# Estimating particle mass from bubble chamber

In an example of a book, it shows how to estimate the $$\pi^-$$ rest mass 139MeV$$/c^2$$ from the trajectory of $$\pi^-$$ in a bubble chamber.

It starts from the charge-to-mass ratio formula $$\tag{1} \frac{q}{m} = \frac{v}{Br}$$ which is derived from the equality of Lorentz force and centripetal force $$qvB = m\frac{v^2}{r}$$ where the uniform magnetic field is assumed to be perpendicular to particle trajectory. To get $$q/m$$, we then need the values of $$v, B, r$$. The value of the uniform magnetic field $$B$$ is chosen by the experimentalist. The radius of curvature $$r$$ can be measured. As for $$v$$, the author substitute it with light speed $$c$$. The author argues that $$\pi^-$$ moves at a very high speed in the bubble chamber and so $$v$$ is approximated by light speed. Therefore, the formula becomes $$\frac{q}{m} = \frac{c}{Br}$$ We can then solve for $$m$$ as $$$$\tag{2} m = \frac{qBr}{c}$$$$ Substitute in values of variables on the right-hand-side, the author estimated the rest mass of $$\pi^-$$ to be 136MeV/$$c^2$$ which is close to the value listed on wiki $$139\text{MeV}/c^2$$.

Questions:

1. The "relativistic correction" done by the author looks fishy for me. Per the comment in this thread, Relativistic centripetal force the relativistic version of the charge-to-mass ratio formula should be $$\tag{3} m_0\gamma^2 v^2 / r = \gamma q v B \quad\text{or}\quad m_0 = \frac{qBr}{\gamma v}$$ where $$m_0$$ is the rest mass of the particle. That is, the relativistic correction should be to replace $$m$$ in equation (1) by $$\gamma m_0$$, i.e. interpret $$m$$ in (1) as relativistic mass (which is rest mass multiplied by the Lorentz factor $$\gamma = 1/\sqrt{1-v^2/c^2}$$).

2. If I estimate the rest mass of $$\pi^-$$ using (3) with the numbers provided by the book and assume $$v = 0.99c$$, I get $$17.7\text{MeV}/c^2$$ which is far less than the value $$139\text{MeV}/c^2$$ on wiki, https://en.wikipedia.org/wiki/Pion

I don't know if the author of the book was sloppy and the numbers given in the example were just made up?

Derivation of charge-to-mass formula: From the covariant form of Lorentz force $$\frac{dp_{\alpha}}{dt} = qF_{\alpha\beta}\frac{dx^\beta}{dt}$$ We choose the spatial component $$\alpha = i$$ and assume magnetic field $$B_i = \epsilon_{ijk}F^{jk}/2$$ is nonzero only in a spatial component, say $$i = 3$$. We then have
$$\frac{dp}{dt} = qvB$$ where we denote $$p_1, v_1, B_3$$ as $$p, v, B$$ respectively. Now, with $$p = m_0 \gamma v$$ and $$dv/dt = v^2/r$$ we get $$\gamma m_0 \frac{v^2}{r} = qvB$$ which is (3).

Actually it is not difficult to see (3) makes more sense than (1) in the relativistic limit $$v \rightarrow c$$. For a particle with nonzero rest mass, the radius $$r$$ goes up. This simply means it is more and more difficult to accelerate the particle to curve if the particle speed approaches light speed.

• Your (3) is wrong, and I'm not sure how you teased it out of that answer. Start by chucking the oxymoronic "relativistic mass". There is only one mass parameter, the relativistic invariant. Commented Jun 8 at 15:16
• The radius of curvature expression involves the momentum, relativistic or not. The velocity involves the ratio p/E. Never use or touch "relativistic mass" ever again! Commented Jun 8 at 15:45
• @CosmasZachos So (3) is correct, isn't it? (3) is consistent with physics.stackexchange.com/questions/327182/… Commented Jun 8 at 15:54
• Viz. Commented Jun 8 at 18:58
• Right, so the (9.153) of the linked book $R = \frac{1}{q}\frac{p}{B}$ is consistent with (3) with the substitution of $p = \gamma m_0 v$. That means (3) is correct and (1) is wrong if $m$ in (1) is rest mass. Agree? @CosmasZachos Commented Jun 9 at 0:59

You certainly have a good point, and I apologize for my confusing/wrong leading comment, a knee-jerk reaction to the mention of the deprecated/taboo term "relativistic mass" in lieu of $$E/c^2=m\gamma$$. (I was expecting you to write your sound answer to your own question, sure to be approved by as many people as liked your question.) Indeed, (3) is right and (1) is wrong. Not knowing the book you are referring to, I can't tell if the author faked it or not... (2) is wrong. The Lorentz factor γ is there.
The key point is that the radius of curvature of the particle trajectory is, as stated in the linked answer and every E&M text, $$R=\frac{p}{qB}=\frac{\gamma mv}{qB},\tag{3}$$ where p is the momentum, at all energies, relativistic or not. As v and thus γ increase, B must increase to prevent the Radius of curvature from blowing up to a straight line, as you properly point out is illustrated by the enormity of the CERN ring.
Just a hypothetical reverse engineering of your numerical example. If your book were applying the correct formula with the γ without telling you, so $$v=c(1-\epsilon)$$, where $$\epsilon = 1/100$$, then the answer would be $$\gamma v = {c(1-\epsilon) \over \sqrt{1-(1-\epsilon)^2}}\approx {c\over \sqrt{2\epsilon}} \approx c 7.07,$$ resulting in something quite close to the physical answer. So maybe your book utilized γ without telling you?
• the example is in a section about introductory collider experiment in a pre college level textbook. The author mentioned that the particles are close to light speed and then directly replace $v$ with $c$... (I don't know why they thought the error would be small where Lorentz factor can be huge) Commented Jun 11 at 16:39