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Pressure in the Vicinity of a Lunar Astronaut Space Suit due to Outgassing of Coolant Water

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Fields, an Heuristic Approach

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The Transcendentality of

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Maxwell's Equations: The Vector and Scalar Potentials

A Possible Scalar Term Describing Energy Density in the Gravitational Field

A Proposed Relativistic, Thermodynamic Four-Vector

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Calculating the Energy from Sunlight over a 12 hour period
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Field Equations and Equations of Motion (General Relativity)
The observer in modern physics
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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' = y(x - vt)
y' = y
z' = z
t' = y(t - vx/c2)

where y = 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

upside down delta . V + differentialS/differentialt = 0

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

differential/differentialx, differential/differentialy, differential/differentialz, and differential/differentialt

from K to K'. We first use the chain rule to write

differential/differentialx = (differentialx'/differentialx)differential/differentialx' + (differentialt'/differentialx)differential/differentialt'
differential/differentialy = differential/differentialy'
differential/differentialz = differential/differentialz'
differential/differentialt = (differentialt'/differentialt)differential/differentialt' + (differentialx'/differentialt)differential/differentialt'

We next use eq. 1 to show that

differentialx'/differentialx = y
differentialt'/differentialx = -yv/c2
differentialt'/differentialt = y
differentialx'/differentialt = -yv

We finally substitute eq. 5 into eq. 4, to obtain

differential/differentialx = y(differential/differentialx' - (v/c2)differential/t')
differential/differentialy = differential/differentialy'
differential/differentialz = differential/differentialz'
differential/differentialt = y(differential/differentialt' - vdifferential/differentialx')

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

upside down delta . V + differentialS/differentialt = 0
or: differentialVx/differentialx + differentialVy/differentialy + differentialVz/differentialz + differentialS/differentialt = 0

We substitute, into eq. 2a, the transformed operators from eq. 6

y(differentialVx/differentialx' - (v/c2)differentialVx/t')
+ differentialVy/differentialy' + differentialVz/differentialz' + y(differentialS/differentialt' - differentialvS/differentialx') = 0

and rearrange terms

differential{y(Vx - vS)}/differentialx' + differentialVy/differentialy' + differentialVz/differentialz'
+ differential{y(S - vVx/c2)}/differentialt' = 0

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

differentialV'x'/differentialx' + differentialV'y'/differentialy' + differentialV'z'/differentialz' + differentialS'/differentialt' = 0

and, (comparing eq. 8 with eq. 9)

V'x' = y(Vx - vS)
V'y' = Vy
V'z' = Vz
S' = y(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 upside down deltaT

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

a upside down delta2T = differentialT/differentialt

where a is the thermal diffusivity of the medium. Taking the divergence of eq. 11, and using eq. 12, we obtain

upside down delta. q = -k upside down delta2T = -(k/a) differentialT/differentialt = -differential[(k/a)T]/differentialt

from which, immediately, follows a continuity equation

upside down delta . q + differential[(k/a)T]/differentialt = 0

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' = y(qx - v(k/a)T)
qy' = qy
qz' = qz
[(k/a)T]' = y((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 upside down deltaT = 0, and, therefore, qx = 0, everywhere along the rod in K. Equations 15 then simplify to

qx' = - yv(k/a)T
[(k/a)T]' = y(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 yv(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 rho be the static charge density in some frame of reference. Then for the frames K and K' being considered above,

jx' = y(jx - vrho)
jy' = jy
jz' = jz
rho' = y(rho - (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, rho, and carry no electric current, so that jx = 0. Equations 17 then simplify to

jx' = - yvrho
rho' = yrho

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, rho, 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, upside down deltaT, to drive the heat flux along the rod, so the observer in K' must infer an electric potential gradient, -upside down deltaflux, 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

if it is seen approaching by the observer in K', and

[(1 - v/c)/(1 + v/c)]1/2

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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