and old universes depleted of energy and matter

and antimatter

does that matter to doctor who?

who knows....

# ALMA MICROPHYSICA

Marți Theory ABYSSUS ABYSSUM INVOCAT String Theory has the potential to describe everything. Even though it has not been proven true, like all Theoretical FIELD

## joi, 31 martie 2016

### WHEN THE SOMATORY OF ALL COEFFICIENTS OF CONTRACTION TENDS TO INFINITY YOU HAVE A PONTUAL UNIVERSE -the expansion of the universe started to accelerate recently (meaning a few billion ...BUT IS ONLY A PONTUAL MEMENTO MORI OF ALL UNIVERSES THE SOMATORY OF ALL EXPANSION IS BIGGER THAN ALL THEPONTUAL CONTRACTIONS IN THE FORM OF BLACK HOLES AND OTHER ASS HOLES DA GAMMA RAY ..... Homogeneity requires the proportionality coefficient to be only a function of time.

Expansion of the Universe
One of the main goals of cosmology is to figure out how the universe expands as a function of time.
1.1 Expansion and Conservation
To describe the evolution of the average universe, one needs only two kinds of equations:
1. The equation that relates the density and pressure of constituents of the universe (such as
baryons, cold dark matter, photons, neutrinos, dark energy) to the expansion of the universe,
and
2. The equation that describes the energy conservation of the constituents.
Consider a line connecting two arbitrary points in space (which is expanding), and call it L. As the
universe expands, L changes with time. As you will derive in homework using

General Relativity, the equation of motion for L is given by

L¨(t) = − 4πG 3 L(t) X i [ρi(t) + 3Pi(t)] ,

(1.1) where ρi(t) and Pi(t) are the energy and pressure of the ith component of the universe, respectively. Here, note that the absolute value of L does not affect the equation of motion for L. Therefore, one may define a dimensionless “scale factor,” a(t), such that L(t) ≡ a(t)x, where x is a timeindependent separation called a “comoving” separation, which is in units of length. In cosmology, 1 we often encounter the Hubble expansion rate, H(t), which is defined by H(t) ≡ a˙(t) a(t) . (1.2) The dimension of this quantity is 1/(time). The age of the universe can be calculated from the above definition of H, which gives H(t)dt = da/a. Now, if we know H as a function of a instead of t, we obtain t = Z da aH(a) . (1.3) Another interpretation of H is found by writing L˙(t) = H(t)L(t), which tells us that H(t) gives a relation between the distance, L, and the recession velocity, L˙ . For this reason, it is often convenient to write H(t) in the following peculiar units: H(t) = 100 h(t) km/s/Mpc, where h is a dimensionless quantity. The current observations suggest that the present-day value of h is h(ttoday) ≈ 0.7.∗ Dividing both sides of equation (1.1) by L and using L(t) = a(t)x, we find one of the key equations connecting the energy density and pressure to the expansion of the universe: a¨(t) a(t) = − 4πG 3 X i [ρi(t) + 3Pi(t)] (1.4) As expected, positive energy density and positive pressure slow down the expansion of the universe.† This equation cannot be solved unless we know how ρi and Pi depend on time. How ρi depends on time is given by the energy conservation equation, while how Pi depends on time is usually given by the equation of state relating Pi to ρi and other quantities. As you will derive in homework, the energy conservation equation is given by X i ρ˙i(t) + 3a˙(t) a(t) X i [ρi(t) + Pi(t)] = 0 (1.5) Equation (1.5) is general and does not assume presence or absence of possible interactions between different components. If we assume that each component is conserved separately, then we have ρ˙i(t) + 3a˙(t) a(t) [ρi(t) + Pi(t)] = 0, (1.6) ∗The most precise value of h(ttoday) to date from the direct measurement using low-z supernovae and Cepheid variable stars is h(ttoday) = 0.742 ± 0.036 (Riess, Macri, et al., ApJ, 699, 539 (2009)). † If we ignore the effect of pressure relative to that of the energy density (which is always a good approximation for non-relativistic matter), and write ρ(t) in terms of the total mass enclosed with a radius L, P i ρi(t) = 3M 4πL3 , then equation (1.1) becomes L¨ = − GM L2 , which is the familiar Newtonian inverse-square law. Although one must not apply the Newtonian mechanics to describe the evolution of space (because Newtonian mechanism assumes static space), this is a convenient way to understand equations (1.1) and (1.4)for each of the ith component. Note that the second term contains the pressure, and thus how the energy density evolves depends on the pressure.‡ Looking at equations (1.4) and (1.5), one might think that we cannot solve for a(t) unless we have the equation of state giving Pi(t) as a function of ρi(t) etc. While in general that would be true, for these equations a little mathematical trick lets us combine equations (1.4) and (1.5) without knowing the evolution of P(t)! First, rewrite equation (1.4) as a¨(t) a(t) = 8πG 3 X i ρi(t) − 4πGX i [ρi(t) + Pi(t)] . (1.7) Using equation (1.5) on the second term of the right hand side, we get a¨(t) a(t) = 8πG 3 X i ρi(t) + 4πG 3 a(t) a˙(t) X i ρ˙i(t) a˙(t)¨a(t) = 8πGa(t)˙a(t) 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t) 1 2 (˙a 2 ) · = 4πG(a 2 ) · 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t). (1.8) As this has the form of A˙ = BC˙ + BC˙ = (BC) · , it is easy to integrate and obtain: a˙ 2 (t) = 8πGa2 (t) 3 X i ρi(t) − κ, (1.9) where κ is an integration constant, which is in units of 1/(time)2 . (A negative sign is for a historical reason.) Dividing both sides by a 2 (t), we finally arrive at the so-called Friedmann equation: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) − κ a 2(t) . (1.10) ‡While it is a wrong explanation, it is useful to compare this equation to the first law of thermodynamics: T dS = dU + P dV, where T, S, U, and V are the temperature, entropy, internal energy, and volume, respectively. To a very good accuracy, the entropy is conserved in the universe, dS = 0. The internal energy is U ∝ ρa3 and the volume is V ∝ a 3 , and thus d(ρa 3 ) + P d(a 3 ) = 0, which gives ρ˙ + 3a˙ a (ρ + P) = 0. This is a wrong explanation because it assumes that the pressure is doing work as a increases. However, in the average universe, the pressure is the same everywhere, and thus there is no under-pressure region against which the pressure can do work. Equation (1.5) must be derived using GR, which you will do in homework, but the above thermodynamic argument is an amusing way to arrive at the same equation. Also, this gives us some confidence that it is not crazy to think that the evolution of ρ depends on P. 3 A beauty of this equation is that it is easy to solve, once a time dependence of ρi(t) is known, which is usually the case. General Relativity tells us that the integration constant, κ, is equal to ±c 2/R2 where R is the curvature radius of the universe (in units of length) and c the speed of light. When the geometry of the universe is flat (as suggested by observations), R → ∞ (giving κ → 0), and thus one can ignore this term. Since we have so much to learn, to save time we will not consider the curvature of the universe throughout (most of) this lecture: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) (1.11) 1.2 Solutions of Friedmann Equation In order to use solve equation (1.11) for a(t), one must know how ρi(t) depends on time. To find solutions for a(t), let us first assume that the universe is dominated by one energy component at a time, i.e., a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) ≈ 8πG 3 ρi(t), (1.12) and further assume that ρi depends on a(t) via a power-law: ρi(t) ∝ 1 a ni (t) . (1.13) 4 Finding the solution is straightforward: a(t) ∝ t 2/ni . (1.14) This is usually an excellent approximation, except for the transition era where two energy components are equally important. There are 3 important cases: 1. Radiation-dominated (RD) era. A radiation component (photons, massless neutrinos, or any other massless particles) has a large pressure, PR = ρR/3,§ which gives ρR(t) ∝ 1/a4 (t), or nR = 4. We thus obtain aRD(t) ∝ t 1/2 . (1.15) The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 1 2t . (1.16) 2. Matter-dominated (MD) era. A matter component (baryons, cold dark matter, or any other non-relativistic particles) has a negligible pressure compared to its energy density, PM ρM, which gives ρM(t) ∝ 1/a3 (t), or nM = 3. We thus obtain aMD(t) ∝ t 2/3 . (1.17) §Again, a “wrong” derivation, but there is an intuitive way to get this result using the equation of state for non-relativistic ideal gas (this is obviously a wrong derivation because we are about to apply non-relativistic equation of state to relativistic gas!): P = nkBT = ρ kBT hEi , where n is the number density, T the temperature of gas, kB the Boltzmann constant, and hEi the mean energy per particle. For relativistic particles in thermal equilibrium, hEi ≈ 3kBT, which gives P ≈ ρ/3. Now, actually, it turns out that the error we are making by using non-relativistic equation of state for relativistic gas cancels out precisely the error we are making by using an approximate relation hEi ≈ 3kBT. This gives us the exact relation, P = ρ/3 for relativistic particles. More precisely, the equation of state for relativistic gas takes on the form P = (1 + )ρ kBT hEi with hEi = 3(1 + )kBT, giving P = ρ/3. Here, ' 0.05 and −0.10 for Fermions and Bosons, respectively. 5 The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 2 3t . (1.18) 3. Constant-energy-density-dominated (ΛD) era. A hypothetical energy component (let’s call it Λ) whose energy density is a constant over time, nΛ = 0. In this case we cannot use equation (1.14). Going back to equation (1.12) and setting ρΛ = constant, we get ˙a/a = constant, whose solution is aΛD(t) ∝ e Ht , (1.19) where an integration constant, H, is the same as the Hubble expansion rate (which is a constant for this model). The expansion of the universe accelerates, which must mean that, according to the acceleration equation (1.4), the pressure of this energy component is negative. The conservation equation (1.5) tells us that such a component indeed has an enormous negative pressure given by PΛ = −ρΛ. (1.20) While this looks quite strange, we now know that something like this may actually exist in our universe, as the current observations suggest that the present-day universe is indeed accelerating. 6 1.3 Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch.

Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch. 7 Now, taking t → t0, we find the present-day expansion rate H2 0 ≡ H2 (t0) = 8πG 3 [ρM(t0) + ρR(t0) + ρDE(t0)] ≡ 8πG 3 ρc(t0), (1.27) which has been determined to be H0 ≈ 70 km/s/Mpc. Here, ρc(t0) is the so-called “critical density” of the universe, which is equal to the total energy density of the universe when the universe is flat. The numerical value of the critical density is ρc(t0) ≡ 3H2 0 8πG = 2.775 × 1011 h 2 M Mpc−3 . (1.28) The critical density provides a natural unit for the energy density of the universe, and thus it is convenient to measure all the energy densities in units of ρc(t0). Defining the so-called density parameters, Ωi , as Ωi ≡ ρi(t0) ρc(t0) , (1.29) one can rewrite the Friedmann equation (1.26) in a compact form: H2 (t) H2 0 = ΩM a 3 (t0) a 3(t) + ΩR a 4 (t0) a 4(t) + ΩDEe −3 R a(t) a(t0) d ln a[1+wDE(a)] (1.30) Basically, most of the literature on cosmology (within the context of General Relativity) use this equation as the starting point.¶ Taking z = 0, one finds that all the density parameters must sum to unity: P i Ωi = 1. In summary, the Friedmann equation is a combination of two key equations: (1) the equation describing how the universe decelerates/accelerates depending on the energy density and pressure of the constituents, and (2) the equation describing the energy conservation of the constituents. Once the Friedmann equation is given with the proper right hand side containing the energy densities of the relevant constituents of the universe, we can find a(t) as a function of time easily. ¶An interesting possibility is that General Relativity may not be valid on cosmological scales. There are scenarios in which the form of the Friedmann equation is modified. One widely-explored example is the so-called DvaliGabadadze-Porrati (DGP) model (Dvali, Gabadadze & Porrati, Phys. Lett. B485, 208 (2000)). In this scenario, the Friedmann equation is modified to: H 2 (t) − H(t) rc = 8πG 3 X i ρi(t), where rc is some length scale below which General Relativity is restored. (For r rc, the potential is given by −GN m/r where GN is the ordinary Newtonian gravitational constant. For r rc, the potential is modified to −G5m/r2 and decays faster. G5 is the gravitational strength in the 5th dimension.) This model has attracted a huge attention of the cosmology community, as it was shown that this modified Friedmann equation gives an accelerating expansion without dark energy. Namely, even when the right hand side contains only matter, the solution for this equation can still exhibit an accelerating expansion. As this is a quadratic equation for H(t), we can solve it and find H(t) = 1 2 1 rc ± r 1 r 2 c + 32πG 3 ρM(t) . At late times when ρ(t) becomes negligible compared to the other term, one of the solutions is given by a(t) ∝ e t/rc , i.e., an exponential, accelerated expansion. 8 At present, the radiation is totally negligible compared to matter, ΩR/ΩM ' 1/3250, and the dark energy density is about 3 times as large as the matter density, ΩDE/ΩM ' 2.7 (with ΩM ' 0.27 and ΩDE ' 0.73). 1.4 Redshift As the universe expands, the wavelength of light, λ, is stretched linearly: λ(t) ∝ a(t), (1.31) which implies that photons lose energy as E(t) ∝ 1/a(t). This is something one can observe, by comparing, for example, the observed wavelength of a hydrogen line to the rest-frame wavelength that we know from the laboratory experiment. We often use the redshift, z, to quantify the stretching of the wavelength: 1 + z ≡ λ(t0) λ(temitted) . (1.32) The present-day corresponds to z = 0. Using equation (1.31), we can relate the observed redshift to the ratio of the scale factors: 1 + z = a(t0) a(temitted) . (1.33) Using this result in the Friedmann equation (1.30), we obtain the most-widely-used form of the Friedmann equation:

H2 (z) H2 0 = ΩM(1 + z) 3 + ΩR(1 + z) 4 + ΩDEe 3 R z 0 d ln(1+z)[1+wDE(z)From this result, it follows that the best way to determine the equation of state of dark energy is to measure H(z) over a wide range of z. If we can only measure the expansion rates at z 1, then Taylor expansion of equation (1.34) with ΩR ΩM and ΩDE ' 1 − ΩM gives H2 (z 1) H2 0 ≈ 1 + 3ΩMz + 3(1 + wDE)(1 − ΩM)z. (1.35) As we know from observations that |1 + wDE| is small (of order 10−1 or less), the third term is tiny compared to other terms, making it difficult to measure wDE. This is why we need to measure H(z) over a wide redshift range

General Relativity, the equation of motion for L is given by

L¨(t) = − 4πG 3 L(t) X i [ρi(t) + 3Pi(t)] ,

(1.1) where ρi(t) and Pi(t) are the energy and pressure of the ith component of the universe, respectively. Here, note that the absolute value of L does not affect the equation of motion for L. Therefore, one may define a dimensionless “scale factor,” a(t), such that L(t) ≡ a(t)x, where x is a timeindependent separation called a “comoving” separation, which is in units of length. In cosmology, 1 we often encounter the Hubble expansion rate, H(t), which is defined by H(t) ≡ a˙(t) a(t) . (1.2) The dimension of this quantity is 1/(time). The age of the universe can be calculated from the above definition of H, which gives H(t)dt = da/a. Now, if we know H as a function of a instead of t, we obtain t = Z da aH(a) . (1.3) Another interpretation of H is found by writing L˙(t) = H(t)L(t), which tells us that H(t) gives a relation between the distance, L, and the recession velocity, L˙ . For this reason, it is often convenient to write H(t) in the following peculiar units: H(t) = 100 h(t) km/s/Mpc, where h is a dimensionless quantity. The current observations suggest that the present-day value of h is h(ttoday) ≈ 0.7.∗ Dividing both sides of equation (1.1) by L and using L(t) = a(t)x, we find one of the key equations connecting the energy density and pressure to the expansion of the universe: a¨(t) a(t) = − 4πG 3 X i [ρi(t) + 3Pi(t)] (1.4) As expected, positive energy density and positive pressure slow down the expansion of the universe.† This equation cannot be solved unless we know how ρi and Pi depend on time. How ρi depends on time is given by the energy conservation equation, while how Pi depends on time is usually given by the equation of state relating Pi to ρi and other quantities. As you will derive in homework, the energy conservation equation is given by X i ρ˙i(t) + 3a˙(t) a(t) X i [ρi(t) + Pi(t)] = 0 (1.5) Equation (1.5) is general and does not assume presence or absence of possible interactions between different components. If we assume that each component is conserved separately, then we have ρ˙i(t) + 3a˙(t) a(t) [ρi(t) + Pi(t)] = 0, (1.6) ∗The most precise value of h(ttoday) to date from the direct measurement using low-z supernovae and Cepheid variable stars is h(ttoday) = 0.742 ± 0.036 (Riess, Macri, et al., ApJ, 699, 539 (2009)). † If we ignore the effect of pressure relative to that of the energy density (which is always a good approximation for non-relativistic matter), and write ρ(t) in terms of the total mass enclosed with a radius L, P i ρi(t) = 3M 4πL3 , then equation (1.1) becomes L¨ = − GM L2 , which is the familiar Newtonian inverse-square law. Although one must not apply the Newtonian mechanics to describe the evolution of space (because Newtonian mechanism assumes static space), this is a convenient way to understand equations (1.1) and (1.4)for each of the ith component. Note that the second term contains the pressure, and thus how the energy density evolves depends on the pressure.‡ Looking at equations (1.4) and (1.5), one might think that we cannot solve for a(t) unless we have the equation of state giving Pi(t) as a function of ρi(t) etc. While in general that would be true, for these equations a little mathematical trick lets us combine equations (1.4) and (1.5) without knowing the evolution of P(t)! First, rewrite equation (1.4) as a¨(t) a(t) = 8πG 3 X i ρi(t) − 4πGX i [ρi(t) + Pi(t)] . (1.7) Using equation (1.5) on the second term of the right hand side, we get a¨(t) a(t) = 8πG 3 X i ρi(t) + 4πG 3 a(t) a˙(t) X i ρ˙i(t) a˙(t)¨a(t) = 8πGa(t)˙a(t) 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t) 1 2 (˙a 2 ) · = 4πG(a 2 ) · 3 X i ρi(t) + 4πGa2 (t) 3 X i ρ˙i(t). (1.8) As this has the form of A˙ = BC˙ + BC˙ = (BC) · , it is easy to integrate and obtain: a˙ 2 (t) = 8πGa2 (t) 3 X i ρi(t) − κ, (1.9) where κ is an integration constant, which is in units of 1/(time)2 . (A negative sign is for a historical reason.) Dividing both sides by a 2 (t), we finally arrive at the so-called Friedmann equation: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) − κ a 2(t) . (1.10) ‡While it is a wrong explanation, it is useful to compare this equation to the first law of thermodynamics: T dS = dU + P dV, where T, S, U, and V are the temperature, entropy, internal energy, and volume, respectively. To a very good accuracy, the entropy is conserved in the universe, dS = 0. The internal energy is U ∝ ρa3 and the volume is V ∝ a 3 , and thus d(ρa 3 ) + P d(a 3 ) = 0, which gives ρ˙ + 3a˙ a (ρ + P) = 0. This is a wrong explanation because it assumes that the pressure is doing work as a increases. However, in the average universe, the pressure is the same everywhere, and thus there is no under-pressure region against which the pressure can do work. Equation (1.5) must be derived using GR, which you will do in homework, but the above thermodynamic argument is an amusing way to arrive at the same equation. Also, this gives us some confidence that it is not crazy to think that the evolution of ρ depends on P. 3 A beauty of this equation is that it is easy to solve, once a time dependence of ρi(t) is known, which is usually the case. General Relativity tells us that the integration constant, κ, is equal to ±c 2/R2 where R is the curvature radius of the universe (in units of length) and c the speed of light. When the geometry of the universe is flat (as suggested by observations), R → ∞ (giving κ → 0), and thus one can ignore this term. Since we have so much to learn, to save time we will not consider the curvature of the universe throughout (most of) this lecture: a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) (1.11) 1.2 Solutions of Friedmann Equation In order to use solve equation (1.11) for a(t), one must know how ρi(t) depends on time. To find solutions for a(t), let us first assume that the universe is dominated by one energy component at a time, i.e., a˙ 2 (t) a 2(t) = 8πG 3 X i ρi(t) ≈ 8πG 3 ρi(t), (1.12) and further assume that ρi depends on a(t) via a power-law: ρi(t) ∝ 1 a ni (t) . (1.13) 4 Finding the solution is straightforward: a(t) ∝ t 2/ni . (1.14) This is usually an excellent approximation, except for the transition era where two energy components are equally important. There are 3 important cases: 1. Radiation-dominated (RD) era. A radiation component (photons, massless neutrinos, or any other massless particles) has a large pressure, PR = ρR/3,§ which gives ρR(t) ∝ 1/a4 (t), or nR = 4. We thus obtain aRD(t) ∝ t 1/2 . (1.15) The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 1 2t . (1.16) 2. Matter-dominated (MD) era. A matter component (baryons, cold dark matter, or any other non-relativistic particles) has a negligible pressure compared to its energy density, PM ρM, which gives ρM(t) ∝ 1/a3 (t), or nM = 3. We thus obtain aMD(t) ∝ t 2/3 . (1.17) §Again, a “wrong” derivation, but there is an intuitive way to get this result using the equation of state for non-relativistic ideal gas (this is obviously a wrong derivation because we are about to apply non-relativistic equation of state to relativistic gas!): P = nkBT = ρ kBT hEi , where n is the number density, T the temperature of gas, kB the Boltzmann constant, and hEi the mean energy per particle. For relativistic particles in thermal equilibrium, hEi ≈ 3kBT, which gives P ≈ ρ/3. Now, actually, it turns out that the error we are making by using non-relativistic equation of state for relativistic gas cancels out precisely the error we are making by using an approximate relation hEi ≈ 3kBT. This gives us the exact relation, P = ρ/3 for relativistic particles. More precisely, the equation of state for relativistic gas takes on the form P = (1 + )ρ kBT hEi with hEi = 3(1 + )kBT, giving P = ρ/3. Here, ' 0.05 and −0.10 for Fermions and Bosons, respectively. 5 The expansion of the universe decelerates. With this solution, we can relate the age of the universe to the Hubble expansion rate: H(t) = a˙(t) a(t) = 2 3t . (1.18) 3. Constant-energy-density-dominated (ΛD) era. A hypothetical energy component (let’s call it Λ) whose energy density is a constant over time, nΛ = 0. In this case we cannot use equation (1.14). Going back to equation (1.12) and setting ρΛ = constant, we get ˙a/a = constant, whose solution is aΛD(t) ∝ e Ht , (1.19) where an integration constant, H, is the same as the Hubble expansion rate (which is a constant for this model). The expansion of the universe accelerates, which must mean that, according to the acceleration equation (1.4), the pressure of this energy component is negative. The conservation equation (1.5) tells us that such a component indeed has an enormous negative pressure given by PΛ = −ρΛ. (1.20) While this looks quite strange, we now know that something like this may actually exist in our universe, as the current observations suggest that the present-day universe is indeed accelerating. 6 1.3 Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch.

Equation of State of “Dark Energy” and Density Parameters The matter has PM ρM; the radiation has PR = ρR/3; and Λ has PΛ = −ρΛ. This motivates our writing the equation of state of the ith component in the following simple form: Pi = wiρi . (1.21) Here, wi is called the “equation of state parameter,” and can depend on time (although it is usually taken to be constant). Why this form? It is important to keep in mind that there is no fundamental reason why we should use this form. This form is often used either just for convenience, or simply for parametrizing something we do not know. At the very least, this form is exact for radiation, wR = 1/3, and for Λ, wΛ = −1. For matter, since wM 1, the exact value does not affect the results very much. The equation of state parameter is almost exclusively used for parametrizing “dark energy,” which is supposed to cause the observed acceleration of the universe. If we assume that w for dark energy, wDE, is constant, then the current observations suggest that (Komatsu, et al., ApJS, 192, 18 (2011)) wDE = −0.98 ± 0.05 (68% CL). (1.22) In other words, the energy density of dark energy is consistent with being a constant (wDE = wΛ = −1). Determining wDE with better accuracy may tell us something about the nature of dark energy, especially if wDE 6= 1 is found with high statistical significance, as it would tell us that dark energy is something dynamical (time-dependent). Ignoring a potential interaction between dark energy and other components in the universe (e.g., dark matter), the energy density of dark energy obeys (see equation (1.6)) ρ˙DE(t) + 3a˙(t) a(t) (1 + wDE) ρDE(t) = 0, (1.23) whose solution is ρDE(t) ∝ [a(t)]−3(1+wDE) . On the other hand, if we do not assume that wDE is a constant, then the energy density of dark energy obeys ρ˙DE(t) + 3a˙(t) a(t) [1 + wDE(t)] ρDE(t) = 0, (1.24) whose solution is ρDE(t) ∝ e −3 R d ln a[1+wDE(a)] . (1.25) Putting these results together, we obtain the Friedmann equation for our Universe containing radiation, matter, and dark energy (but not curvature) as a˙ 2 (t) a 2(t) = H2 (t) = 8πG 3 ρM(t0) a 3 (t0) a 3(t) + ρR(t0) a 4 (t0) a 4(t) + ρDE(t0)e −3 R a(t) a(t0) d ln a[1+wDE(a)] , (1.26) where t0 is some epoch, which is usually taken to be the present epoch. 7 Now, taking t → t0, we find the present-day expansion rate H2 0 ≡ H2 (t0) = 8πG 3 [ρM(t0) + ρR(t0) + ρDE(t0)] ≡ 8πG 3 ρc(t0), (1.27) which has been determined to be H0 ≈ 70 km/s/Mpc. Here, ρc(t0) is the so-called “critical density” of the universe, which is equal to the total energy density of the universe when the universe is flat. The numerical value of the critical density is ρc(t0) ≡ 3H2 0 8πG = 2.775 × 1011 h 2 M Mpc−3 . (1.28) The critical density provides a natural unit for the energy density of the universe, and thus it is convenient to measure all the energy densities in units of ρc(t0). Defining the so-called density parameters, Ωi , as Ωi ≡ ρi(t0) ρc(t0) , (1.29) one can rewrite the Friedmann equation (1.26) in a compact form: H2 (t) H2 0 = ΩM a 3 (t0) a 3(t) + ΩR a 4 (t0) a 4(t) + ΩDEe −3 R a(t) a(t0) d ln a[1+wDE(a)] (1.30) Basically, most of the literature on cosmology (within the context of General Relativity) use this equation as the starting point.¶ Taking z = 0, one finds that all the density parameters must sum to unity: P i Ωi = 1. In summary, the Friedmann equation is a combination of two key equations: (1) the equation describing how the universe decelerates/accelerates depending on the energy density and pressure of the constituents, and (2) the equation describing the energy conservation of the constituents. Once the Friedmann equation is given with the proper right hand side containing the energy densities of the relevant constituents of the universe, we can find a(t) as a function of time easily. ¶An interesting possibility is that General Relativity may not be valid on cosmological scales. There are scenarios in which the form of the Friedmann equation is modified. One widely-explored example is the so-called DvaliGabadadze-Porrati (DGP) model (Dvali, Gabadadze & Porrati, Phys. Lett. B485, 208 (2000)). In this scenario, the Friedmann equation is modified to: H 2 (t) − H(t) rc = 8πG 3 X i ρi(t), where rc is some length scale below which General Relativity is restored. (For r rc, the potential is given by −GN m/r where GN is the ordinary Newtonian gravitational constant. For r rc, the potential is modified to −G5m/r2 and decays faster. G5 is the gravitational strength in the 5th dimension.) This model has attracted a huge attention of the cosmology community, as it was shown that this modified Friedmann equation gives an accelerating expansion without dark energy. Namely, even when the right hand side contains only matter, the solution for this equation can still exhibit an accelerating expansion. As this is a quadratic equation for H(t), we can solve it and find H(t) = 1 2 1 rc ± r 1 r 2 c + 32πG 3 ρM(t) . At late times when ρ(t) becomes negligible compared to the other term, one of the solutions is given by a(t) ∝ e t/rc , i.e., an exponential, accelerated expansion. 8 At present, the radiation is totally negligible compared to matter, ΩR/ΩM ' 1/3250, and the dark energy density is about 3 times as large as the matter density, ΩDE/ΩM ' 2.7 (with ΩM ' 0.27 and ΩDE ' 0.73). 1.4 Redshift As the universe expands, the wavelength of light, λ, is stretched linearly: λ(t) ∝ a(t), (1.31) which implies that photons lose energy as E(t) ∝ 1/a(t). This is something one can observe, by comparing, for example, the observed wavelength of a hydrogen line to the rest-frame wavelength that we know from the laboratory experiment. We often use the redshift, z, to quantify the stretching of the wavelength: 1 + z ≡ λ(t0) λ(temitted) . (1.32) The present-day corresponds to z = 0. Using equation (1.31), we can relate the observed redshift to the ratio of the scale factors: 1 + z = a(t0) a(temitted) . (1.33) Using this result in the Friedmann equation (1.30), we obtain the most-widely-used form of the Friedmann equation:

H2 (z) H2 0 = ΩM(1 + z) 3 + ΩR(1 + z) 4 + ΩDEe 3 R z 0 d ln(1+z)[1+wDE(z)From this result, it follows that the best way to determine the equation of state of dark energy is to measure H(z) over a wide range of z. If we can only measure the expansion rates at z 1, then Taylor expansion of equation (1.34) with ΩR ΩM and ΩDE ' 1 − ΩM gives H2 (z 1) H2 0 ≈ 1 + 3ΩMz + 3(1 + wDE)(1 − ΩM)z. (1.35) As we know from observations that |1 + wDE| is small (of order 10−1 or less), the third term is tiny compared to other terms, making it difficult to measure wDE. This is why we need to measure H(z) over a wide redshift range

## vineri, 16 octombrie 2015

### QUO VADIS CON TEXTUS POLITICUS AB OVUM VIPERA,,,DOS ESTADOS ANORMAIS FAKEFUCKIANUS EM QUE A NORMA É A INCERTEZA DE TAIS ESTADOS POLÍTICOS IN FAKEFUCK QUANTUM MECHANICS UMA PARTÍCULA DE PÓ ATÓMICO PROTON NEUTRON SOCATRON VASCON ETC PARA CADA UMA E TUTTI DELAS THERE ARE SEVERAL POSSIBLE WITH SS IN IT (OR ALLOWED BI THE NATIONAL-SOCIALIST PARTY) STATES OF TO BE OR NOT TO BE ...AND EVERY ATOMIC PARTICLE OR CON OR MORON OR VASCON CAN BE YES IT CAN BE IN SEVERAL OF THESE STATES OR EVEN IN ALL OF THEM SIMULTÂNEA MENTE ASSIS PODEM ESTAR NO GOVERNO E FORA DO GOVERNO E ATÉ JÁ ESTAREM GOVERNANDO-SE

COMO ALTERAR A NATUREZA ATÓMICA

DA POLÍTICA ECONÓMICA MUNDIAL

DESDE 1939 BOMBARDEOU-SE

ECONOMICAMENTE O DEUTÉRIO

COM NEUTRÕES ATÉ OBTER TRÍTIO

SUFICIENTE PARA CRIAR SÓIS

EFÉMEROS NA ECONOMIA TERRESTRE

TAL COMO A BANCA GLOBAL

O TRÍTIO É UM ISÓTOPO INSTÁVEL

QUE SE DESINTEGRA À MENOR CRISE

EMITINDO UM ELECTRÃO SOLITÁRIO

E UM ANTINEUTRINO

TRANSFORMANDO-SE NUM RELES GÁS

DE ENCHER BALÕES OU OUTRAS BOLHAS

ECONÓMICAS CHINESAS ASSIM O Helium 3

DEIXA DE TER AS PROPRIEDADES

FANTÁSTICAS QUE FAZEM DO TRÍTIO

UMA LUZ FULGURANTE NO FIM DE

QUALQUER TÚNEL ECONÓMICO

TAMBÉM SE OBTEM TRÍTIO

IRRADIANDO O LITIUM - 6 COM NEUTRÕES

TARTARUGÕES OU SLOW NEUTRONS

IN ANGLO-SEXON TANTRICUS

OS NEUTRÕES LENTOS SÃO CAPTURADOS

PELO Li- 6 ou Li meia-DÚZIA E

OBVIAMENTE ADMITO-O

DESINTEGRAM-SE COMO A BANCA

MUNDIAL

EM BONS H-3 ou tritium's

E EM MAUS ou MESMO MAU-MAUS

Helium-3

DA POLÍTICA ECONÓMICA MUNDIAL

DESDE 1939 BOMBARDEOU-SE

ECONOMICAMENTE O DEUTÉRIO

COM NEUTRÕES ATÉ OBTER TRÍTIO

SUFICIENTE PARA CRIAR SÓIS

EFÉMEROS NA ECONOMIA TERRESTRE

TAL COMO A BANCA GLOBAL

O TRÍTIO É UM ISÓTOPO INSTÁVEL

QUE SE DESINTEGRA À MENOR CRISE

EMITINDO UM ELECTRÃO SOLITÁRIO

E UM ANTINEUTRINO

TRANSFORMANDO-SE NUM RELES GÁS

DE ENCHER BALÕES OU OUTRAS BOLHAS

ECONÓMICAS CHINESAS ASSIM O Helium 3

DEIXA DE TER AS PROPRIEDADES

FANTÁSTICAS QUE FAZEM DO TRÍTIO

UMA LUZ FULGURANTE NO FIM DE

QUALQUER TÚNEL ECONÓMICO

TAMBÉM SE OBTEM TRÍTIO

IRRADIANDO O LITIUM - 6 COM NEUTRÕES

TARTARUGÕES OU SLOW NEUTRONS

IN ANGLO-SEXON TANTRICUS

OS NEUTRÕES LENTOS SÃO CAPTURADOS

PELO Li- 6 ou Li meia-DÚZIA E

OBVIAMENTE ADMITO-O

DESINTEGRAM-SE COMO A BANCA

MUNDIAL

EM BONS H-3 ou tritium's

E EM MAUS ou MESMO MAU-MAUS

Helium-3

## marți, 29 septembrie 2015

### SE A IGNORÂNCIA É INFINITA PORQUE É QUE EINSTEIN NÃO CONSEGUIU SER AINDA MAIS IGNORANTE? DEUS NÃO JOGA AOS DADOS QUÂNTICOS ? E SE DEUS FOR NEWTONIANO ? EXERCÍCIO DE IGNORÂNCIA PURA TENTANDO CHEGAR AO LIMITE DA DITA CUJA ....DA IGNORÂNCIA PORRA PARA O INFINITO E MAIS ALÉM ...DOS LIMITES DA IGNORÂNCIA ---PROSA RELATIVISTICA A CONSERVAR EM BIFES E BITOQUES ENROLADOS NA PRÓPRIA GORDURA ,,,FACTORES QUE INFLUENCIAM A EFICÁCIA DA IGNORÂNCIA E LIMITAM O SEU DESENVOLVIMENTO ...CONCENTRAÇÃO DO PRODUTO OU DO IGNORANTE TANTO FAZ SE É GORDÃO ESTÁ MAIS CONCENTRADO E LOGO MAIS LIMITADO NA SUA IGNORÂNCIA ...TEMPERATURA A QUE ESTÁ O IGNORANTE A 42º CELSIUS O IGNORANTE DEIXA DE SER FUNCIONAL DEVE MANTER-SE O IGNORANTE E A SUA TEMPERATURA INTERNA A 37º C ABAIXO E ACIMA DISSO O IGNORANTE FICA MUITO LIMITADO NA SUA EXPRESSÃO DA IGNORÂNCIA-----ACHO QUE TEM LIMITES SER IGNORANTE AO INFINITO CANSA BUÉ ..SE ESCREVE A IGNORÂNCIA AGRADECE O AUTO DE FÉ Gosto · Responder · há 2 minutos Ambrosius Catharinus Politus Ambrosius Catharinus Politus OU A SANTA INQUISIÇÃO SE HOUVESSE PINGO DE CIÊNCIA EM TI Ó TORQUEMADA CON SEGUIAS RESPONDER Ó CON COM TERMINAÇÃO EM ÁGORA Gosto · Responder · há cerca de um minuto Ambrosius Catharinus Politus Ambrosius Catharinus Politus EIN STEIN ESTAVA ERRADO FILHA ...Ambrosius Catharinus Politus Dina Nunes SE A IGNORÂNCIA HUMANA NÃO TIVESSE LIMITES CONSEGUIRIAS ESCREVER MAIS Ó SUA IGNORANTEDON'T LOOK A IGNORÂNCIA ESTÁ MESMO AÍ ATRÁS Gosto · Responder · 1 min Ambrosius Catharinus Politus Ambrosius Catharinus Politus SERÁ QUE PHODIAM BLOQUEAR A CIÊNCIA É QUE ESTOU A CHEGAR AOS LIMITES QUOD ERAT DEMONSTRANDUM A IGNORÂNCIA ESTÁ MUITO LIMITADA NEM QUE CHEGASSE AOS MIL ANOS CHEGAVA A INFINITAR ISTO .

a ciência queima tantos hereges

Será que o administrador da página

Isso é tudo muito bonito, mas

poderia bloquear

o senhor Ambrosius Catharinus Politus ?

A Ciência agradece!

A CIÊNCIA FAZ UMA MAMADA?

A GENTE AUTO-BLOQUEIA-SE JÁ

NÃO AINDA NÃO CHEGUEI AO LIMITE

SE CALHAR SOU DEUS

OU O ESPÍRITO SANTO

OU O GRUPO GES ENCARNADO

OU DESCARNADO

UMA MAMADA IGNORANTE DESSAS

Será que o administrador da página

Isso é tudo muito bonito, mas

poderia bloquear

o senhor Ambrosius Catharinus Politus ?

A Ciência agradece!

A CIÊNCIA FAZ UMA MAMADA?

A GENTE AUTO-BLOQUEIA-SE JÁ

NÃO AINDA NÃO CHEGUEI AO LIMITE

SE CALHAR SOU DEUS

OU O ESPÍRITO SANTO

OU O GRUPO GES ENCARNADO

OU DESCARNADO

UMA MAMADA IGNORANTE DESSAS

### DON'T LOOK AT THE WORD TRAIN IN THE TRAIL OF WORDS ....DAS DIRECÇÕES VIRTUAIS NOS UNIVERSOS INEXISTENTES NÃO VÁS POR AÍ DIZEM-ME NENHUNS QUE SE VÃO ÀS DOCES O CAOS NO FAKEFUCK SURGE E DESAPARECE AO ABSERVADOR OU AO ABSURDADOR OU MESMO AO OBSERVADOR ABNORMAL VINDO DE LADOS ADVERSOS E DE DIREÇÕES CONTRÁRIAS E MUITAS VEZES MESMO APARTIDÁRIAS É UMA UCRONIA UTÓPICA O FAKELOOK ,,,LOOK AT THE TRAIL OF WORDS ...NUM UNIVERSO MULTIDIMENSIONAL INEXISTENTE COMO O FACELOOK OU O FAKEFUCK QUEM FICA EMBAIXO PODE SER MAIS IGUAL DO QUE O QUE FICA DESTE LADO OU DAQUELE LADO OU MESMO EM CIMA DO OUTRO LADO É UM NÃO LUGAR ATÍPICO ...DA MULTIDIMENSIONALIDADE NOS UNIVERSOS VIRTUALMENTE INEXISTENTES - ESTUDO Nº69 FEITO COM O APOIO DA FUNDAÇÃO DA UNIVERSIDADE DAS NODOAS DE SANGUE FRESCO - COMO É BOM DE NOTAR O FAXEBOOK APARENTEMENTE SURGE-NOS COM A ILUSÃO DE TER VÁRIOS LADOS NO UNIVERSO FAKEBOOKIANO HÁ OS DESTE LADO OS DAQUELE LADO, OS QUE FICAM POR CIMA E OS QUE FICAM POR BAIXO, OS QUE LEVEM NO VIEGAS POR VIA DA TAL LUZ NO FUNDO DO TÚNEL OU DO WORMHOLE OU DO ASSHOLE INTER-UNIVERSAL É TODO UM CONJUNTO DE DIMENSÕES FICTÍCIAS E ILUSÓRIAS QUE PARECEM APARECER E DESAPARECER AO OBSERVADOR DESATENTO VOTO LIVRE NESTE LADO DA PAPA NESTLÉ DO FAKEBOOK JÁ NAQUELE CANTO DOS LUSÍADAS VOTO PRESO COMO O SOCRATES JÁ NO LADO ONDE A MARGEM SUL COALESCE COM O NORTE MONÁRQUICO E LAZARENTO LEZÍRIALENTO VOTO NA ARQUITECTURA PAISAGISTA DOS TELLES DA ALBERGARIA OU NOS BARROS VERMELHOS DO PSD ALGARVIO FELIZMENTE NO FAKELOOK HÁ URNAS E VOTOS PARA TODOS OS DESGOSTOS Í O PAPA ANDAR EM DIGRESSÃO GLOBAL NUMA CRUZADA CONTRA O ESTADO ISLÂMICO QUE AVANÇA POR TODOS OS LADOS E CANTOS DO FAKELOOK TAL É A GRANDEZA DO FAKELOOK QUE FAZ DE ANÕES GIGANTES E QUE DAS GIGANTESCAS FRAUDES FINANCEIRAS FAZ MICROSCÓPICAS INFLAÇÕES E ATÉ DEFLAÇÕES VIRTUAIS E DE DÍVIDAS E DÚVIDAS COLOSSAIS GERA BITCOIN'S E BIT CON'S VIRTUAIS O ZÉ POVINHO ESTÁ FINO E FLUTUA EM TODAS AS INUNDAÇÕES QUE O VOTO LIVRE NOS TRÁS OU MESMO POR DETRÁS TAL É A VIRTUALIDADE DOS LADOS MULTIDIMENSIONAIS VON FACEFUCK

## vineri, 18 septembrie 2015

### DOS ANÁLOGOS DAS COUSAS VIVAS E MORTAS QUE PULULAM NUM MUNDO CHEIO DE QUIMERAS, NAS QUIMERAS QUE SE ENCHEM DE MUNDOS E FUNDOS A BEM DA NOÇÃO DA NAÇÃO DE QUIMERAS FEITA É TODO UM COLECTIVO DE QUIMERAS QUE SE APROPRIA DAS VIDAS DE OUTROS PARA CONTINUAREM SEMI-VIVAS NOS SEUS MUNDOS QUIMÉRICOS MAIS QUE VIRTUAIS ...DAS QUIMERAS BALOFAS QUE FOSSAM COMO PORCOS E ZURRAM COMO BURRAS E SÃO MONTADAS COMO MULAS. HÁ A IDEIA DE QUE TODAS AS QUIMERAS PODERÃO SER SALVAS DA EXTINÇÃO SE AS CONSERVARMOS EM JARDINS ZOOLÓGICOS VICTORIANOS E SE FOREM MONTADAS REGULARMENTE POR NEONAZIS ESTALINISTAS OU MESMO POR ESTALINISTAS POLPOTIANOS OU POR KIM IL SUNG'S DE SUNGA OU DE CUECAS SUJAS NAS CLOACAS CAPITALISTAS DE DENG XIAO PING MAS AS QUIMERAS TAL COMO AS MULAS NÃO SE REPRODUZEM , CRUZAMENTOS DE BESTAS MUITO BRUTAS DE ESPÉCIES IGNOTAS ESTÃO CONDENADAS À EXTINÇÃO POIS SE CÉSAR ERA UM FASCISTA LAMBE CUS E SE TODOS OS GREGOS ALMEJAM IR AO CU ÀS JUVENTUDES HITLERIANAS O CERTO É QUE SÓ ALUCINADOS PRODUZEM CHORRILHOS DE PALAVRAS DUM PASSADO QUE TERÁ POUCAS SOMBRAS NUM FUTURO DE QUIMERAS MORTAS

TODAS AS QUIMERAS MORTAS OU VIVAS

MARCHAM PARA O PASSADO

TAL É A NATUREZA QUIMÉRICA

DESSAS BRUTAS COUSAS

DE IGNOTOS BRUTUS FEITAS

ANCORADAS EM QUIMÉRICOS FUTUROS

MUITO POUCO PRESENTES

E PESSIMAMENTE PASSADOS

DOS CORNOS OU DOS PIORNOS

TANTO

FAZ

TAL É A NATUREZA QUIMÉRICA

DESSAS BRUTAS COUSAS

DE IGNOTOS BRUTUS FEITAS

ANCORADAS EM QUIMÉRICOS FUTUROS

MUITO POUCO PRESENTES

E PESSIMAMENTE PASSADOS

DOS CORNOS OU DOS PIORNOS

TANTO

FAZ

## luni, 22 iunie 2015

### E A MORTE SAIU À RUA NUM DIA ASSIM OU SE CALHAR ASSAD E VINHA EM MANGAS DE CAMISA E SEM SAMARRA E COM UNS CALÇÕES QUE PARECIAM MESMO UM BIKINI DO ATÓMICO ATOL E A MORTE NEM SEQUER FALAVA SUECO SÓ FALAVA HOLANDÊS E MESMO ASSIM FALAVA MAL PRÓ CARAÇAS ERA UMA LÍNGUA MORTA QUE SE ASSOMAVA EM LÍNGUAS DE PROJECÇÕES SOLARES MUITO MAL PROJECTADAS ...O ESTADO ISLÂMICO A QUE CHEGÁMOS FALHOU REDONDAMENTE EM MESSEJANA ......20 JUNHO FALHA DE MESSEJANA 42º À SOMBRA E AO SOL DAVA PARA ESTRELAR OVOS e mexia mexia o horizontal e fugidio em ondas de calor até as perdizes já vinham cozidas e nenhum manel alegre desatou a chumbá-las alegremente ...a malta do regime estava a fazer a sesta e só caçava no sereno ...já em évora socrates continuava à sombra e refrescava-se com pornographia caseira dum von zeppelintra qualquer a thalassocracia já nã mexe como mexia ...tamém deve tar falhada...

A VIDA TAMBÉM SAI À RUA

MAS PARA CONTINUAR VIVA

ESCONDE-SE EM TODAS AS SOMBRAS

QUE O SOL LANÇA PARA A EXISTÊNCIA

MAS PARA CONTINUAR VIVA

ESCONDE-SE EM TODAS AS SOMBRAS

QUE O SOL LANÇA PARA A EXISTÊNCIA

Abonați-vă la:
Postări (Atom)