Re: Symmetry in Lorentz transformation equations

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Sarma,

Regarding:

> But does this not imply that y<----->t, y'<----->t' and z<----->t,
> and z'<----->t'.
>
> What is the problem?

One of the problems is that you cannot do all of these transformation
replacements simultaneously.  You can do at most one of them (since
there is only one time dimension to exchange the spatial dimensions with).
This means the relationship between the two untouched spatial dimensions
and the 2 exchanged space and time dimensions gets all messed up.  Also
the world lines of massive particles do not respect this "symmetry", and
such a transformation strongly violates causality.  The problem is that
time is simply *not* space--even if it *sometimes* transforms in a
symmetric way with one particular spatial coordinate.

For simplicity of illustration, consider a point particle at rest at the
spatial origin of some coordinate system describing an inertial frame
(and assume that the mass is so small or that G is so small that any
curvature effects of spacetime due to the presence of the particle and
any other gravitational sources is entirely negligible, so that SR is a
good description of nature, and GR does not have to be considered).  Also
assume that the particle was created in some subatomic interaction
at the origin of the time axis.  This particle has a world line which
starts at the origin of Minkowski spacetime and goes along the positive
t-axis as it ages.  Now pick some spatial direction, say the y-axis, to
exchange with the t-axis in spacetime.  Once the exchange is made the
world line of the particle is now along the + y-axis.  The particle at
rest turns into an infinitely fast moving tachyon whose world line is
now *spacelike* along the y-axis and the particle exists everywhere
along the + y-axis simultaneously at the moment t=0.  Both before and
after t=0 the particle has no existence.  Clearly it is not a symmetry of
nature to make this switch. Particles at rest are not the same as
infinitely fast tachyons.

> I think that Lorentz transformation is not
> valid for accelerated frames. Since presence of mass is equivalent
> to accelerated frame, I naturally think that Lorentz transformation
> and its consequences do not apply.

It is true that Lorentz transformations are not used to transform
between accelerated frames.  More complicated transformations do that.
However the problems with the proported x <--> t exchange transformation
show up in the absence of any gravitational effects and the absense
of any accelerated frame as well.  Simple flat inertial frames suffice
to disallow the exchange as a symmetry of nature.

The reason that particles and fields of finite mass cannot exist in a
universe with this symmetry is illustrated in the above example.  Making
the transformation will reciprocate the speed (about c=1) of any motion.
Only (null v = 1) motions that are associated with a massless object can
withstand the exchange and remain a massless (null) motion.  Suppose
we have a noninteracting (quantum) field component A for some particle
of mass m.  Then the relativistic wave equation obeyed by this field
component is:

A_xx + A_yy + A_zz - A_tt = A*m^2 in units where c = h-bar = 1

and the subscripts (..)_xx, etc. indicate second partial derivatives
w.r.t. the subscript parameter.  If we exchange a spatial coordinate,
say, y with t this equation becomes:

A_xx - A_yy + A_zz + A_tt = A*m^2.

Clearly this is a completely different equation and has differently
behaving solutions.

*But* suppose we restrict ourselves to a 1-dimensional (in space)
universe *and* require that m==0 so that any fields present are
massless.  Then the relativistic wave equation boils down to:

A_xx - A_tt = 0 or equivalently, A_xx = A_tt.


Clearly this last equation *is* symmetric under the exchange x <--> t.
If the boundary conditions in space and time are also symmetric under
this exchange then the solution of the equation will be also.  So, if we
have a universe that has only *one* spatial dimension, *and* has only
massless fields and particles, *and* has boundary conditions on the left
and right ends of the spatial line the same as at the beginning and the
end of time, *then* the exchange x <--> t would be a symmetry with
x and t playing equivalent roles.

However, it just so happens that we do not live in such a universe.

David Bowman
dbowman@georgetowncollege.edu