dynamicweb.html

Dynamics of Quantized Space-Time.

Version 3.1

Hontas F. Farmer
University of Illinois at Chicago
Hfarmer@alumni.niu.edu

PACS: 04.60.Gw

Everybody knows that energy is quantized. The General theory of relativity says mass-energy curves space-time. In this composition I postulate a quantum theory of space-time which rest on those facts. Then from mathematical logic I derive novel predictions for future experimental investigation.

This new version of the theory features Heisenberg matrix mechanics.

Statement of the problem at hand.

Gravitation is the most fundamental interaction in the universe.  Absolutely everything interacts
gravitationally with everything else.  Currently we have no internally consistent,experimentally verifiable and
agreed upon fundamental quantum theory of gravity.  There are several contenders for such a theory they all
fail one of the afore mentioned counts.  Notable among them is the theory of Loop Quantum Gravity, and
(M)matrix Theory.  The problem is to postulate a conceptually simple theory of quantum gravity with
associated math's so straight forward that they cannot be argued with.  Furthermore the predicted results of
experiments are of a magnitude such that they can be discerned from the noise of experimentation.

The planck units of measurement.

In terms of standard SI units the Planck units of measurement are as follows.

(1)

In[57]:=

(2)

In[58]:=

(3)

(Let me be perfectly clear that in the context of this paper wherever Δ appears it always means the "rate of change" in the quantity next to it for example Δx=-.) These constitute a compleat set of fundamental units MLT. Units for all other quantities can be derived from these. I will set h and c equal to one this results in a simple and revealing relationship between the units of space, time, and mass. Why am I using these units? Consider the schwarzschild radius of a massless particle.

(4)

If λ is set to zero then the Schwarzschild radius grows to infinity. Suppose we set the wavelength λ to the Planck length.

(5)

The Schwarzschild radius of a photon with a wavelength equal to the planck length is twice the magnitude of the wavelength. This is significant no other unit of measure has this quality. Only a Planck length has the right proportionality to the gravitational constant.

The postulate of quantized space-time

Mass-energy alters the local space-time interval in increments that are integer multiples of the Planck length.

Definitions of the operators

Before this is given mathematical realization definitions are in order.

Definition:

The distribution (Not a density operator) of energy denoted by

(6)

Where is the standard Hamiltonian matrix

(7)

(8)

and

$Δx_μ = G^(1/2) (-{{n_0,, n_1,, n_2,, n_3}})$

(9)

As is the standard practice the coefficients are.

$n_i∈ {1, 2, 3, 4, 5, ... ., n}$

(10)

The subscripts a and b in the expansion of the Hamiltonian are of this type. In practice n will be infinite. μ is a index for the space-time dimensionality of configuration space.

(11)

(12)

The operator in this definition excludes the gravitational interaction. Further down the page a Hamiltonian that accounts for gravity will be defined in terms of these operators.

As defined this is a Hermitian operator which is confirmed by it's symmetry.The energy distribution operator (A) has the units of Energy divided by Planck area (G).

Definition:

The curvature of space time will be represented by the operator .

(13)

$@ _μν = G ( {{-n_0, 0, 0, 0}, {0, n_1, 0, 0}, {0, 0, n_2, 0}, {0, 0, 0, n_3}} )$

(14)

As defined this operator is both Hermitian and unitary. This one has units of Planck area. Given these definitions the postulate of Quantized Space-time can be stated simply. The mass-energy of a quantum state is equal to the curvature of space-time due to the distribution of mass-energy in that particular quantum state.

(15)

The mass tensor defined here is neither Hermitian nor unitary but it is normal. The @ operator is acting as a unitary operator in this equation. These are more than operators these are covariant four tensors of rank two as such they are in full agreement with at least the Special theory of Relativity. They are at least Lorentz covariant.

The quantum Schwarzschild radius formula.

A sure way to test this theory and see if it is at all accurate would be to compute the schwarzschild radius of a object of given mass and see if the result given by this theory is in agreement with the classical result. Consider the operators above defined. Then note that the product of the curvature of space-time with the Hamiltonian operator gives a quantity with units of length. For the sake of simplicity let us consider a very simple situation. A stationary object of Mass M. What would it's Hamiltonian operator look like? For a stationary object with no forces acting on it the relativistic Hamiltonian will reduce to the mass of the object.

(16)

Expanding this quantity in any orthonormal basis gives a diagonal unit four tensor-operator times M.

$M_ij = ( {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}} ) M$

(17)

In this case due to the symmetry of a spherical object the @ operator can be simplified and all of the set equal to one. Thus the operator becomes Newtons gravitational constant times the Minkowski metric. The product of these operators can be simply computed as they both are diagonal. The trace of this operator is.

In[15]:=

Tr[( {{-G M, 0, 0, 0}, {0, G M, 0, 0}, {0, 0, G M, 0}, {0, 0, 0, G M}} )]

(18)

Out[15]=

(19)

In the units chosen for this paper (==1) Schwarzschild radius is well known to be [3].

(20)

This is exactly right. Now put the mass M in units of Planck mass and the expression becomes.

(21)

(22)

In general the Schwarzschild radius of a black hole with an energy equal to that of the state σ that is acting on will be.

(23)

This is proof positive that this theory is worth further investigation. So far all that has been discussed is static equilibrium. Now on to the development of the dynamics of space-time.

The Space of Physical Vectors

Before the dynamics of quantum space-time can be intelligently discussed we must define exactly what space-time is. In this theory space-time is referred to as the space of physical vectors. The @ operator matrix is a form of metric tensor derived from the Minkowski metric. Along with the definition of the displacement operators of the form defines a set of vectors and a metric. This set of vectors along with this metric will define an inner product space If it is closed under addition and scalar multiplication.

Proof that the set of vectors with the form   is a inner product space.

    Let γε

$x^μ = G^(1/2) (-γ {{n_0,, γn_1,, γn_2,, γn_3}})$

(24)

This is just a vector composed of quantum numbers. As such γ can be factored out.

$⇒γG^(1/2) (-{{n_0,, n_1,, n_2,, n_3}})$

(25)

Therefore the set is closed under scalar multiplication. As for linear combination.

$x^μ + x^ν⇒G^(1/2) (-{{n_0,, n_1,, n_2,, n_3}}) + G^(1/2) (-{{m_0,, m_1,, m_2,, m_3}})$

(26)

$⇒G^(1/2) ((-{{n_0,, n_1,, n_2,, n_3}}) + ({{-m_0,, m_1,, m_2,, m_3}}))$

(27)

$⇒G^(1/2) (-{{n_0 - m_0,, n_1 + m_1,, n_2 + m_2,, n_3}} + m_3)$

(28)

Let +=

$⇒G^(1/2) ({{-k_0,, k_1,, k_2,, k_3}})$

(29)

Which is of the same form as the two vectors input.  Therefore the set is closed under linear combination.  Therefore this is s inner product space.  Q.E.D

Furthermore as a metric has been defined on this space by the @ operator this is a compleat metric space and therefore meets the definition of a Hilbert space.  Four vectors denoted by =(-

)would form a subspace of this space.  These would be the coordinates of a single point and not the distance between points.

How about calculus?  is this space time differentiable? Integrable?  Yes and here is proof.

Differentiation is by simple definition..

(30)

Defined if the limit is finite.

In the space defined the change in x is and can be no smaller than n.

(31)

(32)

The denominator is larger than zero and the numerator is larger than zero. Therefore the limit will generally evaluate to a finite quantity. Therefore the derivative exist. Therefore the manifold as defined is differentiable and supports calculus. Q.E.D.

Though derivatives and integrals exist on this manifold they may not be exactly the same as they are in smooth space time. For example the derivative of .

(33)

(34)

So the difference between a derivative evaluated the traditional way and one evaluated on this quantized space time will be on the same order of magnitude as the Planck length. Furthermore this additive constant will not carry over through subsequent differentiations. For most purposes this little uncertainty could be safely ignored in all but the finest calculations.

The Dynamics of Quantized space time.

Gravity is at heart a dynamical force. To solve dynamical problems we need a Hamiltonian operator which takes into account gravitation. In the formula for the A operator above the Hamiltonian H was used. The H used in that section is the standard H used everywhere which does not take into consideration quantum gravitational forces. This is so because no one has known what potential energy to use for the gravitational force. ( even though semiclassical computations have been done using the old standby V=-mgz) One difficulty has been that the gravitational potential would depend on the energies due to the other four forces. So the gravitational potential and the Hamiltonian of the rest of the configuration would be interrelated.

Potential energy

The most rigorous approach to this problem would be to set up and solve a Poisson or Laplace equation with the proper boundary conditions and source terms.

(35)

Where ρ has dimensions of mass-energy per unit volume (volume in four dimensional space). The index μ will be used for vectors in the reference frame of the laboratory observer. The index ν will be used for the reference frame of the particle or field under study. is the velocity of the particle as the laboratory observer sees it. is the momentum of the laboratory as the particle sees it.
Just as is the case in electrodynamics the solution to this equation is the integral of a Greens function.

(36)

Where F[,] has been picked to satisfy whatever boundary conditions are in effect.
The A operator as defined above is also a solution.

Proof:

(37)

(38)

(39)

Because the indices μ are being summed over the covariant and contravariant character of the derivatives is summed away. All there is to do is take two derivatives with respect to .

(40)

ρ is just some arbitrary function of and that describes the distribution of mass-momentum that is generating the gravitational field. The right side of the last equation is such a function so the equality can be set. Therefore A is a solution to the Poisson equation. We also know that any operator that meets the definition of A will satisfy the equation. Consider the following: Anzats.

(41)

We can check this against the Poisson equation.

(42)

This is exactly the form of the function as was defined before.

    Depending on the type of calculation being under taken one of these forms for the quantum gravitational potential will be useful.  The Greens function approach to the solution is good for very theoretical calculations but not very practical.  The second version is good for a pure matrix mechanical approach.  The last potential operator would be good for a Schrodinger type equation.  The potential energy is the curvature operator times the A operator.

(43)

The Hamiltonian Operator.

The gravitational Hamiltonian operator can now be written. In this composition the systems I will consider will all have conservative forces. The Hamiltonian will just be the sum of the kinetic energy and potential energy. The kinetic energy will be

$T = c {p} = c ((- ℏ)^2∂/∂x_μ∂/∂x^μ)^(1/2)$

(44)

(45)

V will be the potential energy as written above. Which one to use will depend on the system under study.
Replace the momentum with it's operator.

$Overscript[H,^] = c {Overscript[p,^]} + V = c ((- ℏ)^2∂/∂x_μ∂/∂x^μ)^(1/2) + V$

(46)

From here a Schrodinger equation that takes the dynamics of quantized space time a.k.a. quantum gravity into account can be written.

(47)

What jumps out at me is that the Hamiltonian is an explicit function of . This relationship can be inverted. These two quantities are interdependent and so this equation is very NON linear. This equation will be very difficult to come up with exactly integrable solutions. This case cries out for a matrix approach. For the matrix approach this type of equation is solved by simply finding a diagonalizing unitary operator. I could not continue any further along this approach without picking a specific example.

Earth's Gravitational field.

The fall of a neutron.

In [Citation reference number 1], Nesvizhevsky et.al. Measured the eigenvalues of a beam of truckload neutrons falling in earth's gravitational field. The following is inspired by this experiment. In the experiment Nesvizhevsky and his team set up a ultra cold neutron source a neutron absorber and a mirror. As figure one shows.

For this part of the composition I will switch back to SI units for these calculations.

From In[25]:=

[Graphics:HTMLFiles/dynamicweb_81.gif]

Figure 1

The only potential acting in this experiment is gravitational and in the vertical direction. In this case the forces are all conservative. With conservative forces the Hamiltonian is equal to the total energy.

$H = c {p_z} - 1/(2 n_zG) U P z^2$

(48)

Remember in this equation p and P are not the same. P is the scalar relative momentum of the source of the gravitational field as seen from the test particle. U is the velocity of the earth as seen from the particle.

Substitute the momentum operator for p but not P. This gives the Hamiltonian operator.

(49)

The problem is to find an orthonormal basis that preferably gives H in a diagonal form. This will make life simpler in the future. Solving the problem in this way should give a full configuration interaction (CI), an exact solution to the problem. I propose the following set of vectors.

In[2]:=

$φ_1 = {2^(1/2)/h^(1/2) Sin[(z π)/h]}$

(50)

In[3]:=

$Overscript[φ_1, _] = {2^(1/2)/h^(1/2) Sin[(z π)/h]}$

(51)

In[4]:=

$φ_2 = {2^(1/2)/h^(1/2) Cos[(z π)/h]}$

(52)

In[5]:=

$Overscript[φ_2, _] = {2^(1/2)/h^(1/2) Cos[(z π)/h]}$

(53)

A quick test of orthonormality.

In[259]:=

(54)

Out[259]=

(55)

In[123]:=

(56)

Out[123]=

(57)

Using this orthonormal basis the matrix elements can be computed.

In[6]:=

(58)

Out[6]//MatrixForm=

$( {{-(h^2 P (2 - 3/π^2) U)/(12 G n), (h^2 P U)/(4 G n π) - ( c π ℏ)/h}, {(h^2 P U)/(4 G n π) + ( c π ℏ)/h, -(h^2 P (2 + 3/π^2) U)/(12 G n)}} )$

(59)

The eigenvalues and eigenvectors of this system will be the compleat exact solution of this system.

In[7]:=

(60)

Out[7]=

(61)

Notice that the spectrum of this system has two distinctive eigenvalues indexed by "n". ''n" would be the de Broglie wavelength of the neutron in a particular state. Taking the postulate of quantized space-time into account, "n" will be an integer multiple of the Planck length. "n" in this formula would be the number of planck length's long the de Broglie wavelength of the particle is at the given height. Obviously the minimum wavelength would be determined by the "rest" or ground state wavelength of the particle. In this experiment which eigenvalues are observed will depend on the altitude the experiment is performed at.

Plugging in integers one through infinity will give the various energies that the neutron would have in the gravitational well of the Earth. The lowest energy level will represent the energy required to lift a neutron out of Earth's gravitational well. That said the Hamiltonian operator solved in this example is not the most general. The potential energy formula used is only good for relatively small distances, for a system where gravity is the dominant interaction.

Numerical analysis

To put numbers to these eigenvalues and eigenvectors raw numbers and physical constants have to be fed to the computer algebra system. I will use a conversion factor between joules and electron volts.

In[11]:=

In[51]:=

I need the number of Planck length's long the neutron's ground state deBroglie wavelength is.

In[52]:=

(62)

In[53]:=

(63)

In[54]:=

(64)

In[55]:=

(65)

In[68]:=

(66)

In[70]:=

(67)

In[58]:=

(68)

In[59]:=

(69)

This will give the energy needed to boost a neutron from its ground state out of the earth's gravitational potential well.

In[71]:=

(70)

Out[71]=

(71)

Using a classical formula I can estimate what altitude this should correspond to. This will give the height the neutron would have to rise in meters.

In[75]:=

(72)

Out[75]=

(73)

For comparison a light year is 9.4605284*meters.

A neutron would have to be 593 Billion kilometers away from the Earth in order to be completely free of earth's influence.  However at that altitude the energy levels will be so finely spaced that the motion of the neutron will be virtually free motion.

I cannot say anything definite about the experiment unless I know the height of the table or workbench that the experiment was performed on.  I could guess that the height is about one meter.

In[227]:=

(74)

In[228]:=

(75)

In[229]:=

(76)

Out[229]=

(77)

Due to the limitations of my computer this is a close as it will come to stating the order of magnitude of the difference between two energy levels at this height. in spite of these limitations I can narrow down how high the table must be. The the n1 I choose corresponds to one meter. The amount added to n1 would be the margin of error for this estimate. So I would put the height of the table that the experiment was done on at about 1 meter give or take meters. Of course the actual strength of gravitations pull on an object would depend on the local energy momentum density. For example the earth's gravitational field is stronger where the ground is denser and weaker a high altitudes. Laboratories wishing to repeat the experiments referred to need to consider such Geological factors when comparing their results.

The energy required to boost a satellite to orbit

Boosting a satellite into orbit is a problem more easily handled using a classical approach. However any quantum theory of gravity must be able to handle classical situations and give results similar to the classical theories. First to derive the Hamiltonian matrix for this type of system. For this example I will use the type of potential represented by equation 38. The symmetry of this problem makes it effectively one dimensional.

In[13]:=

$ψ_1 = {1/r^(1/2) Exp[ (ξ π)/r]}$

(78)

In[14]:=

$Overscript[ψ_1, _] = {1/r^(1/2) Exp[- (ξ π)/r]}$

(79)

In[240]:=

$ψ_2 = {1/r^(1/2) Exp[- (ξ π)/r]}$

(80)

In[241]:=

$Overscript[ψ_2, _] = {1/r^(1/2) Exp[ (ξ π)/r]}$

(81)

Check the orthonormality.

In[15]:=

(82)

Out[242]=

(83)

In[243]:=

(84)

Out[243]=

(85)

The matrix elements of the kinetic energy are.

In[244]:=

(86)

Out[244]//MatrixForm=

(87)

In this basically one dimensional system the Hamiltonian matrix is diagonal.

(88)

The eigenvalues of the system are simply the diagonal entries.

(89)

The eigenvectors of this system are of very simple form.

(90)

Note how the Hamiltonian operator and the eigenvalues do not depend directly on the mass of the particles involved. These eigenvalues would be just as valid for completely massless particles as for massive particles. "n" in this formula would be the number of planck length's long the de Broglie wavelength of the particle is at the given height. Obviously the minimum wavelength would be determined by the "rest" or ground state wavelength of the particle. Notice that in this Hamiltonian operator explicit G has canceled out. Newtons gravitational constant is still a factor however. Keep in mind the postulate of quantized space time. Both the de Broglie wavelength and the altitude r would be integer multiples of the Planck length. The planck length is defined in terms of G.

Numerical analysis

Let me plug in some real numbers and see what size these energy levels are. For this formula I need the ground state deBroglie wavelength of the particle. In terms of planck length's this will have to be.

In[3]:=

(91)

Out[3]=

(92)

In[1]:=

-((299792458meter/s) π (6.58212 * 10^(-16) ev s))/r - ((299792458meter/s) π (6.58212 * 10^(-16) ev s))/(2.29711*10^10 r)

(93)

Out[1]=

(94)

Plug in the altitude of the satellite when it is in geostationary orbit. This equation will tell us the energy in Joules needed to put the satellite into geostationary orbit. This does not take into account the launch vehicle, fuel weight etc. This would have to be the net energy imparted to the satellite.

In[16]:=

(95)

Out[16]=

(96)

The negative sign indicates that this is a binding energy. For comparison the classically computed energy is the Sum of it's kinetic and potential compontents.

(97)

In[92]:=

(98)

Out[92]=

(99)

(100)

In[31]:=

(101)

Out[31]=

(102)

The sum of thses energies.

In[32]:=

(103)

Out[32]=

(104)

The percentage difference between my theories value and the classical value is.

In[35]:=

(105)

Out[35]=

(106)

So the value computed from the theory of quantized space-time is 13.2% smaller than the classically computed value. For a more careful comparison I would need to solve the Full configuration interaction of the satellite with the earth in all four space-time dimensions using the theory of quantized space time. Then I would need to use the Schwarzschild metric to find the energy of the satellite. All in all that would take up more time than I have right now. Stay tuned for further updates.

Final Thoughts

From first principles I have derived for the benefit of you the reader the theory of quantized space time.  This is a theory of canonically quantized space time using matrix-tensor group theory representations of the key quantities A @ and M.  With state space and Space-Time vectors the space of physical vectors has been shown to be a differentiable pseudo Riemannian, Hilbert Manifold.  The quantum mechanically correct hamiltonian of a gravitational field has been shown.

The Hamiltonian operator for a neutron in earth's gravitational field was derived and diagonalized.  The eigenvalues were used to determine an unknown quantity related to the experiment, the elevation the experiment must have been performed at.   The value found was a reasonable one meter.  Two predictions were made with this experiment as a starting point.  Firstly the escape energy of a neutron in earth's gravitational field.  Second the height it would rise to if boosted to this energy.  Lastly the states of a satellite in geostationary orbit is solved for and found to be exactly as one would expect from classical experience.

What I have presented here is a theory of quantum gravity which makes predictions about the real world on a scale that can be tested.  This theory makes predictions that no other theory does.  This theory agree's with classical expectations and quantum measurements.  Only time and experimentation can prove this (or any other) theory.  Experiment is the foundation of science.  With those qualifications I conclude that the theory of quantum space time is a proven as a theory can ever get to be.

    In my next paper I will attack the question of the thermodynamics of quantum black holes. I will address the entropy of a black hole  in light of this alternative  framework provided by Heisenberg matrix mechanics.

References

1 "Quantum States of Neutrons in Earth's gravitational field", V.V.Nesvizhevsky, Nature volume 415, 17 January 2002, Pages 297 to 299.

2 "Study of the neutron quantum states in the gravity field" http://www.arxiv.org/abs/hep-ph/0502081,V.V.Nesvizhevsky,A.K.Petukhov.

My thanks to the Physics department at Northern Illinois Univerisy where I did most of the foundational work for this theory

Created by Mathematica (November 27, 2005)