Simplifying KE/Momentum Relativistic Problem

In summary, the problem involves a D0 meson decaying into a K0 meson and a \pi0 meson. The Kinetic Energy of the \pi0 meson can be found by using the equation E^2 = (pc)^2 + (m_0 c^2)^2 and conservation of energy and momentum. Using four-vectors can also simplify the solution process.
  • #1
Vorde
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Homework Statement



A D0 meson (with a rest mass of 1.86 GeV/c2), initially at rest, decays into a K0 meson (with a rest mass of .51 GeV/c2) and a [itex]\pi[/itex]0 meson (with a rest mass of .12 GeV/c2). What is the Kinetic Energy of the [itex]\pi[/itex]0 meson?


Homework Equations



E=[itex]\gamma[/itex]m0c2

p=[itex]\gamma[/itex]m0v

The Attempt at a Solution



Subtracting final rest energy from intial energy its easy to see that the KE(K0+[itex]\pi[/itex]0)=1.23 GeV

Also: [itex]\gamma[/itex]K0mK0c2+[itex]\gamma[/itex][itex]\pi[/itex]0m[itex]\pi[/itex]0c2 = 1.86 GeV/c^2 (The total energy of the two Mesons must equal the first meson)
Without going through all the horrible steps me and my friend did, eventually we got a equation that solved: VK0=f(V[itex]\pi[/itex]0) (I don't have the final equation we got on me)


When then plugged that back into: [itex]\gamma[/itex]K0mK0c2+[itex]\gamma[/itex][itex]\pi[/itex]0m[itex]\pi[/itex]0c2 = 1.86 GeV/c^2


Now at this point we had a solvable equation, the only variable was V[itex]\pi[/itex]0, but it was a hellish equation. After about 20 minutes working on trying to solve it we ran out of time and had to give up- the algebra was just to hard.
Had we had enough time, I'm completely confident we could have solved it, as it was simply a matter of foiling again and again (I think we needed to do it three levels down).
My question is: Is there a way of solving this problem without having to go in and solve for one of the variables in terms of the other, or doing that in a simpler equation?

Thank you, especially if you made it through this.
 
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  • #2
If you don't need the velocities (i.e. you're not asked for them explicitly), don't take that route to find the energies of the outgoing particles. Instead, use

[tex]E^2 = (pc)^2 + (m_0 c^2)^2[/tex]

which is true for each particle, together with conservation of energy and conservation of momentum. You never have to see a single "v".
 
  • #3
Two suggestions:

1. Stick with energy (total, not kinetic) and momentum.
2. Take advantage of [itex]E^2 - (pc)^2 = (mc^2)^2[/itex].

If you learn how to use four-vectors, you can get the answer in about three or four lines of simple algebra. Read Griffith's chapter on special relativity in his Introduction to Particle Physics.
 
  • #4
Thank you to both of you, while I haven't actually gotten the answer yet, I have gotten numerical values for both the momentum and velocity of one of the particles, I'll be able to get the answer with about 5 more minutes of work.

In addition, I know the momentum energy four vector, but I've never been given a problem requiring it so using it didn't occur to me, I am going to look at its usefulness more carefully.
 
  • #5




I understand your frustration with this problem. Solving equations involving relativistic quantities can be quite complex and time-consuming. However, there are a few tips that may simplify this problem and make it easier to solve.

Firstly, instead of using the full expression for kinetic energy (KE = \gamma m_0 c^2 - m_0 c^2), you can use the simplified form of KE = (\gamma - 1) m_0 c^2. This can save you some algebraic steps.

Secondly, since the total energy of the two mesons must equal the initial energy of the D0 meson, you can set up an equation using the total energy of the K0 and \pi0 mesons, and solve for the velocity of one of the mesons (let's say K0) in terms of the other (in this case, \pi0). This will give you a single equation with one variable, which should be easier to solve.

Lastly, you can also use the fact that the total momentum of the system must be conserved. This means that the initial momentum of the D0 meson (which is zero since it is at rest) must be equal to the final momentum of the K0 and \pi0 mesons. You can use this information to set up another equation and solve for the velocity of one meson in terms of the other.

I hope these tips help you in solving this problem. Remember, as a scientist, it is important to approach problems with patience and perseverance. Good luck!
 

1. What is the equation for calculating relativistic kinetic energy?

The equation for relativistic kinetic energy is KE = (γ-1)mc2, where γ is the Lorentz factor, m is the mass of the object, and c is the speed of light.

2. How does relativistic kinetic energy differ from classical kinetic energy?

Relativistic kinetic energy takes into account the effects of special relativity, such as time dilation and length contraction, while classical kinetic energy does not. This means that as an object approaches the speed of light, its relativistic kinetic energy will increase significantly compared to its classical kinetic energy.

3. Can the equation for relativistic kinetic energy be simplified?

Yes, the equation for relativistic kinetic energy can be simplified by using the approximation γ ≈ 1 + ½(v/c)2, which is valid for velocities much smaller than the speed of light. This results in the simplified equation KE ≈ ½mv2.

4. How is momentum related to relativistic kinetic energy?

Momentum is related to relativistic kinetic energy through the equation p = γmv, where p is momentum, γ is the Lorentz factor, m is the mass of the object, and v is its velocity. This equation shows that as an object's velocity approaches the speed of light, its momentum will increase significantly compared to its classical momentum.

5. What are some real-world applications of relativistic kinetic energy?

Relativistic kinetic energy has important applications in fields such as particle physics and astrophysics, where objects can reach very high velocities and their relativistic effects must be taken into account. It also has practical applications in technologies such as particle accelerators and nuclear reactors.

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