Confined Energy In An Expanding Space Compared With Gravity

CONFINED ENERGY IN AN EXPANDING ELASTIC CONTINUUM COMPARED WITH GRAVITY BJØRN URSIN KARLSEN Abstract. An elastic continuum might be compressed and then released in a controlled manner such that it undergoes a continuous expansion. In this paper I will show that confined disturbance energy in such a space of infinite or nearly infinite extension will create around itself a gradually decreasing compression field that implies a gradually increasing propagation speed of transversal waves with distance from the energy packet. A small test energy packet in the vicinity of the confined energy will be accelerated towards the confined energy, or alternatively feel a pull from it if it is hindered from moving freely in space. In this way packages of confined disturbance energy will influence each other and move around exactly like masses do under the influence of gravity in open space.

1. Confined energy in the spatial continuum In this section I will assume that disturbance energy may be confined in distinct regions of space where the energy for some reason is hindered from spreading in space. First I will show that such confined energy will displace a certain amount of the spatial continuum regardless of how it is distributed. How it comes that it is confined will not be discussed in this paper, but if it is restricted to a certain area of space, it will exert an outward directed pressure that is counteracted by an equally strong grad div u-field. Next I will consider an expanding space and how the energy packets are loosing energy. Finally I will show that confined energy in such a space will create a div u-field that matches the gravitational potential around distributed matter in space. 1.1. Displacement of spatial mass caused by confined energy. In another paper [1] I made a comparison between the Navier-Cauchy equation1, (1.1)

(λs + 2µs ) grad div u − µs curl curl u + b = ρs u ¨,

and Maxwell’s electrodynamic equations. Provided that sinks and sources can be realized in a spatial continuum of infinite extension, I found a complete match between the two sets of equations. The wave speed for transversal and longitudinal waves were found to be r µs (1.2) , c= ρs Date: Revised 05:02:11. 1In order to distinguish the constants of the N-C equation from those in the Maxwell equations, the index s is used to indicate that the property refers to the spatial continuum, i.e. ρs is defined as the mass density of the spatial continuum, and λs and µs are Lam´ e’s elastic constants. 1

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BJØRN URSIN KARLSEN

and s (1.3)

c1 =

λs + 2µs , ρs

respectively. I also found that confined disturbance energy governed by the divergencefree part of the Navier-Cauchy equation, creates a second order tensor T, the stress energy tensor, given by   e Sx /c Sy /c Sz /c  Sx /c −σxx −σxy −σxz   (1.4) Tαβ =   Sy /c −σyx −σyy −σyz  . Sz /c −σzx −σzy −σzz It describes how energy, momentum, and energy flow creates a stress condition represented by a stress tensor known as Maxwell’s stress tensor. Here I will assume that there might be conditions in space that cause the energy to be kept inside ”bodies”, which generally keep their form unchanged over a reasonable span of time. Hence the forces have got to be taken up by the continuum itself, and the principle of conservation of energy is covered by the condition ∇ · T = 0. This condition is fulfilled if the trace of T is like zero so e = σxx + σyy + σzz . It follows that in an isotropic stress field with σxx = σyy = σzz , we acquire σxx = σyy = σzz = 13 e.

(1.5)

Assume that isotropic radiation energy in the spatial continuum is present as a function of position alone, but not of time, i.e. u = u(x, y, z). The energy may then vary from place to place and the resulting radiation pressure will exert a body force b on an infinitesimal volume element of space. The body force in the x-direction can be found by taking the difference of the forces that act on the two opposite surfaces (dy · dz) at (x, y, z) and (x + dx, y, z) bx · dx · dy · dz = 13 e(x, y, z) · dy · dz − 13 e(x + dx, y, z) · dy · dz 1 e(x + dx, y, z) − e(x, y, z) 3 dx 1 ∂e (1.6) =− . 3 ∂x This expression can be expanded to imply the two other spatial directions, and we get ∂e ∂e ´ 1 ³ ∂e i+ j+ k bx i + by j + bz k = − 3 ∂x ∂y ∂z (1.7) b = − 31 grad e. bx = −

Thus confined isotropic radiation that is a function of position alone, will act as a body force which can be inserted into the Navier-Cauchy equation in order to find the deformation it imposes on the spatial continuum. Since the force is expressed as the gradient of a potential, it cannot cause any rotational deformation, and moreover since the energy distribution is presumed to be stationary, we can use

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the irrotational part of Navier-Cauchy equation (1.1) with u ¨ set to zero to find the deformation (1.8)

(λs + 2µs ) grad div u =

1 3

grad e.

If we choose to set the displacement vector, u, like zero in the undeformed space, i.e. the space outside the confined energy, then both div u and e are falling down to zero outside the volume, V , were the radiation energy is confined, so Equation (1.8) can be solved e (1.9) . div u = 3(λs + 2µs ) The physical meaning of (1.8) is that the outwards directed force caused by the confined radiation energy, e, is counteracted by an inwards directed force caused by the gradient of the displacement. The grad div u-field represents the force that keeps the radiation at bay. The deduction above, however, does not prove that the radiation has got to be confined (the reason for that must be sought elsewhere), only that if there is a certain amount of confined energy, then there has got to be a displacement that sets up an exact balance of forces between the expanding force of the confined energy and the gradient of the pressure in space. Moreover, from the divergence theorem I Z (1.10) (A · n) df = div A dV V

V

we can infer that the displacement is independent of how densely the energy is distributed in space; a certain amount of confined energy will always displace the same amount of the spatial continuum regardless of how it is distributed. Hence the total displacement from a volume of space where an amount of disturbance energy, E, is confined is given by E D= (1.11) . 3(λs + 2µs ) 1.2. The expanding spatial continuum. The spatial continuum can be considered to behave like being confined in a huge spherical container with receding walls moving outwards with some constant – or nearly constant – speed V , such that the space it occupies expands in all directions. The radius of the container is given by R = V T where T is the age of the universe since it was born in the Big Bang. If the expansion is uniform, then a nearby point to an observer is moving outwards with a speed given by r v= = rH, T where def 1 (1.12) H = T is known as Hubble’s constant. In this model it is not a constant at all, but in the course of a short time span at a cosmic scale, it can all the same be considered constant. We can also take the time derivative of H and acquire 1 (1.13) H˙ = − 2 = −H 2 , T which is an extremely small quantity.

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A possible acceleration of a spatial point is given by ˙ v˙ = r˙ H + rH. Since r˙ simply is like v we obtain v˙ = rH · H − rH 2 = 0, as expected. This confirms that all points in open space are moving with a constant speed, and can be chosen as the origin of any cartesian coordinate system in uniform rectilinear motion. 1.3. Confined energy in expanding space. Inside a small sphere with radius r there might be an energy 4/3πr3 e. According to Equation (1.5) the pressure inside the sphere is given by p = 13 e. In an expanding space the surface will recede with a speed H · r, and hence the confined energy within the sphere will loose energy at a pace given by the force on the spherical surface times the velocity ∂ 4 3 ( πr e) = − 13 e · 4πr2 · Hr, ∂t 3 ∂e (1.14) = −e · H. ∂t By differentiating e once more with respect on time we obtain: ∂2e ˙ = −eH ˙ − eH. ∂t2 By (1.14) and (1.13) this expression reduces to ∂2e = 2eH 2 . ∂t2 Combined with (1.9) it yields

(1.15)

i ∂2 h e ∂ 2 ( div u) = , ∂t2 ∂t2 3(λs + 2µs ) ∂ 2 ( div u) 2eH 2 = . 2 ∂t 3(λs + 2µs )

Because H is very small (H ≈ 2.3 × 10−18 sec−1 ), the loss of energy is also very small. It amounts to 1% of its initial value in about 1.4 billion years, and even if we waited that long, our standard kilo would have diminished by the same amount at that time so we would not notice the change. When we look into the sky, we also look back in time, so our observations might possibly be affected, but that would probably only affect our interpretation of what we see. 1.4. The div u-field outside an energy concentration in expanding space. In space outside of the confined energy there is no body force, but instead we have got to take into account the dynamic force ρs u ¨ , and since space in this area is supposed to be free of any rotation we can without loss of generality take the divergence of Navier-Cauchy equation (1.1) (λs + 2µs ) div grad ( div u) − ρs

∂ 2 ( div u) + div b = 0, ∂t2

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where b = − 13 grad e is a measure of the effect of the confined energy, see (1.7). If the time derivative of div u is zero, there will be no divergence of u outside the confined energy because the two leftover terms will cancel out. Hence it is necessary to introduce a small additional component, uε , to take care of the effect of the time dependent term. By (1.15) we obtain: (1.16)

2ρs eH 2 . 3(λs + 2µs )2

∇2 ( div uε ) =

I now define some new terms: First a new variable Φ(x) given by: c2 div uε , 2 next the matter density %(x) (without an index) that is related to e(x) through the relation: (1.17)

def

Φ(x) = −

def

e(x) = %(x)c2 , and finally the constant (1.18)

def

G =

ρs c4 H 2 . 12π(λs + 2µs )2

With these properties inserted Equation (1.16) takes the form (1.19)

∇2 Φ = −4πG%.

We notice that G can be interpreted as the gravitational constant2, Φ as the gravitational potential, and % as the mass density of the confined energy. Then Equation (1.19) is an exact match of the classical equation for the Newton potential. This Poisson’s equation can be solved under rather general condition, so if %(r0 ) is a function of the radius vector r0 , then Φ(r) at the position r is given by the well known relation Z %(r0 )dv 0 Φ(r) = G , |r − r0 | where dv 0 is a small volume element around the position r0 and the integration is taken over the whole space. If the total mass equivalence is defined as Z M= % dV, V

and is concentrated in a small area around r0 = 0, then the potential at a distance r from it is given by Φ=

GM . r

The gradient of Φ is given by (1.20)

grad Φ = −

GM r , r2 |r|

2In fact G is not a constant, but a function of H. Both are, however, functions of T −1 where

T is supposed to be very great, hence the variation of G in the course of any time of observation is extremely small.

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BJØRN URSIN KARLSEN

or for later use by reinserting div u from (1.17) we get 2 GM r grad div u = 2 2 (1.21) . c r |r| This shows that outside confined energy with arbitrary distribution there is a div u-field that follows the same laws as the gravitational potential. It is extremely weak, but then the gravitational forces are about 40 orders of magnitude weaker than the strong forces. 2. Movements of confined energy packets in a continuum with varying wave speeds In this section I will show that disturbance energy that moves along filiform paths with velocity c within closed packages, will be accelerated towards bodies of confined energy, or else acted upon by forces like forces in a gravitational field if they are hindered from moving freely in space. 2.1. Confined radiation in a space with varying wave speed. In order to find how mass behaves in a space with varying density, we have got to recapitulate some of the results discussed at the end of the paper Elastodynamics in a Continuum of Infinite Extension [1]. Matter consists of small cells of disturbance energy all moving with the speed c through the spatial continuum. Say that one such cell has an energy E. Then per definition i has a mass given by def E m = 2, (2.1) c and a momentum E p ˆ, c in the direction of the unit vector p ˆ . A material body may consist of k cells, and since momentum is a vector, the momentum of each cell at a given time may be added to obtain the net momentum of the body ¶ k k µ k X X X E P= pn = p ˆ = (mcˆ p)n , c n n=1 n=1 n=1 p=

(2.2)

P = M v. The derivative of P with respect on time is the force needed to give a body with mass M an acceleration. DP D F= = (M v) = M˙ v + M v. ˙ Dt Dt If M˙ ¿ M then the (non relativistic) acceleration with a good approximation is given by DP (2.3) = F = M a. Dt I will also assume that energy packets moving around in free space certainly can exchange momenta and energy, but in between the interactions, they move along their individual strings with constant energy. Now, let us study a single cell, first in a flat space with constant c throughout, and then in a space with varying c.

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We take the time derivative of p: Dp ∂p dp ds = + · , Dt ∂t ds dt where dp/ds = (ˆ p∇)p is the directional derivative of p in the direction of p ˆ , and ds/dt simply is the speed of the energy cell. By (2.2) and the mathematical identity, (A · ∇)A = 12 grad |A|2 − A × curl A, we obtain ´ Dp ∂p ³ c = + p·∇ p·c Dt ∂t E ∂p c2 c2 = + grad |p|2 − p × curl p ∂t 2E µ 2 ¶E ∂p c2 E = + grad − cˆ p × curl p ∂t 2E c2 µ ¶ ∂p c2 E 1 + grad = − cˆ p × curl p ∂t 2 c2 · µ ¶ µ ¶ µ ¶ ¸ ∂p c2 E ∂ 1 ∂c ˆ ∂ 1 ∂c ˆ ∂ 1 ∂c ˆ = + i+ j+ k − cˆ p × curl p ∂t 2 ∂c c2 ∂x ∂c c2 ∂y ∂c c2 ∂z µ ¶ ∂p c2 E −2 = + grad c − cˆ p × curl p ∂t 2 c3 ∂p − m(c grad c) − cˆ p × curl p. = ∂t Finally we sum together all the internal forces and obtain the time derived momentum of the whole body: ¸ k · DP X ∂pn (2.4) = − mn (c grad c) − cˆ pn × curl pn . Dt ∂t n=1 First let the body be in a space with constant c. Since c is no function of time and position, the first and second terms are zero. The third terms represent the windings of the energy packets as they meander around in the body, but if there are no external forces acting on the body, the sum of all the force components are zero. Then let c be a function of position, but not of time, hence c = c(x). The first terms are still zero and the third terms sum up to zero, but the middle terms are no longer zero. According to Equation (2.3) all energy packets get an acceleration component of (2.5)

g = −c grad c,

which becomes the acceleration of the whole body. Notice that the acceleration is negative in the direction of increasing c, and positive in the direction of decreasing c. As a whole the body gets an acceleration towards areas in space with slower speeds of light. We can conclude that a body situated in a field with varying c(x) gets an acceleration, which we can call the acceleration of gravity, that is independent of the mass of the body. 3. Newton’s Gravitational law When Newton formulated his law of gravity some 300 years ago, he thought that the acceleration of a body was caused by a force; the force of gravity. As we have seen above, however, the acceleration is not caused by a force, but by a property of

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BJØRN URSIN KARLSEN

space itself, namely a varying c. The force only enters the picture when a body is hindered from moving freely, so the force of gravity is the force needed to break the acceleration down to zero. Elsewhere I have found that the force needed to give a body an acceleration of a is given by (3.1)

F = Ma

which also follows from Equation (2.4). 3.1. The attraction between two bodies of confined energy. First we have got to find out how the wave speed, c, varies with compression of the spatial continuum. The question is not quite trivial. In another paper, Spatial continuum mechanics [2], I have addressed this question from two points of view. One that the spatial continuum has an intrinsic mass density that only changes when a volume element is compressed to a smaller volume, and the other that the spatial continuum has no initial mass density at all and only gets its inertial properties by the energy that goes into it by compression. Here I will only consider the first alternative. Let a volume element, V0 , be changed to a volume V = V0 + ∆V . Then we can set up the relation: ρs V0 = ρnew V, 1 1 V0 + ∆V = , ρnew ρs V0 and multiply both sides with the property µs µs ¡ ∆V ¢ µs = 1+ . ρnew ρs V0 Then by the identity ∆V = div u, V0 →0 V0 lim

we get the variation of c as a function of div u c2 = c02 (1 + div u), and finally by taking the gradient on both sides of the equation, we obtain (3.2)

c grad c =

1 2 2 c0

grad div u, .

Now let us introduce a mass M1 into space, which generates a potential given by Equation (1.20) grad Φ = −

GM1 r . r2 |r|

In this field at a distance r we introduce another test mass, M2 , which is kept from moving and hence be acted upon with a force given by (3.1) F21 = −M2 · c grad c. Note that the force is acting on M2 and the radius vector is from M1 to M2 , hence the negative sign. From (3.2) we get c grad c, and from (1.21) we get the value of

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grad div u, and with these two properties inserted the mutual attraction between the two bodies above becomes F = M2 12 c02 grad div u c02 2 GM1 2 c02 r2 M1 M2 F =G , r2 This is Newton’s equation for the attraction between two heavenly bodies and needs no additional comments. = M2

3.2. An estimate of the size of µs and ρs . Equation (1.18) puts us in position to estimate the numerical value of µs and ρs . If we set λs = 2µs then according to Equation (1.2) and (1.3) the velocity of longitudinal waves becomes the double of transversal waves, which is a fairly reasonable estimate. By Equation (1.18) we obtain c2 H 2 µs = . 192πG Newton’s gravitational constant G = 6.673 × 10−8 cm3 /g sec2 . The Hubble Space Telescope Key Project Team has measured Hubble’s constant with an uncertainty of 10 percent to3 H = 70 km s−1 Mpc−1 = 2.3 × 10−18 sec−1 . The speed of light c = 3.0 × 1010 cm/sec. With these values inserted we obtain µs ≈ 1.2 × 10−10 g cm−1 sec−2 , and by Equation (1.2) we finally acquire ρs ≈ 1.3 × 10−31 g cm−3 , which is within the same order of magnitude as the estimated mean spatial density of matter in the universe. It is tempting to assume that (ρs c2 ) represents the mysterious dark energy. This becomes even more probable if space only gets its inertia from being compressed [2].

3In a press release from May 9. 1996 at http: oposite.stsci.edu/pubinfo/press-releases/96-21.txt the most probable value is reported to be between 68 and 78 km s−1 Mpc−1 .

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BJØRN URSIN KARLSEN

References 1. Bjørn Ursin Karlsen, Elastodynamics in a Continuum of Infinite Extension. 2. , Spatial Continuum Mechanics, http://home.online.no/˜ukarlsen, 2000. E-mail address: [email protected]