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Exist a physical demonstration why

$$E^2- p^2c^2 =m^2c^4=E'^2- p'^2c^2 $$ is an invariant for transformations of Lorentz?

N.B.: $m$ is mass; $E$ is the energy and $p$ is momentum in the frame $\Sigma$. Similary $E'$ is the energy and $p'$ is momentum in the frame $\Sigma'$.

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  • $\begingroup$ I am sure you meant $E^2-p^2c^2=m^2c^4$. Now, what do you mean by a physical demonstration? In particular, why the given relation is invariant for the transformations of Lorentz is an intrinsically mathematical question--I am not sure it can mean anything to have its physical demonstration. Do you want to ask if there is empirical evidence that this relation holds for experiments related to each other via physical boosts? $\endgroup$ – Dvij Mankad Mar 22 at 22:32
  • $\begingroup$ @DvijMankad I have edited my question. Thank you for your suggestion. I'm a high school teacher and I've never done this before. In addition, if I have asked questions that in my opinion are simply clarifications, it is because I do not have the same skills as those who use these concepts every day. $\endgroup$ – Sebastiano Mar 22 at 22:46
  • $\begingroup$ $E$ here is energy, not electric field. $\endgroup$ – G. Smith Mar 22 at 23:19
  • $\begingroup$ @G.Smith Yes. You have right. Excuse me. Now I edit my question. $\endgroup$ – Sebastiano Mar 22 at 23:21
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    $\begingroup$ Energy & momentum can be combined into the four-momentum vector, which is a scalar multiple of the four-velocity, so both of those vectors Lorentz transform the same way. There are more details in the Wikipedia article on four-momentum, but that article is rather technical. $\endgroup$ – PM 2Ring Mar 24 at 7:21
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The threshold beam energies for particle production experiments depend on the way you set up the experiment (mostly fixed-target or symmetric collider systems but there are asymmetric colliders as well).1

However, those experimental thresholds are deducible on the basis on the invariant mass of the final-state system.

I suppose the simplest possible choice of reactions to discuss here would be $$ e^- + e^+ \to \mu^- \mu^+ \;.$$ If the two muons are at relative rest then the (invariant) mass of the final state is $2m_\mu$.

In symmetric collider mode, you have net zero momentum so you can produce the pair at rest in the lab frame. Accordingly each beam needs to have a kinetic energy $T_\text{collider} = m_\mu - m_e \approx 105\,\mathrm{MeV}$ which is a lot but not a lot.

Let's figure what you need in fixed target mode.2

You need $|\mathbf{p}_{+} + \mathbf{p}_-| = 2m_\mu \approx 211\,\mathrm{MeV}$. But we know $\mathbf{p}_- = (m_e,0)$. We write $\mathbf{p}_+ = (T+m_e,T+m_e)$ because we are very much in the ultra-relativistic regime.3

This leads to \begin{align} (\mathbf{p}_{+} + \mathbf{p}_-)^2 &= (T+2m_e,T+m_e)^2 \\ &= (T+2m_e)^2 - (T+m_e)^2 \\ &= 2Tm_e + 3m_e^2 \;, \end{align} and as you will see $T \gg m_e$ to so a very good approximation $$ \sqrt{2Tm_e} = 211\,\mathrm{MeV} \;.$$ But of course the electron mass is $m_e = 0.511\,\mathrm{MeV}$, so this comes to $$ T_\text{fixed target} \approx 43.6\,\mathrm{GeV} \;,$$ which is a lot!4

And these kinds of predictions are borne out in practice.5


1 There is a wrinkle here in that the production rate at the exact threshold is essentially zero. So we don't find the threshold by slowly dialing up the beam energy until we see the first particle, we do it by measuring the way the rate varies not much above the threshold and then working backwards from the theory.

2 Better use a positron beam as it is hard to find a positron target...

3 The boldface symbols are four-momenta and are broken down $\mathbf{p} = (E,p)$ becase we need care only about the beam direction for this calculation. In principle it ought to be $\mathbf{p} = (E,\vec{p})$. Also note that I'm working in natural units here.

4 You should see right away that conservation of momentum requires that the products be in motion. That's where the extra energy goes: the kinetic energy of the products. The perfectly symmetric collider reaction has the products at rest at threshold.

5 This is also why we build colliders when they are considerable harder to engineer than fixed-target machines.

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  • $\begingroup$ Since I joined StackExchange personally I have always voted in favour of any answer because the user has dedicated his time to me. Your explanation is too complicated for me. But isn't there a simpler way of using either the quadrivers or the Lorentz transformations? $\endgroup$ – Sebastiano Mar 23 at 22:08
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    $\begingroup$ Hmmm. I've tried to describe a physical demonstration (i.e. set of experiments) that shows that the Lorentz invariant mass of a system is a useful quantity. I use a system because I don't know of a way to do it with a single particle. The point is a system of 2 muons at relative rest has a certain (invariant) mass, and that this mass is the thing you need to know in order to predict experimental thresholds for muon pair production. $\endgroup$ – dmckee Mar 23 at 22:37

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