Relationship of Forces

[Little Bang]

I understand that the electromagnetic, strong and weak nuclear forces have been united. Can anyone tell me, with out using math, what is that relationship?

[Qfwfq]

It isn't easy to explain the Standard Model, spontaneous symmetry breaking, Goldstone bosons and the Weinberg angle without using math. Would a spot of linear algebra be too mathematical for you?

[Little Bang]

I can handle the linear algebra Q, mabye that will help.

I guess what I'm looking for is something like the relationship between energy and matter where C is that relationship

[Erasmus00]

I understand that the electromagnetic, strong and weak nuclear forces have been united. Can anyone tell me, with out using math, what is that relationship?

 

I'll try and start with electric and magnetic forces. I'm assuming you know that an electric field acts on a charged particle, while a magnetic field acts on a moving charged particle.

 

Consider the simple case of a uniform magnetic field in all of space and a slow moving particle traveling in that field. It undergoes some acceleration due to the force of a magnetic field.

 

Now, imagine that you are riding along on top of the charge. In this case, it looks to you like the charge isn't moving, but the magnetic field somehow is. This "moving magnetic field" creates a force on your stationary charge, i.e. this field is an electric field. (you must experience the same force, because the principle of relativity tells you that physics cannot be frame dependant).

 

This would be much easier to do with pictures, but I hope you get the idea that electric and magnetic fields change into one another depending on frame of reference. This implies they are just different aspects of the same thing.

-Will

[Qfwfq]

I guess what I'm looking for is something like the relationship between energy and matter where C is that relationship

That seems more like special relativity than the standard model.

 

The crux is that mass is energy. In natural units, c = 1 because a second is a "length" equal to 300 million metres but in a timelike direction, instead of spacelike. In this way the famous formula is really E = m, and it's just the p = 0 case of:

 

m^2 = E^2 - p^2.

 

Mass is a scalar (in Minkowskian geometry), m^2 is the scalar product of the energy-momentum 4-vector (according to the Minkowski metric). This differs from a 4-D Euclidean metric by having the time-time component of opposite sign to the other three diagonal elements.

 

E is called the total energy, it is the sum of rest energy plus kinetic energy where the rest energy is the mass. This is an equation concerning a free particle, no potential energy, it simply gives the relation between energy, 3-momentum and mass.

 

Note that in a Lorentz-covariant treatment, the gamma factor is in the 4-vector components, so the p is actually Newton's momentum divided by the usual root of 1 - v^2. If you write it the way Einstein and Planck did you get the increasing mass that people still talk about but which isn't a good 4-tensor notaion; mass is a scalar and isn't dependant on velocity. It's the total energy that increases.

 

Mass is therefore the energy a body is "made of", at rest with the observer, and comprises any internal kinetic and potential terms so that it isn't exactly equal to the sum of the masses of constituent particles. A piece of iron has a slightly greater mass when it is hot because of the greater kinetic energy of its parts (according to the centre of mass coordinates). Also, it would take energy to evaporate the iron atoms, therefore the mass of the piece of iron isn't exactly equal to that of the same atoms in the vapour state, even at the same temperature. The same goes for atoms (that's how chemical reactions can be exothermic or endothermic) and the same goes for atomic nuclei (that's how atom bombs work).

[Little Bang]

It seems like your say that the equation E = MC^2 is not truely an equality. Tell me if I'm wrong, but I thought that the annhilation of a particle anti-particle exactly fit the equation?

[Qfwfq]

It's an equality for p = 0.

 

For e- e+ anihilation, the nearst to exact you can do, in the c. m. coordinates, is the ground state of the positronium atom (if it can be called the "ground state" as it isn't stable, but you know what I mean). No atom has a mass exactly equal to the sum of masses. It is, however, a difference of few eV. For colliding e- and e+, the cross section is tiny but not exactly zero, anyway the photons would have a greater energy in the c. m. system.

[InfiniteNow]

No atom has a mass exactly equal to the sum of masses. It is, however, a difference of few eV.

Is it that the mass of the whole is greater than the sum of the parts, or that the mass of the whole is less than the sum of the parts?

[sigsfried]

IT has a mass smaller (even if very slightly) than the sum of its parts.

[Qfwfq]

Is it that the mass of the whole is greater than the sum of the parts, or that the mass of the whole is less than the sum of the parts?

For "simple" cases you can argue that a completely stable bound state can't have a greater rest energy than the total of its parts separated but it can be more complicated than. When there are different possible ways of separating into parts it can also depend on which you consider. Some reactions are exothermic and some are endothermic.

[InfiniteNow]

The thought I took away from your answers is that some energy is lost while the "parts" combined, as Q mentioned, heat.