Why are the strong forces inside the nucleus of a atom considered different from gravitational forces? According to F=G Mm/r^2 when the radius approaches zero isn't F going to become incredibly huge just like strong force? Am I missing something here?
Not huge enough. Plug in the numbers and see for yourself. Also consider that the electromagnetic force, the repulsion between protons, increases with decreasing radius in the same way as gravity.
The gravitational force is many many time weaker than the strong and EM force on the atomic scale, even allowing for the fact that r is very small (and EM goes like 1/r^2 too). One way to think of the strong force is like this: Forces act on charges. Electromagnetism acts on particles that have a charge but not on particles that are neutral. Gravity acts on particles that have a charge called 'mass.' The strong force acts on particles that have a charge that we call colour. Instead of the one type of charge in the case of mass and the 2 types of charge (+ and -) in the case of EM, for colour the situation is a bit more complicated. For quarks there are 6 types of charge - red, blue and green and their "anti colours." A proton is made of three quarks - one red, one blue and one green. An anti proton is made of antiquarks. One antired, one antiblue and one antigreen. For gluons (the particles that carry the strong force) things are a bit different. There are actually 8 different charges, which are a combination of a colour and an anticolour. For example you can have a green-antired gluon. Check out the pictures here for more clarification.
My physics book is saying that coulomb's law is not valid for distances lesser than 10^-15 m. What if the nucleons are closer than this minimum distance? Wouldn't gravitational pull prevail then?
Hmmm my book does not say why coulomb law is not valid for less than 10^-15m. Seems that after this range they assume quantum effects.
Welcome to the wonderful world of physics textbooks Please Register or Log in to view the hidden image! They're mostly right, but there are some pretty shitty examples that I've seen. The book is correct, in this instance, but it is completely useless because it just fed you a fact, and didn't give you any explanation. The coulomb force is only an effective description for phenomena that happen at distances longer than 10^-15 m. Below that length, there is a new force, called the electroweak force. We can observe the consequences of this force in radioactive decay, as well as verify the theory at particle accelerators. The differences between the electroweak force and the Coulomb force are pretty esoteric. But, you know that in electromagnetism, the photon interacts with charged particles to mediate the force. Photons can't interact with each other. In the electroweak force, there are four types of photons, and these four types of photons (called B, W1, W2, and W3) can interact with each other. This changes (drastically) the properties of the force. Sure, it's called the Planck Length Please Register or Log in to view the hidden image!
I wouldn't blame my textbook so much. Thing is I have only got my higher secondary textbook which has nothing about quantum physics. It only deals with classical physics.
I don't know the math, so I would appreciate some help... If we assumed the nucleus of an atom is equal to a planck length and the distance from the nucleus to the first electron was a planck length what woudl be the difference between the weak force and the strong force? How would that equation look? Am I way off base with this and completely misunderstanding this? It has been a number of years since I read about the different forces.
Strong interplay Nucleons have gravidynamic interplay see Konovalov V.K. monograph "Fundamental of new physics..."
What is it you want to calculate? The (electrostatic) force between two protons is repulsive, and is much greater in magnitude than gravity. Either way, the proton has mass = 1.7x 10^{-27} kg, and the distance between two nuclei is about one femtometer.
Presume the proton is the size of a planck length and the electron is a planck length away from it. How would that equation work? Also, presume that the weak force is equivalent to the strong force - woudl would that imply about the mass of the proton? I am just trying to recall/figure out the math and how/why it works the way it does.
No, they don't. No. The proton is 1 femtometer is size, and the electron is on the order of 5 nanometers away. The answer is easy, and you CAN calculate the gravitational force between two protons at a separation of 1 fm. Now calculate the electrostatic force between two protons at one femtometer. You should see that the electrostatic force GREATLY outweighs the gravitational attraction. This is how you know that there has to be a strong (i.e. stronger than gravity) force.
I believe a major distinction is that with the gravitational force, the farther apart the two interacting bodies (protons, moons, whatever), the less the attractive force. With the strong force, it's the opposite. The farther apart the two interacting bodies (quarks), the stronger the attractive force. Hence, it is impossible to separate quarks, only cause them to exchange (or be converted into energy, e.g. proton/anti-proton annihilation). Correct me if I'm wrong, Ben.
Actually, the force stays pretty constant. But as W = F times d, if you pull the quark very far, soon you have W > 2mc^2 and you get pair creation, so it's very hard to test at large d, but the residual force (between mesons and/or nucleons) drops off very quickly with distance. Using the pre-Maxwell idea of field lines, the field lines of QCD are sticky and pull together into a tube. Neither this model, nor any particle-based exchange which I can think of would account for a force which uniformly (and without an asymptote) grew larger with distance. QCD does has asymptotic freedom, which means at very short distances, the forces become too weak to measure, which is part and parcel of my field line model and the actual SU(3) gauge theory.
