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Trying to understand reduced mass

by DiracPool
Tags: mass, reduced
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DiracPool
#1
Mar13-13, 08:42 PM
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I understand that the reduced mass is used to simplify a two-body problem into a one body problem, so that instead of calculating two masses rotating around a common center as in the earth and moon, you can treat the earth as stationary, and then used a "reduced" mass for the moon in the calculation.

My question is how is m1m2/m1+m2 calculated to derive the reduced mass. I'm not able to follow the arguments in my readings. Can someone give me a simple, straightforward derivation along with a conceptual "intuition" as to why this works?
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DrClaude
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Mar14-13, 07:11 AM
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Quote Quote by DiracPool View Post
I understand that the reduced mass is used to simplify a two-body problem into a one body problem, so that instead of calculating two masses rotating around a common center as in the earth and moon, you can treat the earth as stationary, and then used a "reduced" mass for the moon in the calculation.
The part in bold is incorrect. What is get is the relative motion of the two bodies, which follows the same equation as is you had a single body of mass "reduced mass" rotating around a point which is the center of mass of the system.

Quote Quote by DiracPool View Post
My question is how is m1m2/m1+m2 calculated to derive the reduced mass. I'm not able to follow the arguments in my readings. Can someone give me a simple, straightforward derivation along with a conceptual "intuition" as to why this works?
The trick is to think of center of mass motion (the two bodies moving as one system) and relative motion.
jtbell
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Mar14-13, 08:02 AM
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Quote Quote by DiracPool View Post
I'm not able to follow the arguments in my readings.
Can you point us to a derivation that you don't understand, and tell us where you get stuck? Then someone may be able to "unstick" you.

If someone simply posts a derivation, they run the risk of repeating something that you've seen already and don't understand.

DiracPool
#4
Mar15-13, 02:28 AM
P: 606
Trying to understand reduced mass

@DrClaude

The part in bold is incorrect. What is get is the relative motion of the two bodies, which follows the same equation as is you had a single body of mass "reduced mass" rotating around a point which is the center of mass of the system.
Here is the reference I used to make the statement you said was incorrect, Dr...

When two bodies revolve around each other, they revolve around the shared center of mass between them. For example, the Earth and Moon revolve around a point between their centers. The Moon causes the Earth to wobble. This complicates equations, for example the solution of orbital period. This same problem comes up with regard to an electron orbiting around a nucleus. The solution to this problem, called the “reduced mass,” therefore applies to both the very large in nature as well as the very small. The solution is to find a system that has the same frequency solution but is simpler to calculate. That simpler solution is to pretend the larger body is stationary in the center and the smaller body orbits with a “reduced mass” at the same distance from the larger object as in the unmodified problem. The two-body problem then reduces to a one-body problem, focused solely on the smaller body’s orbit.
http://www.ehow.com/how_5968910_calc...uced-mass.html

@jtbell

I'm trying to follow the derivation on the wiki page under the "Newtonian mechanics" heading. I get all the math but...why does a-rel = a1-a2 instead of say, a2-a1? Also, why is the result F12/Mred the result we are looking for? I get the derivation, I just don't understand how it works to turn the two body problem into a single body problem. So when they say at the end...

So we conclude that body 1 moves with respect to the position of body 2 as a body of mass equal to the reduced mass
http://en.wikipedia.org/wiki/Reduced_mass


I'm not catching their drift. Any help would be appreciated.
DrClaude
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Mar15-13, 03:49 AM
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The key word in that quote is "pretend." This is not rigourous. But indeed, if one of the objects is much more massive than the other, then the center of mass of the system will be located close to the middle of the massive object, and therefore you can imagine that it is as if you had an object of mass [itex]m_\mathrm{red}[/itex] orbiting in a closer orbit around the massive object.

I've just looked at the Wikipedia page, and I understand completely why you are confused. Maybe this derivation, taken from V. D. Barger and M. G. Olsson, Classical Mechanics: A Modern Perspective, 2nd ed., McGraw-Hill (New York), 1995, will be more helpful.
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DiracPool
#6
Mar15-13, 04:07 AM
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Quote Quote by DrClaude View Post
The key word in that quote is "pretend." This is not rigourous. But indeed, if one of the objects is much more massive than the other, then the center of mass of the system will be located close to the middle of the massive object, and therefore you can imagine that it is as if you had an object of mass [itex]m_\mathrm{red}[/itex] orbiting in a closer orbit around the massive object.

I've just looked at the Wikipedia page, and I understand completely why you are confused. Maybe this derivation, taken from V. D. Barger and M. G. Olsson, Classical Mechanics: A Modern Perspective, 2nd ed., McGraw-Hill (New York), 1995, will be more helpful.
Thank you for that derivation, DrClaude
jtbell
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Mar15-13, 06:26 AM
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Quote Quote by DiracPool View Post
why does a-rel = a1-a2 instead of say, a2-a1?
I expect that it doesn't make any difference in the end. That is, you could just as well start with arel = a2 - a1, and along the way you might have 1 and 2 switched around in some places, or maybe some + and - signs switched, but the differences will cancel out eventually. Try it and see what happens!
DaleSpam
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Mar15-13, 07:45 AM
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Quote Quote by DiracPool View Post
why does a-rel = a1-a2 instead of say, a2-a1?
That just corresponds to a 180º rotation. So if in the inertial frame one object is at (-1,0,0) and the other object is at (3,0,0) then you can either adopt the convention that their relative position is (4,0,0) or (-4,0,0). Either way works fine, but your accelerations will all be 180º "out of phase".


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