Total or kinetic energy in Bethe Bloch stopping power?

  • #1
43
4

Main Question or Discussion Point

The average loss of energy in a material per unit lenght of a particle (in particular an electron, which is stable) is described by the quantity ##dE/dx##.

- for ionization it is given by the Bethe-Bloch formula $$-\left\langle {\frac {dE}{dx}}\right\rangle ={\frac {4\pi }{m_{e}c^{2}}}\cdot {\frac {nz^{2}}{\beta ^{2}}}\cdot \left({\frac {e^{2}}{4\pi \varepsilon _{0}}}\right)^{2}\cdot \left[\ln \left({\frac {2m_{e}c^{2}\beta ^{2}}{I\cdot (1-\beta ^{2})}}\right)-\beta ^{2}\right]$$
- for Bremmstralungh it is given by the Bethe-Heilter formula $${\displaystyle -\left\langle {\frac {dE}{dx}}\right\rangle \approx {\frac {4N_{a}Z^{2}\alpha ^{3}(\hbar c)^{2}}{m_{e}^{2}c^{4}}}E\ln {\frac {183}{Z^{1/3}}}}$$

I can't understand if the "energy ##E##" in the formulas is the total relativistic energy or the kinetic energy ##K## only?

In the first case ##E=K +mc^2##, while in the second case ##E=K##.

It looks like that the first case is the right one, since it's more general, but in that case I cannot understand how the particle, stopping in the material for various processes can change its mass (the electrons are stable so they do not decay after they have stopped). Does this really happen or does it loose before all its kinetic energy and then its rest energy?

I'm confused also because I read that in calorimeters the range of the particle is used to measure its energy: does this energy include the rest energy?
 

Answers and Replies

  • #2
34,056
9,922
For the left side it does not matter as both only differ by a constant which doesn't change the derivative.
The Bethe-Heitler formula is only a good approximation for ##E\gg mc^2## which means it doesn't matter either.
 
  • #3
43
4
For the left side it does not matter as both only differ by a constant which doesn't change the derivative.
The Bethe-Heitler formula is only a good approximation for ##E\gg mc^2## which means it doesn't matter either.
Thank you for the answer! By "differing by a constant" in the left side are you referring to the fact that the rest mass ##mc^2## is constant?
 
  • #4
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
2019 Award
24,040
6,645
It's simpler than that. df/dx = dg/dx is f(x) = g(x) + a constant.
 
Last edited:
  • #5
34,056
9,922
Thank you for the answer! By "differing by a constant" in the left side are you referring to the fact that the rest mass ##mc^2## is constant?
Right. The derivative of the total energy and the derivative of the kinetic energy are the same.
 

Related Threads on Total or kinetic energy in Bethe Bloch stopping power?

  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
1
Views
3K
Replies
10
Views
12K
Replies
1
Views
6K
Replies
2
Views
6K
  • Last Post
Replies
3
Views
787
  • Last Post
Replies
4
Views
1K
Replies
9
Views
7K
  • Last Post
Replies
3
Views
6K
Replies
0
Views
3K
Top