Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Mandelstam variables 1 positive 2 negative

+ 2 like - 0 dislike
2860 views

The three Mandelstam-variables are defined as: $$s=(p_A+p_B)^2=(p_C+p_D)^2,$$$$t=(p_A-p_C)^2=(p_B-p_D)^2$$$$u=(p_A-p_D)^2=(p_B-p_C)^2.$$ Where A and B are the incoming particles and C and D are the outgoing particles, and the second equation follows from the conservation of 4-momentum.

The Mandelstam-variable $s$ gives the center-of-mass energy and is thus always positive, now for the Mandelstam-variables $t$ and $u$ the demands can be found by looking at them in the laboratory-frame, after an easy calculation it can be shown that:$$s\geq \max[m_A^2+m_B^2,m_C^2+m_D^2],$$ $$t\leq \min[m_A^2+m_C^2,m_B^2+m_D^2],$$ $$u\leq \min[m_A^2+m_D^2,m_B^2+m_C^2].$$

This means that there should exist a situation in which the three Mandelstam-variables are positive. I was wondering is such a situation exists, or that I am simply overlooking some facts and that indeed only ONE Mandelstam-variable can be positive?

To elaborate more on the calculation which lead me to this conclusion, I've calculated these variables in the frame in which particle A is at rest (since $s$, $t$ and $u$ are invariant I can do this). This yields for the momenta: $$p_A=(m_A,0),$$$$p_B=(E_B,\vec{p}_B),$$$$p_C=(E_C,\vec{p}_C),$$$$p_D=(E_D,\vec{p}_D).$$ So for my Mandelstam variables I would get that: $$s=(p_A+p_B)^2=p_A^2+p_B^2+2p_A\cdot p_B = m_A^2+m_B^2+2m_AE_B,$$ $$t=(p_A-p_C)^2=p_A^2+p_C^2-2p_A\cdot p_C = m_A^2+m_C^2-2m_AE_C,$$ $$t=(p_A-p_D)^2=p_A^2+p_D^2-2p_A\cdot p_D = m_A^2+m_D^2-2m_AE_D,$$ where I used the first equality in the definition of the Mandelstam variables. I could do the same calculation with the second equality, if I would then discard the terms of the form $2m_AE$ I would get the inequalities above.

$\,$

Edit: I know that the second equality doesn't give simple terms $2mE$, but terms of the form $$E_CE_D-\vec{p_C}\cdot\vec{p_D}=E_CE_D-|\vec{p_C}||\vec{p_D}|\cos(\theta),$$ and since $E^2=m^2+|\vec{p}|^2$ we have that $E>|\vec{p}|$, so these terms are positive.

$\,$

Edit 2: I've calculated these for the non-elastic proces $e^-e^+\rightarrow\mu^-\mu^+$ (since inelastic problems are the source of the problem) and I found that: $$s=2m(m+E_B),$$ $$t=m^2+M^2-2mE_C,$$ $$u=m^2+M^2-2mE_D,$$ where $m$ is the mass of the electron and $M$ is the mass of the muon, particle $A$ (either an electron or positron) is standing still and for particle $B$ (electron or positron) and $C$ and $D$ (the 2 muon particles) the energies are $E_B$, $E_C$ and $E_D$. The variable $s$ is obviously positive, to check if $t$ of $u$ are positive I summed them and used the conservation of energy: $m+E_B=E_C+E_D$, which gives: $$t+u=2(m^2+M^2)-2m(m+E_B)=2(m^2+M^2)-s,$$ now in order to be able to create 2 muons the energy should be high enough, hence $s=(2M)^2$, filling this in yields: $$t+u=2(m^2-M^2).$$ Which brings me back to the question: ''Should only one of the three variables be positive (just as most books claim) or are there special cases''? In my derivation above I used the conservation laws, but as you can see in this example the possibility of $t$ and $u$ being positive (for small enough $E_B$) vanishes by demanding that de center of mass energy $\sqrt{s}$ is large enough, will this always be the case ?

This post imported from StackExchange Physics at 2014-04-14 16:19 (UCT), posted by SE-user Nick
asked Apr 13, 2014 in Theoretical Physics by Nick (30 points) [ no revision ]
From your equations in the first line: $s=(p_A+p_B)^2,\,t=(p_A-p_C)^2 \text{ and }u=(p_A-p_D)^2.$ This states that s, t and u are all positive since they are squares of real numbers.

This post imported from StackExchange Physics at 2014-04-14 16:20 (UCT), posted by SE-user LDC3
@LDC3 The $p$s that appear are 4-momenta. So no, you cannot guarantee that the squares are positive.

This post imported from StackExchange Physics at 2014-04-14 16:20 (UCT), posted by SE-user dmckee
@LDC3, you can see this if you would calculate $s$, $t$ and $u$ in the center of mass- (or momentum)-frame for an elastic collision (so the masses stay conserved). In that case you would get the specific demand that $s\geq (m_1+m_2)^2$, $t\leq 0$ and $u\leq 0$ for the proces to be physical.

This post imported from StackExchange Physics at 2014-04-14 16:20 (UCT), posted by SE-user Nick

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...