Degrees of freedom of a two particle rigid system

We have two particles and the distance between them is fixed, let's suppose we know the coordinates of one particle (2,1) and other particle (x,2). So using distance formula (let's suppose the fixed distance between the particle is 4) we can find $x$,so we don't need four coordinates to specify the system, we just need 3, we can find the other. So no of independent degrees of freedom we have is 3. (let's suppose the motion is in a plane).

But solving the equation we get two solutions, I don't know which one to take?

Here is quote from wiki to clearfy what I mean.

If two particles in space are constrained to maintain a constant distance from each other, such as in the case of a diatomic molecule, then the six coordinates must satisfy a single constraint equation defined by the distance formula. This reduces the degree of freedom of the system to five, because the distance formula can be used to solve for the remaining coordinate once the other five are specified.

but I get two solutions for after solving the equation.

• Can you show some of your work? It seems like you solved the equations for the two particles as if they were independent. If you account for the bound condition you should only have the three degrees of freedom, but is hard to point your mistake without the actual work you did. – rmhleo Jul 25 '15 at 11:06
• @rmhleo :Look at the ans,thats what I meant but given the fixed distance and 3 coordinates I should know exact positions of the particles,but there are two possible situation. – Paul Jul 25 '15 at 11:24

The system has three degrees of freedom, but that doesn't mean that any three numbers necessarily specify the full state of the system. If you have $$x_1$$, $$y_1$$, and $$y_2$$, then as you have noticed there are two possible solutions to $$x_2$$. But you can fully specify the system with $$x_1$$, $$y_1$$ and an angle specifying which direction the second particle is in relative to the first.

You know the other particle has $y = 2$, but you don't know what $x$ is. However you do know the distance, $d$.
The trouble is that there are two possible positions for the particle that have $y = 2$ and a distance $d$ from the first particle. I've drawn these two positions on the graph.
• @dushyanth You can fully describe the system with three numbers: for instance $(x,y)$ of particle 1 and a single angle describing which direction particle 2 is in. – Chris Dec 16 '19 at 5:05
Generally speaking, a two-body system with coordinates $\bf r_1$ and $\bf r_2$ will have the following Hamiltonian $\cal H = \frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2} + $$V(\bf r_1, \bf r_2). And normally you would have eight differential equations to solve (in Hamiltonian formulation, only four in Newtonian mechanics) for k=1,2 and i=x,y:$$\dot p_{ki} = - \frac{\partial \cal H}{\partial r_{ki}}\dot r_{ki} = \frac{\partial \cal H}{\partial p_{ki}}$$When adding the constraint d = \lvert \bf r_2 - r_1 \lvert we can change variables in several different ways, one of them can be:$$r_{x2}=r_{x1}+d \cos \thetar_{y2}=r_{y1}+d \sin \thetap_{x2}=p_{x1}+d \cos \thetap_{y2}=p_{y1}+d \sin \theta$$Where$\theta$is the angle between the positive$x$axis and vector$\bf d = \bf r_2 - \bf r_1$. Now these equations, when put in the system of differential equations above, will reduce to only six, where the unknown variables are the the coordinates and momenta$m_1$and$\theta$and$\dot \theta$in the Hamiltonian formulation (or in Newtonian mechanics just two equations for position of$m_1$and one equation for$\theta\$).