Relativistic Collision: Calculating Mass, Energy & Momentum

In summary: Oops, I always make that mistake. I've come across something puzzling in (c) when finding the kinetic energy of particle R, ##K_R##. I know ##E_R=K_R+m_Rc^2##, so ##K_R=E_R-m_Rc^2=0##, since ##E_R=m_Rc^2##. But from L's perspective, particle R is certainly moving, so how can its kinetic energy vanish? I'm guessing ##E_R=m_Rc^2## is... not quite right?Indeed, ##E_R = m_R c^2## is not quite right. You need to use the formula for the total energy, not the rest energy. You
  • #1
Rubber Ducky
14
0

Homework Statement



Two relativistic particles "L" and "R", each of rest mass ##m_0##, are moving at speed ##v## towards each other (in the frame of an observer). They collide squarely and are stationary afterwards.

(a) From the perspective of one particle, what is the oncoming speed of the other?
(b) What is the rest mass, relativistic mass, total energy, kinetic energy, and momentum of each particle before the collision in the frame of the observer?
(c) Repeat (b) in the rest frame of particle "L"
(d) What is the total momentum and energy before and just after the collision, in the observer's frame?

Homework Equations



##u'=\frac{u-v}{1+\frac{v^2}{c^2}}##

##m=\frac{m_0}{\sqrt{1-\frac{u^2}{c^2}}}##
##E=\frac{m_0c^2}{\sqrt{1-\frac{u^2}{c^2}}}=K+m_0c^2=\sqrt{(pc)^2+(m_0c)^2}##
##p=\frac{m_0u}{{\sqrt{1-\frac{u^2}{c^2}}}}##

The Attempt at a Solution



(a) If the observer is in frame ##S##, and the frame of L is ##S'##, then ##\left |u'_R\right |=\frac{v+v}{1+\frac{v^2}{c^2}}=\frac{2v}{1+\frac{v^2}{c^2}}##

(b) ##m_L=\frac{m_0}{\sqrt{1-\frac{u^2}{c^2}}}=m_R##

The rest of this part is relatively (heh) straight forward, I think. I'm not sure how to apply the same formulas to part (c), though. For example, for the mass of R with respect to L, would I copy the formula from part (b), but replace ##u## by what I found in (a)?

Also, in the frame of L, shouldn't its rest mass be equal to its relativistic mass, since particle L is at rest in this frame, and rest mass is defined as the mass measured by a frame in which the object is at rest?
 
Physics news on Phys.org
  • #2
Rubber Ducky said:
What is the rest mass, relativistic mass, total energy, kinetic energy, and momentum of each particle before the collision in the frame of the observer?

Many people on this forum will cringe from seeing this which is in the formulation of your problem. Relativistic mass is not really used among scientists and with good reason. Instead, total energy (which is exactly the same thing up to a factor of ##c^2##, but we tend to use units where ##c = 1##) will typically be used.

That being said, what you have done so far looks fine. For (c), you can use the same argumentation, but you now have to consider the actual velocity each particle has in the rest frame of L. The answer to:
Rubber Ducky said:
would I copy the formula from part (b), but replace uu by what I found in (a)?
would therefore be "yes".
Rubber Ducky said:
Also, in the frame of L, shouldn't its rest mass be equal to its total energy, since particle L is at rest in this frame, and rest mass is defined as the mass measured by a frame in which the object is at rest?
Yes. Again, note that we would typically talk only about one mass, called the rest mass, invariant mass, or just mass).
 
  • #3
Orodruin said:
Many people on this forum will cringe from seeing this which is in the formulation of your problem. Relativistic mass is not really used among scientists and with good reason. Instead, total energy (which is exactly the same thing up to a factor of ##c^2##, but we tend to use units where ##c = 1##) will typically be used.

That being said, what you have done so far looks fine. For (c), you can use the same argumentation, but you now have to consider the actual velocity each particle has in the rest frame of L. The answer to:

would therefore be "yes".

Yes. Again, note that we would typically talk only about one mass, called the rest mass, invariant mass, or just mass).
Thanks for your reply. My prof feels the same way as you, and my textbook says as much; "we feel that relativistic mass does not lead to any deeper physical understanding, so we will not treat it in this text". So it's weird / frustrating to see a question about it.

For (c), in computing the relativistic mass, I get ##m_R=\frac{m_0}{\sqrt{1-\frac{(u'_R)^2}{c^2}}}##. There's something I'm unsure of - in (a) I needed to find the speed only, but now I believe I'm looking at velocity?

##u'_R=\frac{-v-v}{1+\frac{v^2}{c^2}}=\frac{-2v}{1+\frac{v^2}{c^2}}## since particle L perceives particle R as moving in the negative direction. The simplification for ##m_R## is fine - just wanted to know if I'm using the correct reasoning.

Now for total energy, in (c), isn't it true that ##E_L=\frac{m_0c^2}{\sqrt{1-\frac{u'_Lv}{c^2}}}=m_0c^2##, since particle L perceives itself as being stationary? And ##E_R=m_Rc^2##?
 
  • #4
Rubber Ducky said:
the speed only, but now I believe I'm looking at velocity?

What you used was the expression for relativistic velocity addition (in one dimension, the one in which the Lorentz transformation boosts) and it is sign dependent so for one of the transformations in (a) you technically have to take the absolute value. For the total energy, what appears in the gamma factor should be ##v^2/c^2## (where v is the speed of an object (but the velocity squared is clearly the same as speed squared) and thus only speed is relevant for the gamma factor. This brings me to:

Yes, the total energy of any object in its rest frame is the rest energy. However, watch out with the gamma factor, it should be containing ##u'_L{}^2##, not ##u'_L v##.
 
  • #5
Orodruin said:
What you used was the expression for relativistic velocity addition (in one dimension, the one in which the Lorentz transformation boosts) and it is sign dependent so for one of the transformations in (a) you technically have to take the absolute value. For the total energy, what appears in the gamma factor should be ##v^2/c^2## (where v is the speed of an object (but the velocity squared is clearly the same as speed squared) and thus only speed is relevant for the gamma factor. This brings me to:

Yes, the total energy of any object in its rest frame is the rest energy. However, watch out with the gamma factor, it should be containing ##u'_L{}^2##, not ##u'_L v##.

Oops, I always make that mistake. I've come across something puzzling in (c) when finding the kinetic energy of particle R, ##K_R##. I know ##E_R=K_R+m_Rc^2##, so ##K_R=E_R-m_Rc^2=0##, since ##E_R=m_Rc^2##. But from L's perspective, particle R is certainly moving, so how can its kinetic energy vanish? I'm guessing ##E_R=m_Rc^2## is incorrect?
 
  • #6
This is just one of the drawbacks of relativistic mass, confusion.

Kinetic energy is by definition total energy minus rest energy. The rest energy is proportional to the rest mass (invariant mass), not to the relativistic mass (total energy).
 

1. What is a relativistic collision?

A relativistic collision is a type of collision between two particles where the velocities are close to the speed of light. This type of collision requires the use of the theory of relativity to accurately calculate the resulting mass, energy, and momentum.

2. How do you calculate the mass of a particle in a relativistic collision?

The mass of a particle in a relativistic collision can be calculated using the formula: m = m0 / √(1 - (v/c)^2), where m0 is the rest mass of the particle, v is the velocity of the particle, and c is the speed of light. This formula takes into account the effects of time dilation and length contraction.

3. What is the equation for calculating the energy in a relativistic collision?

The energy in a relativistic collision can be calculated using the formula: E = mc^2, where m is the mass of the particle and c is the speed of light. This formula is based on Einstein's famous equation, E=mc^2, which describes the relationship between mass and energy.

4. How do you find the momentum of a particle in a relativistic collision?

The momentum of a particle in a relativistic collision can be calculated using the formula: p = mv / √(1 - (v/c)^2), where m is the mass of the particle, v is the velocity of the particle, and c is the speed of light. This formula takes into account the effects of time dilation and length contraction.

5. Can the relativistic effects be ignored in low-speed collisions?

In most cases, the relativistic effects can be ignored in low-speed collisions. However, if the velocities are still significant compared to the speed of light, it is important to use the equations for relativistic collisions to accurately calculate the resulting mass, energy, and momentum.

Similar threads

  • Advanced Physics Homework Help
Replies
2
Views
614
  • Advanced Physics Homework Help
Replies
16
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
696
  • Advanced Physics Homework Help
Replies
10
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
0
Views
617
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Special and General Relativity
2
Replies
55
Views
3K
  • Advanced Physics Homework Help
Replies
17
Views
2K
Replies
12
Views
240
Back
Top