Inside the Head of Anatoli Bogorski
I used the Relativistic Calorimetric Equations to predict the temperature inside Anatoli's Head.
Let's review the incident.
Here’s an overview of Anatoli Bugorski's incident, his injuries, and the details surrounding the U-70 Synchrotron accelerator.
The Incident: Anatoli Bugorski’s Accident
- Date: July 13, 1978.
- Location: Institute for High Energy Physics in Protvino, Russia.
- Accelerator: The U-70 Synchrotron was the particle accelerator involved in the accident.
Anatoli Bugorski, a Russian scientist and researcher, was performing maintenance on the U-70 Synchrotron, which was the largest particle accelerator in the Soviet Union at the time. During his work, a malfunction led to an unexpected event where Bugorski accidentally placed his head in the path of a proton beam.
Details of the U-70 Synchrotron
- Type: Proton synchrotron.
- Operational Energy: The U-70 Synchrotron was capable of accelerating protons to 76 GeV (76 billion electron volts), significantly lower than modern accelerators like the Large Hadron Collider (6.5 TeV per proton).
- Circumference: Approximately 1,500 meters.
The U-70 was built in the late 1960s and became operational in 1967. It was primarily used for high-energy physics experiments and was a significant achievement in Soviet particle physics.
Nature of the Exposure
- Proton Beam Energy: The beam that struck Bugorski was at the full energy of the U-70’s capability, 76 GeV per proton.
- Exposure Path: The proton beam entered the back of Bugorski’s head, passed through his skull, and exited near his nose.
- Duration: Although not precisely measured, the exposure likely lasted only a fraction of a second due to the beam’s high speed, but Bugorski was exposed long enough to absorb a substantial dose of energy.
Immediate Effects
- Visual Phenomenon: Bugorski reportedly saw a flash "brighter than a thousand suns" but did not feel any pain.
- Severe Radiation Dose: He received a localized radiation dose estimated between 200,000 to 300,000 rads (2,000 to 3,000 Gy) to the tissue along the beam path, far beyond a lethal dose for whole-body exposure. However, because the beam was highly localized, the damage was confined to the area it traversed.
Medical and Long-Term Effects
1. Initial Injury: The beam severely burned a narrow path of tissue in his head. It destroyed tissue, bone, and nerves along the way, and Bugorski’s face swelled up beyond recognition within days.
2. Partial Paralysis: The right side of his face was paralyzed, and he lost hearing in his left ear. Bugorski also experienced severe seizures in the years following the accident, likely due to damage to brain tissue.
3. Cognitive Effects: Despite the extensive injury, Bugorski’s cognitive functions remained intact, and he was able to continue his work as a scientist after the accident.
4. Long-Term Survival: Bugorski survived against all odds, making his case one of the most unusual in radiation injury history. He continued to live a relatively normal life, albeit with ongoing health issues related to the accident.
Legacy and Scientific Interest
Bugorski’s case remains a unique study of the effects of high-energy proton beams on biological tissue and a testament to human resilience. It is still referenced in studies related to radiation injuries, particle beam therapy, and the biological effects of localized high-dose radiation.
The incident underscores both the risks of working with particle accelerators and the unpredictable nature of radiation injuries, especially when dealing with high-energy particles like protons.
Now, let's do the calorimetric analysis:
This is the Python Code:
import numpy as np
from scipy.integrate import quad
from astropy import units as u
from astropy.constants import e, c, m_p
# Constants and initial parameters
E_0 = 76e9 * u.eV # Initial energy in eV (76 GeV for U-70 Synchrotron protons)
L = 10 * u.cm # Thickness of the material in cm (e.g., water or tissue)
I = 1e-6 * u.A # Estimated current in Amperes (1 microampere)
T = 3 * u.s # Estimated exposure time
# Material properties for water (H2O)
Z_eff = 7.42 # Effective atomic number for water
A_eff = 18.015 # Molar mass of water in g/mol
# Calculate radiation length X_0 for water in g/cm^2 using the provided formula
X_0_g_cm2 = 716.408 * A_eff / (Z_eff * (Z_eff + 1) * np.log(287 / np.sqrt(Z_eff))) * u.g / u.cm**2
# Convert X_0 to meters using the density of water
density_water = 1 * u.g / u.cm**3 # Density of water
X_0 = (X_0_g_cm2 / density_water).to(u.m) # Convert X_0 to meters
# Function for energy as a function of distance x in meters
def energy(x, E_0, X_0):
return E_0 * np.exp(-x / X_0)
# Energy loss function (rate of energy loss) as a function of distance x in meters
def energy_loss(x, E_0, X_0):
return -energy(x, E_0, X_0) / X_0
# Integrate energy loss over the material thickness L for a single proton
E_deposit_per_proton, _ = quad(energy_loss, 0, L.to(u.m).value, args=(E_0.to(u.J).value, X_0.value)) # Result in Joules
E_deposit_per_proton *= u.J # Re-apply units to the result
# Calculate final energy after passing through material for one proton
E_final_single_proton = E_0.to(u.J) + E_deposit_per_proton # Final energy in Joules
# Initial and final momenta
p_0 = np.sqrt((E_0.to(u.J)/c)**2 - (m_p * c)**2).to(u.kg * u.m / u.s) # Initial momentum
p_final = np.sqrt((E_final_single_proton / c)**2 - (m_p * c)**2).to(u.kg * u.m / u.s) # Final momentum
# Momentum transfer per proton
delta_p_per_proton = p_0 - p_final
# Calculate number of protons per second
protons_per_second = (I / e.si).si # Result in protons per second
# Total momentum transfer and energy deposition over time T
total_energy_deposit = E_deposit_per_proton * protons_per_second * T # Total energy in Joules
total_momentum_transfer = delta_p_per_proton * protons_per_second * T # Total momentum transfer in kg*m/s
# Results
print(f"Radiation Length X_0 for water: {X_0:.2e}")
print(f"Effective Current for U-70: {I:.2e}")
print(f"Total Energy Deposited over {L:.2f} in {T:.2f} s: {total_energy_deposit:.2e} J")
print(f"Initial Momentum of a single proton: {p_0:.2e}")
print(f"Final Momentum of a single proton after passing through {L:.2f} cm: {p_final:.2e}")
print(f"Momentum Transfer per proton: {delta_p_per_proton:.2e}")
print(f"Total Momentum Transfer over {T} seconds: {total_momentum_transfer:.2e}")
This code is based on the following:
Here’s a more detailed explanation with the relevant formulas for dE/dx due to ionization losses, especially as they apply to the 75 GeV proton beam that struck Anatoli Bugorski.
Is the dE/dx in the Code Due to Ionization?
Yes, the dE/dx we used in the code to estimate the energy loss of a 75 GeV proton beam passing through tissue or water primarily represents ionization losses.
Why Ionization Dominates for Protons at 75 GeV
For high-energy protons like the 75 GeV beam from the U-70 Synchrotron:
Ionization Energy Loss Formula: Bethe-Bloch Equation
The energy loss due to ionization for heavy charged particles like protons is described by the Bethe-Bloch formula:
where:
This equation describes the ionization losses per unit distance for a proton or other heavy charged particle as it passes through matter.
Radiation Length? X0
In our simplified code, we used an approximation:
where:
The radiation length? X0? is often defined for high-energy electron interactions, where it represents the mean distance over which an electron loses all but? 1/e? of its energy due to Bremsstrahlung. For completeness, the formula for? X0? in terms of material properties is:
where:
However, for heavy particles like protons, ionization losses are dominant, and? X_0? is used as an effective length scale rather than a direct measure of radiative energy loss.
Why We Used This Simplified? dE/dx? in the Code
In our code, we used to approximate the average rate of energy deposition for a proton traveling through 10 cm of material. This simplification works reasonably well because:
Summary Yes, the?dE/dx we used in the code effectively represents ionization losses for the 75 GeV proton beam, as ionization is the dominant energy loss mechanism for protons in this energy range when passing through matter like tissue. The simplified formula, with an effective? X0, provides a practical approximation for estimating energy deposition in this scenario.
THE VERDICT ON ANATOLI BOGORSKI HEAD
Running the code with the Syncroton U-70 protons:
# Constants and initial parameters
E_0 = 76e9 * u.eV # Initial energy in eV (76 GeV for U-70 Synchrotron protons)
L = 10 * u.cm # Thickness of the material in cm (e.g., water or tissue)
I = 1e-6 * u.A # Estimated current in Amperes (1 microampere)
T = 3 * u.s # Estimated exposure time
I considered it would take 3 seconds for Anatoli to get his head into the beam, see some light and no pain, and then move his head out of the beam. It is very likely that it took more since he felt no pain.
Radiation Length X_0 for water: 4.44e-01 m
Effective Current for U-70: 1.00e-06 A
Total Energy Deposited over 10.00 cm in 3.00 s s: -4.60e+04 J
Initial Momentum of a single proton: 4.06e-17 kg m / s
Final Momentum of a single proton after passing through 10.00 cm cm: 3.24e-17 kg m / s
Momentum Transfer per proton: 8.20e-18 kg m / s
Total Momentum Transfer over 3.0 s seconds: 1.54e-04 kg m / s
So, according to Relativity, his head dissipated 46 kilojoules and received an impulse of 1.5E-4 N.s.
In other words, the Relativistic calculation indicates that his head should have exploded.
BELOW ARE THE PREDICTIONS ACCORDING TO MY THEORY:
import numpy as np
from scipy.integrate import quad
from astropy import units as u
from astropy.constants import e, c, m_p
# Constants and initial parameters
E_0 = 0.470e6 * u.eV # Initial energy in eV (76 GeV for U-70 Synchrotron protons)
L = 10 * u.cm # Thickness of the material in cm (e.g., water or tissue)
I = 1e-6 * u.A # Estimated current in Amperes (1 microampere)
T = 3 * u.s # Estimated exposure time
# Convert the initial energy to Joules for calculations
E_0_joules = E_0.to(u.J)
# Material properties for water (H2O)
Z_eff = 7.42 # Effective atomic number for water
A_eff = 18.015 # Molar mass of water in g/mol
# Calculate radiation length X_0 for water in g/cm^2 using the provided formula
X_0_g_cm2 = 716.408 * A_eff / (Z_eff * (Z_eff + 1) * np.log(287 / np.sqrt(Z_eff))) * u.g / u.cm**2
# Convert X_0 to meters using the density of water
density_water = 1 * u.g / u.cm**3 # Density of water
X_0 = (X_0_g_cm2 / density_water).to(u.m) # Convert X_0 to meters
# Function for energy as a function of distance x in meters (exponential decay)
def energy(x, E_0, X_0):
return E_0 * np.exp(-x / X_0)
# Energy loss function (rate of energy loss) as a function of distance x in meters
def energy_loss(x, E_0, X_0):
return -energy(x, E_0, X_0) / X_0
# Integrate energy loss over the material thickness L for a single proton
E_deposit_per_proton, _ = quad(energy_loss, 0, L.to(u.m).value, args=(E_0_joules.value, X_0.value)) # Result in Joules
E_deposit_per_proton *= u.J # Re-apply units to the result
# Calculate final energy after passing through material for one proton (classical approximation)
E_final_single_proton = E_0_joules + E_deposit_per_proton # Final energy in Joules
# Initial velocity from classical kinetic energy: v = sqrt(2 * E / m)
v_0 = np.sqrt((2 * E_0_joules) / m_p).to(u.m / u.s) # Initial velocity
v_final = np.sqrt((2 * E_final_single_proton) / m_p).to(u.m / u.s) # Final velocity after energy loss
# Classical momenta using p = m * v
p_0 = (m_p * v_0).to(u.kg * u.m / u.s) # Initial momentum
p_final = (m_p * v_final).to(u.kg * u.m / u.s) # Final momentum
# Momentum transfer per proton
delta_p_per_proton = p_0 - p_final
# Calculate number of protons per second
protons_per_second = (I / e.si).si # Result in protons per second
# Total momentum transfer and energy deposition over time T
total_energy_deposit = E_deposit_per_proton * protons_per_second * T # Total energy in Joules
total_momentum_transfer
total_momentum_transfer = delta_p_per_proton * protons_per_second * T # Total momentum transfer in kg*m/s
# Results
print(f"Radiation Length X_0 for water: {X_0:.2e}")
print(f"Effective Current for U-70: {I:.2e}")
print(f"Total Energy Deposited over {L:.2f} in {T:.2f} s: {total_energy_deposit:.2e} ")
print(f"Initial Velocity of a single proton: {v_0:.2e}")
print(f"Final Velocity of a single proton after passing through {L:.2f} cm: {v_final:.2e}")
print(f"Initial Momentum of a single proton: {p_0:.2e}")
print(f"Final Momentum of a single proton after passing through {L:.2f} cm: {p_final:.2e}")
print(f"Momentum Transfer per proton: {delta_p_per_proton:.2e}")
print(f"Total Momentum Transfer over {T} seconds: {total_momentum_transfer:.2e}")
Radiation Length X_0 for water: 4.44e-01 m
Effective Current for U-70: 1.00e-06 A
Total Energy Deposited over 10.00 cm in 3.00 s s: -2.85e-01 J
Initial Velocity of a single proton: 9.49e+06 m / s
Final Velocity of a single proton after passing through 10.00 cm cm: 8.48e+06 m / s
Initial Momentum of a single proton: 1.59e-20 kg m / s
Final Momentum of a single proton after passing through 10.00 cm cm: 1.42e-20 kg m / s
Momentum Transfer per proton: 1.69e-21 kg m / s
Total Momentum Transfer over 3.0 s seconds: 3.17e-08 kg m / s
HU predicts a modest 0.285 Joules energy deposition and is consistent with Anatoli Bugorski survival.
Q.E.D.
IN SUMMARY
I created a theory of everything and applied it to everything. The work was visible in 2006 and has been censored since then. That wastes 18 years—a crucial amount of time when Mankind is facing unprecedented danger.
The different ratios in energy prediction between HU and Relativity are basically whatever the Relativistic Energy is divided by 0.47 GeV.
The mismatch increases with energy, as in the case of the Synchrotron, 75 GeV/0.47 GeV or 160. For the Large Hadron Collider, it is 6.5TeV/0.47GeV or 13829.
I believe this analysis of the event of the Russian Scientist Anatoli Bogorski is enough to refute Relativity.