6
$\begingroup$

Firstly I would like to say that despite the fact that you may be thinking that this is a physics question and should probably be asked on that part of the website, I came across reduced mass for the first time in Chemistry - physical and inorganic - and it is used a lot in Chemistry so I thought it suitable for this part of the site.

With that said, I will now elaborate on my question. I hear people say that reduced mass ,$\mu$, is used for two body problems, but what sort of problems? Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

I can follow the derivation (which I shall go through shortly) in a mathematical sense but I can seem to be able to grasp the ideas behind why the steps were taken.

Derivation:

Step [1] - From Newton's laws of motion: $$F_{12}=-F_{21}$$ It follows that: $$m_1a_1=-m_2a_2$$ Step [2] - Simple rearrangement of the result from step 1 to make $a_1$ the subject (it seems a reasonable thing to do but I don't know why it's done). $$a_1=-\frac{m_2}{m_1}a_2$$ Step [3] - The relative acceleration, $a_{rel}$, can be found by subtracting the acceleration of one body from the other (I get how this is done and get that it would give the relative acceleration BUT I don't know why you would need the relative acceleration). $$a_{rel}=a_1-a_2=(1+\frac{m_1}{m_2})a_1=\frac{m_2+m_1}{m_1m_2}m_1a_1=\frac{m_2+m_1}{m_1m_2}F_{12}$$ Step [4] - According to Wikipedia the result from the last step can be used thus (I have no idea what's going on in this step): $$\frac{m_2+m_1}{m_1m_2}F_{12}=\frac{F_{12}}{\mu}$$ Thus: $$\mu=\frac{m_1m_2}{m_2+m_1}$$

Again, I can do the maths but my problem here is conceptual I believe. I have included the derivation in my question because I believe that it's possible that if you were able to shed some light on why particular steps were taken that this may help.

$\endgroup$
  • 1
    $\begingroup$ Reduced mass is purely physical concept, and since you are interested in the concept itself and not its application in some area of chemistry, the question should be asked on Physics.SE. $\endgroup$ – Wildcat Jan 13 '15 at 21:51
  • $\begingroup$ I am very much interested in its applications. As I mentioned in the second paragraph. I will make an edit to make this clearer $\endgroup$ – RobChem Jan 13 '15 at 21:54
2
$\begingroup$

but what sort of problems?

Where two bodies move in the influence of their mutual forces only(that act in pairs and net force on system is zero, so that COM(Centre Of Mass) remains at rest), no external forces! Actually:

"When two bodies are moving in translational,rotational or vibrational motion only under their mutual interaction, then for simplified analysis, we can consider one body at rest and analyze the motion of other body wrt first body and change the inertial mass of moving body to a new value called reduced mass given as."

Is it so that the two body system can be treated as one body or is it to simplify calculations regarding how the two bodies act upon each other?

It is such that we are treating the motional parameters of one fixed(i.e. at rest and at rest forever(until something weird happens)). Suppose you are travelling in a train, when you see an oppositely coming train, it seems faster since when you observe, in your reference frame(i.e., wrt you) the speed of the train becomes $\vec v_{other-train}-\vec v_{your-train}$ Since speed of your train has opposite sign as the speed of other train, it rather becomes: $v_1-(-v_2)=v_1+v_2$ and thus speeds add up and it looks fast. Similarly in case of moving in same direction, it appears slow as speeds subtract.

This often helps because sometimes problems are solved easily since one equation is always reduced.

Vibrational Example

suppose two bodies are connected by spring and vibrating, then both are constantly moving, If they are following SHM (Simple Harmonic Motion), then to calculate time period we need to account both of them, but it could be done easily if we treat one at rest and look at the other body while sitting on the body we wish to have at rest. That turns out to be: $$T=2\pi\sqrt{\frac{\mu}k}=2\pi\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}$$

If we were doing it the traditional way: $$k_1=kl/l_1=kl(m_1+m_2)/(m_2l)\\k_1=k(m_1+m_2)/m_2\\T=2\pi\sqrt{m_1/k_1}=2\pi\sqrt{m_1m_2/(k(m_1+m_2))}$$

Translational example

Try to fin the separation velocity between two planets at a distance $r$ when their distance reduces to $r/2$ using both methods.

Rotational Motion Example

Try to find the total kinetic energy of two particles at a distance l rotating about their COM due to their mutual gravitational attraction without any external force at angular speed $\omega$ by both methods.

$\endgroup$
1
$\begingroup$

You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a single fictitious particle. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass.

In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.

In step 4, you're calculating the mass $\mu$ of the fictitious particle.

Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles.

The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$ This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be $$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$ where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a second fictitious particle with total mass $M=m_1+m_2$ and coordinates $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$).

In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.