Imagine a force whose minimum happens if particles x and y are at a distance 3 *10^8 meters. The particles themselves are very light, say 1 gram each, but you need a gigajoule of energy to change their distance 1 millimeter.
The particle A is in your hand and the particle B is close to the Moon. It takes one second for B to know that you have started pushing on A. You spent 1 GJ to push A just 1 millimeter in 0.1 seconds. The force is 10^12 newtons and the impulse is 10^11 newton seconds.
What happens when B will know that you have spent that much energy and impulse on A? The kinetic energy would imply that A & B start moving at 10^6 m/s, while the impulse would imply a speed of 5 * 10^13 m/s, much faster than light.
These figures do not make sense. A fix to the discrepancy is that the force field of A and B carries an enormous mass-energy. If pushing A 1 millimeter increases the local momentum and the energy of the field by the above numbers, things start making sense.
The minimum possible mass for a harmonic oscillator spring
A /\/\/\____________/\/\/\ B
d
d
The force on the particle A is
F = -kx,
where x is the displacement of A from the lowest potential distance.
Let us push A to the right with a force
F_p = F_0 + kx
for T seconds, where T < 2d/c. We assume that since B will not know about the pushing until d/c seconds later, the field of B will remain static at A during the push. The particle A will accelerate to the right:
a = F_0 / m,
and cover a distance
s = 1/2 F_0 / m * T^2.
The push force as a function of time is
F_p(t) = F_0 + k * 1/2 F_0 / m * t^2
The push impulse is the integral of the above over T seconds:
p = F_0 T + k * 1/2 F_0 / m * T^3 / 3
= F_0 T [1 + k T^2 / (6m)].
At x = 0, the push force is
F_p = F_0,
and at x = s it is
F_p = F_0 + k * 1/2 F_0 / m * T^2.
The work W done by the pushing force is the average of the two forces above, times s:
W = 1/2 [2F_0 + k * 1/2 F_0 / m * T^2]
* 1/2 F_0 / m * T^2
= 1 / (4m) * F_0^2 T^2
* [2 + k T^2 / (2m)].
The kinetic energy of the whole system A & B from the push impulse p is
E = p^2 / (4m)
= 1 / (4m) * F_0^2 T^2
* [1 + k T^2 / (3m) + k^2 T^4 / (36m^2)].
The kinetic energy E must be less than the total work W done by the force:
1 + k T^2 / (3m) + k^2 T^4 / (36m^2)
< 2 + k T^2 / (2m)
<=> (z = kT^2)
z^2 / (36m^2)
< 1 + z / (6m)
<=> (y = z / (6m))
0 < 1 + y - y^2
= -(y - 1/2)^2 + 5/4
<=> (ignore negative values of y)
y < sqrt(5)/2 + 1/2
= 2.24 / 2 + 1/2
= 1.62.
We have
k T^2 / (6m) < 1.62
<=>
m > 0.10 k T^2
If we can assume that the field of B stays static at A until the time 2d/c, then we have
m > 0.40 k d^2/c^2
where m is the mass of a particle, d is the distance of the particles, and c is the speed of light.
A limit on the mass of a space elevator tether
Suppose that we want to lift a payload which weighs F, from the surface of Earth, using a tether attached to the Moon. The length of the tether is L.
We assume that the weight of the tether is negligible.
We let the tether expand a distance e. The strain is then e / L.
We assume that the weight of the tether is negligible.
We let the tether expand a distance e. The strain is then e / L.
Let us assume that the tether consists of Q particles simply chained together under a potential as described above.
Hooke's constant k_0 for the whole tether is
F = k_0 e
<=>
k_0 = F / e
The constant k for an individual pair of particles is k_0 Q. The minimum possible mass for a particle is
m = 0.40 k_0 Q L^2
/ (Q^2 c^2)
The minimum mass M of the whole tether is Q times the above mass m:
M = 0.40 k_0 L^2 / c^2
= 0.40 (F / e) L^2 / c^2.
It does not depend on the number Q of particles.
If the payload is 10,000 N (= 1 metric ton), and we allow an extension e of 10,000 km in the tether from the Moon to Earth, we get
M = 0.40 * 0.001 * 1.27^2 kg
= 0.65 gram
as its minimum possible mass.
https://en.wikipedia.org/wiki/Ultimate_tensile_strength
Graphene has an ultimate tensile strength of 130 gigapascals, and its density is 1,000 kg/m^3. A graphene tether which has a mass of 50 metric tons could extend from the Moon to Earth and could stand the weight of one metric ton of payload.
F = k_0 e
<=>
k_0 = F / e
The constant k for an individual pair of particles is k_0 Q. The minimum possible mass for a particle is
m = 0.40 k_0 Q L^2
/ (Q^2 c^2)
The minimum mass M of the whole tether is Q times the above mass m:
M = 0.40 k_0 L^2 / c^2
= 0.40 (F / e) L^2 / c^2.
It does not depend on the number Q of particles.
If the payload is 10,000 N (= 1 metric ton), and we allow an extension e of 10,000 km in the tether from the Moon to Earth, we get
M = 0.40 * 0.001 * 1.27^2 kg
= 0.65 gram
as its minimum possible mass.
https://en.wikipedia.org/wiki/Ultimate_tensile_strength
Graphene has an ultimate tensile strength of 130 gigapascals, and its density is 1,000 kg/m^3. A graphene tether which has a mass of 50 metric tons could extend from the Moon to Earth and could stand the weight of one metric ton of payload.
If a force is mediated by a virtual particle, what is the minimum mass of the force field?
Our calculation above is based on a simple conservation of momentum and energy argument for a force which is mediated at the speed of light.
If we assume that a force is carried by virtual particles of some kind, can we derive bounds for the energy of the force field?
We have conjectured that the electric field of a charge is an elastic body whose mass-energy density is E^2. Electromagnetic waves are born from the vibration of the elastic body.
A limit on the mass of an anti-gravity device
1 - r_s / r
is roughly 1 - 10^-9 close to the surface of Earth. We conclude that if we want to use a rigid body to force the spatial metric close to Earth to something else than Schwarzschild, the body should stretch much less than 9 millimeters under its own weight.
The gravitational binding energy of Earth is 2.5 * 10^32 J or 3 * 10^15 kg. In an earlier blog post we conjectured that the deformation energy of the spacetime is roughly the same as this number.
If we would have a rigid body several times the size of Earth, we may guess that placing it next to Earth would reduce the deformation energy of the spacetime considerably in the volume which the body occupies.
We may try to calculate the forces on the rigid body from an assumption that the energy would be in the ballpark 10^31 J.
The virial theorem states that the thermal energy of a star is 1/2 of the binding energy. Our conjecture that the deformation energy = the binding energy bears a resemblance to the virial theorem.
The minimal mass for an anti-gravity tether
The tether should stretch much less than 9 millimeters under its own weight. Let us assume that we can hang the tether from a very strong structure at the height of one Earth radius, R = 6,371 km. That is, L = 6,371 km in our formula.
The average gravitational acceleration is the integral
1 / r^2
from 1 to 2, times g = 9.81 m/s^2. It is 1/2 g.
If the mass of the tether is M, then the force is F = 1/2 Mg. We get
M > 0.40 [1/2 Mg / e] (L^2 / c^2)
<=>
e > 0.40 * 1/2 g L^2 / c^2
= 0.9 millimeter.
We were able to solve for the extension e > 0.9 mm, but the mass M can be chosen arbitrarily. It looks like an anti-gravity device might be possible.
Suppose that our tether can stand the force 10 N and it stretches 1 millimeter. What is the minimum mass M of the tether?
M = 0.40 * 10,000 * 4.5 / 10,000 kg
= 1.8 kg.
The tether can lift a payload of 1 newton.
...WORK IN PROGRESS...
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