I am currently reading Introduction to Special Relativity by James H. Smith, and I am attempting to complete the exercise problems for chapter 1 before moving on.

The 4th problem asks

Show analytically, using Equation 1-2 $(u=u'+v)$, that the amount of thermal energy produced in a collision is independent of the frame of reference from which the collision is viewed. Did you have to assume the conservation of Momentum?

I've attempted this problem several times now and am completely stuck. I've been trying to use $\mathit{KE}_{\rm initial} -\mathit{KE}_{\rm post-collision} = E_{\rm thermal}$ (assuming the energy lost only goes into making heat, not sound).

If I am making a horrible error with my set-up please explain the correct set-up. If my set-up is correct, maybe you won't make the many algebra mistakes I inevitably make. Thanks in advance for the help!

UPDATE: I've figured it out and will now try to explain it as thoroughly as possible! So let the ground frame of the collision be O and O' be the frame with a relative velocity v to frame O. in frame O, particle 1 has a mass m1 and a velocity u1, Particle 2 has a mass m2 and a velocity u2. After the collision, the particles should have velocities U1 and U2 respectively. Now the three equations we will need to complete this problem are $$u=u'+v$$ $$m_1u_1+m_2u_2=m_1U_1+m_2U_2$$ $$KE_i-KE_f=Q$$ Where u is the velocity in frame O, u' is the velocity in frame O' and v is the relative velocity between frames. The second equation is merely the expression for conservation of momentum and the third is the way I have introduced the concept of energy loss in the collision. Having the difference between total initial KE and Total post-collision KE gives us the total energy lost in the system to thermal energy.

Remember, my objective was to prove that energy loss in collisions is consistent between frames. So my first Step is to define Q in frame O. After manhandling the algebra a bit, you end up with this $$\frac{m_1(u_1^2-U_1^2)}{2}+\frac{m_2(u_2^2-U_2^2)}{2}=Q$$

We can then define the energy loss in frame O' to be Q' and the description of that would look similar, but with a few more terms. $$\frac{m_1([u']_1^2-[U']_1^2)}{2}+\frac{m_2([u']_2^2-[U']_2^2)}{2}+m_1v(u'_1-U'_1)-m_2v(U'_2-u'_2)=Q'$$

So to prove that $Q=Q'$ We have to prove that the last two terms in the expression for Q' are equivalent and therefore will cancel. We can do so with the law of conservation of momentum $m_1u_1+m_2u_2=m_1U_1+m_2U_2$ rearranging gives us $$m_1(u_1-U_1)=m_2(U_2-u_2)$$

We can now conclude that $m_1v(u'_1-U'_1)=m_2v(U'_2-u'_2)$ and therfore $$Q'=\frac{m_1([u']_1^2-[U']_1^2)}{2}+\frac{m_2([u']_2^2-[U']_2^2)}{2}$$ Meaning that $Q=Q'$ and lost energy during a collision is identical between inertial frames.

If you read this far thank you so much! This is my first post and I learned a lot from thinking through this problem. If you have any advice for me for future posts, please let me know. If There is anything wrong with my explanation above, also let me know.

Here's a picture of all the math I did if you were wondering how I came to any of this.


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    $\begingroup$ Would you please explain why the question suggested to use a non-relativistic equation of $u=u′+v$ ? Do you want to solve the problem within the framework of relativity or Newtonian mechanics? $\endgroup$ – Mohammad Javanshiry Mar 13 '18 at 9:13
  • $\begingroup$ You know the approach you tried is wrong because both of the kinetic energies are frame-dependent and subtracting them will not eliminate the effect. Further, kinetic energies are not linearly dependent upon speed, but momentum is which may give you a clue since the question seems to ask you to look at momentum. $\endgroup$ – honeste_vivere Mar 13 '18 at 14:53
  • $\begingroup$ @MohammadJavanshiry Currently the chapter is understanding reference frames with regard to Newtonian Mechanics. $\endgroup$ – Bad Nat Mar 13 '18 at 15:13
  • $\begingroup$ @honeste_vivere Is it not true that the overall energy lost in the system has to be equivalent? I understand that kinetic energies are different between reference frames, but the thermal energy produced in the collision has to be consistent. If it wasn't I could just point an infrared thermometer at the collision and determine some ultimate frame of reference in the universe. I'll try looking at the question more in terms of momentum conservation, I may have just gotten to caught up in an overly complicated scenario I made for myself. More help or suggestions are always appreciated, thanks! $\endgroup$ – Bad Nat Mar 13 '18 at 15:19
  • $\begingroup$ Yes, total energy must be conserved but I was trying to point out that there must be something else if your approach had not worked. I should point out that temperature is not the same as thermal energy. $\endgroup$ – honeste_vivere Mar 13 '18 at 15:31

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