# How can I calculate the speed of a π+ meson based on its decay distance?

• R3ap3r42
In summary: I can't see my error. :(Hint: You didn't square ##\gamma##. (You can't do a simple distributive multiplication like that.) That's exactly what I mentioned a couple of posts ago.Yes, I used this trick (you mentioned it before on another thread).I am just annoyed I keep getting the exact same result.$$v \gamma \tau = d$$then$$v \gamma = \frac d \tau$$square both sides$$\frac {v^2} {1-v^2} = \frac {d^2} {\tau^2}$$$$\frac {v^2} {1-v^2} = R3ap3r42 Homework Statement Another exercise on Lorentz - particle decay Relevant Equations Lorentz transformations, conservation of momentum and energy A π+ meson is an elementary particle with a mean lifetime, defined in its rest frame, of τ = 2.60×10−8 s. The meson decays to a muon (µ+) and a neutrino (νµ) via the reaction π+ → µ+ + νµ. A π+ traveling in the laboratory decays so that the µ+ travels in the same direction as the original π+ and the νµ travels in the opposite direction. a) At what speed must a typical π+ be traveling if it is to move a distance of 20 m in the laboratory from the point where it is produced to the point where it decays? See my work below. The expected result is 0.932 but I am getting 0.9916, Any help will be much appreciated. R3ap3r42 said: Homework Statement:: Another exercise on Lorentz - particle decay Relevant Equations:: Lorentz transformations, conservation of momentum and energy A π+ meson is an elementary particle with a mean lifetime, defined in its rest frame, of τ = 2.60×10−8 s. The meson decays to a muon (µ+) and a neutrino (νµ) via the reaction π+ → µ+ + νµ. A π+ traveling in the laboratory decays so that the µ+ travels in the same direction as the original π+ and the νµ travels in the opposite direction. a) At what speed must a typical π+ be traveling if it is to move a distance of 20 m in the laboratory from the point where it is produced to the point where it decays? See my work below. The expected result is 0.932 but I am getting 0.9916, Any help will be much appreciated. Let's check your answer:$$\frac v c = 0.9916, \gamma = 7.73, T = \gamma \tau = 2\times 10^{-7}s, d = vT = 59.8m$$Hmm? PS using the book answer:$$\frac v c = 0.932, \gamma = 2.76, T = \gamma \tau = 7.17\times 10^{-8}s, d = vT = 20m$$I did that on a spreadsheet. PPS I don't think you need the Lorentz Transformation for this. You can do everything in the lab frame. I did: Which gives me that answer.PS: What should I use to write the equations the way you guys do? R3ap3r42 said: I did: View attachment 295808 Which gives me that answer. Best to double check. It's hard to read exactly what you did. That's fine, except you should use ##\tau## for the muon's proper time, rather than coordinate time ##t'##. R3ap3r42 said: PS: What should I use to write the equations the way you guys do? Learn Latex: https://www.physicsforums.com/help/latexhelp/ If you reply to any post with Latex in it, you get the raw Latex to copy. Can you explain a little more? This should be really simple as far as I can see and I can't spot my mistake. From what I can see I am using the same equations...$$ d = vT, T = \gamma \tau $$therefore$$ d = v \gamma \tau $$Last edited: R3ap3r42 said: Can you explain a little more? This should be really simple as far as I can see and I can't spot my mistake. From what I can see I am using the same equations...$$ d = vt, T = \gamma \tau $$therefore$$ d = v \gamma \tau $$Yes, that's right. You must have made a mistake extracting ##v## from ##\gamma v##. This is really stressful... I just can't see it. :) R3ap3r42 said: This is really stressful... I just can't see it. :) Try this: Here's a hint and something that doesn't seem to be in textbooks. If you generate the quantity ##\gamma v##, which happens quite often. Note that: PeroK said:$$\gamma v = w \ \Rightarrow \ \frac{v^2}{1 - v^2} = w^2 \ \Rightarrow \ v^2 =\frac{w^2}{1 + w^2}\Rightarrow \ v = \frac{w}{\sqrt{1 + w^2}} \ \text{and} \ \gamma = \sqrt{1 + w^2}$$That's how you extract ##v## and ##\gamma## from ##\gamma v##. Note that I've used ##c=1##. It's just putting numbers into a spreadsheet. PeroK said: Try this: Here's a hint and something that doesn't seem to be in textbooks. If you generate the quantity ##\gamma v##, which happens quite often. Note that: It's just putting numbers into a spreadsheet. Sorry, that assumes ##c = 1##. You'll need to rework that formula first! PeroK said: Try this: Here's a hint and something that doesn't seem to be in textbooks. If you generate the quantity ##\gamma v##, which happens quite often. Note that: It's just putting numbers into a spreadsheet. Yes, I used this trick (you mentioned it before on another thread). I am just annoyed I keep getting the exact same result.$$ v \gamma \tau = d $$then$$ v \gamma = \frac d \tau $$square both sides$$ \frac {v^2} {1-v^2} = \frac {d^2} {\tau^2} $$And so on to give me v = 0.993c I can't see my error. :( R3ap3r42 said: Yes, I used this trick (you mentioned it before on another thread). I am just annoyed I keep getting the exact same result.$$ v \gamma \tau = d $$then$$ v \gamma = \frac d \tau $$square both sides$$ \frac {v^2} {1-v^2} = \frac {d^2} {\tau^2} $$And so on to give me v = 0.993c I can't see my error. :( That was with ##c = 1##. In this case, you have:$$\frac {v^2} {1-v^2/c^2} = \frac {d^2} {\tau^2} \frac {v^2/c^2} {1-v^2/c^2} = \frac {d^2} {c^2\tau^2}$$And, if we let ##\beta = \frac v c##, then:$$ \frac {\beta^2} {1-\beta^2} = \frac {d^2} {c^2\tau^2}$$I got it now. We can only use that trick when the distance is in light-year (or this case light-second), correct? Once again thank you very much. R3ap3r42 said: I got it now. We can only use that trick when the distance is in light-year, correct? Once again thank you very much. That was my fault posting a formula for ##c = 1##. Putting ##c## back in gives:$$\gamma v = w \ \Rightarrow \ \frac {v^2}{1 + v^2/c^2} = w^2$$And, then:$$v = \frac{w}{\sqrt{1 - \frac{w^2}{c^2}}}, \ \gamma = \sqrt{1 + \frac{w^2}{c^2}}$$The trick for reinserting correct factors of ##c## into any expression where ##c = 1## has been used is dimensional analysis. For example, here: R3ap3r42 said:$$ \frac {v^2} {1-v^2} = \frac {d^2} {\tau^2} $$In the denominator on the RHS, ##v^2## must be divided by ##c^2## to have the same dimensions as ##1##. If this is done, then both sides have the same dimensions and the equation is fine. This is rest of the same exercise. b) Determine the energy (in MeV) of the µ+ as determined in the rest frame of the original π+. You should assume that the µ+ has a mass of 106 MeV/c2 , the π+ has a mass of 140 MeV/c2 and the νµ has zero mass. c) What is the speed of the µ+ as determined in the π+ rest frame? I thought it was easier to find c) before b) also... since I could get the energy once I had the speed. I have tried using conservation of energy and conservation of momentum 1 = photon, 2 = muon, 3 = pion E1 + E2 = E3 and p2 - p1 = p3 I used$$ p = \gamma m v $$for the momentum of the particles with mass and$$ E = \frac p c $$for the energy of the photon and$$ E = mc^2 + (1-\gamma)mc^2 $$for the particles with mass That gave me two equations and 2 unknowns p1 and v2. See my working below... Somehow I get the wrong result. What am I doing wrong? Expected results are: b) 110 MeV; c) 0.271c; Last edited: R3ap3r42 said: This is rest of the same exercise. b) Determine the energy (in MeV) of the µ+ as determined in the rest frame of the original π+. You should assume that the µ+ has a mass of 106 MeV/c2 , the π+ has a mass of 140 MeV/c2 and the νµ has zero mass. c) What is the speed of the µ+ as determined in the π+ rest frame? I thought it was easier to find c) before b) also... since I could get the energy once I had the speed. Actually, this makes it much harder. The first thing to know about relativistic collisions and decays is to use energy and mometum, solve for those and derive KE and velocity from those. If you write equations in terms of kinetic energy and velocity things get very messy. R3ap3r42 said: I have tried using conservation of energy and conservation of momentum 1 = photon, 2 = muon, 3 = pion E1 + E2 = E3 and p2 - p1 = p3 I used$$ p = \gamma m v $$for the momentum of the particles with mass and$$ E = \frac p c $$for the energy of the photon and$$ E = mc^2 + (1-\gamma)mc^2 $$for the particles with mass I can't read that attachment well enough to see what's going on. I would try again using total energy rather than KE. You have three equations:$$E_{\pi} = E_{\nu} + E_{\mu}p_{\nu} = p_{\mu}E_{\pi} = m_{\pi}c^2; \ E^2_{\mu} = p^2_{\mu}c^2 + m^2_{\mu}c^4; \ E_{\nu} = p_{\nu}c$$Where ##p## is the magnitide of momentum. Last edited: R3ap3r42 I think I figured it out. I was doing in the rest frame and the exercise asked in the frame of the original pion. I still got the answers in the opposite sequence they asked... I first c) then b). PeroK said: Actually, this makes it much harder. The first thing to know about relativistic collisions and decays is to use energy and mometum, solve for those and derive KE and velocity from those. If you write equations in terms of kinetic energy and velocity things get very messy.I can't read that attachment well enough to see what's going on. I would try again using total energy rather than KE. You have three equations:$$E_{\pi} = E_{\nu} + E_{\mu}p_{\nu} = p_{\mu}E_{\pi} = m_{\pi}c^2; \ E^2_{\mu} = p^2_{\mu}c^2 + m^2_{\mu}c^4; \ E_{\nu} = p_{\nu}c$$Where ##p## is the magnitide of momentum. Yes I will re-work my steps to see if I can do it in the correct order R3ap3r42 said: Yes I will re-work my steps to see if I can do it in the correct order Ultimately, in SR and GR the focus is on vectors (and tensors) and invariant quantities. This means that the energy-momentum four-vector is the key quantity here. Some easier problems may be cranked out using KE and velocity, but in general those quantities do not transform well under the Lorentz Transformation. The sooner you start focusing on energy-momentum and learning the mathematical and algebraic patterns that recur in these problems the better. R3ap3r42 PeroK said: Actually, this makes it much harder. The first thing to know about relativistic collisions and decays is to use energy and mometum, solve for those and derive KE and velocity from those. If you write equations in terms of kinetic energy and velocity things get very messy.I can't read that attachment well enough to see what's going on. I would try again using total energy rather than KE. You have three equations:$$E_{\pi} = E_{\nu} + E_{\mu}p_{\nu} = p_{\mu}E_{\pi} = m_{\pi}c^2; \ E^2_{\mu} = p^2_{\mu}c^2 + m^2_{\mu}c^4; \ E_{\nu} = p_{\nu}c$$Where ##p## is the magnitide of momentum. I did that, except that I used$$ E_{\mu} = \gamma m_{\mu} c^2 $$for the energy of the muon. With the way you mentioned, I was getting 3 unknowns (the moment of the photon and the muon and the gamma/speed of the muon). Getting stuck at this:$$ E_{\mu} = m_{\pi} - \gamma m_{\mu} v_{\mu} $$But going my way gives me first the speed of the muon (and gamma) which I use to find the energy using the same formula above. How did you do it? I still can't see it. R3ap3r42 said: How did you do it? I still can't see it. The way I did it was. Start without equations:$$E_{\pi} = E_{\nu} + E_{\mu} \ \ (1)p_{\nu} = p_{\mu} \ \ (2)E_{\pi} = m_{\pi}c^2 \ \ (3a); \ E^2_{\mu} = p^2_{\mu}c^2 + m^2_{\mu}c^4 \ \ (3b); \ E_{\nu} = p_{\nu}c \ \ (3c)$$First, from (1), (2) and (3c) we have:$$E_{\pi} = E_{\mu} + E_{\nu} = E_{\mu} + p_{\nu}c = E_{\mu} + p_{\mu}c\Rightarrow \ E_{\mu} + p_{\mu}c = E_{\pi} \ \ (4)$$And, from (3b) we have:$$E^2_{\mu} - p^2_{\mu}c^2 = m^2_{\mu}c^4 \ \Rightarrow \ (E_{\mu} - p_{\mu}c)(E_{\mu} + p_{\mu}c) = m^2_{\mu}c^4 \ \ (5)$$Using (4) and (5) gives:$$E_{\mu} - p_{\mu}c = \frac{m^2_{\mu}c^4}{E_{\pi}} \ \ (6)$$Then you can add (4) and (6) and use ##E_{\pi} = m_{\pi}c^2## to get:$$2E_{\mu} = \frac{m^2_{\mu}c^4}{E_{\pi}} + E_{\pi} = \frac{(m^2_{\mu} + m^2_{\pi})c^4}{m_{\pi}c^2}$$That gives you the energy of the muon, from which you can get its momentum and hence the energy-momentum of the neutrino. If the second particle is not a photon or effectively massless neutrino, then there's a bit more algebra. That could be a good exercise. Perhaps with ##M## as the original particle decaying into two particles of mass ##m_1## and ##m_2##. R3ap3r42 PeroK said: The way I did it was. Start without equations:$$E_{\pi} = E_{\nu} + E_{\mu} \ \ (1)p_{\nu} = p_{\mu} \ \ (2)E_{\pi} = m_{\pi}c^2 \ \ (3a); \ E^2_{\mu} = p^2_{\mu}c^2 + m^2_{\mu}c^4 \ \ (3b); \ E_{\nu} = p_{\nu}c \ \ (3c)$$First, from (1), (2) and (3c) we have:$$E_{\pi} = E_{\mu} + E_{\nu} = E_{\mu} + p_{\nu}c = E_{\mu} + p_{\mu}c\Rightarrow \ E_{\mu} + p_{\mu}c = E_{\pi} \ \ (4)$$And, from (3b) we have:$$E^2_{\mu} - p^2_{\mu}c^2 = m^2_{\mu}c^4 \ \Rightarrow \ (E_{\mu} - p_{\mu}c)(E_{\mu} + p_{\mu}c) = m^2_{\mu}c^4 \ \ (5)$$Using (4) and (5) gives:$$E_{\mu} - p_{\mu}c = \frac{m^2_{\mu}c^4}{E_{\pi}} \ \ (6)$$Then you can add (4) and (6) and use ##E_{\pi} = m_{\pi}c^2## to get:$$2E_{\mu} = \frac{m^2_{\mu}c^4}{E_{\pi}} + E_{\pi} = \frac{(m^2_{\mu} + m^2_{\pi})c^4}{m_{\pi}c^2} That gives you the energy of the muon, from which you can get its momentum and hence the energy-momentum of the neutrino.

If the second particle is not a photon or effectively massless neutrino, then there's a bit more algebra. That could be a good exercise. Perhaps with ##M## as the original particle decaying into two particles of mass ##m_1## and ##m_2##.
Yes, it would take me a while to see it. It is more elegant no doubt.
But nothing wrong with the way I did it right?

R3ap3r42 said:
But nothing wrong with the way I did it right?
I wouldn't worry too much if you got the answer out. There's always more than one solution. The approach I posted generalises well. It's worth absorbing the ideas.

## 1. What is a π+ meson and why is its speed important to calculate?

A π+ meson, also known as a pion, is a subatomic particle that is composed of a quark and an antiquark. Its speed is important to calculate because it can provide information about the energy and momentum of the particle, which can in turn help us understand the fundamental properties of matter.

## 2. How is the decay distance of a π+ meson measured?

The decay distance of a π+ meson is typically measured by detecting the decay products of the meson, such as muons or neutrinos, and using their trajectory to determine the distance traveled by the meson before it decayed.

## 3. What factors affect the speed of a π+ meson?

The speed of a π+ meson can be affected by several factors, including its energy, mass, and interactions with other particles. The speed can also vary depending on the reference frame from which it is observed.

## 4. What information is needed to calculate the speed of a π+ meson?

To calculate the speed of a π+ meson, you will need to know its decay distance, which can be measured experimentally, and its mass, which can be obtained from particle data tables. You may also need to consider any other relevant factors that may affect its speed.

## 5. What equations can be used to calculate the speed of a π+ meson?

The speed of a π+ meson can be calculated using the equation v = d/t, where v is the speed, d is the decay distance, and t is the time it takes for the meson to decay. Alternatively, the speed can also be calculated using the equation v = p/E, where p is the momentum and E is the energy of the meson.

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