Wednesday, May 20, 2015

Killing Vectors and Hawking Radiation

Preliminary

We start out with the interval (see equations (6) to (23) in Relativistic Doppler Effect ),

(1) ds2 = -dt2 + dx2 + dy2 + dz2

We define the metric as the coefficient of each of the terms in the above:

(2) η00 = -1, η11 = 1,η22 = 1,η33 = 1,and ηij = 0 for i≠j

We can rewrite equation (1) in the general form,

(3) ds2 = ηαβdxα dxβ

The proper time τ is,

(4) dτ2 = - ds2

This yields,

(5) dτ = dt/γ

(6) where γ = (1 - v2)

We measure the velocity with respect to the proper time τ, not the ordinary time t.

(7) uβ =dxβ/dτ

This gives the important result,

(8) u2 = uu = -1

We define a 4-vector momentum as,

(9) pβ =(p0,pi) = (p0,p)

This gives the following:

(10)p2 = muβmuβ = m2u2 = - m2

And,

(11) E2 = m2 + (p)2.

Putting c into the equation,

(11) E2 = m2c4 + p2c2.

Euler-Lagrange Equations for a free particle in motion

Consider two timelike separated points A and B, and all the timelike worldlines. In fig 1, two such lines are illustrated - a straight line path and a nearby path.

By the variational principle, the world line of a free particle between two timelike separated points extremizes the proper time between them. To see this, each curve will have a value in terms of the proper time,

(12) τAB = ∫AB

Using equations (1) and (4),

(13) τAB = ∫AB {dt2 - dx2 - dy2 - dz2}½

We parametrize this equation by choosing σ such that at point A, σ = 0, and at B, σ =1

(14) τAB = ∫01 dσ {(dt/dσ)2 - (dx/dσ)2 - (dy/dσ)2 - (dz/dσ)2}½

This has the same form as the action of equation (1) in The Essential Quantum Field Theory , repeated below

(15) S = ∫ dt L

By making the correspondence:
the action S → τAB,
the time t → σ
and the Lagrangian L → {(dt/dσ)2 - (dx/dσ)2 - (dy/dσ)2 - (dz/dσ)2}½

We can rewrite the Lagrangian L in terms of the general form (equations 3 and 4),

(16) L = { - ηαβ(dxα/dσ) (dxβ/dσ) }½

Also, another form of the Lagrangian is,

(17) L = dτ/dσ

The corresponding Euler-Lagrange equation ( see paragraph below equation 1 in The Essential Quantum Field Theory )

(18)










Consider a particle freely moving along the x-axis ( x1 = x, x2 = y =0, x3 = z = 0)

(19)Equation (18) becomes (see appendix A),






(20) Using equation (17), substitute for L in the above, we get,






(21) Now multiply both sides by dσ/dτ, we get,






In case you haven't recognized, this is the equation of a straight line. Integrate once,

(22) dx/dτ = c

Integrate a second time,

(23) x = cτ + d

Hence for the extremal proper time, the world line for a particle freely moving from point A to point B is a straight line path (fig 1).

Killing Vectors

Generally speaking, conservation laws are connected to symmetries. For instance, if there is a symmetry under displacement in time, energy is conserved; under displacement in space, momentum is conserved; under rotations, angular momentum is conserved. However, in GR, the metric is often time dependent, angle dependent, position dependent, etc. So how does one tell if there is a symmetry? One clue is if the metric is independent of one of its coordinates. For instance, say the metric is independent of x1. That means, it transforms as,

(24) x1 → x1 + const.

leaving the metric unchanged

The vector ξ with components,

(25) ξα = (0,1,0,0)

lies along the direction the metric doesn't change. This is a Killing vector (in honor of Wilhelm Killing, German mathematician 1847-1923). A Killing vector is a general way of characterizing a symmetry in any coordinate system. For a freely moving particle, one can show,

(26) ξu = constant, (see appendix B)

(27) Also,ξp = constant, where p is the particle momentum.

Schwarzschild Geometry

In GR, the Minkowsky metric ηαβ is replaced by a more general metric gαβ so that equation (3) now reads as,

(28) ds2 = gαβdxα dxβ

Specifically in a Schwarzschild geometry, the metric reads as, (G=c=1)

(29) g00 = -(1 - 2M/r), g11 = (1 - 2M/r)-1, g22 = r2,g33 = r2sin2θ,and gij = 0 for i≠j

For our purposes, we note that the metric is time-independent, and therefore there is a Killing vector, which has the components,

(30)ξα = (1,0,0,0)

Hawking Radiation

Fig 2 shows a rest-mass zero particle-antiparticle pair which has been created by vacuum fluctuations in such a way that the two particles were created on opposite sides of the horizon of a black hole. The components ξp and ξp' must be equal and opposite so that ξ(p +p') = 0, (value of the vacuum). The particle ( ξp > 0) can propagate and can be seen as radiation by an observer at infinity. This also means that the antiparticle ( ξp' < 0) will be absorbed by the black hole, thus decreasing its mass in the process. This is the basis of Hawking's claim that black holes radiate, and in time, will evaporate.

Appendix A







(A3) Equation (18) now reads as,






First calculate,






Putting it altogether, equation (A3) becomes,

(A4)






Appendix B

let α =1, Equation (A3) becomes,

(B1)






(B2) from (A2),






(B3) LHS of (B1) ,






(B4) therefore,






(B5) Now consider,








Using equations (A1) and (17)








Note that we can write,

(B6) η = ηαβ ξα

Substituting in the above,

(B7)








(B8) From (B4), we get ,

ξu = constant

Saturday, December 20, 2014

The Equivalence Principle: BBT or SUT

Einstein made the following reasoning:
he took the Doppler Effect and using the Equivalence Principle, turned it into a Gravitational Shift. And then from there, he derived his eponymous Field Equations.
See The Essential General Relativity.

By reversing this reasoning I've shown the following:
from the Gravitational Shift and using the Equivalence Principle, I've turned that into a Doppler Effect, and from there I then derived the Hubble Equation.
See The Equivalence Principle and the Big Bang Theory.

So one inevitable conclusion is: either the redshift from faraway galaxies is the result of a Doppler Effect and that gives the BBT - or it's a gravitational effect and that gives the SUT. The second inevitable conclusion from the EP is that one interpretation cannot be differentiated from the other, that is, there are no conclusive tests that would favor one over the other. You either accept GR and the EP, or you reject both.

Friday, December 05, 2014

Big Bang Theory Versus Static Universe Theory

In this blog, I will compare the Static Universe Theory (SUT) versus the Big Bang Theory (BBT) in regard to their respective assumptions.

As it has been mentioned in Another argument against the BBT, it is believed that only a Hot Big Bang scenario can explain the CMB, and any other explanation would have to be contrived. We show here quite the contrary that it is the BBT which is contrived in its assumptions to make all the parts fit in.

Static Universe Theory

(1) The Equivalence Principle is valid and the redshift from faraway galaxies is gravitational in nature. See The Equivalence Principle and the Big Bang Theory.

(2) The universe is eternal and infinite.

(3) The CMB can be explained in terms of the surface of infinite redshift. See Olbers' Paradox .

Big Bang Theory

(1) The Equivalence Principle is valid and the redshift is due to a Doppler Effect – the galaxies are moving away from each other.

(2) By extrapolating backward in time, the universe started as a singularity and then expanded.

(3) There is a 4th spatial dimension into which our 3-spatial dimensional world is expanding. See Riemannian Geometry and the Big Bang Theory.

(4) In order to solve the Einstein Field Equations to get to the Friedman Equations, one must assume the universe is homogeneous and isotropic.

(5) To justify (4), one must assume that the universe went through an inflationary period in the early stage.

(6) To justify (5), one must assume that quantum fluctuations popped out of the vacuum some 13.7 billion years.

(7) Since the universe is accelerating, one must assume that the universe is filled with Dark Energy, which must make up 75% of the universe in order to justify a flat space universe (As of now, the Vacuum Energy from (6) is out of step by 122 orders of magnitude with Dark Energy).

(8) To calculate the density of the universe, one must assume the universe is finite in size with its radius equal to its Schwarzschild radius.

(9) In order to tie in the CMB with the BBT, one must assume that the universe must behave like a nearly perfect idealized fluid, so that one can tie in the redshift to the scale factor in (4), which itself is tied in with temperature and time. One can then set a chronology of different reactions that would have happened at different temperatures/times, all of these requiring a number of parameters that can be fine-tuned with observation.

Conclusions

The BBT is a contrived theory which besides the number of assumptions that is needed to support the BBT - a much larger number than the SUT - it nevertheless leaves a certain number of unanswered questions such as: what evidence do we have that a 4th spatial dimension exists? If the universe didn't exist from t = -∞ to t = -13.7 billion years, what caused it to spring out of the vacuum some 13.7 billion years ago? How many more assumptions will the BBT need in order to reconcile the Vacuum Energy with Dark Energy in order to make that fit into the theory?

Monday, December 01, 2014

Olbers' Paradox

Olbers' paradox is the argument that the darkness of the night sky conflicts with the assumption of an infinite and eternal static universe.

Let n be the average number density of galaxies in the universe. Let L be the average stellar luminosity. The flux f(r) received on earth from a galaxy at a distant r is,

(1) f(r) = L/(4πr2)

Consider now a thin spherical shell of galaxies of thickness dr. The intensity of radiation from that shell is,





(2) dJ(r) = flux x number of galaxies in the thin shell.

= L/(4πr2) x n x r2dr

= (nL/4π)dr

We can see that the intensity only depends on the thickness of the shell, not its distance.

The total intensity is found by integrating over shells of all radii.

(3) J = (nL/4π) 0dr = ∞

Accordingly, the night sky should be bombarded by an infinite number of photons.

Explanation

(A) BBT

The primary argument of the Olbers' paradox from the Big Bang Theory is the universe has a finite age (by extrapolating backward in time), and the galaxies beyond a finite distance, called the horizon distance, are invisible to us simply because they are moving faster than lightspeed and therefore their light can’t reach us.

(B) The static model

The paradox is only an apparent contradiction. In this case, if humans had eyes that could see 2mm wavelength (the Cosmic Microwave Background), then one would see the night sky being illuminated from every direction. What's missing in the above calculation is that the wavelength of light travelling immense distance is shifted more and more towards the red. In terms of the wave model, the next peak would take an infinite of time to reach us. This is the surface of infinite redshift. What we see in the CMB are the photons released from a distance slightly less than the surface of infinite redshift. Any photons released from galaxies beyond that distance will not reach us.



Wednesday, November 19, 2014

Another argument against the BBT

In The Equivalence Principle and the Big Bang Theory, we explored the idea that the galaxies are exhibiting a redshift that can be interpreted either as a Doppler effect, that is, they are moving away, or as a gravitational shift, in which case we can safely say they are at rest with each other. An argument for the BBT that every observer in the universe will see every other galaxy moving away is often used to validate the BBT. This argument is misleading. It is only partially true if one ignores that velocity has not only magnitude but it has also a direction.



According to an observer in our galaxy, he sees every other galaxy moving away. (Black Arrows). However, an observer from another galaxy, say the blue one, sees the other galaxies moving away but in different directions than what we see (Blue Arrows).



We will agree with her concerning both of our galaxies - She sees us move away from her, we see her moving away from us along the same line, and since motion is relative, our observation agrees. However, we will disagree about the direction for galaxies Red, Green and any other galaxy not situated along the line joining our two galaxies.

On the other hand, if we agree that instead of a Doppler shift, the galaxies are exhibiting a gravitational shift and are at rest with each other, then we all agree that the net force on any galaxy is zero.



Cosmic Microwave Background

The last remaining argument in favor of the BBT is the Cosmic Microwave Background (CMB). It has been said that only a Hot Big Bang scenario can explain the CMB, and any other explanation would have to be contrived. In future blogs, I will demonstrate that the explanation of the CMB from a Hot Big Bang scenario is itself contrived. Stay tuned.

Saturday, November 01, 2014

The Equivalence Principle and the Big Bang Theory

In this blog I explore the idea that Hubble failed to notice an important aspect of Einstein’s Equivalence Principle.

Consider this thought experiment. Different emitters are placed at different heights from the ground. The earth plays the role of the source of gravity.



They emit light, which from an earth observer, would be blueshifted. According to Einstein’s Equivalence Principle: we can say that the Doppler effect is equal to the gravitational shift ( see equation 2 in The Essential General Relativity , reproduced below)

(1) (Δf/f)gravity = -(Δf/f)doppler = -Δv/c

For emitter 1, we can say,

(2) Δv1 = g(d1) Δt1 (definition of acceleration, where g(d1) is the gravitational potential field at d1.

(3) Define g(d1) = g1

(4) Δv1 = g1 (d1/c) (time = distance/velocity)

(5) However, g1 = (GMsource)/ R21
= (GMsource)/ (Rsource + d1)2
= (GMsource) (Rsource + d1)−2
= (GMsource/ R2source) ( 1 + d1 / Rsource)−2
≈ (GMsource/ R2source) ( 1 − 2d1 / Rsource)
For 2d1 << Rsource

(6) g1 = (GMsource)/ R2source

Substitute (6) into (4),

(7) Δv1 = (GMsource)/ cR2source )(d1)

We can get the same result for emitters 2 and 3,

(8) Δv2 = (GMsource)/ cR2source )(d2)

(9) Δv3 = (GMsource)/ cR2source )(d3)

We can generalize equations 7,8 and 9 as,

(10) Δv = Hd ,

where H = (GMsource)/ (cR2source)

In case you haven’t recognized this, it is Hubble’s equation. When he discovered that all galaxies have a redshifted spectrum, Hubble concluded that all the galaxies were moving away. That is the Doppler Effect. However using Einstein’s Equivalence Principle, we can say that galaxies are at rest, and photons are redshifted( they are moving against gravity). Note that Hubble discovered not a change in velocity but just a velocity. In his days, he did not have the technology to observe such a small change in the galaxies' velocities, and it took nearly 70 years before it was discovered that galaxies are actually accelerating.

In our thought experiment, the emitters are at rest, so one can easily say that such emitters would start to accelerate as they cannot be “nailed” in outer space. What about the galaxies, can they be “nailed” so that we can claim they are at rest with respect to each other? Consider one galaxy against all others.



According to Gauss’ theorem (see fig.3 in Newton's Law of Gravity), a galaxy would be attracted as if all the matter inside the sphere were concentrated at the center of that sphere. One can ignore all the other galaxies outside that sphere. At the same time, one can draw an infinite number of spheres, in which the galaxy would be attracted to the center of each sphere. Here’s a diagram with just three spheres drawn.



If the universe is infinite, we can safely say that the total force on a galaxy is zero, and therefore, the galaxies are nearly at rest.



Does it mean that the Big Bang theory is wrong? No. The Big Bang theory says that for every galaxy, all other galaxies are moving away. But Einstein’s Equivalence Principle also says that we can look at every galaxy at rest with their light being redshifted. Both pictures are equivalent.



APPENDIX

Equation 10 is true for every galaxy, since each one is emitting light from a source of gravity that has an infinite radius. This argument is only valid if the universe is infinite.

Tuesday, October 28, 2014

Riemannian Geometry and the Big Bang Theory



Actually it was Gauss who proposed how to describe the inhabitants on a sphere as if they were unaware of the third dimension. Of course, they would need only two coordinates, why it’s called a 2-sphere. Here’s an example with spherical coordinates φ and θ:



But it was Riemann, Gauss’ student, who extended this idea to higher dimensions, hence why it’s generally known as Riemannian geometry. Notice that we have a 2-sphere embedded into 3D. So if we take our 3-dimensional world then Riemannian geometry tells us that it is embedded into a 4 spatial dimensional world. In General Relativity (GR), time is considered as another coordinate. So for the Big Bang Theory, which is based on GR, we must assume that it's a theory of a universe with 4 coordinates,(3+1), but embedded into 4 spatial dimensions + one temporal dimension.

Notice that if our inhabitants on a 2-D sphere were to observe that their New York City is moving away from their Paris, they would conclude that their universe is expanding. We, the 3-D creatures observing them, would know that someone/something is blowing air or matter into the inside of their sphere, causing their world to act in that way. But what about our 3-D spatial world, which is really treated as a Riemannian 4-D spatial world, what's causing it to being blown away? The question is: is the BBT really the final word on this? Does the fourth spatial Riemannian dimension really exist? Or is it just a fabrication that makes the calculations in the BBT simpler, but as a consequence, it is in fact misleading us? Stay tuned.