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Wednesday, July 20, 2016
Friday, September 04, 2015
Superposition and Quantum States
A lot of confusion in Quantum Mechanics is the result from not being able to differentiate between the real world and the Hilbert Space. Vectors in real space – like velocities, accelerations, forces, etc. – are objects one can actually measure in the real world. On the other hand, quantum states are represented by vectors (more precisely by rays) in a Hilbert space, but these are NOT subjects of measurement. What we measure for a quantum system are probabilities, and those vectors in that Hilbert space are useful mathematical tools to calculate those probabilities.
Illustration
Suppose we have a beam of electrons flowing from right to left:
Notice this is a thought experiment as we really don’t know in what direction the spin of each individual electron points. We can safely say that these directions are at random. Now physicists are interested in measuring these spins. So I need some kind of apparatus, and the good news is that there exists one – a magnetic field. Trouble is that these electrons, with their spin, are tiny magnets, and we know that magnets placed in a magnetic field will align (or anti-align) with the magnetic field. Suppose I place the magnetic field along a certain direction, say the z-axis. Now let’s look at one electron as it approaches the magnetic field.
When that electron penetrates the magnetic field, it will align its spin such that its z-component will yield the value of +ℏ/2 along the z-axis, a spin up, which can be represented as:
Here’s another electron about to penetrate the magnetic field:
This time it will anti-align with the magnetic field, with a spin value of -ℏ/2, a spin down.
On the whole, 50% of the electrons will align with the magnetic field (spin =+ℏ/2, or up), and 50% will anti-align (spin = -ℏ/2, or down).
Comments
(1) Note that before the measurement, the spin of an electron can be in any direction. Passing the electron through the magnetic field forces the electron to change its spin orientation such that it either aligns or anti-aligns with its z-component to be ± ℏ/2. This is what distinguishes quantum physics from classical physics: the act of measuring a quantity will disturb the system.
(2) The other components of the spin are indeterminate: if I were to pass these electrons into another magnetic field, say aligned with the x-axis, again I will find that 50% of the electrons will align with the magnetic field (spin = +ℏ/2), and 50% will anti-align (spin = -ℏ/2), this time along the x-axis. I will no longer know what the spin along the z-axis is.
(3) One way to mathematically represent this quantum system (read, the wave function) is this:
| ψ> = 1/(2^{½}) (| ↑ > + | ↓ >).
Now this is called a superposition of two quantum states, the up and down states. Note that if I want to calculate the probability that the electron has a spin up, I take the product of the vector | ↑ > with the wave function | ψ>, and square that.
P = |< ↑| ψ >|^{2}
= 1/2 [< ↑ | { | ↑ > + | ↓ > }]^{2}
=1/2 [{ < ↑ | ↑ > + < ↑ | ↓ > }]^{2}
Using the orthogonality condition, < ↑ | ↑ > = 1 and < ↑ | ↓ > =0, we get,
P=1/2, or 50%, which is what is observed in the lab.
(4) Now here comes the real crunch. Writing | ψ> = 1/(2^{½}) (| ↑ > + | ↓ >) is called a superposition but it’s not meant to mean that the electron “lives” simultaneously in two states and can’t make up its “mind” in which one it wants to live. Those states do not represent ordinary vectors of real objects - like velocities, acceleration, forces, which was a crucial point that was mentioned above. If it were the case, then since these two vectors are equal in magnitude and opposite in direction I would be able to claim,
| ↑ > = (-1) | ↓ >,
and the orthogonality condition would no longer hold, and P would not equal to 50% - actually it would turn out to be zero!!! What needs to be reminded is that the two vectors, | ↑ > and | ↓ > represent possible states, and the beauty of it all is that they form a complete set of orthogonal unit vectors, in an abstract space called the Hilbert space, which provides a powerful method of calculating probabilities.
Appendix
A word on semantics: note that I used the word "apparatus" when that word description is NOT needed. For instance at the LHC, one thinks of two beam interacting (colliding), and not as one beam interacting with an apparatus - the second beam. Similarly, the beam of electrons described above are interacting with a magnetic field (the "apparatus"). Hence, the whole concept of "wave function collapse" is totally unnecessary. The so-called measurement between a microscopic system and a macroscopic system is illusive as it never happens, it is always a microscopic system interacting with another microscopic system. And the wave function cannot collapse as it is not a function of a real wave. Also, there is no need of hidden variables or "beables". QM can do very well without this extra baggage.
Illustration
Suppose we have a beam of electrons flowing from right to left:
Notice this is a thought experiment as we really don’t know in what direction the spin of each individual electron points. We can safely say that these directions are at random. Now physicists are interested in measuring these spins. So I need some kind of apparatus, and the good news is that there exists one – a magnetic field. Trouble is that these electrons, with their spin, are tiny magnets, and we know that magnets placed in a magnetic field will align (or anti-align) with the magnetic field. Suppose I place the magnetic field along a certain direction, say the z-axis. Now let’s look at one electron as it approaches the magnetic field.
When that electron penetrates the magnetic field, it will align its spin such that its z-component will yield the value of +ℏ/2 along the z-axis, a spin up, which can be represented as:
Here’s another electron about to penetrate the magnetic field:
This time it will anti-align with the magnetic field, with a spin value of -ℏ/2, a spin down.
On the whole, 50% of the electrons will align with the magnetic field (spin =+ℏ/2, or up), and 50% will anti-align (spin = -ℏ/2, or down).
Comments
(1) Note that before the measurement, the spin of an electron can be in any direction. Passing the electron through the magnetic field forces the electron to change its spin orientation such that it either aligns or anti-aligns with its z-component to be ± ℏ/2. This is what distinguishes quantum physics from classical physics: the act of measuring a quantity will disturb the system.
(2) The other components of the spin are indeterminate: if I were to pass these electrons into another magnetic field, say aligned with the x-axis, again I will find that 50% of the electrons will align with the magnetic field (spin = +ℏ/2), and 50% will anti-align (spin = -ℏ/2), this time along the x-axis. I will no longer know what the spin along the z-axis is.
(3) One way to mathematically represent this quantum system (read, the wave function) is this:
| ψ> = 1/(2^{½}) (| ↑ > + | ↓ >).
Now this is called a superposition of two quantum states, the up and down states. Note that if I want to calculate the probability that the electron has a spin up, I take the product of the vector | ↑ > with the wave function | ψ>, and square that.
P = |< ↑| ψ >|^{2}
= 1/2 [< ↑ | { | ↑ > + | ↓ > }]^{2}
=1/2 [{ < ↑ | ↑ > + < ↑ | ↓ > }]^{2}
Using the orthogonality condition, < ↑ | ↑ > = 1 and < ↑ | ↓ > =0, we get,
P=1/2, or 50%, which is what is observed in the lab.
(4) Now here comes the real crunch. Writing | ψ> = 1/(2^{½}) (| ↑ > + | ↓ >) is called a superposition but it’s not meant to mean that the electron “lives” simultaneously in two states and can’t make up its “mind” in which one it wants to live. Those states do not represent ordinary vectors of real objects - like velocities, acceleration, forces, which was a crucial point that was mentioned above. If it were the case, then since these two vectors are equal in magnitude and opposite in direction I would be able to claim,
| ↑ > = (-1) | ↓ >,
and the orthogonality condition would no longer hold, and P would not equal to 50% - actually it would turn out to be zero!!! What needs to be reminded is that the two vectors, | ↑ > and | ↓ > represent possible states, and the beauty of it all is that they form a complete set of orthogonal unit vectors, in an abstract space called the Hilbert space, which provides a powerful method of calculating probabilities.
Appendix
A word on semantics: note that I used the word "apparatus" when that word description is NOT needed. For instance at the LHC, one thinks of two beam interacting (colliding), and not as one beam interacting with an apparatus - the second beam. Similarly, the beam of electrons described above are interacting with a magnetic field (the "apparatus"). Hence, the whole concept of "wave function collapse" is totally unnecessary. The so-called measurement between a microscopic system and a macroscopic system is illusive as it never happens, it is always a microscopic system interacting with another microscopic system. And the wave function cannot collapse as it is not a function of a real wave. Also, there is no need of hidden variables or "beables". QM can do very well without this extra baggage.
Wednesday, August 26, 2015
Entanglement and the Uncertainty Principle
Here's your typical experiment. Suppose these particles are electrons/positrons and they are sent one pair at a time. At O, the two particles are released from rest and sent in opposite directions, and of course because of the conservation of angular momentum, they will have opposite spins as they move toward A and B, where observers are stationed.
A..............................←●O●→...............................B
Let's say those particles are going to be passed through a magnetic field aligned in the direction of the Z-axis, giving each either a spin up (+) or a spin down (-). Alice will record each individual electron as they arrived, along with their spin. Ditto for the positrons on Bob's side.
Here's one entry that Alice might have recorded on day 1.
It is no surprise that after a considerable number of observations, she will find that 50% will be up, which will be denoted by (+), and 50% down, designated by (-).
Case 1. She can verify her results by passing all the electrons along a second magnetic field, also aligned with the Z-axis, and she will find all those which had a spin up(+) will still have spin up(+), and all those with spin down(-) will still have spin down(-). This is what is meant by preparing a quantum system in a given state: the first measurement prepared our quantum system in a given state - it forced all the electrons to have either spin up or down along the Z-axis. The second measurement confirmed that. Mathematically, the system is in a eigenstate.
Of course, because the particles are entangled, Bob if he also measures his positrons along the Z-axis, will record for each positron its opposite spin. The law of conservation of angular momentum demands it. Now, there's no mystery here.
Case 2. But what if Bob had chosen to measure along a different axis, say the X-axis, by putting a magnetic field along that axis. Now again he will measure along the X-axis either spin up(+) or spin down(-).
Say the first electron that Alice actually measured to be up (+) on the Z-axis, and would have been down (-) on the Z-axis for Bob's particle is now up (+) on the X-axis, did Bob overcome the Uncertainty Principle, since he now "knows" the z-component, down (-), and x-components up (+), of that particle? Not really. Should he take that particle into a second magnetic field now aligned with the Z-axis, he will find that there will be a 50% chance that it will be up (+), and a 50% chance down (-). So he doesn't "know" the spin along the Z-axis, only along the X-axis. The act of measuring along the X-axis no longer guarantees that he has a particle with spin down (-) on the Z-axis. And this is at the heart of Quantum Mechanics: we don't know the state of the particle until we make a measurement, and whatever the state of a particle was before the measurement, it can be altered if we pass the particle into a different apparatus(Case 2).
A..............................←●O●→...............................B
Let's say those particles are going to be passed through a magnetic field aligned in the direction of the Z-axis, giving each either a spin up (+) or a spin down (-). Alice will record each individual electron as they arrived, along with their spin. Ditto for the positrons on Bob's side.
Here's one entry that Alice might have recorded on day 1.
It is no surprise that after a considerable number of observations, she will find that 50% will be up, which will be denoted by (+), and 50% down, designated by (-).
Case 1. She can verify her results by passing all the electrons along a second magnetic field, also aligned with the Z-axis, and she will find all those which had a spin up(+) will still have spin up(+), and all those with spin down(-) will still have spin down(-). This is what is meant by preparing a quantum system in a given state: the first measurement prepared our quantum system in a given state - it forced all the electrons to have either spin up or down along the Z-axis. The second measurement confirmed that. Mathematically, the system is in a eigenstate.
Of course, because the particles are entangled, Bob if he also measures his positrons along the Z-axis, will record for each positron its opposite spin. The law of conservation of angular momentum demands it. Now, there's no mystery here.
Case 2. But what if Bob had chosen to measure along a different axis, say the X-axis, by putting a magnetic field along that axis. Now again he will measure along the X-axis either spin up(+) or spin down(-).
Say the first electron that Alice actually measured to be up (+) on the Z-axis, and would have been down (-) on the Z-axis for Bob's particle is now up (+) on the X-axis, did Bob overcome the Uncertainty Principle, since he now "knows" the z-component, down (-), and x-components up (+), of that particle? Not really. Should he take that particle into a second magnetic field now aligned with the Z-axis, he will find that there will be a 50% chance that it will be up (+), and a 50% chance down (-). So he doesn't "know" the spin along the Z-axis, only along the X-axis. The act of measuring along the X-axis no longer guarantees that he has a particle with spin down (-) on the Z-axis. And this is at the heart of Quantum Mechanics: we don't know the state of the particle until we make a measurement, and whatever the state of a particle was before the measurement, it can be altered if we pass the particle into a different apparatus(Case 2).
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) ds^{2} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}
We define the metric as the coefficient of each of the terms in the above:
(2) η_{00} = -1, η_{11} = 1,η_{22} = 1,η_{33} = 1,and η_{ij} = 0 for i≠j
We can rewrite equation (1) in the general form,
(3) ds^{2} = η_{αβ}dx^{α} dx^{β}
The proper time τ is,
(4) dτ^{2} = - ds^{2}
This yields,
(5) dτ = dt/γ
(6) where γ = (1 - v^{2})^{-½}
We measure the velocity with respect to the proper time τ, not the ordinary time t.
(7) u^{β} =dx^{β}/dτ
This gives the important result,
(8) u^{2} = u•u = -1
We define a 4-vector momentum as,
(9) p^{β} =(p^{0},p^{i}) = (p^{0},p)
This gives the following:
(10)p^{2} = mu^{β}mu^{β} = m^{2}u^{2} = - m^{2}
And,
(11) E^{2} = m^{2} + (p)^{2}.
Putting c into the equation,
(11) E^{2} = m^{2}c^{4} + p^{2}c^{2}.
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) τ_{AB} = ∫_{A}^{B} dτ
Using equations (1) and (4),
(13) τ_{AB} = ∫_{A}^{B} {dt^{2} - dx^{2} - dy^{2} - dz^{2}}^{½}
We parametrize this equation by choosing σ such that at point A, σ = 0, and at B, σ =1
(14) τ_{AB} = ∫_{0}^{1} 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 → τ_{AB},
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 = { - η_{αβ}(dx^{α}/dσ) (dx^{β}/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 ( x^{1} = x, x^{2} = y =0, x^{3} = 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 x^{1}. That means, it transforms as,
(24) x^{1} → x^{1} + const.
leaving the metric unchanged
The vector ξ with components,
(25) ξ^{α} = (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) ξ•u = constant, (see appendix B)
(27) Also,ξ•p = constant, where p is the particle momentum.
Schwarzschild Geometry
In GR, the Minkowsky metric η_{αβ} is replaced by a more general metric g_{αβ} so that equation (3) now reads as,
(28) ds^{2} = g_{αβ}dx^{α} dx^{β}
Specifically in a Schwarzschild geometry, the metric reads as, (G=c=1)
(29) g_{00} = -(1 - 2M/r), g_{11} = (1 - 2M/r)^{-1}, g_{22} = r^{2},g_{33} = r^{2}sin^{2}θ,and g_{ij} = 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)ξ^{α} = (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 ξ•p and ξ•p' must be equal and opposite so that ξ•(p +p') = 0, (value of the vacuum). The particle ( ξ•p > 0) can propagate and can be seen as radiation by an observer at infinity. This also means that the antiparticle ( ξ•p' < 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_{1β} = η_{αβ} ξ^{α}
Substituting in the above,
(B7)
(B8) From (B4), we get ,
ξ•u = constant
We start out with the interval (see equations (6) to (23) in Relativistic Doppler Effect ),
(1) ds^{2} = -dt^{2} + dx^{2} + dy^{2} + dz^{2}
We define the metric as the coefficient of each of the terms in the above:
(2) η_{00} = -1, η_{11} = 1,η_{22} = 1,η_{33} = 1,and η_{ij} = 0 for i≠j
We can rewrite equation (1) in the general form,
(3) ds^{2} = η_{αβ}dx^{α} dx^{β}
The proper time τ is,
(4) dτ^{2} = - ds^{2}
This yields,
(5) dτ = dt/γ
(6) where γ = (1 - v^{2})^{-½}
We measure the velocity with respect to the proper time τ, not the ordinary time t.
(7) u^{β} =dx^{β}/dτ
This gives the important result,
(8) u^{2} = u•u = -1
We define a 4-vector momentum as,
(9) p^{β} =(p^{0},p^{i}) = (p^{0},p)
This gives the following:
(10)p^{2} = mu^{β}mu^{β} = m^{2}u^{2} = - m^{2}
And,
(11) E^{2} = m^{2} + (p)^{2}.
Putting c into the equation,
(11) E^{2} = m^{2}c^{4} + p^{2}c^{2}.
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) τ_{AB} = ∫_{A}^{B} dτ
Using equations (1) and (4),
(13) τ_{AB} = ∫_{A}^{B} {dt^{2} - dx^{2} - dy^{2} - dz^{2}}^{½}
We parametrize this equation by choosing σ such that at point A, σ = 0, and at B, σ =1
(14) τ_{AB} = ∫_{0}^{1} 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 → τ_{AB},
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 = { - η_{αβ}(dx^{α}/dσ) (dx^{β}/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 ( x^{1} = x, x^{2} = y =0, x^{3} = 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 x^{1}. That means, it transforms as,
(24) x^{1} → x^{1} + const.
leaving the metric unchanged
The vector ξ with components,
(25) ξ^{α} = (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) ξ•u = constant, (see appendix B)
(27) Also,ξ•p = constant, where p is the particle momentum.
Schwarzschild Geometry
In GR, the Minkowsky metric η_{αβ} is replaced by a more general metric g_{αβ} so that equation (3) now reads as,
(28) ds^{2} = g_{αβ}dx^{α} dx^{β}
Specifically in a Schwarzschild geometry, the metric reads as, (G=c=1)
(29) g_{00} = -(1 - 2M/r), g_{11} = (1 - 2M/r)^{-1}, g_{22} = r^{2},g_{33} = r^{2}sin^{2}θ,and g_{ij} = 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)ξ^{α} = (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 ξ•p and ξ•p' must be equal and opposite so that ξ•(p +p') = 0, (value of the vacuum). The particle ( ξ•p > 0) can propagate and can be seen as radiation by an observer at infinity. This also means that the antiparticle ( ξ•p' < 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_{1β} = η_{αβ} ξ^{α}
Substituting in the above,
(B7)
(B8) From (B4), we get ,
ξ•u = 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. Now note that only facts independent of GR/EP can tip the balance in favor of one of those models.
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. Now note that only facts independent of GR/EP can tip the balance in favor of one of those models.
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 4^{th} 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 4^{th} 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?
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 4^{th} 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 4^{th} 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/(4πr^{2})
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/(4πr^{2}) x n x r^{2}dr
= (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) ∫_{0}^{∞}dr = ∞
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.
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/(4πr^{2})
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/(4πr^{2}) x n x r^{2}dr
= (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) ∫_{0}^{∞}dr = ∞
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.
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