2
$\begingroup$

I am a bit confused about the rotational motion in molecules. Assuming the bond length is constant, the motion can be described as a rigid rotor. In the center of mass frame the energies are given by BJ(J+1) and the wavefunctions are spherical harmonics. However when we measure the energies or the angular momenta, we do it in lab frame. So I am a bit confused. Is the formula for the energy the same both in lab and CM frame? And if not, what is the formula in the lab frame? Also, is the wavefunction the same in both frames or, in other words, is the angular moment of the molecule the same in both frames. Actually I am a bit confused about how is the angular momentum defined in the CM frame. Isn't the molecule stationary in that frame? Yet the wavefunctions in the CM frame (spherical harmonics) do show a clear angular momentum dependence. Can someone help me clarify these things? Thank you!

$\endgroup$
1
$\begingroup$

The molecule is not stationary in the center-of-mass frame. Any rigid body dynamics can be separated into motion of the center of mass, and motion about center of mass. The second includes rotational motion, which is what you are considering. Only translational motion is absent in the CM frame. The question of whether angular momentum, or energy is the same in the lab frame depends on what other kinds of motion the molecule has in the lab frame. Does it have translational degrees of freedom, for example? If not, the energy is the same in both frames. The angular momentum here is intrinsic angular momentum, due to rotations within the molecule, and intrinsic spins of the constituents. This too should be the same in both frames, unless you are considering some odd situation like the molecule revolving around some axis in the lab frame.

$\endgroup$
9
  • $\begingroup$ Thank you for your reply! There is no translation or other weird rotations other than around a fix point (in the lab frame). But I am still confused. Classically if I hold a ball and I rotate with it, the ball has 0 energy according to me, but according to a stationary observed, the energy is Iw^2/2. Why in the case I mentioned the energy is the same but classically it would not be (I assume I have a missunderstanding of the way the frames are defined). Similarly, according to me the ball is not rotating, so the angular momentum is zero, but according to a stationary observer it is not. $\endgroup$ – BillKet Mar 22 '20 at 23:32
  • $\begingroup$ the observer does not rotate, even in the center of mass frame. reference frames should be kept inertial, and a rotating frame of reference is just awkward to work with. the usual expressions for energies like $I\omega^2/2$ are all defined for inertial frames, and need to be modified for a rotating frame $\endgroup$ – NewUser Mar 22 '20 at 23:46
  • $\begingroup$ I am not sure how the CM frame is defined here actually. Here: kth.se/social/upload/5176d9b0f276543c2c2bd4db/CH5.pdf they analyze the problem for nuclei (a bit more complicated but same idea) and they define an " intrinsic axis system". I initially thought it was the CM frame but I am not sure anymore. From what I understand, the frame they define is rotating together with the object and they solve the Schrodinger equation in that frame, yet the energies they get are the same as the one in the lab frame. What am I missing? $\endgroup$ – BillKet Mar 23 '20 at 0:44
  • $\begingroup$ Also, given that the object is just rotating around a fixed point and for molecules and nuclei that point is usually the center of mass, isn't the CM (defined in an inertial way as you mentioned) and the lab frame the same? Why would they make a clear distinction between them in the analysis? That's why I assumed that the body frame is rotating with the body (hence non-inertial). But again if that's the case, I am confused about the energies. Thank you! $\endgroup$ – BillKet Mar 23 '20 at 0:47
  • $\begingroup$ Okay, so the frame here is indeed rotating. The motion of the molecule is divided into two parts, the rotation about its intrinsic axis (K), and the rotation of its intrinsic axis (M). Only the first is seen in the intrinsic frame. The expressions for energy are given in the lab frame, since it is an inertial frame. The molecule is definitely at rest in the intrinsic frame, but the energy would be different from the lab frame, since it is a non-inertial frame. $\endgroup$ – NewUser Mar 23 '20 at 1:40
0
$\begingroup$

Like NewUser mentioned, the motion can always be broken down into motion of centre of mass and motion about centre of mass. In our case, we interact with the system using light. And this causes transitions in the energy levels.

The system factors into one continuous energy spectrum corresponding to the kinetic energy of the free centre of mass and one discrete energy spectrum corresponding to the rigid rotor. The former causes light to scatter and the latter causes absorption. So if you look at absorption spectra, you'll find the signature of the discrete rotational energy levels.

$\endgroup$
3
  • $\begingroup$ Thank you for your reply! What I am confused about is this CM frame. If the CM frame was fixed and inertial, I would understand. But the body frame they use is not the CM, but a frame rotating with the body (see the link I posted in one of the comments). And if it is rotating with the body, it shouldn't have any rotational energy associated to it i.e. if 2 objects rotates at the same time, in the same way, they will appear stationary to each other, hence their energy according to each other is 0. This is how I understand this body frame, too, as something rotating with the body. $\endgroup$ – BillKet Mar 24 '20 at 7:14
  • $\begingroup$ You shouldn’t look at the bodies. Rather, they are rotating about the centre of mass. In the case of two equal bound masses, CM is at the mid point. The rotation is about this point. $\endgroup$ – Superfast Jellyfish Mar 24 '20 at 7:21
  • $\begingroup$ I know that, but this is not what they use in the derivation I am reading (it is for a nucleus not a molecule, but the idea is the same): kth.se/social/upload/5176d9b0f276543c2c2bd4db/CH5.pdf As you can see, they define a frame that rotates with the body, (which obviously can't be the CM frame) and they solve the equations in that frame. $\endgroup$ – BillKet Mar 24 '20 at 7:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.