Proving Massless Particle's Speed: 0.5c

In summary, according to the homework statement, a neutral pion traveling along the x-axis decays into two photons, one being ejected exactly forward and the other exactly backward. The first photon has three times the energy of the second. The equation for the relativistic doppler effect can be used to find the frequency of both photons in a frame that moves with a speed v. By squaring and using P^2 = m^2 c^4 , so that P_1^2 = P_2^2 = 0, it is found that m_\pi^2 c^4 = 2 E_1 E_2 - c^2. Finally, using the fact that one photon has three times
  • #1
im_slow
1
0

Homework Statement


Massless Particles
A neutral pion traveling along the x-axis decays into two photons, one being ejected exactly forward and the other exactly backward. The first photon has three times the energy of the second. Prove that the original pion had speed 0.5c.

Homework Equations


for m=0, E=p*c
conservation of Energy E^2=(c*p)^2+(m*c^2)^2
gamma=1/sqrt(1-Beta^2)
Beta = v/c
p=gamma*m*v
E=gamma*m*c^2

The Attempt at a Solution



sum(momentum photons) = sum (momentum pion)
 
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  • #2
The photon energy is easy to compute in a frame where the pion is at rest.
Use the formula for the relativistic doppler effect to find the frequency of both fotons in a frame that moves with a speed v. The frequency of the forward photon must be 3 times the
frequency of the backwards photon.
 
  • #3
im_slow said:

Homework Statement


Massless Particles
A neutral pion traveling along the x-axis decays into two photons, one being ejected exactly forward and the other exactly backward. The first photon has three times the energy of the second. Prove that the original pion had speed 0.5c.

Homework Equations


for m=0, E=p*c
conservation of Energy E^2=(c*p)^2+(m*c^2)^2
gamma=1/sqrt(1-Beta^2)
Beta = v/c
p=gamma*m*v
E=gamma*m*c^2

The Attempt at a Solution



sum(momentum photons) = sum (momentum pion)

Are you familiar with four-momentum conservation?


You may use [tex] P_\pi = P_1 + P_2 [/tex] where 1 and 2 refer to the photons. After squaring and using P^2 = m^2 c^4 , so that P_1^2 = P_2^2 = 0, you get

[tex] m_\pi^2 c^4 = 2 E_1 E_2 - c^2 {\vec p}_1 \cdot \vec{p_2} [/tex]

now, using the fact that the two photons move in oppposite directions, you find that [tex] 4 E_1 E_2 = m_\pi^2 c^4 [/tex]. Using the fact that one photon has three times the energy of the other one, you then have the energy of each photon in terms of the pion mass.

Then use [tex] E_1 + E_2 = \gamma m_\pi c^2 [/tex] to find the speed.
 
  • #4
I don't understand this. can you try explaining it in more detail?
 
  • #5
= 0
p1*cos(theta1) - p2*cos(theta2) = 0
p1*sin(theta1) + p2*sin(theta2) = 0

Using p1=3*p2 and theta1=0 and theta2=pi, we get p1=3*p2=3*p2*cos(theta1) and p2*sin(theta2) = 0. This means that p1=p2=0, which means that the photons have no momentum. This also means that the pion had no momentum before decaying, since momentum is conserved. Therefore, the pion had no velocity and was at rest, which means that its speed was 0c. This contradicts the statement that the pion was traveling along the x-axis, so we can conclude that the pion must have had some velocity before decaying.

Since the pion is massless, we can use the equation E=p*c to relate its energy to its momentum. Plugging in the values for the two photons, we get E1=3*E2 and p1=3*p2. Using the conservation of energy equation, we can solve for the pion's energy before decaying:

E^2=(c*p)^2+(m*c^2)^2
E^2=(c*p1)^2+(0)^2
E^2=(c*3*p2)^2
E=3*c*p2

Substituting this into the equation E=p*c, we get:

3*c*p2=p*c
3*p2=p
p2=p/3

Using the equation p=gamma*m*v and solving for v, we get:

v=p/(gamma*m)
v=p/(1/sqrt(1-Beta^2)*m)
v=p*m/sqrt(1-Beta^2)

Since the pion is massless, its mass is equal to 0, so we can simplify the equation to:

v=p/sqrt(1-Beta^2)

Substituting in the value for p2, we get:

v=p/3/sqrt(1-Beta^2)

Since the photons have no momentum, we can set p=0 and solve for the velocity of the pion:

v=0/3/sqrt(1-Beta^2)
v=0

Therefore, the pion had a velocity of 0 before decaying, which means its
 

Related to Proving Massless Particle's Speed: 0.5c

1. How can we prove that a particle is massless?

To prove that a particle is massless, we can measure its momentum and energy. If the momentum is non-zero but the energy is zero, then the particle is considered to be massless. This is known as the mass-energy equivalence, as described by Einstein's famous equation E=mc^2.

2. What is the significance of a particle having a speed of 0.5c?

A speed of 0.5c means that the particle is moving at half the speed of light. This is a significant speed as it is the maximum attainable speed in the universe, according to Einstein's theory of relativity. It also means that the particle has a considerable amount of energy, even though it is massless.

3. How do we measure the speed of a massless particle?

Since massless particles have no rest mass, they cannot be accelerated or decelerated. Therefore, the speed of a massless particle can only be measured indirectly by measuring its energy and momentum. This can be done using sophisticated equipment such as particle accelerators.

4. Can a particle be both massless and have a speed of 0.5c?

Yes, a particle can be both massless and have a speed of 0.5c. In fact, there are several known massless particles, such as photons (particles of light) and gluons (particles that mediate the strong nuclear force). These particles travel at the speed of light, which is 1c, but can also have slower speeds, such as 0.5c.

5. What implications does proving a massless particle's speed of 0.5c have in physics?

If a massless particle's speed of 0.5c is proven, it would confirm the predictions of Einstein's theory of relativity and the concept of mass-energy equivalence. It would also have implications for our understanding of the fundamental properties of matter and the behavior of particles at high energies. Additionally, this discovery could have practical applications in fields such as particle physics, astrophysics, and telecommunications.

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