I'm making no assumption, I am going by what you did : trolling and no contribution. This thread contains exactly 9 posts that are physics (1,2,3, 15,22,23,35,36,37), the rest are trolling by arfa brane, Trooper, brucep , OnlyMe and whining / droning by rpenner. OK, I'll give you and all the other barkers a chance to prove me wrong: The solution to the exercise gives only the angle between the two particles after collision, no other useful information. So, the particles describe two arbitrary trajectories, embedded into a (non axially symmetric) cone. Not very useful for an experimenter trying to detect the particles after collision. Actually, not useful at all. Find the exact trajectories of the two particles after collision (or prove that it is impossible to find them). Stop the barking, let's see the physics. Any more barking with no physics will only prove my point, that you are trolls, incompetent to perform any physics. To further shut you up, I will give you a head start: Let the angles made by the two particles with \(\vec{p_1}\) be \(\alpha\) and \(\beta\) , respectively. Then, from post 35, you have: \(p_3 p_4 cos (\alpha+\beta)=\frac{(E_3-m_2c^2)(E_4-m_2c^2)}{c^2}\) Find the other relevant equations, the rest of the relevant information stares you in the face in post 35, feel free to use the provided solution. Post the physics or keep your mouth shut.
First.., is xyx really just a different spelling of Tach? And then... You realize that your list of 9 posts, contains exactly 2 of your 30 posts?... Which implies that since you label the rest as trolling, you have been trolling from the start. The list does not even contain your initial post in this thread! And Post #37 (quoted), the last post in your above list, was mine.., and true, it was not about the physics. Instead it was an appeal to you, to start a new thread instead of derailing this one! One good thing about the new software is that if this really is Tach reincarnated, edits have to be completed within the first 90 minutes after the initial post!
Posts 35,36 solve the general problem, persistent troll.Actually post 35 is sufficient, post 36 contains only some discussion. And no. 1 troll OnlyMe doesn't disappoint: incapable of doing any physics he continues to pollute the thread with his trolling.
Really? Because posts #5 (as edited) and #8 tell a different story. Most of your problems would not have happened if you actually read the thread and worked with the same facts as the rest of us. Instead you disparage posts that you later appear to obliviously repeat and misread the thread in basic ways.
Well, your post 11 shows you to be a real asshole: preachy, dismissive, arrogant <shrug> You, of all people, need to stop droning. See if you can work out the challenge posted in #62 (your barking dogs amply demonstrated that they can't). Give whining and pettiness a rest and see if you can do some physics for a change. There hasn't been any physics in this thread since post 36, let's see if we can return to physics. It is up to you.
Since none of you is doing any physics, here is some more: From the momentum conservation in 35 \(\vec{p_1}=\vec{p_3}+\vec{p_4}\) one gets: \(p_1=p_3 cos \alpha+p_4 cos \beta\) \(0=p_3 sin \alpha-p_4 sin \beta\) Add this to \(p_3 p_4 cos (\alpha+\beta)=\frac{(E_3-m_2c^2)(E_4-m_2c^2)}{c^2}\) Your next move, rpenner.
It was answered exactly in post #29. The basis for the answer was given in post #1. To solve for four variables, one usually needs four relationships. Post #1: In 2 dimensional space, we have 4 unknown momentum variables and 2 relations. The energy relation is a third relation which means there remains a degree of freedom. In 3 dimensional space, we have two degrees of freedom remaining. Post #29: (emphasis added)
So, where is the 4-th relationship? I gave you 3, can you write down the 4-th one? You do not know that the particles are in 2D. There is nothing stopping them from being 3D, actually, their trajectories are embedded in a cone. So, where should an experimenter place his detectors in order to capture the particles after collision?
Everywhere practical. LHC detectors are large cylindrical affairs because there is no preferred direction in the center-of-mass frame but it's impractical have the detector and the collider beams occupy the same space. http://cms.web.cern.ch/news/why-cms-so-big
This is a waffle, not an answer, since the particles can be found ANYWHERE covering a whole plane. So, where would the exercise formalism instruct for detector placement? You realize that the exercise predicts only the angle at the tip of the cone, nothing more, right? You realize what this means, right?
Don't be obtuse. Conservation of momentum means in both the center of mass frame and the stationary target frame the two outgoing trajectories are co-planer which allows the decomposition in the stationary target frame into components parallel and orthogonal to the incoming momentum. Thus (as is the case in posts #1-#62) we only care about the relation between those outgoing trajectories and thus the geometry of that 2-dimensional subspace.
You are becoming patronizing again. But this plane can be ROTATED at ANY angle around the axis going thru \(\vec{p_1}\). You keep missing the basic fact that you are dealing with a FAMILY of planes, not ONE plane. Any of the planes is this INFINITY is equally probable.So the two particles can land ANYWHERE on the transverse section through the cone ("diametrically" opposite). Err, this is false, we care about much more, see above. How about answering the other question, the exercise gives only the angle between the particles after collision, their directions are indeterminate, so how does the exercise help in placing the detectors?
That's the second degree of freedom in 3 dimensions which I already described. But since we are only dealing with one collision, we need only considered one plane for the outgoing trajectories.
WHICH "one plane" would that be? You have an INFINITY of planes to choose from (all the planes passing through \(\vec{p_1}\) are equally valid). How does your exercise pick the plane? Hint: it can't. To make matters worse, in the plane (that you cannot pick), you have no idea of the directions of the two particles, your exercise is capable of only finding the angle between the two directions. How does your exercise pick the respective directions of the particles? Hint: it can't. So, you do not know the plane and you do not know the directions within the plane.
You seem close to understanding what was written in posts #1, #29 and #70, but your question suggests you want a fortune teller, not a physicist.
Resorting to personal attacks doesn't change the fact that you can't determine the trajectories of the two particles because you cannot determine their respective angles with respect to \(\vec{p_1}\). You had all the time to think about a clever way of finding those two angles but you came up empty. You cannot even determine the plane that contains the two trajectories because any plane that embeds \(\vec{p_1}\) is equally valid. Based on your quip above, you seem unable to grok this simple fact. The bottom line is that your exercise is absolutely worthless, it is just a freshman exercise with no application whatsoever. An experimental physicist cannot use its conclusion for anything. But, if it makes you happy and it garners you admirers like the toadies that elbow each other eager to brownose you , like brucep, OnlyMe and arfa brane, then so be it. None of them is capable to contribute any physics if their lives depended on it.
