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  What is the differential cross section of Møller scattering in square meters

+ 1 like - 0 dislike
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Can anyone provide the formula for the differential cross section in square meters (sic) for Møller scattering? And maybe even give an authoritative reference?

I need the formula both for checking my math and for testing some computer programs.

I am interested in a formula which uses the International System of Units (SI) in general and m (meter), kg (kilogram), s (second) and A (Ampere) in particular.

Most sources can agree that the differential cross section in the Center of Mass (CM) coordinate system for a scattering angle $ \theta \in ( 0 , \pi ) $ and for incomming electrons which each have momentum $ p \in \mathbb{R}_+ $ is given by

\[ \frac { d \sigma } { d \Omega } = \beta \frac { 4 ( ( m c ) ^ 2 + 2 p ^ 2 ) ^ 2 + ( 4 p ^ 4 - 3 ( ( m c ) ^ 2 + 2 p ^ 2 ) ^ 2 ) ( \sin \theta ) ^ 2 + p ^ 4 ( \sin \theta ) ^ 4 } { p ^ 4 ( \sin \theta ) ^ 4 } \]

for some constant $ \beta \in \mathbb{R}_+ $ which is measured in square meters in the SI system. Unfortunately, the sources I have found are not very explicit about $ \beta $ and may even disagree with one another.

So the question is: can anyone provide a formula for $ \beta $ using the following physical constants?

Physics Constants

asked Feb 22 in Experimental Physics by grue (10 points) [ no revision ]
recategorized Feb 27 by Dilaton

can you give a link for where this β appears?

$\beta$ is my name for the quantity I ask for. So it probably does not appear anywhere outside this question.

Wikipedia https://en.wikipedia.org/wiki/M%C3%B8ller_scattering says:

So Wikipedia implicitly says $ \beta = \alpha ^ 2 / E ^ 2 _ { C M } $. Unfortunately, that does not answer the question since Wikipedia implicitly sets the speed of light to 1 and thus does not use SI units.

Could this lecture page 3 helo you?https://www2.ph.ed.ac.uk/~vjm/Lectures/ParticlePhysics2007_files/Lecture2.pdf

You can get the whole crossection in SI units , and you know the values of the constants in your β in SI units.

Page 3 (Slide 5) of https://www2.ph.ed.ac.uk/~vjm/Lectures/ParticlePhysics2007_files/Lecture2.pdf indeed says how to restore from natural units to SI units by dividing energy by $ c \hbar $ as is also done in the answer by Vladimir Kalitvianski. So if https://www2.ph.ed.ac.uk/~vjm/Lectures/ParticlePhysics2007_files/Lecture2.pdf and https://en.wikipedia.org/wiki/M%C3%B8ller_scattering agree on what natural units mean, and if the latter uses natural units, and if the latter is correct, then we have $ \beta = c ^ 2 \hbar ^ 2 \alpha ^ 2 / E _ { \mathrm { CM } } ^ 2 $ and my question is answered. But as mentioned in my comment to Kalitvianski's answer, nagging doubts remain.

1 Answer

+ 0 like - 0 dislike

One can see in the grue answer that $d\sigma/d\Omega$ is proportional to $1/E_{\text{CM}}^2$ (the rest is dimensionless and thus unit-independent).

Now, there is a center of mass exponential: $\text{e}^{ {-\text{i}E_{\text{CM}}\cdot t/\hbar}}$ in the scattering problem. By multiplying the numerator and denominator in it by $c$, we get $\text{e}^{ {-\text{i}{E_\text{CM}}\cdot ct/c\hbar}}$.

The product $ct$ can be expressed in meters, so the ratio $c\hbar/E_{\text{CM}}$ can also be expressed in meters. From here one can easily restore the dimension of the cross section in square meters (knowing that $m_e c^2 \approx 0.5\;MeV$ and $r _ e =
\frac { \hbar \alpha } { m c } \approx 2.81794032 \cdot 10 ^ { - 15 } \mbox{meter}$.

answered Mar 2 by Vladimir Kalitvianski (92 points) [ revision history ]
edited Mar 8 by Vladimir Kalitvianski

I read @VladimirKalitvianski 's answer like this: Define $ m = m _ e $ so that we can use $ m $ for the mass of the electron. Now $ c \hbar / E _ { \mathrm { CM } } $ has units of meter and Wikipedia says $ \beta = \alpha ^ 2 / E _ { \mathrm { CM } } ^ 2 $ so if we restore units using $ c = 1 $ and $ \hbar = 1 $ we get

\[ \beta = \frac { c ^ 2 \hbar ^ 2 \alpha ^ 2 } { E _ { \mathrm { CM } } ^ 2 } \]

As suggested by Vladimir Kalitvianski we can compute $ E _ { \mathrm { CM } } $ from the momentum $ p \in \mathbb { R } _ + $. The formula would be

\[ E _ { \mathrm { CM } } = 2 \sqrt { p ^ 2 c ^ 2 + m ^ 2 c ^ 4 } \]

where we can make use of $ m c ^ 2 \equiv 0.5 MeV $. That gives

\[ \beta = \frac { \hbar ^ 2 \alpha ^ 2 } { 4 ( p ^ 2 + m ^ 2 c ^ 2 ) } \]

That answer could easily be correct. But now problems start. Wikipedia is silent about what unit system is used. One can easily see that Wikipedia uses $ c = 1 $. A qualified guess would be that Wikipedia also uses $ \hbar = 1 $. But what about vacuum permittivity $ \varepsilon $? Is it eg $ 1 $ or $ 1 / ( 4 \pi ) $? A factor of $ 4 \pi $ is dimensionless and thus escapes dimensional analysis.

If I start from David J Griffiths, Introduction to elementary particles, 2nd rev. version, and if I do a long derivation then I end up with

\[ \beta = \frac { 4 \pi ^ 2 \varepsilon ^ 2 c ^ 2 \hbar ^ 2 \alpha ^ 2 } { E _ { \mathrm { CM } } ^ 2 } \]

Since Griffiths uses Gaussian units we have $ 4 \pi \varepsilon = 1 $ so

\[ \beta = \frac { c ^ 2 \hbar ^ 2 \alpha ^ 2 } { 4 E _ { \mathrm { CM } } ^ 2 } \]

That is off by a factor 4 from Wikipedia. Since 4 is dimensionless that factor also escapes dimensional analysis.

So I end up with the question: who is wrong? Griffiths? Wikipedia? Me? Wikipedia is not an authoritative source and my derivation based on Griffiths could be flawed. So I look for an answer which is independent of both Wikipedia and my own derivations.

But what about vacuum permittivity ε? Is it eg 1 or 1/(4π)? A factor of 4π is dimensionless and thus escapes dimensional analysis.

Indeed, it should escape the dimentional analysics, but it cannot escape the numerical value. Whatever is used (I guess $\varepsilon=1$), it contributes anyway to dimentionless factors in the cross section.

@VladimirKalitvianski Ok, but how can I then know if $ \beta = \frac { c ^ 2 \hbar ^ 2 \alpha ^ 2 } { E _ { \mathrm { CM } } } $ or $ \beta = \frac { c ^ 2 \hbar ^ 2 \alpha ^ 2 } { 4 E _ { \mathrm { CM } } } $ or both are incorrect?

I guess the first expression is right, i.e., without 4 in the denominator, but with $E$ squared.

Thanks for pointing out $ E $ squared. Also, I made a flaw when copying my results based on Griffiths to this thread. The $ \beta $ based on Wikipedia is still $ \beta = c ^ 2 \hbar ^ 2 \alpha ^ 2 / E _ { \mathrm { CM } } ^ 2 $ but the one based on Griffiths should have been $ \beta = c ^ 2 \hbar ^ 2 \alpha ^ 2 / ( 2 E _ { \mathrm { CM } } ^ 2 ) $.

It turns out both of these $ \beta $ are correct. The difference is that Griffiths defines the differential cross section in an unorthodox way:

\[ \frac { d \sigma } { d \Omega } = \left( \frac { \hbar c } { 8 \pi } \right) ^ 2 \frac { S | { \cal M } | ^ 2 } { ( E _ 1 + E _ 2 ) ^ 2 } \frac { | \mathbf { p } _ f | } { | \mathbf { p } _ i | } \]

The peculiarity in Griffiths' definition is that it includes $ S $ which is $ 1 $ if the two particles in the final state are different and $ 1 / 2 $ if they are identical. That makes the differential cross section of Møller scattering according to Griffiths half as big as the differential cross section according to other sources. As an example, Peskin and Schroeder, An introduction to quantum field theory, 1995, page 108 very explicitly says that Pescin and Schroeder includes $ S $ in the total cross section but not in the differential one, which makes perfect sense. What Pescin and Schroeder does seems to be the standard.

By the way, I have been able to derive $ \varepsilon \hbar c = 1 $ from one of the equations in Wikipedia. So there are strong reasons to believe that Wikipedia uses $ \varepsilon = \hbar = c = 1 $ as you say.

There is further evidence that Wikipedia is right in http://www-heaf.astro.hiroshima-u.ac.jp/thesis/ogata2001.pdf which states

\[ \frac { d \sigma } { d \Omega } = \frac { r _ e ^ 2 } { 4 } \left( \frac { m _ e c } { p } \right) ^ 2 \frac { ( 3 + \cos ^ 2 \theta ) ^ 2 } { \sin ^ 4 \theta } \]

in the ultrarelativistic limit where $ m $ is the rest mass of the electron, $ p $ is the momentum of each particle in the CM system and $ r _ e $ is the classical electron radius.

@VladimirKalitvianski  Could I persuade you to edit you answer to include that

\[\begin{array}{l} \frac { d \sigma } { d \Omega } = \frac { r _ e ^ 2 } { 4 } \frac { ( m c ) ^ 2 } { p ^ 2 + ( m c ) ^ 2 } \frac { 4 ( ( m c ) ^ 2 + 2 p ^ 2 ) ^ 2 + ( 4 p ^ 4 - 3 ( ( m c ) ^ 2 + 2 p ^ 2 ) ^ 2 ) ( \sin \theta ) ^ 2 + p ^ 4 ( \sin \theta ) ^ 4 } { p ^ 4 ( \sin \theta ) ^ 4 } \end{array}\]

where the classical electron radius $ r _ e $ is given by

\[ r _ e =
\frac { \hbar \alpha } { m c } \approx 2.81794032 \cdot 10 ^ { - 15 } \mbox{meter} \]

That would very directly answer the original question.

It seems one cannot mark an answer as correct, but at least I can upvote you answer (if my limited 10 point reputation permits). In any case, thanks for your help on sorting all this out.

Vladimir I wonder how come you mention only one digit after the dot in the value of the mass of electron, while for r_e you seem to mention more than one digit after the dot. Why is that?

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