) The coordinate origin is preserved. I.e., the original and boosted observers' spacetime origins coincide.

3) Satisfy spatial isotropy. (Strictly speaking, this is not a separate assumption, since preservation of the zero-acceleration equation of motion $d^2 x^i/dt^2 = 0$ already implies spatial isotropy.)

4) Boosts along any given (fixed) spatial direction form a 1-parameter Lie group, with parameter $v$ denoted velocity. If the original velocity (at the spacetime origin) is 0, then the boosted velocity (still at the spacetime origin) is $-v$. (The choice of $-v$ means the transformation represents an active boost of an observer from $0$ to $v$, which corresponds to a passive boost of the coordinates from $0$ to $-v$.)

Then, transformations belonging to a 1-parameter Lie group must commute. This allows the form of the transformations to be simplified, and reveals the existence of a constant with dimensions of inverse velocity squared.

5) Since neither the original observer, nor the boosted observer are in any way special, there should exist an inverse transformation. Analysis of the equations reveals that the parameter for the inverse transformation is $-v$. (I.e., this does not need to be assumed, contrary to statements in many textbooks).

6) The group transitivity property requires that successive boost transformations with parameters $v$ and $v'$ must be equivalent to a single boost with parameter $v'' = v''(v,v') = v''(v',v)$. Analysis of this requirement reveals the existence of another constant with dimensions of inverse time, which I'll call "$H$".

The final form of these generalized Lorentz transformations are given in the Kerner and Manida references in my post linked above. (My "$H$" constant corresponds to Manida's "$R$", which seems to be identifiable as a Hubble-like constant.) The velocity-squared constant satisfies the familiar properties of $c^2$.

Taking a limit as $H\to 0$ recovers the usual linear form of the Lorentz transformations.

"Linearity of the Lorentz transformations"

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