# Conversion Gev -> barn?

1. Aug 10, 2009

### maani

Hi I have computed a cross section to 10-6 GeV2. Now I have to convert it to barn, but don't know how. Can anybody help me? Is it ok to have a cross section in units of GeV2 or is my result completely wrong?
Thanks!

2. Aug 10, 2009

Staff Emeritus
You made a mistake somewhere. It should be 1/GeV2.

The conversion factor you need is $\hbar c$ = 200 MeV fm.

3. Aug 10, 2009

### humanino

Equivalently, $\left(\hbar c\right)^{2}=0.389\text{ GeV}^2\text{mbarn}$

4. Aug 11, 2009

### Count Iblis

If you're doing quantum gravity, then:

GeV^2 = 1.752*10^(-80) barn

5. Aug 11, 2009

### humanino

It seems to me confusing in the context of the original question to hide $\hbar$ and c. Vanadium 50's and my result are on the first page of the particle data group booklet or review, so I think we were justified not to give further details (unless requested). I'd like to request a clarification about your formula. It seems to me, energy and length have inverse dimension for instance. Can you please re-establish the proper $\hbar$, c and (probably) G factors ?

6. Aug 11, 2009

### Count Iblis

Well, I agree that the OP really meant GeV^(-2) and agree with your answers. Now, if you put G = 1, then of course, any power of GeV could be a cross section (because you've made physics dimensionless).

Now, I don't work in particle physics so, I don't have the conversion factors in my head. So, what I always do is use a few well known formulae that contain hbar, c and G to do the conversion.

To convert GeV^2 to a cross section, you can use that in General Relativity, mass and length have the same dimensions (if you put c = G = 1). So, GeV^2 is already a cross section and no additional conversion using hbar needs to be performed.

To restore G and c, we just hijack the formula for gravitational potential energy, so:

m^2 G/r = energy = m c^2

this is a dimensionally correct expression, that doesn't need to make sense. So, we have:

m G/(c^2 r) = dimensionless

Or:

E G/(c^4) = length

where E is an energy. So, we see that:

cross section = E^2 G^2/c^8

If you know the formulas for Planck length, Planck energy etc. etc., you can do the conversion directly. To convert GeV^n to a cross section, you simply divide this by the Planck energy to the power n and multiply by the Planck length squared.

7. Aug 12, 2009

Staff Emeritus
I think this is singularly unhelpful.

We have someone who is doing their first calculation - we know this, because they are having unit problems. Suggesting they start popping Planck masses in until the units come out right is not going to help them get the right answer. It's simply the wrong thing to do.

8. Aug 12, 2009

### maani

Thanks to all of you for the fast answers. This was really my first computation and it was completely wrong. I am still working on it. But i have learned at least how to convert the units.

9. Aug 12, 2009

Staff Emeritus
It's very smart to carry along the units, at least until you gain more facility with these calculations. People will say, "who cares if you drop an hbar or a c", but I would respond, "but who knows what else you dropped?" Then later when you get more experience, taking the shortcut becomes more reasonable.

10. Aug 12, 2009

### Count Iblis

It is very unhelpful to keep students indoctrinated in the wrongful use of units. Students learn units the wrong way in high school and even at university, the false myth of the meaning of units is promoted, to the detriment of science. Then what you see is that advanced university level students are struggeling with what should be a trivial high school physics exercise.

Even many professional physicisist do not understand units as
http://arxiv.org/abs/hep-th/0208093" [Broken].

Last edited by a moderator: May 4, 2017
11. Aug 12, 2009

### Count Iblis

It's not just a shortcut, natural units are a bona fide unit system. The conversion back to SI units is very simple. With some minor practice, the chances of making mistakes are much less likely if you use natural units than if you use SI units.

The constants c, hbar, G, k_b, etc. etc. are nothing more than irrelevant conversion factor. Then, if we intend to use SI units when actually inserting numbers in the equation, we should make sure the correct conversion factors are present in the final result. But it can be extremely cumbersome to derive the equation with the conversion factors already present in the intermediary steps.

It is a bit like the complicated formulas frequently used by engineers were e.g. pressure appears in different ways, e.g. in atmospheres and in mm Hg. Then the formula also contains a conversion factor which has exactly the same interpretation as c, hbar, G, k_b etc.

12. Aug 12, 2009

### humanino

Not really. Keeping track of the powers of mass is quite a useful consistency check. I'd say it helps to avoid errors. Anyway, unless the original posters intends to go into quantum gravity, in which case this is not the appropriate sub-forum, advising him to overlook all dimensions together is certainly not very pedagogical. First he should learn the dimensions of the various fields, like spinors/vectors etc... at the very least.

13. Aug 12, 2009

### Count Iblis

Yes, I agree that it's useful to keep track of the powers of the mass. But then you can put hbar = c = 1. The fact that inverse mass is a length should be common knowledge. Even I know that and I don't work with this stuff on a daily basis.

14. Aug 12, 2009

### humanino

When I said power, I meant both positive and negative. That's indeed what we do all the time. The problem comes about with m^2 G/r=E and setting G=1. With an arbitrary number of hidden G factors, you loose the ability to check the number of powers of mass (or energy, or length, or time) on both sides of the equation. For instance, with
cross section = 1/(E^2)

15. Aug 12, 2009

### Count Iblis

Indeed, but then, you're not going to set G = 1 in ordinary QFT computations. And if one contemplates a fundamental theory, then one has to be reasonble and accept the fact that Nature may be fundamentally dimensionless.

If you formulate some lattice statistical mechanics model, like the Ising model, you only have pure numbers. But close to the critical temperature you can look at some scaling limit in which you can formulate hte model as some effective field theory. The correlation length then enters the effective theory as an inverse mass.

Cardy writes in one of his books that the Renormalization Group is simply a sophisticated way of doing dimensional analysis.