Absence of gravity at atomic level at all?

[heartless]

Hello,
I'm not familiar with what scientists working at accelerators say, but almost every book, and documentary movie say that Quantum Mechanics don't cover the gravity. I just gave a quick thought, well, maybe the gravity isn't there at all? Atomic level simply lacks the gravity and is only based upon 3 forces, electro-magnetism, strong and weak force? If the gravity, exists there, it must be very weak, but wouldn't atoms fall into the gravitational field of one another and start circling around each other? Hey, so how is it with this gravity at atomic level? And well, maybe I'm just out of time, and every one knows the answer except for me :)

P.S, just a quick question, what keeps the electrons tied up to an atom?

Thanks for all the help :shy:

 

[chroot]

Gravity certainly exists at atomic scales, but its strength is negligible when compared to the other three forces.

- Warren

 

[jtbell]

Let's calculate the gravitational potential energy of two protons or neutrons (mass about $m = 2.7 \times 10^{-27}$ kg) in a nucleus, separated by about $r = 10^{-15}$ m. The gravitational constant is about $G = 6.7 \times 10^{-11}$N-m^2/kg^2. The potential energy is

[tex]U = - \frac{Gm^2}{r}[/itex]

which gives about $-4.9 \times 10^{-49}$ J, or about $-3.0 \times 10^{-30}$ eV.

To put this in perspective, the binding energy per proton or neutron in most nuclei is in the general ballpark of a few million ($10^{+6}$) eV. So, unless gravity behaves wildly differently on a nuclear scale than on a macroscopic scale, its effects are insignificant compared to those of the other forces.

 

[heartless]

Thanks,
now let me ask you another question, suppose the gravity at atomic level is stronger than nuclear strong force. How would if affect atoms, and all the particles?

 

[-Job-]

Let's calculate the gravitational potential energy of two protons or neutrons (mass about $m = 2.7 \times 10^{-27}$ kg) in a nucleus, separated by about $r = 10^{-15}$ m. The gravitational constant is about $G = 6.7 \times 10^{-11}$N-m^2/kg^2. The potential energy is

[tex]U = - \frac{Gm^2}{r}[/itex]

which gives about $-4.9 \times 10^{-49}$ J, or about $-3.0 \times 10^{-30}$ eV.

To put this in perspective, the binding energy per proton or neutron in most nuclei is in the general ballpark of a few million ($10^{+6}$) eV. So, unless gravity behaves wildly differently on a nuclear scale than on a macroscopic scale, its effects are insignificant compared to those of the other forces.

You used $r = 10^{-15}$. I don't know if, from a particle perspective, that is close or not, but the closer the better. If we were to take r arbitrarily close to 0 we would get a significant value. I'm not suggesting we take the limit of the gravitational potential as r goes to 0, but i wonder why you used the value of r that you used. How small can r be, in theory? Naturally, at this small scale things would have to be extremely close in order for gravity to have a chance, but in the event that they did, it sounds like it would be hard to get them apart. What are some of the main reasons why the nuclear forces are not caused by gravity?

 

[dextercioby]

Well, the smaller "r" gets in gravity, the more you'd have to use GR...

Daniel.

 

[jtbell]

i wonder why you used the value of r that you used.

Atomic nuclei have diameters in the range of $10^{-15}$ to $10^{-14}$ m, so $10^{-15}$ is a reasonable order of magnitude estimate for the separation of two "neighboring" nucleons in a nucleus.

 

[arivero]

More than "absence of gravity", I like to think of "presence of gauge forces". The point is, suppose you put a general cut-off asking that no force between two particles can be greater than the Force of Planck $$F_P \propto {\hbar c \over l_P^2$$ . If we had only gravity, this cut-off happens to be milder than the Planck Length cut-off because two electrons have a gravity force

$$F_G=\hbar c ({m_e\over m_P})^2 {1\over r^2}$$

so they can approach a lot without violating the Planck Force.

Fortunately they also happen to have an electromagnetic force

$$F_E=\alpha(r) \hbar c {1\over r^2}$$

so that at distances of order Planck Length, the *electromagnetic* force is the one causing a force of order Planck Force, and then hitting the cut-off. Thus the gauge forces are needed to make the Force cutoff and the Distance cutoff compatible. Best said, Force cut-off + Gauge forces imply the length cut-off. Without gauge forces, the Force cut-off and the length (area, if you prefer) cutoff are two separate hypothesis.