Degrees of freedom

Jul 2013
9
0
I am a mathematician refreshing on my classical mechanics.

This is not a homework problem and I am not taking a class in classical mechanics right now.

I am using L Hand and J Finch, a textbook on analytical mechanics (based on a series of lectures from Cornell).


Question: Show that a rigid body with three or more mass points has six degrees of freedom.

Attempt: of course we are talking about a 3 dimensional eucldiean space.
The formula that the book provides: N = 3M - j
Where M is the number of the particles. j is the number of constraints. N is the number of degrees of freedom.

For now the coordinates only represent the position. They do not mention momentum yet. So for instance, a rigid body composed of two point masses has 5 degrees of freedom. since there is one constraint connecting the two point masses and there are 6 coordinates in total.

I attempted to use combinatorics to generate a formula for M point masses (M is greater than or equal to 3). N = 3M -MC2 (M choose 2).
It clearly fails.

I am guessing that some constraints are dependent on others.
I tried an example of 5 point masses. there are 5C2 = 10 constraints. And I calculated 3*5 - 10 = 5 degrees of freedom. There is something wrong.

I need some help to PHYSICALLY understand how constraints are dependent on each other and how to mathematically derive a formula.

Thank you
 

ChipB

PHF Helper
Jun 2010
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Morristown, NJ USA
I believe the issue is that for a rigid body in 3D space once you consider more than 3 points in that body all constraints are redundant. If you know the positions of points A, B and C, then the positions of all other points is on teh rigid bosy already known. Consider an example: suppose you have a cube and you label three of the corners (pick any three) as A, B and C. You can move the cube laterally, forward and back, and up and down so that point A is defined by it's (x,y,z) coordinates. Then you can rotate the cube about the x and y axes to fix point B - that's two additional degrees of freedom for it - then finally rotate about the z-axis to fix point C. Hence there are a total of 6 degrees of freedom for the cube. Once you have defined values for those 6 degreees of freedom all points of the cube are locked in - there is no additional flexibility to move some other point D without disturbing the positions of A, B or C. Hence even though the cube has an infinite number of points it has only 6 degrees of freedom. The value for J in your formuka is:

J=0 for M=1
J=1 for M=2
J=3 for M=3
J=3(M-2) for M >= 3.
 
Jul 2013
9
0
I believe the issue is that for a rigid body in 3D space once you consider more than 3 points in that body all constraints are redundant. If you know the positions of points A, B and C, then the positions of all other points is on teh rigid bosy already known. Consider an example: suppose you have a cube and you label three of the corners (pick any three) as A, B and C. You can move the cube laterally, forward and back, and up and down so that point A is defined by it's (x,y,z) coordinates. Then you can rotate the cube about the x and y axes to fix point B - that's two additional degrees of freedom for it - then finally rotate about the z-axis to fix point C. Hence there are a total of 6 degrees of freedom for the cube. Once you have defined values for those 6 degreees of freedom all points of the cube are locked in - there is no additional flexibility to move some other point D without disturbing the positions of A, B or C. Hence even though the cube has an infinite number of points it has only 6 degrees of freedom. The value for J in your formuka is:

J=0 for M=1
J=1 for M=2
J=3 for M=3
J=3(M-2) for M >= 3.

Thank you for your prompt reply.

Please confirm if I have a clear understanding:

The way I see it is that we have a 3 dimensional euclidean space and we need three (linearly independent) vectors to define a basis (points A, B and C) and any vector D can be expressed as a combination of A,B and C. Therefore the motion of the particle D is uniquely determined by the motion of those 3 particles (which, by the way, happen not to be unique, I believe.)

* Can you please tell me how did you derive the formula for J for m >= 3?
For instance for m = 3 there are three constraints. The way I visualize them are the rigid bonds that connect pairs of point masses.
Now for m =5 we get 9 constraints. But we have 10 rigid bonds. How is that explained? How is that translated "geometrically"?

Thanks again
 

ChipB

PHF Helper
Jun 2010
2,369
294
Morristown, NJ USA
I derived my formuila for J empirically, so that M = 6 for any value of N>=3.

I agree that for M=3 there are three constraints. Perhaps its best to think of these constraints not as distances but rather using spherical coordinates for the direction and magnitude of the relationships between the points. Thus given the (x,y,z) position of point A, point B's distance from A is set by the nature of the rigid body (ie the length of the edge of the cube in my previous example) and its direction from A can be defined by two angles in the spherical coordinate system. Thus point A and B together have 5 degrees of freedom. Point C's position can then be defined as its distance from A and B (already pre-defined) and the angle of the ABC plane as it is rotated about the AB line. Thus it adds one more degree of freedom, for a total of 6. For M = 4 there are an additional 3 constraints, as the direction of point D from A is locked by angles relative to the AB line and the AC line and its distance from A. Of course point D could also be considered to be locked to point B instead of A, but there is no need to have values as measured from both and B. Thus if you have the measurements of D from A then getting measurements of D from B is redundant, as this extra infomation is not needed to define where point D is in the solid. For M = 5 point E is also locked to point A by its angles and distance, so it too has three constraints. Again, you could also use angles and distance from points B, C, or D, but you only need one set of values to define point E. Thus moving from 4 points to 5 points adds exactly three independent constraints. Stated another way, using your concept of rigid bonds between points - when you add point E you only need to know the length of bonds to three of the other four points to define its position; the length to the 4th point is redundant.
 
Last edited:
Jul 2013
9
0
I derived my formuila for J empirically, so that M = 6 for any value of N>=3.

I agree that for M=3 there are three constraints. Perhaps its best to think of these constraints not as distances but rather using spherical coordinates for the direction and magnitude of the relationships between the points. Thus given the (x,y,z) position of point A, point B's distance from A is set by the nature of the rigid body (ie the length of the edge of the cube in my previous example) and its direction from A can be defined by two angles in the spherical coordinate system. Thus point A and B together have 5 degrees of freedom. Point C's position can then be defined as its distance from A and B (already pre-defined) and the angle of the ABC plane as it is rotated about the AB line. Thus it adds one more defree of freedom, for a total of 6. For M = 4 there are an additional 3 constraints, as the direction of point D from A is locked by angles relative to the AB line and the AC line and its distance from A. Of course point D could also be considered to be locked to point B instead of A, but there is no need to have values as measured from both and B. Thus if you have the measurements of D from A then getting measurements of D from B is redundant, as this extra infomation is not needed to define where point D is in the solid. For M = 5 point E is also locked to point A by its angles and distance, so it too has three constraints. Again, you could also use angles and distance from points B, C, or D, but you only need one set of values to define point E. Thus moving from 4 points to 5 points adds exactly three independent constraints. Stated another way, using your concept of rigid bonds between points - when you add point E you only need to know the length of bonds to three of the other four points to define its position; the length to the 4th point is redundant.
It is perfectly understood.
I just need to give it some time to get used to it.

Thank you very much.
 
Oct 2013
1
0
Nicely presented information in this post, I prefer to read this kind of stuff. The quality of content is fine and the conclusion is good. Thanks for the post.
 

Pmb

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