Return to Question

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

• $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

• Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

• $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

• Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

• $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

• Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

• $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

• Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle around this line as axis, $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.

The dumbbell consists of two point masses at $(x_1, y_1, z_1)$ and $(x_2,y_2,z_2)$ separated by a distance $l$. The constraint equation is: \begin{equation} (x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2=l^2 \end{equation}

Had there been no constraint equation, the number of degrees of freedom would have been $3N = 3 \times 2 = 6$. But now because of the constraint equation, only 5 out of the 6 coordinates are independent, the last coordinate is determined by the constraint equation.

$\mathbf{1}.$ But can we really know the $6$th coordinate? The constraint equation will give two values for the last coordinate, because there will be quadratic equation in the unknown coordinate. Suppose $(x_1, y_1, z_1) = (1, 2, 3)$ and $(x_2, y_2, z_2) = (x_2, 3, 4)$ and $l = \sqrt6$, then we have two values for $x_2 = 3, -1$. The last coordinate is not uniquely determined - so is the number of degrees really $5$?

$\mathbf{2}.$ There is a new suggestion: $(X_c, Y_c, Z_c, \theta, \phi)$ where $(X_c, Y_c, Z_c)$ are the coordinates of the center-of-mass and $\theta$ and $\phi$ are the polar and the azimuthal angles. If this is the effect of the constraint relation, then it doesn't really eliminate one redundant coordinate, but replaces the entire set, with one less coordinate. Am I right?

Edit 1: May be I should modify my question after getting the answers. I see why $(X_c, Y_c, Z_c, \theta, \phi)$ are legitimate Dof's of the dumb-bell, I never had any problem with that.

Please consider now any rigid body in general. It has $6$ Dof's, two examples of which are mentioned below:

• $(X_c, Y_c, Z_c, \alpha, \beta, \gamma)$: $(X_c, Y_c, Z_c)$ are CoM co-ordinates, and $(\alpha, \beta, \gamma)$ are the Eulerian angles (or, the pitch, roll and the yaw) of the rigid body.

• Take a line through any point $(x_1, y_1, z_1)$ inside the rigid body, fix a line through the point using two angles $\theta, \phi$ (or fix a unit vector along the line, using two components). The last $Dof$ is obtained by specifying an angle $\theta'$ that a point not on this line makes with, say the horizontal.

My question is more about the constraint relations. There are $3N$ co-ordinates to begin with, and each constraint relation eliminates one redundant co-ordinate. If this is true, because a rigid body has $6$ Dof's, after considering all the constraint relations (minus all the redundant constraints), I will have only $6$ (out of the set of $3N$ co-ordinates to start with). So can I have the Dof's to be any $6$ co-ordinates out of the initial $3N$, for example $(x_1, y_2, z_3, x_4, x_5, z_6)$ - is that sufficient to locate the rigid body?

If not, the constraint relations only lowers the number of co-ordinates, doesn't eliminate (remove) though. The reduced set of co-ordinates doesn't have to be any of those before reduction, it could be any, just one less in number.