Count the degrees of freedom using the minimum number of coordinates required to specify a position.

Let's say there are 3 atoms, 3 coordinates for center of mass, and 3 for an axis. But we need 6 coordinates to specify a line in a three dimensional space: $(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$. Which line do we choose? Even if we have a line, how do we use the remaining three coordinates to specify the atoms?

my reference are this image from Chemical Reactor Analysis and Design Fundamentals 2ed Edition by JB Rawlings, page195

  • $\begingroup$ So the "line"? Are you talking about internal coordinates (Z-matrix)? $\endgroup$
    – DSVA
    Jun 14, 2017 at 20:19
  • 4
    $\begingroup$ N atoms need 3N coordinates to specify their positions as x, y, z. But in a molecule atoms are attached to one another so 3 coordinates are needed for translational movement of the whole molecule and 3 for whole molecule rotation (about x, y, z axes) leaving 3N-6 degrees of freedom to describe vibrations. Is this what you are asking about? $\endgroup$
    – porphyrin
    Jun 14, 2017 at 21:50

1 Answer 1


You have N atoms, each atom can be described by a set of cartesian coordinates (x,y,z). So for each atom we have 3 degrees of freedom which means we end up with 3N degrees of freedom. If we look at an isolated structure without any external field or anything like that then only the relative position of each atom to the others matters. So we can get rid of some degrees of freedom. Translating every atom in the same way into x, y or z direction won't change anything, so we can reduce the degrees of freedom by 3. Rotating it around x, y or z doesn't change anything either, so we are reducing by another 3, so we end up with 3N-6.

  • $\begingroup$ Is there any chance you could edit the question to make it clearer what it's asking? Right now I don't really get what it's trying to say. $\endgroup$ Jun 14, 2017 at 20:15
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    $\begingroup$ @orthocresol comment in the wrong place? $\endgroup$
    – DSVA
    Jun 14, 2017 at 20:15
  • $\begingroup$ Nope, I wanted to ask you because you answered, so I assumed you understood it. $\endgroup$ Jun 14, 2017 at 20:15
  • $\begingroup$ @orthocresol ah got it. I don't think I'm able to do that. All I can "understand" that he is trying to determine the degrees of freedom in a way I don't understand and I don't even know why he's trying this way. I basically answered to the topic and explained how to get to the 3N-6 in (one) correct way. $\endgroup$
    – DSVA
    Jun 14, 2017 at 20:18
  • $\begingroup$ Heh, fair enough. Hopefully OP clarifies. $\endgroup$ Jun 14, 2017 at 22:13

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