BOB LAZAR: ELEMENT 115
Bob explained that half a pound of stable isotope of element 115 was used as the fuel. In about 10 years after his story, this element, isotope named Moscovium, was synthesized and its half-life was 0.65 seconds. Bob suggests that natural isotope could be different (in number of neutrons).
Another explanation is possible. When particles and atoms are formed, they encapsulate time of the environment. Some of them can have very slow time canned inside, see chapter 11 for example of billion to trillion times slower time in particles. But even better, check Extraordinary Stabilizing Power of Time Dilation: if time within an atom slows down only 100 times, then its lifetime increases by 100⁵, which is 10 billion times. And if atomic clock slows down 1,000 times, then its lifetime increases by 1,000⁵, which is 1,000 trillion times. Thus, if time in “where and when” this element 115 was created was 1000 times slower than our time, then lifetime of it increases from 0.65 seconds to 20 million years. Where is such time and place — with 100–1000 times slower time? Whatever we see around us has/had 12 times slower time at most, and it comes from the past, and currently slow time in galactic centers and around heavy stars called “nearly black holes”. But that’s about it in our neighborhood. But CMB (Cosmic Microwave Background, from the outskirts of our Universe, see chapters 11 and 5) corresponds to billion-trillion times redshift — time dilation.
Next puzzle about element 115: half pound of it was packing an astronomical amount of energy — that alone contradicts with Einstein’s E = mc², is it? Before checking that, let’s look at something else called flywheel (ancient and well-known human invention — even Leonardo DaVinci was improving it). It is puzzling too, and energy calculations there relate to the element 115 energy problem.
Flywheel is a mechanical energy accumulator. Here is one used in U.S. school buses since the previous century. Nowadays, modern ring-shaped flywheels made of carbon fiber rolls and running in vacuum are installed on top-notch racing cars and more:
Let’s start far away: say we have an object of mass M, moving at a speed V. It has energy
E(V) = M*V²/2
If the speed V is high, it is difficult to change it, even for a small change of speed Δ, because change of the energy ΔE will be big:
E(V+Δ) = M*(V+Δ)²/2 = M*V²/2 + M*V*Δ + Δ²/2
ΔE = E(V+Δ) — E(V) = M*V*Δ + Δ²/2 > M*V*Δ
ΔE is a big number even for a moderate change of speed Δ when V is a large. Let’s explore similar problem for a bus with a flywheel installed on it — changing speed or direction of the bus when flywheel is packed with energy and rotates at a very high speed:
Let say all points A1, A2, B1, B2, C1, C2 have same mass M. Before bus start moving, energy of these points is 6*M*V²/2, but we will see that it is easier to track energy in pairs, opposite one another with the wheel center in between. When bus was not moving, energy formulas were:
E(A1)+E(A2) = M*V²/2+ M*V²/2
E(B1)+E(B2) = M*V²/2+ M*V²/2
E(C1)+E(C2) = M*V²/2+ M*V²/2
When bus has speed Δ, energy changes:
E(A1)+E(A2) = M*(V+Δ)²/2+ M*(V-Δ)²/2
ΔE(A1)+ΔE(A2) = M*(V+Δ)²/2+ M*(V-Δ)²/2-M*V²/2-M*V²/2 = M*Δ²
E(B1)+E(B2) = M*(V²+Δ²)/2+ M*(V²+Δ²)/2
ΔE(B1)+ΔE(B2) = M*(V²+Δ²)/2+ M*(V²+Δ²)/2-M*V²/2-M*V²/2 = M*Δ²
To calculate general case for points C1 and C2, I moved velocities at these points to the right on the picture above. This created a parallelogram, so the longer blue arrow’s length by power of 2, by Pythagorean theorem, is: h²+ (d+Δ)². And the shorter blue arrow’s length by power of 2 is: h²+ (d-Δ)². Initial velocity V in power of 2 is: V² = d² + h². So:
E(C1)+E(C2) = M*(h²+(d+Δ)²)/2 + M*(h²+(d-Δ)²)/2 = M*h²+M*d²+M*Δ²
ΔE(C1)+ΔE(C2) = M*h²+M*d²+M*Δ²–2M*(d² + h²)/2 = M*Δ²
Energy of all pairs change by M*Δ², and for the whole wheel that results in
ΔE = MW*Δ²/2
where MW is the total mass of the wheel, and the division by 2 comes from “pairs” — number of which is “half” of number op points, each has mass M.
We have just confirmed that the energy canned in high RPM flywheel (high speed V) does not contribute to the bus inertia (energy change MW*Δ²/2), otherwise it would be a collision hazard. The wheel weight does contribute to the bus inertia. This problem is solved in racing cars installations, where light carbon fiber flywheels are used instead of heavy steel.
Back to the element 115 with time inside slowed thousand times. In “frozen candle” puzzle, we saw that for an external observer everything inside a slow timezone is hardly moving (even the candle flame neither flickers nor waves). For extremely slow time that means even electrons “move slowly” and other particles “vibrate” slowly (non-relativistically). I wonder if quantum physics (which is hundred percent statistical, because it describes >10¹⁰ velocities in <10⁻¹⁰ space) still works when time slows down dramatically. For us, it is important that these particles’ “trajectories” have central (point) symmetry:
So, we can apply the same math we applied for the flywheel to understand that these atoms and particles inertia is not impacted by amount of energy packed inside. That explains why half a pound of element 115 with slow time inside — meaning packed with energy (see chapters 4 and 8 for more) — can be carried around — literally weighing or with inertia of half a pound and not tons. Classic “bus with flywheel” model shows that a bus at rest can encapsulate various energies. Similarly, an atom at rest can have various amount of energy: it can pack very different “time”, and particles inside the atom can pack various “times”. That challenges E = mc² paradigm of rest mass and energy equivalence (there is no rest in nature, in theory only).
To my knowledge, physics of slow time has not been developed yet. Let’s try figuring energy formula for a slow time out of two well-known formulas:
Relativistic energy formula: E = M*C²*(1-(V/C)²)⁻⁰·⁵ ……(1)
Doppler shift formula: Z+1 = (1+V/C)⁰·⁵/(1-V/C)⁰·⁵ ……(2)
From (2) we can get V/C = ((Z+1)²–1) / ((Z+1)²+1) …..(3)
Substituting V/C in (1) with the right side of (3) gives us:
E = M*C²* ( 1 — ((Z+1)²–1)² / ((Z+1)²+1))² )⁻⁰·⁵
E = M*C²*( ((Z+1)²+1)² — ((Z+1)²–1)² )⁻⁰·⁵ * ((Z+1)²+1)
E = M*C²*( 4*(Z+1)² )⁻⁰·⁵ * ((Z+1)²+1)
E = M*C²* ((Z+1)²+1)/ (2*(Z+1)) ………………………(4)
For brevity, I will use symbol D instead of Z+1. Then (4) shortens to:
E = M*C² * (D²+1) / (2D) = M*C² * (D+1/D) / 2 ……….(5)
where D is time dilation. Formulas above are for an object moving in a straight line only, and they are not about a non-moving object at rest (like ΔE > M*V*Δ formula was for a straight movement, and not for a flywheel).
Formula E = M*C² should be more like
E = M*C²*D² …………………………………………………(6)
when we talk about energy encapsulated in an atom at rest, because measurements are in sec⁻², and time in the atom can be D-times dilated compared to the current time. Astronomical factor D² explains half a pound of stuff containing an astronomical amount of energy. Check E = mc²D² proof here.
Check Simple quantum explanation of gravity without mass or math solving another Bob Lazar’s anecdote/riddle:
And Second Layer in Bob Lazar’s Craft Shielding! shows incredibly smart engineering of that craft.
