Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A short lived particle is created at the interaction point and then decaying a distance $d$ away, in some detector. My question is what is the smallest distance $d$ that can be measured experimentally? Of course it will be different from a detector to another.

Any links that explain how this analysis, of discovering a particle, is done experimentally would be appreciated.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

See Tao Han's Collider Phenomenology: Basic Knowledge and Techniques, Page 22, and the referenced articles.

As a rough guide, for the ATLAS detector the resolution is of the order $10\mu m$. It's also possible to resolve the vertex displacement in the longtitudinal direction alone, in the case that there's only one charged particle track to extrapolate backwards. In this case the resolution is about an order of magnitude worse.

The full formula for the resultion also depends on the transverse momentum and the angle between the particle track and the beam line.

share|improve this answer
    
You should say what the answer is, so that it is available without digging through the reference. –  Ron Maimon Nov 28 '11 at 6:38
    
That is an excellent reference. But still it does not answer the question what is the smallest distance for a displaced vertex that can be measured. –  Revo Nov 28 '11 at 16:57
    
@Ron Maimon: OK, I've added some details. –  felix Nov 28 '11 at 17:02
1  
@Revo: I guess the shortest distance should be comparable to the resolution quoted in the article. But if you are really asking for the shortest distance that can be distinguished from zero at a statistical significance, I don't know the answer. –  felix Nov 28 '11 at 17:04
    
@felix Great, thanks alot. –  Revo Nov 28 '11 at 17:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.