Tom Roberts wrote:
I think that playing with the Lorentz transformation (LT) and the relativistic velocity composition equation (RVCE) are futile, because
these are relationships among DIFFERENT COORDINATE SYSTEMS. But nature clearly uses no coordinates whatsoever, so coordinate relationships are irrelevant at the fundamental level [#]. That's why all modern physical theories are expressed in terms of tensors, which are naturally
independent of coordinates [@]. For the LT and RVCE the relevant tensor
is the 4-velocity of an object -- the LT and RVCE are formulated
specifically so when an object's 4-velocity is projected onto the
different inertial coordinate systems, the tensor itself remains
unchanged (aka invariant).
[#] Not to mention the artificial requirement of inertial
coordinates.
[@] There are other mathematical methods to ensure coordinate
independence. Tensors are the simplest, due to their
requirement of multi-linearity. Nonlinear approaches are MUCH
more complicated, and to date no need for them has been
demonstrated.
Specifically, for a given inertial coordinate system (x^i) and an object
with 4-velocity U, the components of U are:
U^i = dx^i/dtau. {i=0,1,2,3}
where tau is the proper time of the object, and d is
partial derivative.
We also have:
U = U^i d/dx^i d is still partial derivative
Note that dx^1, dx^2, and dx^3 need standard rulers for their
definition, and dx^0 needs standard clocks, all of which must be at rest
in the inertial frame. So it is impossible to define these relationships
for coordinates moving faster than c relative to any inertial frame. IOW
a tachyon cannot have a rest frame.
Hi Tom,
Thanks for reponding. Yes, tachyons always travel c < u < \infty
Essential exercise for the reader:
Given two inertial frames A and B, calculate the components
of a given tachyon's 4-velocity relative to each; check
whether the LT and RVCE hold for the tachyon. The essential
question is: what does the norm of its 4-velocity mean?
As A and B are inertial frames, the metric components for
each are diag(-1,1,1,1).
If you can't/won't do this exercise, you'll never really understand
tachyons.
Tom Roberts
I have done the 4-momentum work and found that there is a problem for
tachyons. Transformation yields P' = [γ(E/c - pv/c),γ(p - Ev/c^2)],
yielding the belief that tachyon energy becomes negative for certain
observers in other inertial frames. This occurs because |pc| > E for
tachyons, contrary to the case for bradyons.
The LT is the nuts and bolts of tensor transformations, so sometimes
it's necessary to check that they still work in a new domain.
A philosopher wrote:
"Civilization advances by extending the number of important
operations which we can perform without thinking of them."
― Alfred North Whitehead
But that can cause problems when your only tool is a hammer and the
new operation involves turning a screw. This is the situation with
tachyons: The 4-vector approach is Whitehead, thinking like Bilaniuk
et al. did in their Meta Relativity paper where they concluded that
negative energy meant the tachyon would be observed to reverse direction
and go back in time to boot (no one checked that the 3-momentum
component didn't reverse direction).
Tachyon energy is never observed to be negative, however, at the nuts
and bolts level:
For one thing, if E = mc^2/sqrt(u^2/c^2 - 1) can be considered as a law
of physics, then E' = mc^2/sqrt(u'^2/c^2 - 1) must also be a law of
physics by the Principle of Relativity. Looking at the transformation
of E to E':
E' = γmc^2 sqrt[(1 - uv/c^2)^2]/sqrt(u^2/c^2 - 1)
E' = γE[(1 - uv/c^2)^2]^(1/2)
we can see where the problem is: The 4-vector approach canceled the ^2
exponent with the ^(1/2) exponent and left the (1 - uv/c^2) term in the transformation. This didn't cause problems for bradyons and luxons, but
it does for tachyons when u > c^2/v. What is the square of a negative
number? Positive, of course. What is the square root of a positive
number? Well, it has two roots, one positive, one negative, but the
positive root is the principle one. Furthermore, the fact that
E' = mc^2/sqrt(u'^2/c^2 - 1)
NEVER goes negative over the range c < u' < \infty signifies that we
eschew the negative root.
Sorry for the amount written here, but it seems to me that there is a
lot of misinformation about tachyons that should be cleared up. All of
this is in DOI: 10.13189/ujpa.2023.170101. Have you read it?
Gary
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