Difficulty with relativistic units, eV/c^2

In summary, the conversation discusses the difficulty in working with units of mass at the quantum level, specifically in a Compton scattering problem. The confusion arises in the denominator and the solution is to convert from mass to length using the equation \Delta \lambda =\frac{h}{m_{e}c}(1-cos(\theta )). It is recommended to think in terms of mc^2 rather than just m, and to use the value of hc in calculations.
  • #1
yoshtov
29
0
I'm having some difficulty in working with units of mass at the quantum level. This difficulty most clearly manifests itself when I'm doing a Compton scattering problem.

Recall that Compton scattering is given by
[tex]\Delta \lambda =\frac{h}{m_{e}c}(1-cos(\theta ))[/tex]
and that the rest mass of an electron is 5.11 x 10^5 eV/c^2

My confusion seems to come about in the denominator. I find myself unsure of whether to multiply by c, c^2, or maybe something else entirely. In any case, for my problem, I keep getting the wrong answer, regardless of my method of attack.

If anyone could shed any insight on my problem, and also give me a big picture of relativistic units (i.e., here is how to handle relativistic units in calculations), it would be very much appreciated.
 
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  • #2
To convert from mass to length, we use

[tex]\hbar c \sim 197~ \mathrm{MeV\cdot fm}.[/tex]

So your dimensionful quantity is

[tex] \frac{h}{m_e c} = \frac{2\pi \hbar c}{m_e c^2}, [/tex]

which is easily computed from the information given.
 
  • #3
I like to think not in terms of ##m = 5.11 \times 10^5## eV/c2, but rather ##mc^2 = 5.11 \times 10^5## eV. Before plugging this into an equation, I insert c's so that I have this combination to substitute for. In your example:

$$\Delta \lambda =\frac{h}{m_{e}c}(1- \cos \theta) = \frac{hc}{m_e c^2}(1 - \cos \theta)$$
$$\Delta \lambda = \frac{1.24 \times 10^{-6} eV \cdot m}{5.11 \times 10^5 eV} (1 - \cos \theta)$$

Most tables of constants in textbooks list values for hc in various units, or of course you can construct them yourself from values for h and c. hc comes up often enough that I've memorized it naturally.

Similarly for momentum: p = 100 keV/c means pc = 100 keV. An electron with that momentum has energy

$$ E = \sqrt{(mc^2)^2 + (pc)^2} = \sqrt{(511 keV)^2 + (100 keV)^2} = 520.7 keV$$
 
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Related to Difficulty with relativistic units, eV/c^2

1. What are relativistic units?

Relativistic units are a system of measurement used in physics to describe the behavior of objects at high speeds or in strong gravitational fields. These units take into account the effects of special relativity, a theory proposed by Albert Einstein.

2. What is the significance of eV/c^2 in relativistic units?

eV/c^2 is a unit of mass in relativistic units, where c is the speed of light and eV is the unit of energy known as electronvolt. This unit is commonly used in particle physics to describe the mass of subatomic particles, such as electrons and protons.

3. How are eV/c^2 and kilograms related?

eV/c^2 is equivalent to approximately 1.78 x 10^-36 kilograms. This conversion factor is used to translate between relativistic units and the more commonly used SI units, such as kilograms and meters.

4. Why is it difficult to work with relativistic units?

Relativistic units can be difficult to work with because they often involve very small or very large numbers, making calculations and comparisons challenging. Additionally, these units may not be familiar to those who are more accustomed to working with traditional units of measurement.

5. How can I convert between eV/c^2 and other units of mass?

To convert between eV/c^2 and other units of mass, you can use the conversion factor of 1 eV/c^2 = 1.78 x 10^-36 kg. You can also use online conversion calculators or reference tables to assist with the conversion process.

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