# Strength of fundamental forces

1. Dec 26, 2009

### Ranku

What is the unit of 'strength' of the fundamental forces?
When it is said that gravitational force is weaker than the strong nuclear force, how are they compared?
What is meant when it is said that the fundamental forces approach each other in strength at higher energies in the early universe?

2. Dec 26, 2009

### arunma

As you know, the electromagnetic and gravitational forces, when described classically, each have inverse square laws with force constants G and $k_e$. But since they have different units, they can't really be compared. Instead we compare their dimensionless coupling constants. For the electromagnetic force it's the fine structure constant:

$$\alpha = \dfrac{e^2}{4\pi \epsilon_0 \hbar c} \approx \frac{1}{137} = 0.00729...$$

We can also define a gravitational coupling constant:

$$\alpha_G = G\dfrac{m_e^2}{\hbar c} \approx 1.7 \times 10^{-45}$$

As you can see, the definition of the $\alpha_G$ depends on what fundamental particles you're considering. You'll find different values of $\alpha_G$ depending on where you look, and the above value is for two electrons. But no matter which elementary particles are chosen, the gravitational coupling constant will always be many orders of magnitude smaller than its electromagnetic counterpart.

As for the forces approaching each other, my particle physics is a bit rusty on this one, but I believe this has to do with the running of the coupling constants. This is why the electromagnetic and weak interactions are said to be somewhat unified.

Last edited: Dec 26, 2009
3. Dec 26, 2009

### Pythagorean

I never quite understood that either from a analytical point of view. How do you compare mass to charge? Through the inertia? F = ma, But then, the gravitational force need not come into effect just because there's a mass involved does it?

This is perplexing to me that mass has a multiplying effect on all forces while also playing "charge" to the gravitational field. Charge itself doesn't seem to have this fundamental connection to force or motion of all particles in this view. I'm fairly ignorant about the other two forces.

4. Dec 26, 2009

### Ranku

I found something called weak interaction constant: 1.2 x 10^-5. Would that be the coupling constant of the weak interaction?
Also I found this http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/couple.html" [Broken] that gives quite a different value for the gravitational coupling constant. What do you have to say on that?

Last edited by a moderator: May 4, 2017
5. Dec 26, 2009

### espen180

Asking "why is gravity so weak" might be asking the wrong question from a particle physics standpoint. By adopting Planck Units (See http://en.wikipedia.org/wiki/Planck_units ), the coefficients in front of the common force expressions disappear, and both the Coloumb force and Gravity become

$$\frac{x_1x_2}{r^2}$$ where x is mass or charge.

The question should then be "Why is the mass of particles so small?", not "Why is G so small?".

6. Dec 26, 2009

### Staff: Mentor

Well said!

7. Dec 26, 2009

### arunma

The value I've seen for the weak coupling constant is $3.22 \times 10^{-31}$. Too bad I don't have my old particle physics notes with me, because I had the expression in my notebook.

As for the different values of $\alpha_G$, as I mentioned in my earlier post, the value of the constant depends on the two fundamental particles you're considering. The value I've cited is for two electrons. You can also compare two protons, an electron and a proton, or any two charged particles that are well-understood.

Last edited by a moderator: May 4, 2017
8. Dec 26, 2009

### Galap

I don't remember this explanation very well, as I heard it a couple of years ago, but I think string theory's explanation as to the weakness of gravity is that gravitons aren't confined to our brane (our universe) and therefore escape off into other dimensions, lessening the force felt here. Can someone who knows more about this say it better?

9. Dec 26, 2009

### Ranku

Could you clarify what is "running of the coupling constants" ?

Appreciate your conceptually clear explanation of the other issues.

10. Dec 26, 2009

### arunma

No problem. Let me just add the disclaimer that this is going to be (at best) a sketchy explanation, since it's been a couple years since I took particle physics, and my research doesn't really go much beyond electrons and muons.

In any field theory that describes one of the fundamental forces, our Lagrangian will have an interaction term which includes a dimensionless coupling constant that describes the strength of the interaction. In the case of EM and gravitational terms, these constants are related to the constants in front of Coulomb's and Newton's Gravitational Laws. It turns out that these constants are dependent on the energy scale that your instrument is probing the coupling at. I think the reason for this is because of the virtual particles that mediate the interactions (i.e. the virtual photon for EM processes, the graviton for gravity, etc.). Basically you get loops in the Feynman diagrams, and integrating over these loops gives divergent integrals. You can take care of these divergent integrals with a trick called renormalization, and one of the side-effects of this is that the coupling constants become energy dependent.

11. Dec 26, 2009

### diazona

Yep, I think that's it (at least, based on what I know of the subject in conjunction with Wikipedia). Basically the idea is that the value of the coupling turns out to depend on the energy of the particles you're using in your experiment (specifically, the energy of the exchanged particle that mediates the force).

12. Dec 26, 2009

### arunma

Thanks for confirming. I should know this, since I just finished QFT this Fall semester. Unfortunately I had to do an observing shift at my collaboration's telescope the week we were learning about renormalization, so everything I know about that came from reading the book at 4 am (while monitoring trigger rates with my peripheral vision), and yes, also Wikipedia.

13. Dec 26, 2009

### diazona

lol... isn't grad school fun? (I hope I'm assuming correctly that you're a grad student too) Actually I'm in kind of the same situation, I also just finished taking a QFT class this past semester. In my case it's part 1 of a 2-part sequence, and all the renormalization stuff is being put off for the spring, so most of my knowledge about running couplings comes from the HEP project I've been working on and a few partially-comprehensible "teaser" lectures.

14. Dec 27, 2009

### Ranku

Thanks a lot mate. Sometimes grad students make ideal expositors, cause they are still learning and at the same time know enough to explain. :)

15. Dec 27, 2009

### arunma

Your assumption is quite correct. Good to hear you're taking the second semester. Unfortunately the professor who teaches our QFT class got bored of university life and took a job at a national lab, so we're stuck with no QFT teacher for this Spring (and I guess it's not the type of class that someone can just substitute for on such a short notice). Also being an experimentalist in particle astrophysics, I don't really need QFT for anything, so I don't know if I can let it slide past my advisor. But hey, there are always ways to get around advisors. :)

So, are you in HEP?

I know what you mean. The way one of the professors in my department puts it, grad students are at that funny point where we know a fair amount of stuff, but can still remember what it was like not knowing it. Maybe we should start writing all the introductory textbooks.

16. Dec 27, 2009

### Ranku

Intellectual idealism is the fuel of knowledge.