Difference between Generalised co-ordinates and degrees of freedom

[Aniket1]

What is the difference between Generalised co-ordinates and Degrees of freedom in classical mechanics? I know that they are not equal when we have non-holonomic equations of constraints. But I don't know why.

 

[Studiot]

In general for holonomic constraints, g = 3N-c

Where g is the number of independent generalised coordinates sufficient to specify (mechanically) the system of N particles and c is the number of independent constraint equations.

If the constraints are not specified as equations then the above is non holonomic and the above does not hold. Further conditions of constraint are needed to specify the system.

A simple example is a system of particles constrained to the surface of a sphere

R²=x²+y²+z²

This is holonomic.

But if the particle is constrained to the atmosphere and space above the sphere the system is not.

R²<x²+y²+z²

 

[ZealScience]

Firstly, degrees of freedom is the NUMBER of general coordinates required to describe the EOM. Simply saying "Generalised coordinates" is a bit vague.

Secondly, not all the coordinates are required. Under most instances, we assume that bodies are rigid, meaning that particles are constrained to each other, and we only need to consider the body as a whole. Otherwise we would need a set of coordinates to describe each molecule (if the molecules are rigid of course).

In addition to that, there are constraint forces which furthermore eliminates degree of freedom. (e.g if a particle is on the table, with no extra vertical forces other than gravity, then you only need two coordinates rather than three, when the particle is free to move everywhere).

Usually, just try to count how many general coordinates you need at least to describe the free body, then take away the holonomic constraints, as you have mentioned, so that you don't have to deal with number on the order of Avogadro constant.

 

[Aniket1]

In general for holonomic constraints, g = 3N-c

Where g is the number of independent generalised coordinates sufficient to specify (mechanically) the system of N particles and c is the number of independent constraint equations.

If the constraints are not specified as equations then the above is non holonomic and the above does not hold. Further conditions of constraint are needed to specify the system.

A simple example is a system of particles constrained to the surface of a sphere

R²=x²+y²+z²

This is holonomic.

But if the particle is constrained to the atmosphere and space above the sphere the system is not.

R²<x²+y²+z²

I think the example you have given is for bilateral and unilateral constraint relations (repsectively) . Nonholonomic relations are those which do not depend on velocity explicitly.

 

[Aniket1]

Firstly, degrees of freedom is the NUMBER of general coordinates required to describe the EOM. Simply saying "Generalised coordinates" is a bit vague.

Secondly, not all the coordinates are required. Under most instances, we assume that bodies are rigid, meaning that particles are constrained to each other, and we only need to consider the body as a whole. Otherwise we would need a set of coordinates to describe each molecule (if the molecules are rigid of course).

In addition to that, there are constraint forces which furthermore eliminates degree of freedom. (e.g if a particle is on the table, with no extra vertical forces other than gravity, then you only need two coordinates rather than three, when the particle is free to move everywhere).

Usually, just try to count how many general coordinates you need at least to describe the free body, then take away the holonomic constraints, as you have mentioned, so that you don't have to deal with number on the order of Avogadro constant.

What I wanted to know is : Why is the number of "Generalised co-ordinates" different than the number of degrees of freedom for systems having non-holonomic constraints even when they have the same definition?
Books say: For holonomic cases both are equal to 3N-k (N=number of particles and k=number of constraint relations)
For non-holonomic cases Gen co-ordinates=3N-k
DOF= 3N - k - k' (k'=number of nonholonomic constraints.

 

[Studiot]

That is a slightly different question from the one you originally asked. I did wonder if that was what you really meant.

Start with the 3N degress of freedom (in 3D), which I assume you understand.

That means you need 3N pieces of information (the value of 3N variables) to fully determine the system.

If you can write 3N simultaneous equations or otherwise specify these values eg by the old standby x₃ = 0 (given) then you can solve the system.

If you can introduce other equations from other sources then you can reduce the numberrequired from 3N.

But they have to be solvable equations, simultaneous with the rest of the set.

An inequality, such as I gave in my second previous example, does not satisfy this requirement and so does not qualify in the reducing set.
It is, nevertheless, a constraint condition.

So the equation that I originally offered could be rephrased:

The number of coordinates required = 3N minus the number of other equations you can dig up.

 

[Aniket1]

That is a slightly different question from the one you originally asked. I did wonder if that was what you really meant.

Start with the 3N degress of freedom (in 3D), which I assume you understand.

That means you need 3N pieces of information (the value of 3N variables) to fully determine the system.

If you can write 3N simultaneous equations or otherwise specify these values eg by the old standby x₃ = 0 (given) then you can solve the system.

If you can introduce other equations from other sources then you can reduce the numberrequired from 3N.

But they have to be solvable equations, simultaneous with the rest of the set.

An inequality, such as I gave in my second previous example, does not satisfy this requirement and so does not qualify in the reducing set.
It is, nevertheless, a constraint condition.

So the equation that I originally offered could be rephrased:

The number of coordinates required = 3N minus the number of other equations you can dig up.

I haven't changed my question. I think you are getting confused between non-holonomic constraint relation and unilateral constraint relation. Non-holonomic constraint relations, too ARE equalities (with velocity terms). But still they don't reduce the number of generalised co-ordinates. They however, do decrease the degrees of freedom. I want to know why..
http://en.wikipedia.org/wiki/Generalized_coordinates#Holonomic_constraints

 

[Studiot]

Pity you aren't paying any attention to what I was saying.

 

[andrien]

here in this thread I have explained some issues regarding non holonomicity and non integrable relation
https://www.physicsforums.com/showthread.php?t=647529&highlight=holonomic
non holonomic system contain atleast one non integrable relation.These relations reduces the degree of freedom owing to non integrability.The example of a sphere on rough surface in the thread provided above will illustrate it.

 

[Aniket1]

here in this thread I have explained some issues regarding non holonomicity and non integrable relation
https://www.physicsforums.com/showthread.php?t=647529&highlight=holonomic
non holonomic system contain atleast one non integrable relation.These relations reduces the degree of freedom owing to non integrability.The example of a sphere on rough surface in the thread provided above will illustrate it.

Thanks !