# Degrees of freedom

ajayguhan
A particle has 3 degrees of freedom, therefore N particles have 3N degrees of freedom. But a two particle whose distance between them is constant has 5 degrees of freedom instead of 6. I know that the fixed distance is constarin so it reduced the degrees of freedom, but why does the constarin reduces the degrees of freedom?

## Answers and Replies

Mentor
Because if I tell you the coordinates of one particle, I only need to tell you two bits of information (usually two angles) to be able to know where the other particle is, as the distance is fixed.

So the total information needed is 3 coordinates and 2 angles, hence 5 degrees of freedom.

ajayguhan
3 co ordinates are cartesian while other two is polar doesn't it bother? I mean the co ordinates should of the same type or can they vary?

Mentor
3 co ordinates are cartesian while other two is polar doesn't it bother? I mean the co ordinates should of the same type or can they vary?
I doens't matter. You don't even have to have cartesian coordinates to start with. The same situation can be described using spherical coordinates for the first particle. You just usually use the simplest representation for the problem: the physical equations are equally valid whichever system you use. It just happens that most of the time, some representations lead to simpler equations.

It is common for n-body problems to separate the centre of mass motion from the relative motion of the bodies. If the centre of mass is rotating, it makes more sense to describe it using spherical coordinates, while the relative motion can be described with cartesian coordinates or spherical coordinates or cylindrical coordinates or anything else.

Also, some constraints are difficult to work with in certain coordinate systems. In your example, keeping the distance fixed between the two particles, I wouldn't even know how to write the equations of motion preserving that distance in cartesian coordinates, whereas they are very simple in spherical coordinates.

ajayguhan
Got it, thanks.