Stellar Splitting and pairing NEW Black holes foun

What would this mean in terms of our cosine shaped particles energy profile? Walk to the edge of the basic w shape and your angular momentum is h, walk down a little further, on the cosine curve, and youre accelerating but theres a skew. You simply cannot walk straight down then up the spike in the middle. When you spiral your way to the bottom, your angular momentum will be one. That gives you the energy to spiral your way up that near vertical central spike to the top, where your angular momentum will again be h.

Ho hum, its boring up here, back down again and out. This time you would spiral in the opposite direction. If you were right handed when you went in you would come out left handed. This would be a mobius ball.

Sloat, Time gets lost in the math trick when you cancel it by using time and units of time. The equation e=hf has both the unit "second" and time itself without any units. This is why we get a bundle of energy particles with more energy as the frequency increases. It time is considered the photon energy is always the same no matter what the frequency. Its much less messy.

Sloat, Working on the details you have provided: the cosine profile is the blackbody profile of a Planck bundle of 511kev? Is this just a mistake on my part? Can you explain this in more detail? thanks

Hi Jim, I would argue that all particles have the same energy shape and not just the electron. If you want to graph it use y = cos (n ln x) The only problem is that computer graph drawing programs can only handle about a hundred steps. So any graph drawn will only give you a rough idea of that wo way propagation of the wave for change in n.

Hi Jim, on the fermi velocity question. Ive got a speed of gravity of h = c^2 / b^2 the speed of gravity being b. Ive also got h = e^2 mu / 2a epsilon
e being the electron charge
mu being the permeability of free space
a being the fine structure constant
epsilon being the permitivity of free space

That 2a alters the permeability of free space to the reciprocal of the speed of light, which I think is fascinating.

So whats v when we have 2ah = v^2 / b^2 the answer comes out at
36.21749159E 6 metres per second (note here the engineering notation, as fermi velocities are given in millions of metres)

The equation for the fermi velocity is V = c sqrt(2 E / mc^2)
V is the fermi velocity
E is the fermi energy in electron volts (for copper 7 eV)
mc^2 is the energy of an electron in electron volts

For the moment lets say that the square root part is correct. Then we can divide that into 36.21749159E 6 to get a new value for c of 6.919337855E 09 Thats about 23 times the speed of light. Its natural log is just over 22 so I think there is scope to look further at this. After all the idea that gravity plays a part in electricity is what the whole unified field theory is about.

Have we got any whizz kids here who can look at how the fermi energy is worked out? The thing about gravitational cores of tiny radius is that they will have a greater mean free path in a metal than the outer electromagnetic electron radius.

Sloat, Why I'm asking about the blackbody issue is that the Planck bundle is a collection of photons in all wavelenghts. As you know the bundle is a dimensionless point so how do all waves fit on the same point being that they have some dimension. The part of the bundle at high frequency would arrive first right? Or do all photons arrive at the same time or zero time? You need to examine the real meaning of h rather than use it in any old way don't you think?