physics 141. lecture 13. -...

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 13,

Physics 141.Lecture 13.

There are multiple-particle systems in your future.


I drove to Toronto in August to fly my favorite airplane.

PH-BFT at YYZ (Toronto)


And look at this one …. My favorite plane and no smog in Beijing.

PH-BFT at PEK (Beijing)


If you ever wonder where PH-BFT is?Use the link on the PHY 141 webpage.


Physics 141.Lecture 13

• Course Information:• Homework set # 5• Laboratory experiment # 3

• Complete our discussion of Chapter 8:• A quick review of topics discussed on last time:

• Emission and absorption spectra.• Vibrational energy levels.• Rotational energy levels.

• Incoherent and coherent emission of light – the laser.

• Start our discussion of Chapter 9:• Center of mass.• Motion of complex objects.


Physics 141.Course information.

• Homework:• Homework set # 5 is due on Friday October 14 at 12 pm.• Homework set 6 will be WebWork only and will be due Friday 10/21

at 12 pm.

• Laboratory:• Laboratory report # 3 will count twice as much as reports # 2, which

counted twice as much as report # 1. Make sure you put into practicewhat you learned from reports # 1 and # 2.

• Exam # 2:• Exam # 2 will take place on Tuesday October 25.• The material covered on this exam is Chapter 5 – 9.• The format of the exam will be similar to the format of exam # 1.


Quantization of energy.A quick review: emission patterns.

• Light is emitted when an excitedatom makes a transition to alower energy level.

• Since the light emitted duringthese transitions have discretewavelengths, the energy levels ofatoms must be quantized.

• The energy levels serve as asignature (finger print) for theatom, and the emission patterncan be used to identify the atom.


Quantization of energy.A quick review: absorption patterns.

• When an atom is in its groundstate, it can only absorb photonsof specific frequencies.

• Only photons with an energy thatexactly match possible transitionsbetween energy levels in the atomare absorbed by the atom.

• The absorption spectrum canalso be used as a signature of theatoms.


Quantization of energy. A quick review: molecular vibrational energy.

• When we measure the vibrationalenergy levels for a two-atomicmolecule we find that at lowenergies the vibrational modelworks fine (nearly uniformenergy spacing).

• At higher energies, the potentialwell starts to deviate from aharmonic oscillator well, and thevibrational energy levels are nolonger uniformly spaced.


Quantization of energy. A quick review: molecular rotational energy.

• Molecules can also carryrotational energy.

• The rotational energy of amolecule is also quantized, butthe spacing between levelsincreases with increasingexcitation energy.

• The rotational energy is found tobe equal to

where l is an integer (0, 1, … )and I is the moment of inertia(depends on mass and shape).

El =

12I

l l +1( )2


Completing Chapter 8.Applications: the laser.

• An important application ofenergy quantization in atoms isthe laser.

• Since light emitted when an atomtransitions between states has awell defined energy, the lightemitted is well defined (theuncertainty in the wavelength isdominated by the thermal motionof the atoms).

• The mono-energetic nature oflaser light is important for manyoptical applications.



• The atoms used in lasers havelong lived excited states.

• Although eventually the atomwill make a transition to theground state, and emit the light aphoton with energy E, thistransition can be "stimulated"when a photon with exactly thesame energy E interact with theexcited atom.

• This type of emission is calledstimulated emission, and thesecond photon is in phase withthe first photon (coherentphotons).



• If most of the atoms in the systemare in their ground state, it is verylikely that the emitted photons willbe reabsorbed.

• In order for the photons to escape,a mechanism needs to bedeveloped which:

• Makes it unlikely to find atoms intheir ground state.

• Ensures that most atoms are in thelong-lived excited state.

• The mechanism that is used toreduce the number of atoms intheir ground state is calledpumping.



• The pumping mechanism worksin the following way:

• The atoms are excited to a high-energy state (this can be done invarious way, such as an electricdischarge).

• The atoms in the high-energyexcited state decay quickly to along-lived intermediate state (thisprocess is spontaneous emission).

• At any given time, most atomswill be in the long-livedintermediate state.

• As soon as the atoms arestimulated to decay to the groundstate, the pumping mechanismwill remove them from that state.


Lasers.Principle of operation.

http://www.repairfaq.org/sam/laserfaq.htm#faqwil


The Personal Response System (PRS).Quiz Lecture 13.


3 Minute 56 Second Intermission.

• Since paying attention for 1 hourand 15 minutes is hard when thetopic is physics, let’s take a 3minute 56 second intermission.

• You can:• Stretch out.• Talk to your neighbors.• Ask me a quick question.• Enjoy the fantastic music.• Go asleep, as long as you wake up

in 3 minutes and 56 seconds.


The center of mass.

• Up to now we have ignored theshape of the objects we arestudying.

• Objects that are not point-likeappear to carry out morecomplicated motions than point-like objects (e.g. the object maybe rotating during its motion).

• We will find that we can usewhatever we have learned aboutmotion of point-like objects if weconsider the motion of the center-of-mass of the extended object.


But what is the center of mass and where is it located?

• We will start with considering one-dimensional objects. For an objectconsisting out of two point masses, the center of mass is defined as

d

Xcm

m1 m2

xcm = m1x1 + m2x2

m1 + m2= 1

Mmixi

i∑



• Let’s look at this particularsystem.

• Since we are free to choose ourcoordinate system in a wayconvenient to us, we choose itsuch that the origin coincideswith the location of mass m1.

• The center of mass is located at

Note: the center of mass does notneed to be located at a positionwhere there is mass!

d

Xcm

m1 m2

xcm = m2d

m1 + m2



• In two or three dimensions thecalculation of the center of massis very similar, except that weneed to use vectors.

• If we are not dealing withdiscreet point masses we need toreplace the sum with an integral.

rcm = 1Mrdm

V∫



• We can also calculate the positionof the center of mass of a two orthree dimensional object bycalculating its componentsseparately:

• Note: the center of mass may belocated outside the object.

xcm = 1M

mixii∑

ycm = 1M

mi yii∑

zcm = 1M

mizii∑


Calculating the position of the center of mass.An example.

• Consider a circular metal plate ofradius 2R from which a disk ifradius R has been removed. Letus call it object X. Locate thecenter of mass of object X.

• Since this object is complicatedwe can simplify our life by usingthe principle of superposition:

• If we add a disk of radius R,located at (-R/2, 0) we obtain asolid disk of radius R.

• The center of mass of this disk islocated at (0, 0).

x-axis

y-axis(b)

x-axis

y-axis(a)

Mass M (solid disk)

Mass M/4

Mass 3M/4


Calculating the position of the center of mass.An example.

• The center of mass of the soliddisk can be expressed in terms ofthe disk X and the disk D weused to fill the hole in disk X:

• This equation can be rewritten as

• Where we have used the fact thatthe center of mass of disk D islocated at (-R/2, 0).

• Thus, xcm,X = R/3.

x-axis

y-axis(b)

x-axis

y-axis(a)

Mass M (solid disk)

Mass M/4

Mass 3M/4

xcm,C =

xcm,X mX + xcm,DmD

mX + mD= 0

xcm,X = −

xcm,DmD

mX= RmD

mX

X

D


Motion of the center of mass.Non-relativistic limit.

• To examine the motion of the center of mass we start withits position and then determine its velocity and acceleration:

Mrcm = miri

i∑

Mvcm = mivi

i∑

Macm = miai

i∑


Motion of the center of mass.Non-relativistic limit.

• The expression for Macm can be rewritten in terms of theforces on the individual components:

• We conclude that the motion of the center of mass is onlydetermined by the external forces. Forces exerted by onepart of the system on other parts of the system are calledinternal forces. According to Newton’s third law, the sumof all internal forces cancel out (for each interaction thereare two forces acting on two parts: they are equal inmagnitude but pointing in an opposite direction and cancel ifwe take the vector sum of all internal forces).

Macm =

ddt

Mvcm( ) = dPcm

dt=

Fi

i∑ =

Fnet ,ext


Motion of the center of mass.Linear momentum.

• Now consider the special case where there are no externalforces acting on the system:

• This equations tells us that the total linear momentum of thesystem is constant.

• In the case of an extended object, we find the total linearmomentum by adding the linear momenta of all of itscomponents:

dPtot

dt= 0

Ptot = Mvcm = mi

vii∑ = pi

i∑


Conservation of linear momentum.Applications.

Internal forces are responsible for the breakup.


Conservation of linear momentum.Applications.


Next: more about multi-particle systems.

physics 141. lecture 13. -...

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