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A Proposed Relativistic, Thermodynamic Four-Vector
It
is customary to begin a discussion, in special relativity, by first
choosing two Cartesian frames of reference, K and K', which are oriented
with their three spatial axes coinciding. These frames are then put
into relative, uniform translation, with velocity, v, in the x direction.
For two such frames of reference in relative motion, with K' moving
in the positive x direction relative to K, the Lorentz transformations
take the familiar form:
x'
= (x
- vt) |
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y'
= y
z' = z
t' = (t
- vx/c2)
where
= 1/(1 - v2/c2)1/2, and c = 3 X 108
m/sec is the speed of light, a constant in all frames of reference.
It is also customary to define any set of four quantities, which transform
according to eq. 1, as components of a relativistic
four-vector. Familiar examples of such four-vectors include the velocity
four-vector, the energy-momentum four-vector, and the current density
four-vector.
In
the calculation that follows, it will be shown that any vector, V
= (Vx, Vy, Vz), and scalar, S, which
are related through a continuity equation, i.e., an equation of the
form
in
all frames of reference, transform according to eq. 1,
and, therefore, comprise components of a four-vector; i.e., that V
and S satisfying eq. 2 is a sufficient condition for
their Lorentz-transformability. It will also be proposed that certain
thermodynamic quantities, which are shown to be related through a continuity
equation, be investigated as components of a relativistic, thermodynamic
four-vector.
We
begin by using eq. 1 to transform the differential
operators
from
K to K'. We first use the chain rule to write
/x
= (x'/x)/x'
+ (t'/x)/t'
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/y
= /y'
/z
= /z'
/t
= (t'/t)/t'
+ (x'/t)/t'
We
next use eq. 1 to show that
t'/x
= -v/c2
t'/t
=
x'/t
= -v
We
finally substitute eq. 5 into eq. 4,
to obtain
/y
= /y'
/z
= /z'
/t
= (/t'
- v/x')
Let
us now consider a set of quantities comprising a vector, V, and
a scalar, S, in the frame of reference K, which are related through
a continuity equation; i.e., for which
or:
Vx/x
+ Vy/y
+ Vz/z
+ S/t
= 0 |
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We
substitute, into eq. 2a, the transformed operators
from eq. 6
+
Vy/y'
+ Vz/z'
+ (S/t'
- vS/x')
= 0 |
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and
rearrange terms
{(Vx
- vS)}/x'
+ Vy/y'
+ Vz/z'
If
we now require that eq. 8 have the same form in K'
as in K; i.e., that eq. 8 be a continuity equation
in K', then, there must be a vector, V' = (V'x', V'y',
V'z'), and a scalar, S', in K', such that
V'x'/x'
+ V'y'/y'
+ V'z'/z'
+ S'/t'
= 0 |
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and,
(comparing eq. 8 with eq. 9)
V'x'
= (Vx
- vS) |
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V'y'
= Vy
V'z' = Vz
S' = (S
- vVx/c2)
Eq.
10 has a form identical to eq. 1, and we are forced
to conclude that the vector, V, and the scalar, S, together form
components of a relativistic four-vector. (Q. E. D.).
The
above argument tells us that relativistic four-vectors may be identified
from the continuity equations of physics. The remainder of this discussion
is devoted to a continuity equation whose terms may not have received
much attention as a four-vector; i.e., one involving thermal heat flux,
and thermal energy density.
The
heat conduction equation is
q
= -k T
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where
q = (qx, qy, qz) is heat flux
in cal/(m2 sec), k is the thermal conductivity of the medium,
and T is the absolute temperature. The thermal diffusion equation is
where
a is the thermal diffusivity of the medium. Taking the divergence of
eq. 11, and using eq. 12, we obtain
.
q = -k 2T
= -(k/a) T/t
= -[(k/a)T]/t |
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from
which, immediately, follows a continuity equation
The
quantity (k/a)T is, dimensionally, a thermal energy density. Together,
the quantities, qx, qy, qz, and (k/a)T,
must form components of a relativistic four-vector. Thus, they must
satisfy the system
qx'
= (qx
- v(k/a)T) |
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qy'
= qy
qz' = qz
[(k/a)T]' = ((k/a)T
- (v/c2)qx)
To
date, the author and his colleagues have not seen this particular system
of equations in any of the literature on special relativity. Eq.
15 appears to provide interesting insights into the behavior of
thermodynamic systems, as seen by observers in different states of relative
uniform motion.
Consider
a one dimensional problem in which a long [infinite], solid
rod, a black body, is lying along the x-axis, at rest in the frame of
reference K. Let the rod have a uniform temperature, so that T
= 0, and, therefore, qx = 0, everywhere along the rod in
K. Equations 15 then simplify to
qx'
= - v(k/a)T
|
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[(k/a)T]'
= (k/a)T
The
first equation in 16 suggests that an observer in
a frame of reference, K', in relative motion to the observer in K, will
measure a non-zero heat flux, qx', of magnitude v(k/a)T.
He will also observe that the thermal energy flow is in the same direction
as the motion of the rod in his frame of reference. This result might
be understood by invoking a Doppler shift in the thermal radiation emitted
by the rod. The portion of the rod which is approaching him (i.e., which
lies in the positive x' direction) will be blue shifted relative to
the portion of the rod receding from him. Since, for a black body, the
peak in the observed radiation frequency is proportional to the temperature
at which the radiation is emitted (Wein's law), the observer must conclude
that the portion of the rod ahead of him is hotter than the portion
behind; i.e., that a thermal gradient exists along the rod, driving
the heat flux which he observes.
Alternately,
let two points, A and B, be marked on the rod such that the observer
in K is situated midway between them. Let clocks, and thermometers,
be placed at each of the points, and let the clocks be synchronized
in K. Finally, let the observer in K' be seen, by the observer in K,
as moving toward point B, and away from point A. At a time t0,
when the observer in K' is adjacent to the observer in K, let both observers
record the temperatures at A and at B by observing the thermometers
placed at each of the points. They might do so by sending out light
pulses to illuminate the thermometers. In each frame of reference, the
light pulses must travel out to the thermometers and be reflected back
in order for the thermometers to be read. Following this procedure,
the observer in K measures equal temperatures. The observer in K', however,
does not. The observer in K' does not see the clocks at A and B as reading
the same. He sees the clock at B as reading an earlier time than the
clock at A. Since the rod is radiating, it is losing thermal energy
with time, and cooling. Observations made at different times must therefore
record different temperatures. More specifically, for observations made
at two different times, the earlier observation will record a higher
temperature than the later one. Again, qualitatively at least, we must
conclude that the observer in K' will record a higher temperature at
B than at A.
The
second equation in 16 suggests that the observer
in K' will detect a lower thermal energy density, [(k/a)T]',
than the observer in K. This result might be understood by invoking
the time dilatation. The thermal energy density is made up of a summation
of the individual random thermal oscillations of the molecules making
up the rod. These motions must exhibit a distribution of frequencies
which correlate with the frequencies of the observed thermal radiation.
In K', these frequencies must have a lower value than in K, and hence
must indicate a lower thermal energy density.
Additional
insight may be gained, at this point, by a comparison with the current
density four-vector. Let jx, jy, and jz
be components of the current density, and
be the static charge density in some frame of reference. Then for the
frames K and K' being considered above,
jx'
= (jx
- v)
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jy'
= jy
jz' = jz
'
= (
- (v/c2)jx)
As
previously, we consider a one dimensional problem in which a long [infinite]
wire is lying along the x-axis, at rest in the frame of reference K.
Let the wire have a uniform surface charge density, ,
and carry no electric current, so that jx = 0. Equations
17 then simplify to
jx'
= - v
|
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This
case is analogous to the thermodynamic case we have been discussing.
A uniform, cylindrical electric field exists everywhere along the wire
in K. This electric field corresponds to the radiation field of the
rod; the static charge density, ,
to the thermal energy density, (k/a)T; and the non-zero current density,
jx', to the non-zero heat flux, qx', in K'. (A
difference exists between the two cases, in that the observer in K'
will detect a magnetic field in the electric case, which has no counterpart
in the thermodynamic case.) Eq. 18 tells us that,
while the observer in K detects a uniform static charge distribution
everywhere along the wire, and no electric current, the observer in
K' detects a reduced static charge distribution, and a non-zero electric
current, moving in the same direction as the motion of the wire in his
frame of reference.
As,
in the thermodynamic case, the observer in K' required a thermal gradient,
T,
to drive the heat flux along the rod, so the observer in K' must infer
an electric potential gradient, -,
along the wire to drive the electric current. This gradient arises from
the nature of the charge distribution, along the wire, as seen by the
observer in K'. Recall that the charge distribution is uniform in K.
In K', it is relatively lower seen approaching than receding. If unit
charges are placed at regular intervals along the wire in K, (i.e.,
at x = 0, 1, 2, etc.), and the charge density is defined as the [average]
charge per unit length of wire, then, due to the finite propagation
speed of light, and taking into account the length contraction, the
separation between the regular intervals in K will appear larger, seen
approaching, and smaller, seen receding, by the observer in K'. In fact,
a unit length in K will appear as the length
[(1
+ v/c)/(1 - v/c)]1/2 |
19. |
if
it is seen approaching by the observer in K', and
[(1
- v/c)/(1 + v/c)]1/2 |
20. |
if
it is seen receding17. The linear charge
densities will appear, respectively, lower, if seen approaching, and
smaller, if seen receding, in K'. The observer in K' will infer, from
this difference in charge densities, the required electric potential
gradient.
These
results are preliminary. Perhaps they will shed some new light on the
relativistic four-vector in general, and its possible application to
thermodynamics in particular.
17Consider
the case of a unit "train" approaching the observer at velocity, v.
Its Lorentz contracted length must be
1/=
(1 - v2/c2)1/2
Imagine
that an observer uses a camera with a rapid shutter to photograph the
approaching train. A wave front, which left the front of the train at
the time t, and another, which left the rear of the train at a time
1/[(c-v)]
earlier, arrive simultaneously at the film plane, and are recorded.
Thus, the image of the approaching unit train appears, not contracted,
but stretched out. In fact, the train appears to have length
v/[(c-v)]
+ 1/
= {(1 + v/c)/(1 - v/c)}1/2 > 1.
A similar
argument may be made for the receding train, with the result that the
observed length is now
{(1 - v/c)/(1 + v/c)}1/2 < 1/.
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