Physics Joke ;)

[Tweety.Abd]

Some nuclear physics geek humor. Like it if you get it. If not, just comment. :P

[rudrax]

Well, I don't get it. Like charges use to repel but how does they combine at last?

[Tweety.Abd]

Well, I don't get it. Like charges use to repel but how does they combine at last?

You got it right, two positive protons will repel via electrical charge. However, when they touch, the strong (nuclear) force is much, much stronger and will hold the two protons together. The electrical force is still there, but the strong nuclear force is 10x (or more) stronger. If you think about it, without the strong force, the nucleus shouldn't exist, since the protons repel each other electrically.

[calguyhunk]

Well, I don't get it. Like charges use to repel but how does they combine at last?

two positive protons will repel via electrical charge. However, when they touch, the strong (nuclear) force is much, much stronger and will hold the two protons together. The electrical force is still there, but the strong nuclear force is 10x (or more) stronger.

That's exactly what happens in a Nuclear Fusion. For fusion (to fuse is to come together) between two similarly charged nuclei to take place, extremely high energies are needed to overcome the repulsive electrostatic Coulomb force (coulomb repulsion) :yes:

They don't combine easy. But when they do, you can't take 'em apart. That's when you get the Hydrogen bomb ya'll :P

EDIT: Pardon moi for the over geekification of what was supposedly a 'joke' thread :P :hehe:

[rudrax]

Well, I don't get it. Like charges use to repel but how does they combine at last?

You got it right, two positive protons will repel via electrical charge. However, when they touch, the strong (nuclear) force is much, much stronger and will hold the two protons together. The electrical force is still there, but the strong nuclear force is 10x (or more) stronger. If you think about it, without the strong force, the nucleus shouldn't exist, since the protons repel each other electrically.

That means, it requires -10x force than the electrical repulsion force to hold two protons together?

[calguyhunk]

That means, it requires -10x force than the electrical repulsion force to hold two protons together?

Depends. ;)

To break the electric potential energy of two point charges (e.g., protons), the energy required to reach a separation r, (at which the nuclear attractive force becomes dominant, negating the electrostatic force) is given by U=ke²/r, where k = Coulomb's constant and e is the proton charge. :)

[rudrax]

That means, it requires -10x force than the electrical repulsion force to hold two protons together?

Depends. ;)

To break the electric potential energy of two point charges (e.g., protons), the energy required to reach a separation r, (at which the nuclear attractive force becomes dominant, negating the electrostatic force) is given by U=ke²/r, where k = Coulomb's constant and e is the proton charge. :)

Can you please explain the situation when r=0?

[ande]

At first I was thinking two ions - engineers way of thinking. :P

@rudax

Ones r is small enough nuclear force comes into action and will pull protons together hence making literally unbreakable bond.

At that time electrostatic force is of much smaller scale and can be disregarded in practical calculations.

Note that nuclear force is effective only on small distances.

That is why valence electrons are capable to become free when hit by very small amount of energy, photons, without being exposed to external electric field.

[cruelsister]

One molecule overheard another molecule saying, "I'm positive that a free electron once stripped me of an electron after he lepton me. You gotta keep your ion them"

[calguyhunk]

Can you please explain the situation when r=0?

I went through 2 of my old text and reference books on particle physics and they couldn't answer me. :( Google couldn't answer me :(

If I remember correctly, for theoretical calculations, we'll have to assume r-->0 rather than r=0. In reality, when r=0, the equation is undefined. Therefore, on the Cartesian coordinate system, the equation would have a vertical asymptote at r=0. ;)

[calguyhunk]

At first I was thinking two ions - engineers way of thinking. :P

You're an engineer? Sheldon would have hated you. "You may be good at doing your job, but your job ain't worth doing in the first place" he'd have said. :P :hehe:

Damn you theoretical Physicists :lol:

[rudrax]

Can you please explain the situation when r=0?

I went through 2 of my old text and reference books on particle physics and they couldn't answer me. :( Google couldn't answer me :(

If I remember correctly, for theoretical calculations, we'll have to assume r-->0 rather than r=0. In reality, when r=0, the equation is undefined. Therefore, on the Cartesian coordinate system, the equation would have a vertical asymptote at r=0. ;)

Well, if we put r = 0 in your equation, then the equation becomes, U = ke²(1/0) => U = ke².∞

What after this?

[calguyhunk]

Well, if we put r = 0 in your equation, then the equation becomes, U = ke²(1/0) => U = ke².∞

What after this?

As I said -

I went through 2 of my old text and reference books on particle physics and they couldn't answer me. :( Google couldn't answer me :(

If I remember correctly, for theoretical calculations, we'll have to assume r-->0 rather than r=0. In reality, when r=0, the equation is undefined. Therefore, on the Cartesian coordinate system, the equation would have a vertical asymptote at r=0

Beyond that, you better get hold of Stephen Hawking :P

[dcs18]

Somebody invite Ambrocious, here - he will unravel the conspiracy hatched by them protons (using an Alex Jones video.) :P

[rudrax]

Well, if we put r = 0 in your equation, then the equation becomes, U = ke²(1/0) => U = ke².∞

What after this?

As I said -

>>I went through 2 of my old text and reference books on particle physics and they couldn't answer me. :( Google couldn't answer me :(

If I remember correctly, for theoretical calculations, we'll have to assume r-->0 rather than r=0. In reality, when r=0, the equation is undefined. Therefore, on the Cartesian coordinate system, the equation would have a vertical asymptote at r=0

Beyond that, you better get hold of Stephen Hawking :P

So, r can not be zero. That proves that two protons can not overlap with each other, unless, infinite amount of nuclear force is applied; Since, U = ke².∞ => U = ∞ (anything multiplied with infinity, gets dissolved in infiinity) and applying infinite amount of nuclear force, makes no sense :dunno:

Hello, Mr. Tweety.abd, where are you?

[dMog]

what is this ..big bang theory?

[avmad]

[Tweety.Abd]

That means, it requires -10x force than the electrical repulsion force to hold two protons together?

Technically, it requires 1 plus a tiny amount times the repulsion force to hold two protons together. But since they fly around, bumping into one another, if the forces are closely balanced, they can get knocked off easily. In order to have stable nuclei, the strong force has to be much higher...100x or more the repulsion energy. For small nuclei, it's like 100x more. For bigger nuclei, it's much less...2 or 3 or 10x. That's part of the reason some heavy nuclei are unstable.

Hello, Mr. Tweety.abd, where are you?

Sorry for the delayed response, I were busy for the past few days and didn't get time required to answer your question. I didn't know it would go from being a joke to serious physics discussion.

[rudrax]

Hello, Mr. Tweety.abd, where are you?

Sorry for the delayed response, I were busy for the past few days and didn't get time required to answer your question. I didn't know it would go from being a joke to serious physics discussion.

Not a problem bro.

That means, it requires -10x force than the electrical repulsion force to hold two protons together?

Technically, it requires 1 plus a tiny amount times the repulsion force to hold two protons together. But since they fly around, bumping into one another, if the forces are closely balanced, they can get knocked off easily. In order to have stable nuclei, the strong force has to be much higher...100x or more the repulsion energy. For small nuclei, it's like 100x more. For bigger nuclei, it's much less...2 or 3 or 10x. That's part of the reason some heavy nuclei are unstable.

The underlined part is a nice info to get. I'd like to draw your attention to the above #15 post expecting some arguments/idea from you.

[Tweety.Abd]

Well, if we put r = 0 in your equation, then the equation becomes, U = ke²(1/0) => U = ke².∞

What after this?

As I said -

>>I went through 2 of my old text and reference books on particle physics and they couldn't answer me. :( Google couldn't answer me :(

If I remember correctly, for theoretical calculations, we'll have to assume r-->0 rather than r=0. In reality, when r=0, the equation is undefined. Therefore, on the Cartesian coordinate system, the equation would have a vertical asymptote at r=0

Beyond that, you better get hold of Stephen Hawking :P

So, r can not be zero. That proves that two protons can not overlap with each other, unless, infinite amount of nuclear force is applied; Since, U = ke².∞ => U = ∞ (anything multiplied with infinity, gets dissolved in infiinity) and applying infinite amount of nuclear force, makes no sense :dunno:

Hello, Mr. Tweety.abd, where are you?

The thing to keep in mind is that equations like that have a region of applicability. Let me give you an easier-to-understand example. The same logic works between two objects with mass. In that case, the potential energy is U = G M m/r. Same problem. Suppose that we are talking about two planets, one with radius r1 and one with radius r2. That equation only works for radii > r1 + r2. Bring the planets so they are close enough to touch (or closer), and the equation fails.

Similarly in electrostatics, what you said works just fine when two objects are far away from one another. Get them closer and this equation fails. The distance at which this equation fails depends on the details...for a proton it fails when two objects get closer than a femtometer apart (the size of a proton). For electrons, it's trickier, but there comes a point where quantum field theory effects come into effect. At that point, the equation fails. That's the bottom line to remember. That equation doesn't apply all the way down to r = 0. This solves your problem easily.