What would be the (relativistic) mass of a proton, in grams, as it is traveling at the maximum possible speed in the LHC?
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$\begingroup$ 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. $\endgroup$– CuriousOneCommented Sep 21, 2015 at 5:51
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$\begingroup$ profmattstrassler.com/articles-and-posts/… $\endgroup$– user81619Commented Sep 21, 2015 at 8:52
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$\begingroup$ The mass is about $1.67\times10^{-24}$ g. $\endgroup$– Kyle KanosCommented Sep 21, 2015 at 10:33
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1 Answer
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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.
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$\begingroup$ 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? $\endgroup$ Commented Sep 22, 2015 at 6:50
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$\begingroup$ 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. $\endgroup$ Commented Oct 4, 2015 at 15:57