For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### The Rigid Rotator I

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Possible Confusion from the Previous Discussion
- Rotation of a Single Mass Around a Fixed Center
- Rotation of a Single Mass Around a Fixed Center
- Angular Velocity
- Rotational Inertia
- Rotational Frequency
- Kinetic Energy for a Linear System
- Kinetic Energy for a Rotational System
- Rotating Diatomic Molecule
- Rotating Diatomic Molecule: Part 1
- Rotating Diatomic Molecule: Part 2
- Rotating Diatomic Molecule: Part 3
- Hamiltonian of the Rigid Rotor

- Intro 0:00
- Possible Confusion from the Previous Discussion 0:07
- Possible Confusion from the Previous Discussion
- Rotation of a Single Mass Around a Fixed Center 8:17
- Rotation of a Single Mass Around a Fixed Center
- Angular Velocity
- Rotational Inertia
- Rotational Frequency
- Kinetic Energy for a Linear System
- Kinetic Energy for a Rotational System
- Rotating Diatomic Molecule 19:40
- Rotating Diatomic Molecule: Part 1
- Rotating Diatomic Molecule: Part 2
- Rotating Diatomic Molecule: Part 3
- Hamiltonian of the Rigid Rotor 36:48
- Hamiltonian of the Rigid Rotor

### Physical Chemistry Online Course

### Transcription: The Rigid Rotator I

*Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.*0000

*Today, we are going to talk about the rigid rotator.*0004

*Let us just go ahead and get started.*0007

*Now, before I actually start discussing the rigid rotator, I want to go back and*0010

*actually clear up a possible fuzzy point from our previous discussions.*0014

*Let us go ahead and write that down.*0020

*Before I discuss the rigid rotator, I would like to go back and clear up the possible confusion,*0028

*there may not be any confusion but just in case, I like to make sure that it is pretty clear.*0051

*A possible confusion from the previous discussion being the harmonic oscillator.*0056

*Let me go ahead and draw out this parabolic thing for the harmonic oscillator, the energy levels.*0069

*Just put a couple of them here.*0084

*We have got R = 1, R = 2, R = 3, R = 4, R being the quantum number for the harmonic oscillator and*0087

*these being the particular energy levels.*0096

*I’m going to have a 0 point energy so let us go ahead and this is going to be R=0, 1, 2, and 3.*0100

*Some students are a little fuzzy about what this parabola is and*0111

*why are we using this parabola to represent these different energy levels, about what this parabola is.*0132

*They sometimes think the motion of the particle itself, of the mass of the particle or the mass is somehow parabolic.*0144

*And that is not true.*0168

*It is really easy to lose your way in what it is that is going on in the big picture.*0169

*There is so much math going on, there are so much derivation of equations, manipulation of equations,*0174

*that oftentimes it is very simple to lose your way in terms of remembering what are we trying to do and*0180

*what is it that we are actually trying to elucidate, what are we looking for.*0186

*Let us talk about what is going on.*0191

*They think that the motion of the particle is parabolic.*0193

*The answer is no, the motion of the particle is not parabolic.*0195

*Let me go ahead draw and a line here.*0199

*This is a 0 point, the equilibrium position of the harmonic oscillator.*0203

*You remember the harmonic oscillator is just a particle that moves back and forth like this.*0206

*Let us go ahead and put A and A, this is A1, A1, or – A1.*0211

*For the first energy, the amplitude is that far.*0220

*We put a little bit more energy into the system, the amplitude is going to increase.*0225

*We know that, it will just go a little bit farther .*0229

*A3 and – A3, they correspond to these energy levels.*0232

*The more energy you put in, the bigger the amplitude.*0238

*That is what is actually happening here.*0242

*The motion of the particle is harmonic, it just goes back and forth like this.*0246

*As it deviates from the equilibrium position, what this parabola describes*0249

*is the amount of potential energy that is being stored in the system as making its way out to its farthest point.*0255

*As it goes here, the potential energy goes to here.*0262

*As it goes here, the potential energy goes to here.*0267

*As it comes back and goes to 0, the potential energy falls and all of it becomes kinetic.*0272

*If it goes the opposite way, either here or here, the potential energy of the system rises to here or to here.*0277

*You remember that the potential energy function of this was ½ KX².*0284

*If I want the potential energy of the harmonic oscillator, it is ½ KX².*0292

*This is the parabola ½ KX² is the parabola.*0296

*It is describing how much potential energy this system is experiencing as it deviates at distance X from that.*0300

*That is why we are using this parabola to describe it and these of course are the energy levels of higher energy levels.*0310

*You have bigger amplitudes and the potential energy goes off into infinity.*0316

*Do you remember when we are talking about the particle in a box?*0320

*The particle in a box, it was just something like this.*0323

*This particle is moving back and forth, there is no potential energy in there.*0326

*But the minute it hits the wall, the potential energy shoots up to infinity.*0330

*In other words, it cannot escape that box.*0334

*Here, the potential energy rises parabolically, it does not just boom rise up to infinity that way.*0338

*It rises parabolically.*0345

*As you get farther from the center, there is a potential energy that is building up in the system.*0347

*That is what the parabola represents.*0355

*The motion itself is harmonic.*0357

*It is just back and forth represented by a sin or cos curve.*0359

*This just represents the potential energy function.*0363

*I hope that clears up a little bit.*0367

*Let me actually write that down so we have it.*0369

*The motion of a particle is harmonic which is just a fancy word for back and forth, repeating, periodic, whatever you want to call it.*0372

*The further the mass goes from equilibrium position, which is this position right there, the parabola describes its potential energy.*0395

*It is a potential energy that is important in all of this.*0433

*It is the potential energy function that actually changes the Schrӧdinger equation.*0437

*Recall that V of X = ½ KX².*0444

*As the particle or the mass deviates from equilibrium position at distance X its potential energy increases parabolically.*0454

*Increases as ½ KX².*0487

*I hope that clears a little bit up, if there was confusion.*0494

*If not, no worries.*0496

*Let us start talking about the rigid rotator.*0499

*Now I diatomic molecule oscillates along this bond.*0503

*We already discussed the harmonic oscillator but that is not the only thing that it does.*0507

*A diatomic molecule also rotates in space, rotates this way.*0510

*It is a rotating molecule also, so we have rotational motion.*0518

*We can model that rotational motion by something called the rigid rotator.*0522

*A molecule not only oscillates along its bond, it also rotates about its center of mass.*0531

*Let us say you have a big ball here and a little ball here, the center of the mass is not going to be in the center.*0565

*It is going to be a little closer to over here.*0571

*It can actually end up rotating about that thing like that .*0572

*That is all, nothing strange.*0576

*Let us start by looking at the rotation of a single mass about a fixed center and then we will talk about 2 masses, the actual diatomic molecule.*0582

*Let us start by looking at the rotation of a single mass around a fixed center.*0595

*We have a center like this and we have maybe a mass like that and it is moving a circle.*0624

*That is it, we have a nice, fixed circle that way and the radius of the rotation we will just go ahead and call it R.*0633

*I’m going to start with the basic equation for circular motion and that is going to be S= R θ.*0641

*S is the distance that the particle actually makes.*0650

*It is the distance along the circle.*0654

*R is the radius and θ is the angle and radians.*0657

*If it travels 90° or π/2, let us say something like this.*0660

*From here to here, S would be this distance and θ would be this fact that there.*0665

*Basic fundamental definition of radian measure.*0671

*S= R θ, let us go ahead and differentiate this with respect to θ.*0677

*We have DS D θ = DD θ of R θ.*0682

*R is a constant so it comes out.*0689

*I’m getting a little confuse here.*0699

*We are going to differentiate this with respect to time, my apologies.*0705

*S is a distance and θ is an angle.*0708

*We are going to do DS DT = DDT of R θ.*0712

*The R is constant so it comes out and what we are left with is R D θ DT.*0721

*DS DT is the time derivative of a distance which is the velocity.*0727

*What we get is velocity = R and the time derivative with respect to angle rotation gives me the angular velocity RO.*0732

*Velocity = R × the angular velocity.*0741

*The linear velocity of the particle is equal to R × the angular velocity or the angular velocity is equal to the linear velocity ÷ the radius.*0746

*Here, ω is the angular velocity and we have seen it before.*0754

*Its unit is radians per second.*0764

*Let us go ahead and go.*0772

*We know that the equation for the kinetic energy of a particle is ½ its mass × velocity².*0774

*Let us go ahead and actually put this for V into here.*0785

*We are going to get ½ mass × R ω² = ½ M R² ω².*0793

*This thing right here, the ½ M R², we refer to that as I.*0806

*It is called the moment inertia or more commonly known as it is called the rotational inertia of the system.*0812

*This is called the rotational inertia of the system and it plays the same role that mass does for a particle moving in a straight line.*0824

*We just call that inertia, linear inertia, that is the mass.*0839

*This I is the same thing for bodies that rotate.*0843

*The rotational inertia and we also refer to as the moment of inertia of the system.*0848

*So far so good.*0863

*I, and then we have O².*0868

*The kinetic energy = the rotational inertia × the angular velocity².*0870

*That is an equation for the kinetic energy of a rotating body.*0883

*I want to freshly do it on this page.*0892

*That is fine, I can go ahead and do it on this page.*0894

*We said that this is in radians per second.*0897

*I'm going to go ahead and multiply it by one cycle.*0902

*It means one complete cycle, one complete circle, that is 2 π radians.*0905

*It is 2 π radians, radians cancels with radians and we are left with ω / 2 π cycles per second which is the same as the Hz.*0911

*ν which is equal to the angular velocity ÷ two π is the rotational frequency.*0927

*Ω is the angular velocity.*0936

*ν which is the ω ÷ two π gives me the rotational frequency in cycles per second.*0941

*We are just setting up some variables here, that is all.*0956

*This right here is the rotational inertia, I’m going to keep the ½ separate.*0982

*We will do it that way so the kinetic energy is equal to ½ I ω².*0988

*We are ready to move on.*1000

*For a linear system, we have the kinetic energy is equal to ½ the mass × the velocity² = M² V² / 2M.*1001

*Just do the mathematical manipulation.*1026

*= MV² / 2M.*1030

*What we have is the kinetic energy is equal to the linear momentum ÷ twice the mass.*1034

*This is just kinetic energy in terms of momentum.*1042

*That is all, nothing strange about it.*1047

*Momentum is mass × velocity so the kinetic energy is the square of the linear momentum ÷ twice the mass.*1057

*Now for a rotational system, we have the kinetic energy = ½ the rotational inertial × the angular velocity².*1063

*Again, notice the correspondents ½, ½ mass rotational inertia, linear velocity, angular velocity.*1082

*That is going to equal I² O² / 2 I which is equal to IO² / 2 I.*1089

*Therefore, for the rotation we have something analogous, the kinetic energy is equal to L²*1103

*which is the angular momentum of the rotational system ÷ twice the rotational inertia.*1112

*Again, the rotational inertia was the M R², the mass × the R.*1120

*I is equal to M R².*1125

*Here where L is the angular momentum.*1129

*Notice the correspondence between linear system, and rotational systems, the angular momentum of the system.*1136

*Nice and basic discussion of simple rotational motion, a single mass rotating about a fixed center.*1151

*I can talk about its linear velocity and momentum.*1158

*I can talk about its angular velocity and its angular momentum.*1162

*I can talk about its rotational inertia.*1165

*These are all the equations that represent it.*1167

*I can talk about its rotational frequency and things like that.*1170

*Let us go ahead and talk about a rotating molecule.*1175

*We have a rotating diatomic molecule.*1182

*Let us go ahead and set up the system here.*1195

*We have one of the atoms here, let us make it slightly bigger.*1197

*Let us go ahead and make this one over here a little bit smaller.*1202

*Let us go ahead and call this mass one over here.*1208

*Let us call this mass two and the rotation is going to be in this direction.*1211

*Let us go ahead and pick a counterclockwise rotation.*1217

*And its center of mass is going to be let us say somewhere around right there.*1222

*Let us go ahead and call the distance from the center of mass to mass one, we will call that R1.*1227

*And from the center of mass to mass 2, we will go ahead and we will call that R2.*1232

*We will call a total distance between the masses R.*1238

*This total distance is going to be R.*1243

*R is going to equal R1 + R2.*1245

*That is our basic setup of a rotating diatomic molecule.*1251

*Atom, atom, it is going to rotate about its center of mass.*1254

*If they happen to be the same mass, the same atom, then it is going to rotate right down in the middle.*1257

*That is all, basic physics.*1262

*What is interesting here is the angular rotation is actually going to be the same for both.*1266

*I will go to the next page.*1280

*Let me draw my molecule here, the center of mass, R1, mass 1, R2, mass 2 and this is going to be R.*1282

*Now the velocity 1 is equal to R1 ω 1.*1301

*We also have the velocity of mass 2,the linear velocity is going to equal R2 × ω 2.*1317

*But the angular velocities are the same, they are rotating at the same rate.*1330

*Our O1 is equal to ω 2.*1336

*Velocity 1 = R1 ω and velocity 2 = R2 ω.*1344

*We just call it ω, it is an ω 1 and ω 2.*1351

*I have got the kinetic energy = ½.*1356

*The kinetic energy of the systems is just the kinetic energy of the first mass + the kinetic energy*1363

*of the second mass 1/2 of M1 V1² + ½ M2 V2², that is going to equal ½ M1 R1² ω².*1367

*I hope can keep all of my variables straight here.*1392

*I hope that you will be vigilant and make sure I do not make too many mistakes.*1395

*½ M2, we have R2² and ω² so we can rearrange this.*1399

*It becomes ½ M1 R1² + M2 R2².*1408

*I’m going to pull out the ω² right there and this is going to equal ½.*1424

*I'm going to call this thing ,the M1 R1² + M2 R2² I ω².*1432

*Where I is equal to the M1 R1² + M2 R2².*1440

*This is the rotational inertia of the two body system.*1457

*I just manipulated some equations from the same thing.*1462

*I started with the kinetic energy and I work my way and I wrote it, I expressed it in terms of something × the angular velocity².*1466

*That something is what we call the rotational inertia.*1474

*The rotational inertia of this 2 body system is this right here.*1478

*It is the rotational inertia of the 2 body system.*1483

*Let us see what are we going to do now.*1495

*We count what we have.*1501

*We have the kinetic energy = ½ the rotational inertia × the angular velocity.*1504

*And we have this equation right here.*1512

*Sorry, I tend to rewrite things several times.*1514

*M1 R1² + M2 R2².*1518

*When we look up here, we also have an equation for the center of mass.*1527

*We know where the center of mass is.*1531

*The center of mass is right here and there is a relationship that exists for the center of mass from classical physics.*1533

*It is going to be this mass × its distance from the center of mass and that is going to equal this mass × its distance from the center of mass.*1539

*We have another relationship here.*1549

*We have the center of mass relationship which is going to be M2 R2 = M1 R1.*1553

*We also have the relation that R is equal to R1 + R2.*1570

*The total R is equal to the sum of the radii.*1576

*These are the relationships that we have and I want to play with this a little bit.*1579

*I’m going to start with this one.*1582

*I’m going to solve for R1 and R2.*1585

*R1 is equal to R - R2 and R2 is equal to R - R1.*1590

*I’m going to start fiddling with these equations and see what it is that I can come up with.*1599

*I have the center of mass relation M1 R1 = M2 R2.*1605

*I'm going to substitute one of those other equations that I just created from R.*1614

*It is going to be M1 and we said that R1 is equal to R - R2 is equal to M2 R2.*1621

*I get M1 R- M1 R2 = M2 R2.*1632

*I'm going to move this / here and then factor out.*1641

*I get M1 R1 is equal to M1 + M2 × R2.*1645

*I’m just going to solve for R2.*1655

*I get R2 is equal to M1 R1 / M1 + R1.*1657

*Interesting, let us do the other one.*1666

*For all practical purposes, I do not really need to go through all these derivations for you.*1672

*Basically, what I can do is I can just present the Schrӧdinger equation for the rigid rotator*1676

*and I can present the wave functions and the energy functions for you.*1682

*It is important to actually go through this.*1688

*The stuff that we are going through, please know you do not necessarily have to actually know this or*1689

*even recreate it unless your teacher asks you to.*1696

*That is just not going to happen.*1701

*This is just part and parcel of your basic scientific literacy.*1702

*It is important to see these derivations to at least have seen what a derivation is,*1706

*what kind of mathematical manipulation is involved in this.*1711

*Again, this is a part of your scientific training.*1715

*You need to see these techniques over and over again, that is why we are going through these.*1719

*You do not have to know this, as long as you can follow it,*1725

*that is all that matters and it is all that you going to be expected to do, I promise.*1727

*Let us do the other one.*1732

*We started again with the M1 R1= M2 R2.*1734

*This time we are going to go M1 R1 = M × R – R1.*1745

*We are going to solve this one for R1.*1753

*We get M1 R1 = M2 R – M2 R1.*1756

*I’m going to bring this over here and factor.*1762

*Again, I'm going to get M1 + M2.*1764

*This time it is going to be × R1 = M2 R.*1769

*When I solve for R1, I end up with.*1774

*It is going to be M2 R ÷ M1 + M2.*1780

*I made a little mistake here.*1787

*We always have M, R, and things floating around.*1789

*Let me take a second and make sure we got everything straight.*1793

*M1 R1/ M1 + M2, M2 R ÷ M1 + M2.*1796

*So far so good.*1805

*We said that I, the rotational inertia, is equal to M1 R1² + M2 R2².*1808

*I’m going to take these values that I calculated.*1820

*This is R, I have to make sure that all of these things are correct here.*1823

*I do not want to make any crazy mistakes.*1839

*So far so good.*1840

*We are going to plug in R2 and R1 into the equation for the rotational inertia.*1841

*We are going to end up with M1 R1² R1.*1847

*I’m going to square this.*1855

*It is going to be M2² R² / M1 + M2², that × that, + M2.*1858

*For R2, I’m going to plug this value in.*1875

*It is going to become M1² R² ÷ M1 + M2².*1878

*I’m going to go ahead and pull out the R².*1901

*I’m going to express this as M1 M2.*1904

*M2, I’m going to separate that out, / M1 + M2.*1915

*Again, this is just some mathematical manipulation here.*1922

*+ M2 × M1 × M1.*1926

*I’m expressing the M1² / M1 + M2².*1931

*And all of that I'm going to multiply by R².*1943

*I'm going to express this in a slightly different way.*1949

*This is going to be M1 M2 ÷ M1 + M2 × M2/ M1 + M2 + M1 M2/ M1*1955

*+ M2 × M1/ M1 + M2, all of this × R².*1982

*This is I, right?*1999

*Well guess what, this right here M1 M2/ M1 + M2.*2001

*M1 M2/ M1 + M2 that is equal to μ, the reduced mass.*2009

*I put this and this together and I get M2 + M1/ M1 + M2 × R², this is just 1.*2025

*I have the rotational inertia = μ × R².*2042

*Where again, μ is the reduced mass.*2050

*We saw this already when we discussed the harmonic oscillator.*2058

*We ended up taking this 2 body problem and reducing it to a 1 body problem.*2063

*Where now, we are no longer talking about the mass of 1 or mass of 2.*2067

*We actually combined it in this thing called the reduced mass.*2071

*Μ = M1 × M2/ M1 + M2.*2075

*We combine the two masses and now we can treat this as a single particle of mass μ rotating about a fixed center.*2087

*This is what we have done.*2098

*Let me go back to black here.*2102

*Where μ is the reduced mass and R = R1 + R2.*2105

*It is just the distance between those two masses.*2120

*What we have done, we have again turned a 2 body problem into a 1 body problem.*2123

*In other words, we can treat 2 masses rotating about their center of mass as a single mass,*2150

*which we call μ rotating in a circle of radius R.*2180

*Where R is given by this and μ is given by that.*2196

*Let us go ahead and move over to the next page.*2207

*Earlier, we found that the angular momentum is equal to the rotational inertia × the angular velocity.*2209

*Which is just like the linear momentum is analogous to linear momentum equaling mass × linear velocity.*2228

*Angular momentum = rotational inertia × angular velocity*2235

*We also found that the rotational inertia is the μ × R².*2241

*We also found that kinetic energy is equal to angular momentum² / twice the rotational inertia.*2249

*This is very important relationship for rotating systems.*2263

*The angular momentum is given by rotational inertia × angular velocity.*2266

*The rotational inertia is actually given by the reduced mass × the R², the distance between the two masses.*2270

*Again, we treat this as if it is a single mass.*2277

*A reduced mass rotating about a fixed center.*2281

*The kinetic energy of this rotational system is equal to the square of the angular momentum ÷ twice the rotational inertia.*2283

*Again, the rotational inertia is given by this μ R².*2291

*There is no potential term here.*2296

*There is no potential energy term, notice.*2300

*For some diatomic molecule that is rotating in free space, in any orientation actually does not matter,*2310

*there is no potential energy term for this.*2317

*There is no potential energy term because there are no external forces on the masses.*2319

*If we want to, we can certainly put these molecules in a magnetic field and create some external forces.*2337

*But as it is, it is just a rotating molecule, there no external forces.*2342

*The orientation of the rigid rotator.*2355

*When we say rigid rotator we are just talking about the thing that is actually turning,*2366

*the diatomic molecule or whatever it is that we happen to be discussing.*2369

*The rigid rotator which is just a molecule.*2373

*The orientation of the rigid rotator does not affect the energy because there is no external force.*2379

*It does not matter how it is rotating, it does not change the energy of the rotating system.*2392

*Since there is no potential energy term, the hamiltonian of the rigid rotator is just the kinetic energy operator.*2400

*It is just K, the kinetic energy operator.*2428

*The kinetic energy operator which is going to equal –H ̅/ 2M but now it is twice the reduced mass and δ².*2435

*This is just the initial discussion.*2461

*In the next lesson we will continue on and talk about the Schrӧdinger equation.*2463

*Thank you so much for joining us here at www.educator.com.*2468

*We will see you next time, bye.*2469

1 answer

Last reply by: Professor Hovasapian

Sat Apr 23, 2016 6:56 PM

Post by Tram T on April 20, 2016

Hello prof. Hovasapian,

-Since Harmonic Oscilators has Potential Energy V(x), what is the external force applied on the oscillating diatomic molecule?

-Is it because the IR photon has more energy than microwave which cause the external force to pull atom and make it vibrate?

1 answer

Last reply by: Anthony Salamanca

Sat Apr 9, 2016 2:28 AM

Post by Anthony Salamanca on April 9, 2016

HOW DID U GO FROM r1 to r and back to r1

1 answer

Last reply by: Professor Hovasapian

Mon Feb 16, 2015 3:04 PM

Post by Ahmed Thabet on February 14, 2015

I would like about determining LS coupling? Finding Ms & Ml & degeneracies? Can you give some examples?

Or give the link if its here?