Sunday, July 24, 2022
Saturday, July 23, 2022
Thursday, November 25, 2010
Brooklyn's Finest
Brooklyn's Finest
[Blu-ray]
DVD — Richard Gere
Fighting failed government policy with more police, guns, corruption, and endless loss of lives., March 8, 2010 I assume the title New York's Fr-lest" was avoided for trademark conflict. The "finest" motto is the subject of ridicule, as those poor cops have nothing [fine] going for them on any front.
The Ex-NYC Police Chief Bernard Kerik's imprisonment is an example of the ugliness of working in a bureaucratically corrupt system. The charisma, ultimate sacrifice in securing, protecting, and helping society to conform to law and order, are rewarded with the ultimate despair of each policeman's family, poor health, contempt to their exceptional commitments by government officials, and suspicion and mistrust of the public. On its façade, the policemen are given figurative lip service and glorified for political gains. In reality, the policemen are left to the perils of excessive and outmatching fire power of criminals and drug dealers, the shear number of unemployed, uneducated, and troubled citizens, and the endless and complicated social decay. The profiteers are the career-thirsty state-officials, sitting on their Mahogany, shinning desks, isolated from the reality of urban warfare. Every fine cop portrayed in the movie, and fits perfectly real life scenario, is exposed to the ultimate death, yet compensate hourly and on overtime basis.
Most cops work irregular shifts, smoke, and stressed to the end, could not support their family, and resort to prostitution to fill their missing human needs.
Furthermore, the perfect kids of this great Country of ours, who were raised to glorify the saintship of being on the right-side of the law, such as cops, are faced with the reality of political corruption that serves state officials. On the other side, the movie displays the endless blood battles that claim the lives of many innocent blacks, whose only crime was possessing, dealing, or using drugs. Not as if the government cares about eradicating drugs from society, since that requires creating economic situation conducive to financial security, but because state officials could play the PR game of staying in office and gaining more power.
Most of the dead and dying, poor blacks, have no access to opportunities or any glimpse of hope that could brighten their future. The movie succeeded in questioning the rationale of trusting the government to solve any social or economic problem without the creative and proactive role of the public in holding its governments accountable in making and enforcing laws that could do good to the people.
Monday, November 22, 2010
Nuclear Fusion Query
Dealing with beam-based nuclear fusion compelled me to reexamine the values used in the following four conventional equations:
The following values are used to check the meaning of the terms in the above equations:
electron mass, kg: 9.10938E-31
neutron mass, kg: 1.67493E-27
proton mass, kg: 1.67262E-27
deuterium mass, kg: 3.34449E-27
tritium mass, kg: 5.00827E-27
eV in Joule: 6.24219725343321E+18
unit of mass eV 8.37185E+25
speed of light, m/sec: 299792458
helium 3 mass, kg: 5.00823E-27
helium 4 mass, kg: 6.64648E-27
Mass Balance of Equation (1)
Mass difference = 3.34449E-27+ 5.00827E-27-6.64648E-27-1.67493E-27
= 3.13563E-29 kg
Energy difference, J = 3.13563E-29* sq(299792458) = 2.25318E-12 Joules
Energy difference, MeV = 2.25318E-12*6.24E+18/E+06= 17.6 MeV
In the above equation, it was assumed that the fused nucleus was at zero velocity such that:
0 = (mass of He-4)* (velocity of He-4) + (mass of neutron)* (velocity of neutron)
The above momentum conservation equation yields:
Squaring and multiplying by (mass of neutron)/ (mass of He-4) gives"
(Energy of neutron) = 17.6(3.968213597/(1+3.968213597))
= 14.0574792 MeV
(Energy of He-4) = 17.6/(1+3.968213597)
= 3.542520799 MeV
As we see, we did not include the kinetic energies
of the reactants D and T. We only assumed that the binding energy
is the net energy output. Recall that this reaction has peek fusion cross section
at kinetic energy 100 KeV. The proper way to write equation (1) is therefore:
Where E(D-T) is the kinetic energy of the original projectiles D and T. This raises this question:
For example, in beam-based fusion, one could either make both D and T have equal energies in the lab reference or make D possess greater speed, such that the net energy of D and T remains the same but the proportion of energies of D to T is different. I the projectiles D and T collide with net zero velocity after fusion, then:
How would we retrieve E(D-T) that was spent in energizing the projectiles?
Let us leave this question for now and turn to the next equation.
Mass Balance of Equation (2)
From the outset we must conserve the electronic charge by adding an electron on the right hand side.
Mass difference = 2* 3.34449E-27-5.00827E-27- 1.67262E-27 - 9.10938E-31 = 7.17906E-30 kg
Energy difference, J = 7.17906E-30* sq(299792458) = 6.45222E-13 Joules
Energy difference, MeV = 6.45222E-13*6.24E+18/E+06= 4.03 MeV
(Energy of proton) = 1.010647969 MeV
(Energy of T) = 3.0219763 Mev
(Energy of electron) = 0.000550 MeV
Let us now modify Equations (3) and (4) so as to represent the real reactants:
All help, suggestions, and questions are warmly appreciated.
| (1) | 21D | + | 31T | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) | ||||||
| (2) | 21D | + | 21D | → | 31T | ( | 1.01 MeV | ) | + | p+ | ( | 3.02 MeV | ) | 50% | |||||
| (3) | 21D | + | 21D | → | 32He | ( | 0.82 MeV | ) | + | n0 | ( | 2.45 MeV | ) | 50% | |||||
| (4) | 21D | + | 32He | → | 42He | ( | 3.6 MeV | ) | + | p+ | ( | 14.7 MeV | ) | ||||||
The following values are used to check the meaning of the terms in the above equations:
electron mass, kg: 9.10938E-31
neutron mass, kg: 1.67493E-27
proton mass, kg: 1.67262E-27
deuterium mass, kg: 3.34449E-27
tritium mass, kg: 5.00827E-27
eV in Joule: 6.24219725343321E+18
unit of mass eV 8.37185E+25
speed of light, m/sec: 299792458
helium 3 mass, kg: 5.00823E-27
helium 4 mass, kg: 6.64648E-27
Mass Balance of Equation (1)
| (1) | 21D | + | 31T | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) |
= 3.13563E-29 kg
Energy difference, J = 3.13563E-29* sq(299792458) = 2.25318E-12 Joules
Energy difference, MeV = 2.25318E-12*6.24E+18/E+06= 17.6 MeV
In the above equation, it was assumed that the fused nucleus was at zero velocity such that:
0 = (mass of He-4)* (velocity of He-4) + (mass of neutron)* (velocity of neutron)
The above momentum conservation equation yields:
(velocity of neutron)/(velocity of He-4) = - (mass of He-4)/(mass of neutron)
Squaring and multiplying by (mass of neutron)/ (mass of He-4) gives"
(Energy of neutron)/(Energy of He-4) = (mass of He-4)/(mass of neutron)
= 6.64648E-27/1.67493E-27 = 3.968213597(Energy of neutron) = 17.6(3.968213597/(1+3.968213597))
= 14.0574792 MeV
(Energy of He-4) = 17.6/(1+3.968213597)
= 3.542520799 MeV
As we see, we did not include the kinetic energies
of the reactants D and T. We only assumed that the binding energy is the net energy output. Recall that this reaction has peek fusion cross section at kinetic energy 100 KeV. The proper way to write equation (1) is therefore:| 21D | + | 31T | + E(D-T) | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) | + E(D-T) |
Where E(D-T) is the kinetic energy of the original projectiles D and T. This raises this question:
How to determine E(D-T) in the case of beam-based fusion when either of the energies of D and T projectiles could be manipulated precisely?
For example, in beam-based fusion, one could either make both D and T have equal energies in the lab reference or make D possess greater speed, such that the net energy of D and T remains the same but the proportion of energies of D to T is different. I the projectiles D and T collide with net zero velocity after fusion, then:
How would we retrieve E(D-T) that was spent in energizing the projectiles?
Let us leave this question for now and turn to the next equation.
Mass Balance of Equation (2)
From the outset we must conserve the electronic charge by adding an electron on the right hand side.
| (2) | 21D | + | 21D | + | E | (D-D) | → | 31T | ( | 1.01 MeV | ) | + | p+ | ( | 3.02 MeV | ) |
|
Mass difference = 2* 3.34449E-27-5.00827E-27- 1.67262E-27 - 9.10938E-31 = 7.17906E-30 kg
Energy difference, J = 7.17906E-30* sq(299792458) = 6.45222E-13 Joules
Energy difference, MeV = 6.45222E-13*6.24E+18/E+06= 4.03 MeV
(Energy of proton) = 1.010647969 MeV
(Energy of T) = 3.0219763 Mev
(Energy of electron) = 0.000550 MeV
Again, to determine E(D-D), in beam-based nuclear fusion, the kinetic energy for the projectiles D and D is subjected to manipulating the individual energies of each projectile.
Let us now modify Equations (3) and (4) so as to represent the real reactants:
| (3) | 21D | + | 21D | → | 32He | ( | 0.82 MeV | ) | + | n0 | ( | 2.45 MeV | ) + E(D-D) | ||
| (4) | 21D | + | 32He | → | 42He | ( | 3.6 MeV | ) | + | p+ | ( | 14.7 MeV |
|
Query
What is the proper proportion of energies, of deuterium and tritium projectiles, in the laboratory frame of reference, that could peak the fusion cross-section, in beam nuclear fusion?
(Hint: beam-based nuclear fusion refers to using beams of ions of D and or T to induce nuclear fusion. This differs from bulk plasma-based fusion where the temperatures of the two projectiles are dependent on the temperature of the bulk plasma.)
Mohamed F. El-Hewie
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