Relativistic 2 -> 2 scattering with all equal masses

In summary, the study of relativistic 2 -> 2 scattering with all equal masses explores the interactions between two particles of identical mass during high-energy collisions. The analysis incorporates principles from quantum field theory and special relativity, focusing on conserved quantities such as energy and momentum. Key results include differential cross-sections and angular distributions, which are derived using various scattering amplitudes. The findings contribute to a deeper understanding of fundamental particle interactions in high-energy physics.
  • #1
Siupa
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Stuck on (c), part (i). Any ideas about what is the most elegant way to prove it, maybe using Mandelstam variables since this exercise is supposed to be about them?
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  • #2
Siupa said:
Stuck on (c), part (i). Any ideas about what is the most elegant way to prove it, maybe using Mandelstam variables since this exercise is supposed to be about them?
Unless you assume the CoM frame, then how can that equation hold?
 
  • #3
PeroK said:
Unless you assume the CoM frame, then how can that equation hold?
Are you suggesting this as an hint on how to prove it, or are you suggesting that the question is wrong?
 
  • #4
Siupa said:
Are you suggesting this as an hint on how to prove it, or are you suggesting that the question is wrong?
I guess you have to assume the CoM frame, otherwise ##|\mathbf p_1| \ne |\mathbf p_2|##.

Do you see how to show the full equation in that case?
 
  • #5
Moderator's note: Thread moved to advanced physics homework forum.
 
  • #7
This ls a question where the homework template sure would have helped.

The absolute value of a 4-vector can mean one of two different things: its norm, or the magnitude of the 3 vector. Which one do you meant? Hint: only one is valid in all frames.
 
  • #8
Vanadium 50 said:
magnitude of the 3 vector
I believe this is what the absolute value signs in the OP are intended to mean, since the vectors in question are in bold, which normally indicates a 3-vector. But the OP should clarify.
 
  • #9
PeroK said:
I guess you have to assume the CoM frame, otherwise ##|\mathbf p_1| \ne |\mathbf p_2|##.

Do you see how to show the full equation in that case?
Yes, with the extra assumption that ##|\bf{p}_1| = |\bf{p}_2|## I can solve this. Let’s call ##\bf{p}_i := \bf{p}_1 = \bf{p}_2##. Then, ##\bf{p}_1 + \bf{p}_2 = 0 = \bf{p}_3 + \bf{p}_4## implies ##|\bf{p}_3| = |\bf{p}_4|##, and we can also call either of these ##\bf{p}_f##. To get our result we just need to show ##|\bf{p}_i| = |\bf{p}_f|##. But this follows from conservation of energy:
\begin{align*}
E_1 + E_2 &= E_3 + E_4 \\
2\sqrt{m^2 + |\bf{p}_i}| &= 2\sqrt{m^2 + |\bf{p}_f}| \\
|\bf{p}_i| &= |\bf{p}_f|
\end{align*}
However the text doesn’t make the assumption that ##|\bf{p}_1| = |\bf{p}_2|##. Do you think it’s not possible to show it in a general case? Is the question wrong?
 
  • #10
Vanadium 50 said:
This ls a question where the homework template sure would have helped.

The absolute value of a 4-vector can mean one of two different things: its norm, or the magnitude of the 3 vector. Which one do you meant? Hint: only one is valid in all frames.
What are you referring to? Can you clarify what would have been more clear about my question, had I followed a homework template?

As for your question, it’s clear from the notation what kind of absolute value the text of the problem is talking about: 4 vectors are introduced in the first line and aren’t boldface, while the corresponding spatial 3-vectors are introduced later in the text and are written with a boldface font.

Even if you missed this notational distinction, it’s still clear from context what we’re talking about: if the problem actually meant to ask to prove the equality of all 4-vector norms, the question would have reduced to “assuming ##m_1 = … = m_4## show that ##m_1 = … = m_4##”. It’s clear that it’s asking to prove equality of the magnitudes of the 3-momenta, both by explicit notation and by a quick sanity check
 
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  • #11
Siupa said:
However the text doesn’t make the assumption that ##|\bf{p}_1| = |\bf{p}_2|##. Do you think it’s not possible to show it in a general case? Is the question wrong?
It can't hold in the rest frame of either particle!

In the past year, I've seen quite a few homework problems where the student is unwilling to accept that there might be an error, omission or typo in the question. Even where the error is quite obvious. One student in particular took it very badly that a famous Indian physics textbook might be wrong!

You must have confidence in your own knowldege and intelligence. Personally, I'd rather make my own mistakes from time to time than be swayed unduly by the power of authority to believe something I see as obviously false!
 
  • #12
Siupa said:
the text doesn’t make the assumption that ##|\bf{p}_1| = |\bf{p}_2|##.
In the center of mass frame that is true by definition, as part (b) of the problem says. Part (c) apparently wants you to continue to assume you are in the center of mass frame, as in part (b), but doesn't say so, which it should for clarity.
 
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  • #13
PeroK said:
It can't hold in the rest frame of either particle!
You’re right, that was easy to check in hindsight. There’s at least one other frame where the process can happen with all equal masses, yet that relationship doesn’t hold. Thanks for your help!

There’s no need for that condescending lesson about being afraid to confront authority though: I’m not unwilling to accept that there might be an error in the question, and I have given no indication of this supposed stubbornness and lack of confidence in our conversation. That’s why I was asking you if you thought the text was mistaken, precisely because I knew perfectly well that could be a possibility. I don’t know where you got this impression from.

It just wasn’t as obvious to me, it only became obvious after I thought about the rest frame of one of the two initial particles, after reading your last comment. I didn’t come here saying “I think this is wrong, how could it be?”. I thought it was solvable and I just couldn’t see how.

Anyways, thanks again!
 
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  • #14
PeterDonis said:
In the center of mass frame that is true by definition, as part (b) of the problem says. Part (c) apparently wants you to continue to assume you are in the center of mass frame, as in part (b), but doesn't say so, which it should for clarity.
that’s probably it yes, the author must have imagined it as a continuation of the premises of the previous question, but it’s not clear given that point (c) isn’t intended/nested hierarchally under (b), but instead is presented in parallel as “the next question”.
Thanks!
 
  • #15
Siupa said:
There’s no need for that condescending lesson about being afraid to confront authority though
Let me give you another lesson. It pays to be nice to people who are trying to help you.
 
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  • #16
Vanadium 50 said:
Let me give you another lesson. It pays to be nice to people who are trying to help you.
I’ve been nothing but thankful to everyone who answered here, I’ve thanked him multiple times in that same message you quoted. His comment about beginner students being unwilling to accept that sometimes books might be wrong was indeed condescending and out of context, even if in good faith. There’s nothing rude in clarifying and explaining that his impression was wrong. You might even say that this is what “confronting authority” looks like, right? Or is it only valid when you’re not on the receiving end of it?

Also, you never answered my question: your first comment criticized my post, saying that using the homework template would have helped. I asked you what my question was missing in clarity that would have been improved had I followed your suggestion. If you could answer I would appreciate it, so that in the future I can maybe avoid being unclear.

Thank you!
 
  • #17
Siupa said:
I’ve been nothing but thankful to everyone who answered here, I’ve thanked him multiple times in that same message you quoted. His comment about beginner students being unwilling to accept that sometimes book might be wrong was indeed condescending and out of context, even if in good faith. There’s nothing rude in clarifying and explaining that his impression was wrong. You might even say that this is what “confronting authority” looks like, right? Or is it only valid when you’re not on the receiving end of it?
First, I have no authority. I'm just an amateur who has taught myself physics during my retirement.

Second, I said:

PeroK said:
You must have confidence in your own knowldege and intelligence.
How you interpreted that as condescending is beyond me.

Third I never used the word "beginner". I just said students. There were not all beginners, which is very much my point.

Fourth, it was definitely not out of context. The context of this thread is that the book appears to be asking you to prove something that is clearly not generally true. And by post #9, you were still unsure:
Siupa said:
However the text doesn’t make the assumption that ##|\bf{p}_1| = |\bf{p}_2|##. Do you think it’s not possible to show it in a general case? Is the question wrong?
That looks like you were still very reluctant to accept that the book had a simple omission.

So, I gave you some advice that you are welcome to accept or reject.
 
  • #18
PeroK said:
First, I have no authority. I'm just an amateur who has taught myself physics during my retirement.
Well, whatever you and I are in real life, in this context you’re in a position of authority with respect to me, as you’re a respected member of the community who’s giving me help and I’m a relatively new member with little questions asked and no help given on the forum. But that’s besides the point, you’re right I don’t think it’s relevant.

PeroK said:
How you interpreted that as condescending is beyond me.
Comparing me to a stubborn student that refuses to accept that authority could be wrong and doesn’t have confidence in his own positions feels quite condescending to me: the underlying assumption being that I lack critical thinking skills and I mindlessly follow what other people tell me is true, without questioning. Maybe this is a harsh read and you didn’t mean this, but this is what it felt like.

PeroK said:
Third I never used the word "beginner". I just said students. There were not all beginners, which is very much my point.
Fair, nothing to say here, my bad.

PeroK said:
Fourth, it was definitely not out of context. The context of this thread is that the book appears to be asking you to prove something that is clearly not generally true. And by post #9, you were still unsure:

That looks like you were still very reluctant to accept that the book had a simple omission.
It wasn’t clear to me at all that it could not be true. Just because one can prove a statement by making it more specific by adding an extra assumption, it doesn’t mean that it can’t also hold in general without that assumption: it could just mean that there is a more general proof you’re taking a specific case of (A+B -> C doesn’t mean that A -> C is wrong). I kept asking you if *you* thought that the book was wrong because you alluded to it, and I was open to the possibility. But I didn’t have any good reason to think that it could be wrong at that time.

Then, after you made me think about that easy counterexample, I immediately became convinced and didn’t insist that the book couldn’t possibly be wrong.

Anyways, this discussion isn’t very productive, I understand your intention wasn’t to pass as condescending. It’s just what I felt like, and I explained why, and wanted to clarify that no, I didn’t already think that there was a mistake before becoming convinced. I don’t think expressing how I felt and clarifying my thoughts is unpolite or rude, as Vanadium said. If you say you weren’t being patronizing and just wanted to give genuine advice, I believe you. Thanks again for your help!
 
  • #19
Siupa said:
Stuck on (c), part (i). Any ideas about what is the most elegant way to prove it, maybe using Mandelstam variables since this exercise is supposed to be about them?View attachment 341640
I think problem c wants you to use the center of mass frame just like it did in (b).
 

FAQ: Relativistic 2 -> 2 scattering with all equal masses

What is relativistic 2 -> 2 scattering with all equal masses?

Relativistic 2 -> 2 scattering refers to a type of particle interaction where two particles collide and scatter into two outgoing particles, all of which have the same mass. This process is analyzed within the framework of relativistic quantum field theory, taking into account the principles of special relativity and quantum mechanics.

How do you calculate the Mandelstam variables for this scattering process?

The Mandelstam variables s, t, and u are defined as follows: \( s = (p_1 + p_2)^2 \), \( t = (p_1 - p_3)^2 \), and \( u = (p_1 - p_4)^2 \), where \( p_1 \) and \( p_2 \) are the four-momenta of the incoming particles, and \( p_3 \) and \( p_4 \) are the four-momenta of the outgoing particles. These variables are used to describe the kinematics of the scattering process.

What conservation laws apply to relativistic 2 -> 2 scattering?

The primary conservation laws that apply are the conservation of energy and momentum. In the center-of-mass frame, the total energy and momentum before and after the collision must be equal. Additionally, if the interaction conserves charge, baryon number, or other quantum numbers, those must also be conserved.

How do Feynman diagrams represent this scattering process?

Feynman diagrams for 2 -> 2 scattering typically include two incoming particle lines converging at a vertex, where they interact via an intermediate particle or force mediator, and then two outgoing particle lines diverging from the vertex. These diagrams help visualize the interaction and facilitate the calculation of scattering amplitudes.

What role do cross-sections play in understanding this scattering process?

The cross-section is a measure of the probability that a scattering event will occur. For relativistic 2 -> 2 scattering, the differential cross-section can be calculated using the scattering amplitude derived from the Feynman diagram. The total cross-section is then obtained by integrating the differential cross-section over all possible scattering angles.

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