Relativity - energies of particles in circular motion

In summary, the conversation discusses the energy of a proton, deuteron, and alpha particle in circular motion with the same radius. The classical approach suggests that the energy for the deuteron would be half that of the proton, and for the alpha particle it would be equal to the proton. However, the relativity approach complicates things by introducing the concept of relativistic momentum and its relationship to energy. Further exploration is needed to determine the correct approach for tackling this problem.
  • #1
Ciumko
4
0
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.]
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
You didn't explain your reasoning. You just said it was a guess.
 
  • #3
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?
 
Last edited:
  • #4
Ciumko said:
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.
 

1. What is the theory of relativity?

The theory of relativity is a fundamental concept in physics that describes how the laws of physics are the same for all observers in uniform motion. It consists of two main theories: special relativity, which deals with objects moving at constant speeds, and general relativity, which takes into account the effects of gravity.

2. How does relativity affect the energies of particles in circular motion?

According to special relativity, the energy of a particle in circular motion is dependent on its speed and mass. As the speed of the particle approaches the speed of light, its energy increases significantly. This is known as relativistic energy, and it is a key concept in understanding the behavior of particles in circular motion.

3. Can relativity explain the behavior of particles in circular accelerators?

Yes, relativity plays a crucial role in understanding the behavior of particles in circular accelerators, such as the Large Hadron Collider. These accelerators use magnetic fields to keep particles in circular motion at high speeds, and relativity helps to explain the increase in energy and mass of these particles as they approach the speed of light.

4. How does relativity affect the concept of mass?

According to special relativity, mass is not a constant quantity but instead depends on the speed of an object. As an object's speed increases, its mass also increases, and this effect becomes more significant as the speed approaches the speed of light. This is known as relativistic mass and is a key concept in understanding the behavior of particles in circular motion.

5. Can relativity be applied to everyday situations?

Yes, while relativity is often associated with complex theories and advanced physics, it has practical applications in everyday life. For example, GPS technology uses the principles of relativity to accurately calculate the positions of satellites and receivers on Earth. Additionally, the understanding of relativistic energy and mass has led to advancements in nuclear energy and particle physics.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
912
Replies
5
Views
869
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
993
  • Introductory Physics Homework Help
Replies
3
Views
757
  • Introductory Physics Homework Help
Replies
28
Views
1K
Replies
35
Views
2K
  • Quantum Physics
2
Replies
36
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
Back
Top