Relativity - energies of particles in circular motion

[Ciumko]

Summary: What is energy of proton, deuteron and alpha particle in circular motion of the same radius.

Hello, I have a problem.

Here is the content of an exercise:

In some experiment, proton with energy of 1MeV is in circular motion in isotropic magnetic field. What energies would have deuteron and alpha particle to orbit in a circle of the exact same radius r?Wouldn't be much of a problem if it wouldn't be "relativity" problem. In classical example I would guess that for deuteron energy would be twice as big as for single proton (because of additional neutron), and for alpha particle it would be four times the energy of single proton. I am not 100% familliar with ale the relativity nuances, so I'd like to ask: Is my reasoning correct? Or there are additional effects taking place?

[Moderator's note: Moved from a technical forum and thus no template.]

 

[Herman Trivilino]

You didn't explain your reasoning. You just said it was a guess.

 

[Ciumko]

Yes and I would like to know the answer - just to know that my reasoning is right or wrong.

$$E_{k}=\frac{1}{2}mv^{2}$$

so, as the radius have to remain the same:

$$F_{c}=\frac{mv^{2}}{r}$$

As the mass grows up to 2 or 4 the times of single proton you get energy two or four times the initial.But from the other side we know that $F=\frac{mv^{2}}{r}=Bqv$, taking velocity from that and applying in classical approach -> $$E_{k}=\frac{1}{2}mv^{2}=\frac{1}{2}\frac{B^{2}q^{2}r^{2}}{m}$$ we have dependence of type $E_{k}=E_{k}(\frac{q^{2}}{m})$ so:

If for proton $E_{k}=1MeV$ then for deuteron (which is one proton and one neutron, so mass is twice as big as single proton) $E_{k}=0.5MeV$, and for alpha particle (two protons and two neutrons) $E_{k}=1MeV$

Which one is correct for classical approach and how to tackle it in relativity?

 

[PeroK]

Yes and I would like to know the answer - just to know that my reasoning is right or wrong.

$$E_{k}=\frac{1}{2}mv^{2}$$

so, as the radius have to remain the same:

$$F_{c}=\frac{mv^{2}}{r}$$

As the mass grows up to 2 or 4 the times of single proton you.But from the other side we know that $F=\frac{mv^{2}}{r}=Bqv$, taking velocity from that and applying in classical approach -> $$E_{k}=\frac{1}{2}mv^{2}=\frac{1}{2}\frac{B^{2}q^{2}r^{2}}{m}$$ we have dependence of type $E_{k}=E_{k}(\frac{q^{2}}{m})$ so:

If for proton $E_{k}=1MeV$ then for deuteron (which is one proton and one neutron, so mass is twice as big as single proton) $E_{k}=0.5MeV$, and for alpha particle (two protons and two neutrons) $E_{k}=1MeV$

Which one is correct for classical approach and how to tackle it in relativity?

This all looks correct, except the notation at the end is confusing.

You need something like

$E_p = \frac{(qBr)^2}{2m} = 1MeV$

Then:

$E_d = \frac{(qBr)^2}{4m} = \frac12 E_p$

And,

$E_a = \frac{(2qBr)^2}{8m} = E_p$

Regarding relativity, the same formula applies for momentum:

$p = qBr$

But now $p = \gamma mv$ is the relativistic momentum and is related to the (total) energy of the particle $\gamma mc^2$ by:

$E^2 = p^2c^2 + m^2c^4$

You could see what you can do with that.