The Lorentz transformation is a mapping of the $x', y',
z',$ and $t'$ coordinates relative to $x, y, z, t$ that preserves the speed of
light:
(1) $x^2 + y^2 + z^2 – c^2t^2 = x'^2 + y'^2 + z'^2 – c^2t'^2 = 0$
If $K'$ is moving with speed $v$ in the $x$ direction relative to $K$ (so that $K$ is moving with speed $-v$ in the $x$ direction relative to $K'$), and we use the Galilean transformation, we get an erroneous
(2) $x'^2 + y'^2 + z'^2 – c^2t'^2 = -2vxt + v^2t^2$
which is NOT equal to zero, and hence does not describe a sphere of
radius $ct$ emanating from the origin of $K'$, but something else (I'm
pretty sure it's a sphere that's been shifted negatively in the $x$
direction by a small amount, relative to $K'$). The Lorentz
transformation says we modify both the $x' = x – vt$ and the $t' = t$
relationships in order to ensure that we end up getting 0.
However, as a naive mathematician, I look at equation (2) and think,
"Ok, so we've got to modify the rule for $x'$, or $t'$, or both, to get
rid of that nasty $-2xvt + v^2t^2$ term in the $x'$ rule. Why not just
set $t'$ up so that instead of coming out to $-c^2t^2$, it actually comes
out to $-c^2t^2 + 2vxt – v^2t^2$, so that the extra terms cancel out?" I
naively went ahead with this idea and I got the following coordinate
transformation:
$$
x' = x – vt\\
y' = y\\
z' = z\\
t' = \frac{1}{c} \sqrt{(v^2 + c^2)t^2 – 2vxt}
$$
If you plug this into (1), you get 0 like you're supposed to.
Now, I haven't been able to find an explanation as to why this transformation is unsatisfactory. Of course, it's no longer linear – but is there some basic reason why the transformation must be linear? Obviously this transformation satisfies Einstein's postulate about the constancy of the speed of light for all inertial observers.
Presumably, then, it must violate the first postulate, regarding the equivalence of all physical phenomena for inertial observers – otherwise we'd need more information in order to derive the Lorentz transformation, which I think is supposed to be derivable from the two postulates alone.
So, in the end, I have one question I'm trying to answer:
- Does this transformation violate equivalence for inertial observers? Why? (ideally, someone could give me an example of a force that suddenly appears in my coordinate system that didn't already exist in the other coordinate system)
Best Answer
Flat spacetime in special relativity is the same everywhere, so we must have translational symmetry in time and space. If we start with two events separated by $\Delta x^\mu$ in one frame, the separation $\Delta x'^\mu$ in a boosted frame should only depend on $\Delta x^\mu$, and not on $x^\mu$ itself. This is only satisfied for linear transformations.
Using your transformations, observers would be able to identify a privileged point. This violates the principle of relativity even for observers at rest relative to each other, since it depends on where they put their origins.