Chapter 23
A Particle Solution—The Inertia of
Energy
We will now turn to the solution q = q
I
(x, t), our kink solution (111), that according
to (98a), (91) and (99) gave us our natural, stationary measuring-rods L o and also our
moving measuring rods L
and the including, all so important Lorentz contraction
(112). Here the following questions will be considered: Do these internal measuringrods, on which the internal observers base their experiences, mechanically behave
as we outside observers expect our outside measuring-rods to behave? Are we in the
position of attributing these measuring-rods as a whole a single mass and a single
velocity? Is an internal observer’s single measuring-rod L
a single object in the sense
of Newtonian mechanics? In order to answer these questions we need to calculate
the tensor (290) for function (111) and to control whether we receive a mathematical
expression that fulfils (279). We will now calculate this:
q
I
(x, t) =
2a
π
arctan exp
π(x − vt)
L o γ
, γ =
1 −
v 2
c 2
o
.
Using
cos α =
1
√
1 + tan 2 α
, cos(4α) − 1 = −8
tan
2
α
(1 + tan 2 α) 2 ,
the following is derived,
cos
2π
a
q
I
− 1 = cos
4 arctan exp
π(x − vt)
L o γ
− 1 =
© The Editor(s) (if applicable) and The Author(s), under exclusive
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H. Günther, Elementary Approach to Special Relativity,
https://doi.org/10.1007/978-981-15-3168-2_23
247