granpa

that is not a rounding error and your math checks out perfectly Energy doesnt work either Kinetic energy = (gamma-1)*mc^2 500-1 + 500-1 = 998 999.5-1 = 998.5

Thank you Elfmotat So for circular motion proper accelleration = (proper velocity)^2/r works for all velocities! F = m*(proper acceleration) So centripetal force should equal m*(proper velocity)^2/r This result makes perfect sense though I don't understand why for an object moving in a straight line proper acceleration = acceleration*gamma^3 And I don't see any way of reconciling it with the concept of transverse mass This is what I am trying to understand https://en.wikipedia.org/wiki/Relativistic_mechanics#Force I'll keep working on it I was expecting everything to reduce to F = m*(proper acceleration) https://en.wikipedia.org/wiki/Proper_acceleration#Acceleration_in_.281.2B1.29D According to https://en.m.wikipedia.org/wiki/Relativistic_mechanics#Force F=dp/dt p = m*gamma*v So I assume F=m*d(gamma*v)/dt It then says that the result is that F = gamma^3*ma for straight line acceleration F = gamma*ma for acceleration at a right angle (like a circle) a= v^2/r for a circle ???? Therefore F = gamma^1*m*v^2/r for a circle

The electron in a hydrogen atom is bound by the electromagnetic force. The electron and a neutron is bound by the strong force

Thank you very much

What is the magnitude of the proper acceleration of an object moving in a circle at a given (unchanging) relativistic speed? For an object moving in a straight line proper acceleration = acceleration*gamma^3 My question has to do with the relationship between transverse mass and longitudinal Mass

We were discusiing electron orbits

Velocity of B (2 particles) in A's frame = 0.999cVelocity of C (rocket) in A's frame = 0.999c Velocity of B in C's frame = velocity of the two particles from the point of view of the rocket = 0.999998c What are you not getting?

Well I'm not going to keep repeating myself but I will say this the centrifugal force follows an inverse Cube law until the particle reaches relativistic speed at which point it then starts following an inverse Square law so at that point the centrifugal force increases exactly the same rate as the electrostatic force would increase It follows an inverse Cube law if angular momentum is conserved that is

Yes thats the whole point. From the point of view of the rocket the particle is stationary There isn't a stationary particle in the center to begin with but if there were then that particle from the Rockets point of you would be moving very close to the speed of light and the two particles moving in would, at every point in time, be perfectly in line with it therefore those two particles cannot be moving in at a 45 degree angle. I don't know what the angle is but it must be very very tiny. Microscopic even. This means that momentum must equal gamma*mv

There are 2 particles output One is stationary The stationary particle in the second image is the same particle in the first image that is going off to the left

Yes the speed of the 2 particles from the point of view of the rocket. I assumed 45 degree angle but I'm no longer sure about that If the angle were very much shallower than that would explain how momentum is conserved and would not require momentum to be the square root of gamma

the electron is moving very fast by adding a neutrino of course

In the second image each particle has sqrt(500)/sqrt(2) = 15.8 momentum in the east west direction The final particle has sqrt(1000) = 31.6 I didnt calculate vy I used the formula for total velocity given here http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html see the section "relative speeds" since the velocities were orthoginal it simplified greatly

Two particles moving very close to the speed of light in the north-south direction collide and end up moving in opposite directions due east and west. Each starts with gamma=22.37 and end up with exactly gamma=22.37. Total momentum is zero at all times. v = 0.999 gamma = 22.37 Now from the point of view of a rocket moving at 0.999c the two particles come together Collide and then one particle becomes stationary while the other flies away with all of the momentum. It should be possible from this thought experiment to determine the equation for relativistic momentum velocity of final nonstationary particle from the point of view the the rocket is calculated by velocity addition formula = (u+v)/(1+uv) (0.999+0.999)/(1+0.999*0.999) = 0.9999994995 v = 0.9999994995 gamma = 999.5 velocity of 2 particles before collision from the point of view of the rocket is calculated by http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html ux = 0.999 uy = 0 vx = 0 vy = 0.999 v=sqrt((0.999-0)^2 + (0-0.999)^2 - (0.999^2)(0.999^2)) v=0.9999980019975039930129749248984603081269824945912709 gamma = 500 This makes sense in terms of energy conservation since kinetic energy equals (gamma - 1)*mc^2 500 + 500 = 1000 But it doesn't make any sense in terms of momentum if momentum = gamma*mv However, everything works fine if Momentum = sqrt(kinetic energy) So my guess is momentum = sqrt(gamma)*mv corrected the image

1) The model I used already accounts for the Heisenberg uncertainty principle. 2) proton

according to this site http://www.sjsu.edu/faculty/watkins/relamomentum.html Relativistic angular momentum = γ^3mvr if so then gamma = 3.38 3.38*(electron mass)*(velocity of light)^2/(10^-14 m) Wolfram says force = 27.6 newtons and the force is 12 times stronger than electromagnetism would be at that distance

https://en.wikipedia.org/wiki/Neutron#Structure_and_geometry_of_charge_distribution Structure and geometry of charge distribution An article published in 2007 featuring a model-independent analysis concluded that the neutron has a negatively charged exterior, a positively charged middle, and a negative core.%5B67%5D In a simplified classical view, the negative "skin" of the neutron assists it to be attracted to the protons with which it interacts in the nucleus. (However, the main attraction between neutrons and protons is via the nuclear force, which does not involve charge.) The simplified classical view of the neutron's charge distribution also "explains" the fact that the neutron magnetic dipole points in the opposite direction from its spin angular momentum vector (as compared to the proton). This gives the neutron, in effect, a magnetic moment which resembles a negatively charged particle. This can be reconciled classically with a neutral neutron composed of a charge distribution in which the negative sub-parts of the neutron have a larger average radius of distribution, and therefore contribute more to the particle's magnetic dipole moment, than do the positive parts that are, on average, nearer the core. https://www.google.com/search?q=neutron+charge+distribution

I just showed that they do You can't argue with math

you might want to get your hearing checked someday Angular momentum of electron = (planks constant)/(2pi) Relativistic angular momentum = γmvr Relativistic centripetal force = γmv^2/r Gamma*(electron mass)*c*(10^-14 m)=(planks constant)/(2pi) solve for x Wolfram says gamma = 38.6 38.6*(electron mass)*(velocity of light)^2/(10^-14 m) Wolfram says force = 316 newtons The force between 2 electrons at that distane is ( Coulomb's constant )*(electron charge)^2/(10^-14 m)^2 Wolfram says 2.3 newtons According to those equations the force is 137 times stronger than electromagnetism would be at that distance (exactly as swansont said above) and that is sufficient for the electron to fit inside a neutron

What else would I have been saying? In order to know whether it would fit inside a neutron you first have to calculate the Bohr radius

That's true for all reactions. Not just nuclear. Even chemical reactions

https://en.wikipedia.org/wiki/Nuclear_fusion Energy released in most nuclear reactions is much larger than in chemical reactions, because the binding energy that holds a nucleus together is far greater than the energy that holds electrons to a nucleus. For example, the ionization energy gained by adding an electron to a hydrogen nucleus is 13.6 eV. Less than one-millionth of the 17.6 MeV released in the deuterium-tritium (DT) reaction So in other words the strong force is actually billions of times stronger than electromagnetism. Which is exactly what I was saying in the op. So if the strong force is that strong then that could explain how the electron fits inside a neutron. What is the bohr radius for an electron orbiting around a proton under that kind of force

The strong force doesn't follow an inverse Square law and is virtually non-existent outside of the nucleus

a chemical reaction might release 10 ev of energy. a nuclear reaction releases millions of times that. we are told that that is because the nuclear force is much stronger. At first that seems to make perfect sense but upon closer inspection the numbers clearly dont work out. The strong force is only 100 times as strong as electromagnetism and it only acts over a very short distance (about the size of the nucleus or 1/10,000 of an angstrom) now energy = force * distance so 100 * (1/10,000) = 0.01 even though the force is stronger it should release much less energy.

Doesnt change much plot 3.3*x+((5.5-3.3)*(5/6)^3)/x^2, x=5/6 to 1 https://www.wolframalpha.com/input/?i=plot+3.3*x%2B%28%285.5-3.3%29*%285%2F6%29^3%29%2Fx^2,+x%3D5%2F6+to+1