PGRE | Positronium Question

  • #1

Homework Statement


Hello PF, I'm studying for the PGRE and I am having trouble understanding why exactly we use the reduced mass to calculate the energy levels of the positronium atom. Well I think I understand why but I just can't visualize, conceptually, what exactly using the reduced mass is doing for us.


Homework Equations


The energy levels of the Hydrogen atom are given by

[itex]E= -\frac{13.6eV}{n^{2}}[/itex]

where the value 13.6eV is the Rydeberg constant and equal to:

[itex]-13.6eV= -\frac{m_{e}q^{4}_{e}}{8h^{2}\epsilon_{o}}[/itex]

Here, the subscript e denotes electron values.

The Attempt at a Solution


In the case of Hydrogen, the proton in the nucleus is so much more massive than the electron that we can approximate it to be stationary while the electron orbits around it. In the case of positronium however, the electron and positron have the same mass and so it is not a very good model to take the positron as being stationary. Then the positron and electron orbit each other. What I don't understand is, how does replacing the mass of the electron in the equation for the Rydeberg constant with the reduced mass of the system account for this fact? Can someone paint me a picture, and explain what exactly this reduced mass system looks like physically? I guess I just never really understood reduced mass problems.

By replacing the mass of the electron with the reduced mass of the positron-electron system, are we in affect changing the problem such that the positron is now a stationary object around which the reduced mass that we calculated is orbiting?

Thanks in advance!
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,909
The two-body problem discusses the motion of two interacting point masses m1 and m2. There are no external forces.
The position vectors of the bodies are [itex]\vec r_1[/itex] and [itex]\vec r_2[/itex], and the force of interaction is F. We assume central force of interaction so [itex]\vec F_{12}=-\vec F_{21}=\vec F[/itex].
The equations of motion for both bodies are


[tex]m_1 \ddot {\vec r_1}=\vec F[/tex]
[tex]m_2 \ddot {\vec r_2}=-\vec F[/tex]

The sum of the equations results in
[tex]m_1 \ddot {\vec r_1}+m_2 \ddot {\vec r_2}=0[/tex].

In the lack of external forces, the CM does not accelerates. We can choose the centre-of mass frame of reference, where the CM is in rest in the origin.

[tex]m_1 \vec r_1+m_2 \vec r_2 =0[/tex]

The equations of motion can be rewritten as

[tex]\ddot {\vec r_1}=\frac{\vec F}{m_1 }[/tex],
[tex]\ddot {\vec r_2}=-\frac{\vec F}{m_2 }[/tex],

and subtracted:
[tex]\ddot {\vec r_1}-\ddot {\vec r_2}=\vec F(\frac{1}{m_1 }+\frac{1}{m_2})[/tex].


Let be [itex]\vec r[/itex] a new variable defined as

[tex]\vec r = \vec r_1- \vec r_2 [/tex] and 1/m1+1/m2 is the reciprocal of the effective mass μ

[tex]\mu=\frac{1}{1/m_1+1/m_2}[/tex]

With these notations we got the equation

[tex] \mu \ddot {\vec r}=\vec F [/tex]

Which is the same as the equation of motion of a single point mass μ around the the CM as origin.

If r is known, the motion of both bodies around the CM are obtained as

[tex]\vec r_1=\frac{m_2 \vec r }{m_1+m_2}[/tex],

[tex]\vec r_2=-\frac{m_1 \vec r }{m_1+m_2}[/tex].

ehild
 
  • #3
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,883
1,460
With these notations we got the equation

[tex] \mu \ddot {\vec r}=\vec F [/tex]

Which is the same as the equation of motion of a single point mass μ around the the CM as origin.
Slight correction. That's the equation for when you fix one of the bodies at the origin, and the other one, with reduced mass μ, moves around it.
 
  • #4
ehild
Homework Helper
15,543
1,909
Slight correction. That's the equation for when you fix one of the bodies at the origin, and the other one, with reduced mass μ, moves around it.
The vector r means the difference r1-r2. The equation [itex] \mu \ddot {\vec r}=\vec F [/itex] is for the position r of an imaginary body of mass μ moving around the origin, according to the force of interaction. The origin was chosen at the CM of the two bodies.

Using relative position does not mean that the position of the reference particle is fixed. If it were so, the CM would orbit around the fixed particle which is impossible.

This method solves the two-body problem in terms of new variables r=r1-r2 and rcm =(m1r1+m2r2)/(m1++m2) The equation of motion of the bodies transform into two other equations, one for the CM

[itex](m_1+m_2)\ddot {\vec r}_{CM}=0[/itex]

and the other one for the body with effective mass

[tex] \mu \ddot {\vec r}=\vec F [/tex].

It is worth seeing: http://en.wikipedia.org/wiki/Two-body_problem


ehild
 
  • #5
Thank you very much, this helps a lot.
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,883
1,460
The vector r means the difference r1-r2. The equation [itex] \mu \ddot {\vec r}=\vec F [/itex] is for the position r of an imaginary body of mass μ moving around the origin, according to the force of interaction. The origin was chosen at the CM of the two bodies.
The way I look at it is when you say [itex]\vec{r} = \vec{r}_1 - \vec{r}_2[/itex], you're effectively making the position of the second body the origin because [itex]\vec{r}[/itex] is the position of body 1 relative to body 2. The effect of considering body 2 fixed is that body 1 no longer appears to have mass m1; instead, it has the reduced mass μ. Obviously, neither body is really fixed, but the equation of motion you would get if you pretended body 2 was fixed, body 1 had mass μ, and the force between then was F is [itex] \mu \ddot {\vec r}=\vec F [/itex].

I guess it doesn't really make a difference in the end. It just seems unnatural to me to think of the vector r extending from the center-of-mass to the position of a hypothetical body because r really has nothing to do with where the center of mass is but only on the relative positions of the two bodies.
 
  • #7
ehild
Homework Helper
15,543
1,909
The way I look at it is when you say [itex]\vec{r} = \vec{r}_1 - \vec{r}_2[/itex], you're effectively making the position of the second body the origin because [itex]\vec{r}[/itex] is the position of body 1 relative to body 2.
Working with r and rcm is transformation of the variables. If you have a function f(x,y) and transform the variables to u=x-y and v=x+y, you do not mean that y is fixed.

I guess it doesn't really make a difference in the end. It just seems unnatural to me to think of the vector r extending from the center-of-mass to the position of a hypothetical body because r really has nothing to do with where the center of mass is but only on the relative positions of the two bodies.
The vector r extends from the origin to the position of a hypothetical body. But the origin happens to be at the CM of the real bodies.
It makes a difference if you take r1 fixed or the CM fixed. With fixed r1, the final solution is r1=const and r2(t)=r(t) which is wrong.

Are Quantum Mechanics or Relativity Theory not unnatural? There are strange mathematical models we apply in Physics. Only the result counts.

There is no imaginary body, but the equations of motion transform to separated ones in the new variables, what makes the solution easy. If you forget about the effective mass and try to solve the system of equations for the interacting bodies, how will you do it?

As a simple example, there are two masses m1 and m2 fixed to the ends of a spring with spring constant k with unstrecthed length L0. At t=0, the spring is stretched to length Li. Describe the motion of both bodies.

Choosing a frame of reference with axis along the spring and the variables x1 and x2.
The equations of motion are

[tex]m_1\ddot {x}_1=k(x_2-x_1-L_0)[/tex]
[tex]m_2\ddot {x}_2=-k(x_2-x_1-L_0)[/tex].

How would you solve?

I would introduce the new variables
[tex]r=\frac{x_1m_1+x_2m_2}{m_1+m_2}[/tex]
and
[tex]s=x_2-x_1-L_0[/tex]

to obtain the transformed equations

[tex](m_1+m_2)\ddot {r}=0[/tex]

[tex]\ddot {s}=-(k/m_1+k/m_2)s[/tex],

Without any bad feelings that r and s are unnatural, I solve the problem as a blind Mathematician, and I get r=at+b, and s=Acos(ωt +φ) with

[itex]\omega=\sqrt{k(1/m_1+1/m_2)}[/itex].

Now I turn back to be a Physicist and recognise that r is the position of the CM, and as the whole spring was in rest initially, a=0.
As s= length of the spring - unstretched length, I recognize that the spring oscillates with the frequency as if one end would be fixed, and a mass of [itex]\mu=\frac{1}{1/m_1+1/m_2}[/itex] would be attached to the other end.

ehild
 

Attachments

Last edited:

Related Threads on PGRE | Positronium Question

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
947
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
954
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
10
Views
3K
Replies
7
Views
2K
Replies
1
Views
1K
Top