Thunderbolts Forum v3.0

Maybe this thread can be used for requesting calculations for various things.

Calculate Pressure in Imploding Plasma Filament

Let's say the mass is 1 solar mass = 2×10^30 kg.
The two halves of the filament each contain half a solar mass = 10^30 kg.
The two halves collide at 0.85c (c = light speed) over an area of 1.5x10^12 km^2.
What is the maximum pressure during impact?
Can anyone show how to calculate that?

Is this correct?
PxV=E?
If so, E=.5mv^2;
P=E/V=.5mv^2/V
(.5 x 10^30 kg x .85c x .85c) / 0.7 x 10^27 m^3
= 4.11 x 10^6 kg m/m^2 ?

P=F/A; A = … m^2; F = … kg m?
F=ma; a = v/t; ...

I would not use Volume(V) here.
More treat it like two colliding balls.
(Inelastic collision)

http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html
Use airplane & duck example

Okay, Zyxzevn.
Airplane & Duck Collision Force
http://hyperphysics.phy-astr.gsu.edu/hb ... se.html#c4
Mass of bird m= ___ kg, weight = ___ N = ___ lb.
Velocity of aircraft = ___ m/s = ___ km/h = ___ mi/h, which for bird length = ___ cm gives an estimated impact time of ___ seconds. Under these conditions, the impact force is
Average impact force = ___ N = ___ lb. = ___ tons.

F.av = m d.v/d.t [d is delta or change]

I know the mass, 10^30 kg, and the velocity, 0.85 c. But gravity & weight would not apply, would they? I'd use a range of estimates for the length, from about 1.4x10^6 km up to several AU's.

F = 10^30 kg x .85c / ? sec
s/v=t; s = 700,000 km minimum.
v = .85 c = 255,000 km/sec
700/255 = 2.75 sec.
F = 10^30 kg x 255,000 km/sec / 2.75 sec
= ...
Will that get the Force? After I calculate the av. Force, I'll get the pressure by dividing the Force by the area. Right?

I wanted to see if an imploding filament would produce enough pressure to cause fusion of heavy elements from light elements, esp. hydrogen. Lightning seems to cause a small amount of fusion sometimes, esp. in the Sun's photosphere, I guess. One website says "Pressure squeezes the hydrogen atoms together. They must be within 1x10^-15 meters of each other to fuse." This site https://physics.stackexchange.com/quest ... emperature says 10^21 or 10^28 Pa is needed to produce fusion.

F = (solar mass) 10^30 kg x (velocity) .85c / (time) ? sec
F = 10^30 kg x 2.55x10^8 m/sec / 2.75 sec
= 9.3 x 10^37 kg m/sec^2 = 9.3 x 10^37 N (newtons)
1 Pascal, SI unit of pressure, is 1 newton acting on an area of 1 square meter.
Sun surface area: 6.088 x 10^12 km^2
Surface Area of a Sphere with radius r equals 4πr^2
Area of Circle equals πr^2
Sun's area = .25 x 6.1 x 10^12 km^2 = 1.5 x 10^12 km^2
Pressure = F/A = 9.3x10^37 N / 1.5 x 10^12 km^2
= 6.2 x 10^25 Pa
If 10^21 Pa is enough for fusion, then the imploding filament would indeed produce fusion.
But if the filament implosion impact took longer than 3 seconds, the pressure would have been less, so less fusion.

Did not have much time to react.

Lloyd wrote: ↑Fri Mar 20, 2020 4:15 pm I wanted to see if an imploding filament would produce enough pressure to cause fusion of heavy elements from light elements, esp. hydrogen. Lightning seems to cause a small amount of fusion sometimes, esp. in the Sun's photosphere, I guess. One website says "Pressure squeezes the hydrogen atoms together. They must be within 1x10^-15 meters of each other to fuse." This site https://physics.stackexchange.com/quest ... emperature says 10^21 or 10^28 Pa is needed to produce fusion.

Ok that is clearer.
In my experience the pressure varies a lot when electric fields/currents are involved.
Fusion shows up in lightning and Fusors with ease.

Anyway, here is how I would calculate it:

First try to find the average force that most of the object will encounter.
I will criticize this approach below.

Average force and pressure of 2 Colliding objects:

If two objects collide and merge, you have a distance over which they merge.
Let's say that is R (radius), and they approach with speed V.
The de-acceleration (a) stops the movement.
This happens with a short time (T).
R= 0.5 * a * T^2
In the same time (T) the de-acceleration reduces V to zero.
V= a * T
(so: T= V/a)

We try to get acceleration (a) from the equation.
R = 0.5 * a * (V/a)^2
R = 0.5 * (V^2) / a
a = 0.5 * (V^2) / R

From acceleration (a) we can get the force (F) and the pressure (P)

F = m*a ,where m is mass of object.
P = F / Area ,where Area is surface area where objects touch each other.
Area goes from 0 to (PI * R^2)

For the average force, we may get PI*R^2

So:
Average P = ( m / Area ) * (0.5) * (V^2) / R
Average P = ( 0.5 * m * V^2 ) / ( Area * R )

With your data:
Using c= 3E8 [m/s] and converting [km] to [m].
Estimate R= 0.66E8 [m]
P= 0.5 * 2E30 [kg] * (0.85* 3E8 [m/s] )^2 / ( 1.5E12 *1E6 [m^2] * 0.66E8 [m] )
P= 1E30 * ( 1.7^2 ) * 1E16 / ( 1E26 )
P= (~3) *1E46 / 1E26 = 3E20 [Pascal]

Criticism and correction:

This pressure is certainly not constant during the collision.
The Area varies during the merging of the 2 objects.
The front ends of the objects get a sudden shock, because the front end
has a very small R, and small A.
The way the material compresses is also important, as gas will
compress a lot easier than liquid or solid. This reduces the mass
in the equation.
But you could try to calculate how a square meter of the front surface would behave when
it collides onto the other surface.
Now you get more the pressure equation that you described in your first post,
where the gas gets compressed into a very small volume.
But this only becomes a lot more than the above equation, when
there is both a gas and liquid/solid state.

But in reality gets far more complex.
The shock will cause a strong shock-wave, and that will cause an explosion.
Also the compression and resistance will suddenly increase the heat of of the front of the object.
These super-sonic shocks will have far higher pressure than shown.

It will work like a meteoroid that explodes in the atmosphere,
before it reaches the surface.

This explosive collision is the most interesting part of the collision
and is far too difficult to calculate.
Instead simulators are used to calculate the effects of such collisions.

How do we calculate the magnetic field locations/directions (Ampere's Law) of a 3D finite spherical sheet current which encapsulates another smaller (anode or cathode) sphere?



Blue Sphere = radius 1
Orange Sphere = radius 2
Red Lines = r_1 - r_2 (current sheet thickness)
Yellow Line = r_2
Blue Sphere Surface Area = 4πr_1^2
Orange Sphere Surface Area = 4πr_2^2

The closed loop for the system is from the outer surface of the orange sphere to the outer surface of the blue sphere.

I have no idea how one even defines the vectors for the dl segments for this kind of geometry.

And yes, what I am trying to work out is the magnetic field locations/directions for the current sheet that extends from the solar chromosphere to the heliosheath, ignoring the planets, etc, for now.

Feel free to fill in the actual radius of the sun for the inner sphere, and the heliosheath for the boundary of the outer sphere (~120 AU), and the current density K of the solar wind.

I'd just like to point out that magnetic fields produced by a spherical current does not seem to have been considered before.

This site lists only wires, loops and solenoids.

JP Michael wrote: ↑Mon Mar 23, 2020 2:10 am I'd just like to point out that magnetic fields produced by a spherical current does not seem to have been considered before.

Because they are equal to zero. See 18-2 here.

Cheers @paladin17.

Higgsy actually posted a good reply before the rollback/forum crash, saying that B = j = 0 in this arrangement. He also did make the point that a filament (=wire) could form if some instability occurred in the plasma of the current sheet of this spherical arrangement.

Thus if we are going to find magnetic fields in this kind of current sheet, they will tend to be in regions of plasma instability which have formed defined Birkeland current filaments.

JP Michael wrote: ↑Thu Apr 02, 2020 7:18 pm Cheers @paladin17.

Higgsy actually posted a good reply before the rollback/forum crash, saying that B = j = 0 in this arrangement.

Actually, I didn't say that j = 0 because it isn't; since j is current density, j cannot be zero as a consequence of how the problem is set up - a spherically symmetrical current. In spherical co-ordinates, since the current density is spherically symmetrical ∂j/∂φ = ∂j/∂θ = 0 and |j|(r) = i/4r^2 where i is total current and r goes between r_1 and r_2, the radii of the inner and outer spherical surfaces. I then pointed out that the magnetic field also has to be spherically symmetrical and B must therefore be radial between the spherical surfaces and since the magnetic flux through a closed surface is zero (Gauss's law for magnetism), B must be zero.

paladin's link actually formalises my statement that B must be purely radial (∇ x B = 0). I set the problem up as purely steady state. But it can't strictly be both steady state and spherically symmetrical, as the only way create a spherically symmetrical current is to have some charge flowing from the inner to the outer spherical surface and therefore ∂Q/∂t on each surface is not zero. The charge cannot be replenished from a battery or DC power supply for example as the current flow in the connections would destroy the spherical symmetry. The link then goes on to show that ∇ x B is zero (the two terms, Ampere's law and the displacement current term cancel). Therefore, B = 0