Challenging problem involving special relativity

In summary: Ei=E1+E2 (initial energy)pi^2=p^2 (momentum conservation)Ei^2=E1^2+E2^2+2E1E2 (energy conservation)Ei^2=p^2+m^2+m^2+2m^2=2p^2+3m^2 (substituting values)sqrt(Ei^2)=sqrt(2p^2+3m^2) (taking square root)Ef=sqrt(M^2+pf^2)=sqrt(2p^2+3m^2) (
  • #1
Kyle.Nemeth
41
0

Homework Statement



The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass M requires an energy Mc2. With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass mp, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a product particle. (a) Show that the energy available to create a product particle is given by Mc2 = 2mpc2√1 + K/2mpc2. This result shows that when the kinetic energy K of the incident proton is large compared with its rest energy mpc2, then M approaches (2mpK)1/2/c. Therefore, if the energy of the incoming proton is increased by a factor of 9, the mass you can create increases only by a factor of 3, not by a factor of 9 as would be expected.

Homework Equations



E2 = (pc)2 + (mc2)2

E = K + mc2

The Attempt at a Solution



I started this problem numerous times by modeling the two protons and the product particle as an isolated system and stating that pi = pf by the conservation of momentum. I then proceeded to use the energy-momentum relationship to find expressions for pi and pf. From there, I figured I'd be able to find the fraction of energy that goes into creating the product particle but after MANY many attempts, it seems I've just been running in circles. There seems to be something that I'm missing and I can't pinpoint what it is.

Here's what I did:

Ei2 = pi2c2 + (2mpc2)2 is the initial energy of the proton-proton system.

Ef2 = pf2c2 + (Mc2)2 is the final energy of the proton-proton system.

I started by solving for pi and pf and then used pi = pf. From there I attempted to solve for Mc2 but it seems that this has led me awry.
 
Last edited:
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  • #2
Use units with c=1.
Given:
E1=sqrt(p^2+m^2)
p1=p
E2=m
p2=0
Use energy and momentum conservation to determine Ef and pf of a product particle with mass M:
Ef=sqrt(M^2+pf^2)
 
Last edited:

Related to Challenging problem involving special relativity

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that explains the behavior of objects in motion, particularly at high speeds. It states that the laws of physics are the same for all observers in uniform motion, and that the speed of light is constant for all observers.

2. What are the key concepts of special relativity?

The key concepts of special relativity are the principle of relativity, which states that the laws of physics are the same for all observers in uniform motion, and the constancy of the speed of light, which states that the speed of light is the same for all observers regardless of their relative motion.

3. How does special relativity affect our understanding of time and space?

Special relativity introduces the concept of time dilation, which means that time passes slower for objects in motion than for objects at rest. It also explains the phenomenon of length contraction, where objects in motion appear shorter in the direction of their motion. These concepts challenge our traditional understanding of time and space as fixed and absolute.

4. What are some real-world applications of special relativity?

Special relativity has been used to develop technologies such as GPS and particle accelerators. It also plays a crucial role in the field of nuclear energy, as it explains the relationship between mass and energy in Einstein's famous equation, E=mc^2.

5. What are some common misconceptions about special relativity?

One common misconception is that special relativity only applies to objects moving at very high speeds. In reality, the principles of special relativity apply to all objects in motion, regardless of their speed. Another misconception is that special relativity contradicts the laws of physics, when in fact it is a consistent and well-tested theory that has been verified through numerous experiments.

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