Why not find mass increase in a simple way?

[Dale]

I make no bones of whatever you wish me to call it, since may be very soon it will change. Since your objection is only to words, any terms suits me fine.

Good, then I think the only substantive issue is that $p\ne mv$, but it seems like you understand that now also.

Now, after so many posts, can you please at last show me what is the 'correct' way in which you determine the radius , step by step, and show me why my way is not simpler and better than yours?

I don't know your formula so I can't comment on it, but whatever it is just replace any occurences of "relativistic mass" with "total energy" and make sure that you don't substitute mv for momentum. The complexity would be identical.

I believe that the correct formula is $r=p/qB$ where $p$ is the momentum (use the correct relativistic formula, not mv), $q$ is the charge, and $B$ is the magnetic field.

 

[vela]

At LHC they run protons at 7TeV = 7000 GeV ( rest mass is .93827 GeV),so they are adding the equivalent of 7000/.938 = 7461 proton rest masses (= 1/cosθ (=0.000134)) both if you please to consider it mass or energy : it is not a name or a procedure that shapes or changes reality.

Its speed will be sinθ= sqrt(1-1/7461^2) * c : v = .99 99 99 99 101*2.99*10^10 , its momentum (tanθ) will increase by 7461 times and so will the radius of its circle: in a 5.5-T MF the orbiting radius will increase from 57 cm to r = 57*7461= 4.243 Km, the actual radius of LHC ring (26.6 km).

So you're calculating $\gamma$ and saying $\gamma = 1/\cos\theta$ for some parameter $\theta$. This allows you to hide the square root in the trig functions to find the speed $v= \sin\theta= \sqrt{1-1/\gamma^2}$ and then the momentum is $p = m\gamma v = m(\sin\theta/\cos\theta) = m\tan\theta$. That's fine, but I don't see the advantage here.

the simple and logic way of finding the radius of the particle is to multiply the regular radius of any proton (57) by 7 T eV / .938 GeV, that is 57*7461 cm.,you haven't yet shown how you figure that out.

The radius of curvature is given by (both relativistically and non-relativistically) $r = \frac{p}{qB}$ where $p$ is the momentum of the particle. The difference between the two momenta $p_{rel} = m\gamma v$ and $p_{non} = mv$ is just a factor of $\gamma$, so as soon as you find $\gamma = E/(mc^2)$, you know the factor by which $r$ increases between the relativistic and non-relativistic cases. There's no need for the stuff you did before.

Typically, the information you start with is the energy of the particle and its mass, and what you want is the momentum. Calculating $p = \sqrt{(E/c)^2-(mc)^2}$ and then plugging the result into $r = \frac{p}{qB}$ is pretty straightforward. There's no need to find the speed of the particle, the non-relativistic momentum, etc.

 

[Dale]

All right, let's make something clear :
rest mass is equivalent to energy
toenergy of any kind bound to massive particles is equivalent to mass, it applies r thermal energy, KE, gluons etc...is that right?

Again the relationship (in natural units) is $m^2=E^2-p^2$. So for a point particle you cannot increase E without increasing p. For multiple particle systems it is possible to increase E without increasing p using internal degrees of freedom, such as rotation.

 

[jtbell]

I believe that the correct formula is $r=p/qB$ where $p$ is the momentum (use the correct relativistic formula, not mv), $q$ is the charge, and $B$ is the magnetic field.

Yes, but you have to be careful with the units when calculating. Taking my electron from post #13, its momentum in energy units is pc = 1422 keV = 2.275 x 10^-13 J. Dividing by c gives p = 7.583 x 10^-22 kg·m/s. Put it in a magnetic field of 1 mT = 0.001 T.

In SI units, $$r = \frac {p}{qB} = \frac {7.583 \times 10^{-22}~\rm{kg \cdot m/s}} {(1.60 \times 10^{-19}~\rm{C})(0.001~\rm{T})} = 4.74~\rm{m}.$$ In "eV units", the unit of q is the magnitude of the electron charge, e, so $$r = \frac {p}{qB} = \frac {pc}{qBc} = \frac {1422 \times 10^3~\rm{eV}} {(1~\rm{e})(0.001~\rm{T})(3.00 \times 10^8~\rm{m/s})} = 4.74~\rm{m}.$$ Or you can think of p as having units of keV/c, so $$p = \frac{1422 \times 10^3~\rm{eV}}{3.00 \times 10^8~\rm{m/s}}$$ and you can use the original form of the equation. $$r = \frac {p}{qB} = \frac {\left( \frac{1422 \times 10^3~\rm{eV}}{3.00 \times 10^8~\rm{m/s}} \right)} {(1~\rm{e})(0.001~\rm{T})} = 4.74~\rm{m}.$$ A factor of c has to come in somewhere, if you use "eV units" for momentum.

 

[Dale]

Yes, but you have to be careful with the units when calculating. Taking my electron from post #13, its momentum in energy units is pc = 1422 keV = 2.275 x 10^-13 J.

Doesn't the particle physics use "eV units" for everything? So B = 1 mT = 0.6925 eV^2 and r = 4.74 m = 2.402 10^7 1/eV. I must have something wrong because I get
$$r=\frac{p}{q \; B} = \frac{1422 \; keV}{1 \; 0.6925 \; eV^2} = 2.05 \; 10^6 \; 1/eV = 0.405 \; m$$

 

[jtbell]

I don't remember anybody using eV² for magnetic field strength, but that was 35-40 years ago. Maybe things are different now.