What would be the (relativistic) mass of a proton, in grams, as it is traveling at the maximum possible speed in the LHC?

  • $\begingroup$ The same as for a proton at rest. Relativistic mass is an old concept that isn't being used anymore. If you want to do a quick and dirty one, the max. proton energy is 7TeV, which is about 7000GeV/0.938GeV or approx. 7460 times the rest-mass energy of the proton. $\endgroup$ – CuriousOne Sep 21 '15 at 5:51
  • $\begingroup$ profmattstrassler.com/articles-and-posts/… $\endgroup$ – user81619 Sep 21 '15 at 8:52
  • $\begingroup$ The mass is about $1.67\times10^{-24}$ g. $\endgroup$ – Kyle Kanos Sep 21 '15 at 10:33

Let's consider two separate things:

  • The mass (i.e. rest mass) of anything doesn't depend on its relative motion to an observer (i.e. is Lorentz invariant). For a proton, $m_p\simeq 1\,\text{GeV}/c^2$.

  • The energy (occasionally egregiously called mass or relativistic mass in old-fashioned sources) of an object isn't Lorentz invariant. In the future, the LHC will collide protons with $$E / c^2 =\gamma m_p \simeq 6.5\,\text{TeV}/c^2$$ energy each, in the laboratory frame.

Note that the conventional unit for particle masses is $\text{eV}/c^2$. You can convert between units with this conversion factor $$ 1.782 661 907 \times 10^{-36}\,\text{kg} = 1\,\text{eV}/c^2 $$ and this table of SI suffixes for T, G etc.

  • $\begingroup$ Okay, I understand the difference between invariant mass and relativistic mass. Now, is the radial force required to keep a proton in its orbit in the LHC a function of its invariant mass or its relativistic mass? $\endgroup$ – John Fistere Sep 22 '15 at 6:50
  • $\begingroup$ John: The best way to think of it is as a function of momentum. Because $\vec{F} = \frac{\mathrm{d}\mathbf{\vec{p}}}{\mathrm{d}t}$ holds in both Newtonian and Einsteinian mechanics. Then you use $\vec{p} = \gamma m \vec{u}$. In the old way of talking you would have folded the $\gamma$ with the $m$ and called the result the "relativistic mass", but due to the nature of the Lorentz transform the transverse inertia is different from longitudinal inertia. $\endgroup$ – dmckee --- ex-moderator kitten Oct 4 '15 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.