Energy-momentum density and pressure relations for a relativistic ideal gas with a bulk motion
aa r X i v : . [ phy s i c s . g e n - ph ] J un Energy-momentum density and pressure relationsfor a relativistic ideal gas with a bulk motion
Ashok K. Singal
Astronomy and Astrophysics Division, Physical Research Laboratory, Navrangpura,Ahmedabad - 380 009, IndiaE-mail: [email protected]
18 June 2020
Abstract.
We derive here, from first principles, the energy-momentum densities ofa perfect fluid, in the form of an ideal molecular gas, in an inertial frame where thefluid possesses a bulk motion. We begin from the simple expressions for the energydensity and pressure of a perfect fluid in the rest frame of the fluid, where the fluidconstituents (gas molecules) may possess a random motion, but no bulk motion. Froma Lorentz transformation of the velocity vectors of molecules, moving along differentdirections in the rest frame of the fluid, we compute their energy-momentum vectorsand number densities in an inertial frame moving with respect to the rest frame ofthe liquid. From that we arrive at the energy-momentum density of the fluid in aframe where it has a bulk motion. This way we explicitly demonstrate how a couple ofcurious pressure-dependent terms make appearance in the energy-momentum densityof a perfect fluid having a bulk motion. In addition to an ideal molecular gas, wecompute the energy-momentum density for a photon gas also, which of course matcheswith the energy-momentum density expression obtained for a molecular gas havingultra-relativistic random motion.
1. Introduction
One of the simplest examples of a perfect fluid is an ideal gas, where the constituentmolecules of the gas are assumed to be structureless finite masses of negligible dimensionsthat may have head-on elastic collisions with each other as well as with the walls of thecontainer but have otherwise no interaction. Thus there are no side-ways forces, therebyno drag or viscosity as well as no sheer stresses and there is no conduction of heat. Sucha fluid is described completely by the energy density and pressure in an inertial frame,called the rest frame, where the pressure is isotropic, and the stress-energy tensor hasa simple, diagonal form. Perfect fluids are used in making relativistic models of theinterior of a star where an idealized distributions of matter may be assumed [1, 2].Perfect fluids are also employed in studying the evolution of isotropic universe modelsusing the general theory of relativity [1, 3]. nergy-momentum density and pressure in a relativistic ideal gas S , to another frame, say S , where a fluid element thus attains a bulkmotion, work gets done against pressure during an increasing Lorentz contraction of thefluid element. (ii) During the movement of a fluid element, work is continuously beingdone on the fluid element at its trailing end against pressure, while an equal and oppositework is being done by the fluid element at its leading end. Thus the energy gained bythe fluid element at its trailing end is being delivered by the fluid element at its leadingend. Even though there is no net gain or loss of energy, yet due to this, there is acontinuous flow of energy across the fluid element between its two ends, which results inan additional contribution to the momentum of the fluid element. This latter term wasin fact shown to have given rise to the intriguing, century old, 4/3 problem [5, 6, 7, 8] inthe electromagnetic momentum of a moving charge. Further, in the apparent paradoxof there being nil electromagnetic momentum in a moving electrically charged parallel-plate capacitor, carrying a finite electromagnetic field energy during its motion alongthe plate separation, added to the fact that the field energy paradoxically decreases withan increase in the system velocity, it was the contribution of these very terms to theenergy-momentum of the electromagnetic system that was shown to be responsible fora successful resolution of these paradoxes [4, 8, 9].Even though the work done during a Lorentz contraction as well as the momentumcontribution of differential work being done on the ends of a moving fluid elementexplained in an unambiguous manner how during the acceleration of the fluid to achievea bulk motion, the energy and momentum densities get contributions from the pressurewithin the fluid, it would nevertheless be desirable if one was able to derive thesepressure-dependent terms in the energy and momentum densities, from first principles,using Lorentz transformations, directly in the frame S itself, where the fluid has achieveda bulk motion. It is true that using the tensor notation leads to results in a rather quickand an easy manner, still a somewhat laborious exercise using velocity transformationsfrom one Lorentz frame to another and the thereby computation of energy-momentumdensities could be quite instructive.It seems that for this purpose one might require some specific models of the fluid. nergy-momentum density and pressure in a relativistic ideal gas
2. Energy-momentum density of a perfect fluid with a bulk motion
Let S be the rest frame of the perfect fluid, then the energy-momentum tensor in S isof a diagonal form [1, 4, 10, 11, 12] T ′ = ρ , T ′ = T ′ = T ′ = p , (1)with all non-diagonal terms zero, implying an energy density ρ and an isotropic pressure p , but a nil momentum density (by definition). It is important to note that ρ hereincludes not only the rest mass energy of the fluid molecules, but also their kineticenergy, which for instance, in the ideal gas model of the fluid, arises due to the randommotion of the gas molecules that also generates pressure in S .In the lab frame S , where the perfect fluid has a bulk motion v , say along the x-axis,with a Lorentz factor γ = (1 − v /c ) − / , the energy and momentum densities are givenby the components T and T = T of the energy-momentum tensor, [1, 4, 10, 11, 12]as ρ = T = γ ρ + p v c ! , (2) σ = T /c = γ ( ρ + p ) v c , (3)Here a perplexing question could be raised. Why does the energy density ρ in frame S depend upon pressure p (Eq. (2)), when the random motion of molecules in frame S is already accounted for by the kinetic energy, which is included in ρ . How comethe random motion enters twice in this expression, once through the kinetic energy ofthe random motion and the second time through the pressure term which again resultsfrom the random motion? In fact, one would intuitively expect ρ = γ ρ (the extra γ factor simply due to the Lorentz contraction of the liquid volume, as seen in frame S ).Equally intriguing is the presence of p in σ , the momentum density (Eq. (3)). With ρ /c being the mass density of the system in rest frame S of the fluid, one would haveexpected the momentum density to be simply σ = γ ρ v/c . Why does an additionalterm ( γ pv /c ), proportional to pressure, appear in the expression for σ ?
3. An ideal molecular gas as a perfect fluid
Let us assume an ideal gas, enclosed inside a container of volume V , having a uniformnumber density n of molecules, each of rest mass m and moving with a velocity v in a nergy-momentum density and pressure in a relativistic ideal gas S . Let γ = (1 − v /c ) − / be the Lorentzfactor of the random motion of molecules. The pressure for the gas is related to theenergy density as (Appendix A, Eq. (A.3)) p = n m γv ρ v c , (4)where ρ = n m c γ is the relativistic energy density of the gas that includes the restmass energy of molecules as well as the kinetic energy due to their random motion inframe S . If the motion of molecules in the gas is relativistic with v ∼ c , the pressure p would be comparable to the energy density ρ .As seen in the lab frame S , the container of fluid moves with a velocity v along thex-axis, with γ = (1 − v /c ) − / as the corresponding Lorentz factor. Now a moleculemoving along θ with respect to the x -axis, as seen in S , will be moving with a velocitycomponent, V x , parallel to the x-axis in the lab frame S , given by [12, 13, 14] V x = v x + v v x v /c , (5)Writing V x = V cos θ and v x = v cos θ , we can write Eq. (5), as V cos θ = v cos θ + v v v cos θ /c , (6)from which we get V cos θ − v = v cos θ γ (1 + v v cos θ /c ) . (7)Now even if the motion of molecules in the frame S has an isotropic distribution,it would not be so in the frame S (Fig. 2). This curious fact arises because all thosemolecules that may have velocity component towards right (that is along positive x direction) in frame S , will also move towards right in frame S too and with somewhatlarger magnitudes. In addition, all those molecules will also be moving towards right inthe frame S which were moving in frame S towards left but with velocity magnitudessmaller that v , the velocity of the frame S with respect to the frame S . One has toalso consider that the container walls, AB and CD , though at rest in frame S , are inmotion in S . Therefore, successive molecules bouncing off the wall CD , and movingtowards left after that, have larger separations, due to the meanwhile movement of thewall towards right, as compared to the separation between molecules that are bouncedoff towards right by the wall AB . As a result there may be, on the average, a largernumber of molecules moving towards right than those moving towards left, as seen inframe S . The relative densities for molecules moving at different angles θ with respectto the x-axis, can be determined in the following way.The time τ that a molecule moving with velocity V x in frame S would take betweensuccessive collision with two opposite container walls AB and CD , that are normal tothe x-axis and moving towards right with speed v (Fig. 2), is given by V x τ = lγ + v τ , (8) nergy-momentum density and pressure in a relativistic ideal gas Figure 1.
A box bound by walls, AB , CD , AC and BD (the depth along the z-axissuppressed in the diagram), containing ideal gas, is stationary in frame S . The gasconstituents may have a random motion with an isotropic distribution, and thus nobulk motion, in frame S . EF (dashed line) is a cross section of the container, at somearbitrary position along x direction. where length l of the arm AC of the container in frame S is Lorentz contracted in frame S , along the x direction, by a factor γ = (1 − v /c ) − / , and the wall CD moves adistance v τ towards right in time τ . Thus the time interval between successive collisionswith the container walls τ = lγ ( V x − v ) , (9)in frame S will be longer than τ = l/v x in frame S and the corresponding moleculewill be spending relatively more time after bouncing off the container wall AB by afactor, v x /γ ( V x − v ), before a change in V x takes place because of a collision with thecontainer wall CD . Therefore the number density of molecules moving with V x , as seenin frame S , will be relatively higher than those seen moving with v x in frame S , by afactor ττ = v x γ ( V x − v ) . (10)Therefore the number density in frame S will not be isotropic, even though in frame S the number densities of molecules with velocity components in all directions wereuniformly distributed. nergy-momentum density and pressure in a relativistic ideal gas Figure 2.
The box
ABDC , at test in frame S , is moving with velocity v alongthe x-axis with respect to the lab frame S . Even the cross section, EF (dashed line),of the container is moving with velocity v in frame S . Figure shows in frame S , theLorentz transformed velocities of the gas molecules, which had an isotropic distributionin frame S . With the help of Eq. (7), we get ττ = v cos θ γ ( V cos θ − v ) = γ v v cos θ c ! = D (say) , (11) D is akin to the Doppler factor δ for photons (see the next Section). Thus, the numberdensities of molecules within a given liquid element in frame S , will be D times that inframe S , which is different from the mere Lorentz contraction factor, γ , of the volumeof liquid element.Here we should point out that for molecules moving towards − x direction in theframe S , i.e., for molecules with θ > π/ V x < v and both τ and τ are negative.What really happens here is that instead of the time interval between ( n )th and ( n +1)thcollision, one gets in such cases the time interval between ( n )th and ( n − τ /τ still correctly gives D , the ratio of number densities in frames S and S for the corresponding molecules.Let Γ = (1 − V /c ) − / be the Lorentz factor of the motion of the molecule in thelab frame S . Then from a Lorentz transformation of the velocity 4-vector γ ( c, v x , v y , v z )in S to Γ( c, V x , V y , V z ) in S , we haveΓ = γ γ (cid:18) v v x c (cid:19) = γ D , (12) nergy-momentum density and pressure in a relativistic ideal gas S will be D times that in frame S .Here a question could be still baffling – While it could be readily understood thatthe energy of a molecule moving along θ in frame S , could be different in frame S fromthat in frame S (Eq. (12)), how come the number densities of molecules in differentdirections within a given liquid element vary in frame S , beyond the Lorentz contractionfactor, γ , of the volume of liquid element? After all they remain uniform in frame S ,not only in all directions but also at all times. For instance, we could select a crosssection, say EF normal to the x-axis (Fig. 1) within the fluid, then in frame S , withan equal number of molecules crossing EF from left to right as from right to left, wewill see a zero net number-flux through EF , i.e. n v π Z π cos θ π sin θ d θ = n v Z π cos θ sin θ d θ = 0 , (13)The same set of molecules, with a nil net number flux, as seen in S , should be seen inframe S too, crossing EF in equal numbers in opposite directions simultaneously. Howcould it be possible if the number density of molecules moving towards right is differentfrom that of the ones moving towards left?Actually the cross section EF , as seen in S , will be moving towards right with aspeed v (Fig. 2), and the molecules moving towards right, with number densities higherby factor D , will have to catch up with the receding away cross section EF , while formolecules moving towards left, EF also will be moving towards them, and in spite oftheir lesser densities, the number of molecules encountering EF head on and crossingit from right to left need not be at a proportionally lower rate. Also, even if a moleculelying on left of EF is moving towards right, but with a velocity V x < v , it would not beable to catch up with EF to cross it from left to right. However, if a similar molecule,with a velocity V x < v , lies on the right side of EF , then it could be overtaken by EF moving towards right with velocity v , and such a molecule will be seen to pass throughcross-section EF from right to left, even though its velocity in S is towards right. As aresult of these effects, the total number of molecules crossing EF in opposite directionscould be still equal, even in S .To verify this explicitly, we calculate the net number of molecules, N , crossing perunit of time, through a moving cross section EF that is moving towards right withspeed v , as seen in S . N = n π Z π D ( V cos θ − v ) 2 π sin θ d θ , = n v γ Z π cos θ sin θ d θ = 0 , (14)where we have used Eq. (11). The appearance of γ in the denominator on the righthand side in Eq. (14), as compared to that in Eq. (13), merely indicates the time dilation(d t = γ d t ) for the number of molecules through the moving cross-section EF , per unittime, observed in frame S . In any case, the net number crossing EF is nil, implying thenumber of molecules crossing EF in opposite directions being equal, in frame S too. nergy-momentum density and pressure in a relativistic ideal gas S , we can make use ofthe fact that the isotropic distribution of the rest frame S has got modified, becausefor the molecules moving within an element of solid angle dΩ = sin θ d θ d φ around θ in frame S , not only the number density gets modified by a factor D , even the energyof individual molecules ( m c γ ) in S , is higher by a factor D in S . Then an integrationover the solid angle yields for the energy density, ρ , as ρ = n m c γ π Z π Z π D sin θ d θ d φ = n m c γ π Z π D π sin θ d θ , = ρ Z π γ v v cos θ c ! sin θ d θ = ρ γ v v c ! = γ (cid:16) ρ + pv (cid:17) , (15)where we have used Eq. (4) for p , the pressure. Here ρ = n m c γ is the relativisticenergy density of the gas, that includes the rest mass energy density as well as thekinetic energy density of random motion of molecules in rest frame S .The momentum of a molecule moving along θ in frame S is m Γ V cos θ = m γ D V cos θ in frame S . Then proceeding in the same way as for the energy density,the momentum density is computed as σ = n m γ π Z π D V cos θ π sin θ d θ , = ρ γ c Z π v v cos θ c ! V cos θ sin θ d θ = ρ γ c Z π v v cos θ c ! ( v cos θ + v ) sin θ d θ = γ ρ v c ! v c = γ ( ρ + p ) v c . (16)where we have made use of Eq. (6).The derived Eqs. (15) and (16) are the desired expressions for the energy andmomentum densities in the lab frame S (Eqs. (2) and (3)). As was mentioned in Section2, one would intuitively expect ρ = γ ρ and σ = γ ρ v /c . In fact this would indeedbe the case if only the energy of individual molecules had gone up by a factor D withoutan accompanying increase in the number density by another D factor (which is muchmore than just a Lorentz contraction factor γ of the volume element). It is the factor D in the energy density that results in a term ∝ v that survives when integrated over θ from 0 to π , leading to the pressure term in the energy-momentum density expressions.We can compare the effects of the magnitude of random motion in S and thatof the bulk motion of the liquid seen in S . If the random motion in S is non-relativistic ( v ≪ c ), then pressure plays only a minimal part in the expressions forenergy-momentum densities (Eqs. (15) and (16)). However, if the random motion isrelativistic with v comparable to c , then pressure too is comparable to ρ . But eventhen pressure plays a substantial role in the energy density (Eq. (15)) only if the bulk nergy-momentum density and pressure in a relativistic ideal gas v ∼ c ). On the other hand, pressure plays a significantrole in the momentum density (Eqs. (13)) even if the bulk motion happens to be non-relativistic.We can, of course, relax the restriction of the gas containing only one categoryof molecules and that too, with only a single value of speed v . For that we make asummation of the energy density in frame S over all categories of molecules in the gas, ρ = Σ( n m γ ) i c , and p = Σ n i m i γ i v /
3, however, with a caveat that each i th categoryhas an isotropic velocity distribution in frame S . Then we still obtain expressions for ρ and σ , the energy and momentum densities of the gas in the lab frame S , in completeagreement with Eqs. (2) and (3).
4. A photon gas as a perfect fluid
Let us assume that the volume, V , inside the container is filled with radiation, madeup of photons of energy hν , having an isotropic distribution, as seen in the rest frame S . As seen in the lab frame S , the container of the fluid has a bulk motion v , say,along the x-axis.Now, the pressure p for the radiation is related to the energy density ρ as (seeAppendix A, Eq. (A.7)) p = ρ / . (17)We can then write from Eqs. (2) and (3), the energy and momentum densities for thefluid elements in the lab frame S as ρ = γ ρ v c ! , (18) σ = 43 γ ρ v c . (19)Here we see that the pressure, being comparable to the energy density ρ , even incase of a non-relativistic bulk motion of the fluid, makes a one-third contribution tothe momentum density (Eq. (19)), although its contribution to the energy density ( ρ ,Eq. (18)), could be ignored for a non-relativistic motion, with v ≪ c . We want toderive Eqs. (18) and (19) in the lab frame S, with respect to which the fluid elementhas a bulk motion v .A photon moving along θ with respect to the x-axis, as seen in S , will be movingalong θ in the lab frame S , with the velocity components parallel to the x-axis, givenby [14] cos θ = cos θ + v /c v cos θ /c , (20)cos θ − v c = cos θ γ (1 + v cos θ/c ) . (21)Now a photon of energy hν in frame S , will have in the lab frame S an energy hν = hν δ , where δ is the Doppler factor given by [12, 14] δ = γ (1 + v cos θ/c ) . (22) nergy-momentum density and pressure in a relativistic ideal gas δ can be obtained from D (Eqs. (11)) by substitutionof V = v = c .Now as in the case of molecular gas in the Section 3, even though the motion ofphotons in the frame S has an isotropic distribution, in frame S it would not be so.This happens because a photon with a velocity component c cos θ in frame S would takea time τ between successive collisions with two opposite container walls AB and CD that are normal to the x-axis, given by cτ cos θ = lγ + v τ , (23)where length l of the container is Lorentz contracted along the x direction by a factor γ . Thus the time interval between successive collisions τ = lγ ( c cos θ − v ) , (24)in frame S will be longer than τ = l/c cos θ in frame S and the corresponding photonwill be spending more time between successive collision with the walls by the factor,cos θ /γ (cos θ − v /c ), before a change in cos θ takes place because of a collision witha wall normal to the x-axis. Therefore the number density of photons moving alongcos θ will be relatively higher by a factor cos θ /γ (cos θ − v /c ) in frame S , even thoughin frame S the number densities of photons with components in all directions wereuniform. ττ = cos θ γ (cos θ − v /c ) , (25)Using Eqs. (21) and (22), we can write ττ = δ , (26)Accordingly, the number density of photons in frame S is higher by a factor δ . This isin addition to the fact that the energy of individual photons will also be higher by afactor δ (Eqs. (22)).The energy density of photon gas in S can be obtained by an integration over thesolid angle, as ρ = n hν π Z π δ π sin θ d θ (27) ρ = n hν Z π γ (1 + v cos θ /c ) sin θ d θ = γ ρ v c ! , (28)where ρ = n hν is the energy density of the radiation.The momentum density is computed in the same way to get σ = n hν πc Z π δ cos θ π sin θ d θ , = γ ρ c Z π v cos θ c ! (cos θ + v /c ) sin θ d θ , nergy-momentum density and pressure in a relativistic ideal gas γ c ρ (cid:18) v c (cid:19) = 43 γ ρ v c . (29)where we have used Eq. (20).The derived Eqs. (28) and (29) are the desired expressions for the energy andmomentum densities of the radiation in the lab frame S (Eqs. (18) and (19)). Like inthe case of molecular gas in Section 3, here too not only the energy of individual photonsgoes up by a factor δ , there is an accompanying increase in the number density by thesame factor δ . It is this δ factor in the energy density that results in the pressure termin the energy-momentum density expressions.In the case of radiation, pressure is comparable to the energy density in S ( p = ρ / v ≪ c ). However,pressure does play a significant role in the momentum density (Eqs. (19)) even if thebulk motion happens to be non-relativistic. The expressions (Eqs. (15) and (16)) for theenergy and momentum densities in the case of an ultra-relativistic gas ( v ≈ c ) reduceto those of the radiation (Eqs. (28) and (29)).In the case the photon gas has a frequency distribution, e.g., in a Planckiandistribution, then ρ = Σ( n ν ) i h , again assuming that photons of each i th frequencyhave an isotropic velocity distribution. Then we still obtain expressions for ρ and σ ,the energy and momentum densities of the radiation in the lab frame S , in completeagreement with Eqs. (18) and (19).
5. Conclusions
We derived, from first principles, expressions for the energy and momentum densitiesof an ideal gas, which is a simple example of a perfect fluid, in an inertial frame inwhich the fluid moves, with a constant bulk velocity. Our Starting point was the restframe, where the random motion of the gas constituents, giving rise to the pressure, isisotropic and the fluid accordingly has no bulk motion, and where stress-energy tensor isof a diagonal form, described by only the energy density and pressure. From a Lorentztransformation of energy and momentum of individual molecules, moving along differentdirections in the rest frame of the fluid, and the transformation of their number densities,we computed the energy-momentum density of the fluid in the frame where it has a bulkmotion. We thereby showed how a bulk motion of the fluid, in the presence of randommotion in the rest frame within the fluid, gives rise to a couple of pressure-dependentterms in the energy and momentum densities of a perfect fluid system, which is an idealgas in our case. We also derived relations for the energy-momentum density of radiationin terms of the energy density and pressure of the photon gas in the frame where thegas has an isotropic distribution. nergy-momentum density and pressure in a relativistic ideal gas Appendix A. Pressure in an ideal gas model of a perfect fluid
We assume that the perfect fluid comprises an ideal gas containing, for simplicity, ahomogeneous number density n of identical, non-interacting, structureless molecules,each of rest mass m , and moving with the same speed v , though in random directions.The fluid may be confined within the walls of a container, supposed to be stationaryin an inertial frame, say S . The distribution of velocities of molecules is thus assumedto be isotropic in the rest frame S . We further assume that the molecules of the gas are“point masses” that may undergo only elastic collisions with each other and the wallsof the container in which both linear momentum and kinetic energy are conserved. Theelastic collisions may give rise to a finite pressure, but no exchange of energy, thus noheat conduction. We also assume the molecules to be otherwise non-interacting, so no“sideways” interactions with other molecules in the neighbourhood thus no sheer stress,no drag or no viscosity.A molecule, initially moving along an angle θ with respect to the x-axis, onencountering a container wall, say, CD (Fig. 1), normal to the x-axis, impinges upon itwith a velocity component v x = v cos θ , rebounds with a velocity component − v x , andthereby impart a momentum to the wall along x direction∆ P = 2 γm v cos θ . (A.1)where v may be relativistic with a Lorentz factor γ = (1 − v /c ) − / .The number of molecules that hit a unit area of the wall CD , while moving along θ , per unit solid angle per second, is n v cos θ / π . The pressure is force per unit area,exerted by the molecules of the gas on the walls of the stationary container. Assumingan isotropic distribution of velocity vectors, the pressure of the gas on the wall CD is p = 2 n m γv π Z π/ cos θ π sin θ d θ . (A.2)It should be noted that molecules moving only within 0 ≤ θ < π/ CD .The pressure can thus be written as p = n m γv ρ v c . (A.3)Here ρ = n m c γ is the relativistic energy density of the gas, that includes the restmass energy as well as the kinetic energy of molecules per unit volume.We can relax the restriction of the gas containing only one category of moleculesand that too, with only a single value of speed v . In such cases, we sum the energydensity over all categories of molecules in the gas, having even some spread in theenergy distribution, ρ = Σ n i m i γ i c , and p = Σ n i m i γ i v /
3, where each i th category hasan isotropic velocity distribution.The expression for pressure in ultra-relativistic gas ( v ≈ c ) simply becomes p = ρ . (A.4) nergy-momentum density and pressure in a relativistic ideal gas v ≪ c and γ ≈ p = n m v ρ k . (A.5)where ρ k = n m v / ρ k = Σ n i m i v /
2, again assuming each i th category has an isotropic velocity distribution.On the other hand, if the container comprises radiation, then we can replace thegas molecules by photons with a number density n , each of energy, say hν , moving inrandom directions. The change in the momentum of a photon, initially moving along θ with respect to the x-axis, after collision with the container wall, is∆ P = 2 hν cos θ c , (A.6)while the number of photons moving along θ , per unit solid angle, reaching the wall persecond is n c cos θ / π , implying a pressure due to an isotropic distribution of photonvelocities p = 2 n hν π Z π/ cos θ π sin θ d θ = ρ , (A.7)where ρ = n hν is the energy density of the radiation. In the case the photons are witha spread in the frequency distribution, e.g., in a Planck distribution, then ρ = Σ n i hν i ,again assuming photons of each i th frequency having an isotropic velocity distribution.Of course, the expression for the radiation pressure is the same as for the ultra-relativistic gas (Eq (A.4)). Appendix B. Pressure of an ideal gas model of perfect fluid with a bulkmotion
Conventionally, pressure is defined as force per unit area on a fluid cross-section, in therest frame S , where the fluid element has no bulk motion. We assume that the fluidmoves with a bulk velocity v with respect to the frame S , implying the rest frame S is moving with velocity v with respect to the frame S . Then pressure, defined as forceper unit area, can be calculated in the frame S too, where the fluid has a bulk motion,however, for that we have to choose a cross-section, say, CD , that is stationary in S (Fig. 1), and has a motion with respect to the frame S (Fig. 2).Now for the ideal, relativistic, gas model of Appendix A, we want to calculate theforce exerted per unit area on the wall CD that is moving with velocity v along x-axis, as seen in frame S (Fig. 2). A molecule moving with velocity, v , and a velocitycomponent v k = v cos θ , along the x -axis, in S , after an elastic collision with the wall CD , bounces off with a velocity component − v k along the x -axis. This molecule, asseen in the lab frame S , will be moving, before and after the collision, respectively with nergy-momentum density and pressure in a relativistic ideal gas V + k , V −k , as V + k = v k + v v k v /c ,V −k = − v k + v − v k v /c , (B.1)and the corresponding Lorentz factorsΓ + = γ γ (cid:18) v v k c (cid:19) , Γ − = γ γ (cid:18) − v v k c (cid:19) . (B.2)Therefore, the impact on the wall CD , due to a collision of an individual molecule, willbe ∆ P = m (Γ + V + k − Γ − V −k ) , = 2 m γ γv k . (B.3)The number of molecules with velocity V + k = V cos θ hitting the wall CD (moving withvelocity v in S ), per unit of time, in the frame S will be proportional to∆ N ∝ D ( V cos θ − v ) , (B.4)the factor D here is due to the increase in number density in the frame S .The net force per unit area or pressure, measured in frame S , in a direction parallelto the bulk motion of the fluid, then is obtained from an integration over the solid angle. p = 2 m γn π Z π/ γ v v cos θ c ! v k ( V cos θ − v ) 2 π sin θ d θ = ρ v c Z π/ cos θ sin θ d θ = ρ v c , (B.5)where we have used Eq. (7), and ρ = n m c γ is the relativistic energy density of thegas. The value for pressure p in frame S calculated this way is consistent with that inthe rest frame S (Eq (A.3)).For calculating the pressure in a direction normal to the bulk flow, we consider amolecule moving with velocity, v , and velocity components v k , v ⊥ in S . This moleculewill be moving in the lab frame S with a velocity components V ⊥ [13] V ⊥ = v ⊥ γ (1 + v k v /c ) = v ⊥ D . (B.6)After the collision, v ⊥ will be reversed in sign, and so will be V ⊥ . Therefore themomentum imparted to a wall normal to the x direction by this molecule will be∆ P = 2 m V ⊥ Γ + = 2 m v ⊥ γ . (B.7)The number of molecules with velocity V ⊥ , hitting a wall normal to the x direction, perunit of time, in the frame S will be proportional to∆ N ∝ D V ⊥ , (B.8) nergy-momentum density and pressure in a relativistic ideal gas D is due to the increase in number density in the frame S . It should be notedthat molecules in only one fourth of azimuth angle range, i.e., π/
2, will be hitting anyone particular wall, say, BD (Fig. 2), normal to the direction of the bulk motion.Integrating over the solid angle, the net force per unit area or pressure measuredin frame S , in a direction normal to the bulk motion of the fluid, then is p = 2 m γn π Z π D V ⊥ v ⊥ π θ d θ , = m n γ Z π v ⊥ sin θ d θ , = m n γ v Z π sin θ sin θ d θ = ρ v c (B.9)which is the same as in the frame S (Eq (A.3)). Thus the pressure is an invariantquantity.In the case of radiation, we can put v = V = c in the above formulation, which willalso make D = δ , and we get p = ρ . (B.10)in all directions.It is important to note that in Eq. (B.10) (or even in (B.9)), pressure p is notderived in terms of ρ , the energy density in the frame S where the fluid element has abulk motion, i.e., p = ρ/ p = ρV / c for gas comprisingmolecules), and that the pressure instead is defined properly in terms of the energydensity ρ in frame S , where the random motion of the gas constituents is of an isotropicdistribution. References [1] C. W. Misner, K. S. Thorne and J. A, Wheeler,
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