New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Dr. Simeon Wiza


Dr. Simeon Wiza
GPA 3.96

Stephen Sharpe

Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Stephen Sharpe
Class Notes
25 ?




Popular in Course

Popular in Physics 2

This 16 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 505 at University of Washington taught by Stephen Sharpe in Fall. Since its upload, it has received 13 views. For similar materials see /class/192439/phys-505-university-of-washington in Physics 2 at University of Washington.


Reviews for MECHANICS


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/09/15
Lecture 9 Oscillations in Large N and Continuous Systems The remainder of Chapter 4 in FampW Now let s think about what happens when we have N coupled variables say N identical point masses and NH identical springs longitudinal motion as in FampW Fig 241 or N identical point masses equally spaced along a massless string transverse motion as in FampW Fig 242 This is again a normal mode problem but now we eXpect to find N modes The former is described by the Lagrangian with 7 representing the longitudinal displacement from equilibrium L TU 227712EZ77117712 91 j1 ji with boundary conditions 170 mm 0 The second situation transverse displacements is described by identical mathematics if the transverse displacements are small If the spacing between masses is a the tension in the string is rand the transverse displacement is represented by u the restoring force on a given mass in the small angle limit ie the transverse restoring force is given by T sin6 T ua is r yj yjlayj1 yja The corresponding Lagrangian is E N z L N 2 L T U 2 EM Za gm 5 92 with boundary conditions 0 HNH 0 Thus by the rule of Feynman that the same equations have the same solutions we need only solve one of the problems and then substitute uk 17 k and k c ra So let s focus on the latter and its equation of motion r r r r m j 39 j 39 H 39 j1 1 m j 2 1 j1 171 Z 039 93 a a a a As in previous discussions we have a linear differential equation with constant coefficients and we are encouraged to try an exponential form as an Ansatz Jt MUM W Ae lka 94 Physics 505 Lecture 9 1 Autumn 2005 where a is an angular frequency and k is the wave number k 27rl where A is the wavelength Note that looking ahead we have written the solution as a function of the continuous coordinate x which we sample only at the discrete values xj qf Substituting this Ansatz into the equation of motion and canceling all common factors yields 21 1 ma 2 e39k 6quot O a a 21 41 ka 9 5 3 a2 1 cos ka 2 s1n2 ma This sort of relationship between a and k is called a dispersion relation It corresponds to the more familiar relationship between energy and momentum for a free particle E Z p2 m2 To the extent that a is not linear in k waves of different wavelengths will travel with different velocities and disperse To turn this into an eigenvalue problem we need to consider the boundary conditions for the problem Before considering the fixed end case discussed above we first consider what are called periodic boundary conditions corresponding to a closed loop of string Thus we set the transverse displacement at one of the string equal to the displacement at the other end u 0 u Na or more generally u ja u ja Na translational invariance when translated by a distance Na the circumference of the loop This means Agraaw 2 Aeikwkmiwt 2 1 W 2 W 2 27m N odd gtkn Na N 2 N 2 41 gtan sm ma N For either even or odd N there are as expected N possible eigenvalues of k and a Other values of n simply reproduce the same N solutions Note that the solutions 96 Physics 505 Lecture 9 2 Autumn 2005 tend to come in pairs with opposite signs for k corresponding to motion in both directions in x We can think about this motion as the motion of points in the wave i k x7 139 form With constant phase 0 as in e quot W e 1 0 Re gt cos Q50 2 cos knx cont So we can define a phase velocity I 39 N cquot 0 kn 4 Tsin M zJi sm n 97 ma N 27m m nN This expression first tells us that the direction of motion changes with the sign of n we have eigenmodes moving in both directions We also see that the characteristic magnitude of the phase velocity is W This is the actual magnitude for n gt 0 when the second factor approaches unity However for gt N 2 we find cnN2 gt ZWIr lt M This n dependence will lead to dispersion The kn range AM gt co to Adm 2a ie the spacing between the masses can be at most 12 corresponding wavelength of the oscillations in 27r Na varies in the of a cycle This reminds us of the essential feature of a chain of discrete masses it cannot support arbitrarily short wavelengths Note as eXpected that the n l eigenmode has wavelength A Na the translational invariance length Now we return to consider the case of xed ends u0 uNa a 0 To satisfy this condition we make use of the opposite moving solutions noted in the previous paragraph and write recall that linear equations always allow linear superposition uxt 2 A64 e k c e i39 98 This will still satisfy the equation of motion if k and a satisfy the dispersion relation in Eq 95 and by construction this form satisfies the boundary condition at x 0 Now the eigenvalue problem arises from the other boundary condition Physics 505 Lecture 9 3 Autumn 2005 Na gar Aeim eikaNl 67mm Z 0 gt sinkaN 1 0 gtkaN 1 2 mt k 217 3 quot aNl 4r 7m gtw 1l s1n quot ma 2Nl Note the differences from the periodic boundary condition case of Eq 96 especially the factor of 2 in the numerator and the absence of solutions of both signs This latter point just reminds us that with xed ends the eigenmodes are standing waves rather than propagating waves ie there are nodes or zeros in the wave shape that occur at fixed positions in x Thus the nth normal mode looks like nl2N 99 mrxj 7W uxjt21An s1naN 1 e 910 gt Real Part 2 2Aquot sin sinwnt N 1 This expression illustrates that the x and t dependences have factored into separate functions Instead of moving waves as in the periodic boundary condition example where the x and t dependence stay coupled when we take the real part we have standing waves Also in contrast to the periodic case the n l eigenmode here has wavelength 1 2aN 1 2 21 or twice the length of the string ie 12 a cycle corresponds to the distance between nodes which must be fixed at the endpoints Finally we want to consider what happens in the continuum limit ie a gt 0 N gt 00 with length aN 1 2 l fixed This is the limit that we use to study an actual matter system even though matter is really composed of a discrete set or atoms We can basically take the corresponding limit of the above expressions where we replace ma by a linear mass density a which could be a function of x in which case the tension will also typically be a function of x Following FampW we also change the label of the displacement from equilibrium to remind us of the Physics 505 Lecture 9 4 Autumn 2005 continuum limit Ll xt gt uxt For periodic boundary conditions now a loop of continuous string the dispersion relation for any fixed mode n as N gt 00 yields Periodic Boundary Conditions k gt2 n 0ili2 4r 7rn f2nn a gt gt quot ma N a l 911 w c gt n k a n 2 1 kn n lnl Thus all finite n modes have the same phase velocity ie no dispersion and now we see arbitrarily short wavelength modes present The n 1 case still corresponds to a full cycle in the distance of periodicity I For the fixed end case there remains a factor of 2 difference in various places Fixed Ends Boundary Conditions k gt 1 n 12 4r 7rn a gtquot gt ma 2Nl a l 912 w c gt quot kn a k n n The case n 1 still corresponds to just 12 cycle along the length of the string Physics 505 Lecture 9 5 Autumn 2005 xl 916 The Lagrangian density plays a central role in the application of these ideas to field theory Physics 505 Lecture 9 6 Autumn 2005 The general solution to the excitation of a system string described by such a Lagrangian can as expected be written as a linear combination of the normal modes For the fixed end boundary conditions we can write the normal coordinates as following the notation of FampW gquot t Cquot cosant 15quot 2 mt mt r 917 gn Z Qngn can 2 02 l l 7 where the constants Cquot Q5quot are fixed by the initial conditions the temporal boundary conditions The corresponding orthonormal eigenfunctions matched to the spatial boundary conditions are pxsmkxa p39 kpok quot 32 21 c l n 918 idxapmxpnx 5mquot The last line expresses the usual orthonormality of the eigenfunctions with respect to the diagonalized metric m a Thus the general solution has the form note again the separation of variables termbyterm 00 xtZPnx nf 919 nl Expressed in terms of the normal modes we can perform the x integral in the Lagrangian see Eq 915 either using the explicit form of pquot or by performing an integrations by parts and perform one of the sums using the orthogonality of the modes to find the diagonal form Physics 505 Lecture 9 7 Autumn 2005 L w 2 920 nl In terms of the normal modes the nite length continuum problem corresponds to an infinite but still discrete set of decoupled harmonic oscillators As a final comment we note that Hamilton s principle of least action applied to the continuum form of the action in Eq 9 16 yields the continuum form of Lagrange s Equation 3 L 3 L 0 a 66u6t ax 66u6x au 93921 Here we have applied the calculus of variations to variations with respect to 3 functions 14 361 altax u altat ll all of which are functions of the 2 independent variables x and t Applying this result to the string Lagrangian density in Eq 915 L 02112 T2 12 yields the wave equation Eq 914 Physics 505 Lecture 9 8 Autumn 2005 Lecture 3 Inertial Reference Frames Intro to Chapter 2 in FampW We have been explicitly making assumptions about the frames of reference in which we have been defining equations and now we want to make those issues more explicit We start with Newton s second equation in either vector ie linear algebra notation d d F 31 or tensor ie component notation F 32 These equations implicitly define a class of reference frames the inertial frames simply by the fact that these equations are true in those frames with no extra contributions We are also explicitly using Cartesian coordinates in the second equation ie rectilinear axes that extend to infinity We also typically assume that these axes are embedded in a 3 dimensional Euclidean vector space with a metric defined so that we know how to construct scalar and vector products By assumption we take the Cartesian coordinates axes to be orthogonal ie to have vanishing scalar products and to be specified by a complete set of 3 unit vectors By definition any vector in this space can be represented by a linear combination of the unit vectors with appropriate coefficients the components We start by considering an abstract inertial frame in some empty Euclidean space with enormous symmetry ie invariant under possibly time dependent translations and rotations In practice we have in mind a frame at rest with respect to the distant low number density stars In this lecture we want to consider the range of other reference frames which we are transported to by some transformation such that the form of Newton s equation is in some sense unchanged When we say unchanged by some coordinate transformation there are actually two relevant levels of constancy The weaker level is covariance meaning that in the new primed coordinates the equation has the same form Physics 505 Lecture 3 1 Autumn 2005 JF7c rgt Fir lc 33 although the righthand side of the new equation is allowed to have a new functional form in the sense that the individual components are different 1117 75 71 This change in detail still allows the new frame to be inertial What we are excluding is the appearance of new terms in the equations and we will provide examples below Actually certain types of new terms can appear in the component version of Newton if we transform to curvilinear coordinates still in an underlying inertial frame If the new frame is an inertial frame the vector version Eq 31 of Newton remains unchanged On the other hand it is also possible that in the new reference frame even the details of the component version of the righthandside are unchanged ie that F 739 VI In this case we have an invariance of the dynamics a stronger statement than covariance and as noted in Lecture 1 there will be an associated conserved quantity constant of the motion as specified by Noether s Theorem Examples are translational invariance when Ff 0 leading to momentum conservation and rotational invariance in the central force problem leading to angular momentum conservation Returning to the issue of covariance we consider first the simplest inhomogeneous transformations translations We simply move the origin of the reference frame by a specific vector distance 170 Thus in the new frame the coordinates are given in terms of the old components by V rk rk r0k rk rk39r0k 3394 We can rewrite Newton in terms of the coordinates in the new frame as d mr d mf7mf zwz l E kEETV 35 It is clear that Newton s equation is form covariant and still valid in the new frame if it was valid in the old frame as long as there is no new term on the lefthandside of Physics 505 Lecture 3 2 Autumn 2005 Eq 35 Thus the constrain of covariance for translations ie that the new frame is still an inertial frame is that m 0 36 Thus we are allowed to make constant translations and allowed to transform to a frame moving with a constant relative velocity and still be in an inertial frame where the form of Newton s equation is unchanged d m J r r F r 3 7 dt 1k lt gt In such an inertial frame we cannot determine by performing a physics experiment for example throwing a ball whether the frame is fixed with respect to the distant stars or moving with a constant velocity other than looking at the stars themselves On the other hand if we transform to an accelerating frame we find d mf m u Fj39rk39 38 Observed in this primed frame forcefree particles will accelerate past a primed observer ie an observer fixed in the frame with acceleration just as if there were a uniform gravitational field Similarly the observer herself will experience a force in order to stay at a fixed point in the frame just as if there were gravity Thus the acceleration of a reference frame is experimentally detectable and acts locally just like a gravitational eld The fact that the identical mass factor acts both as the coupling strength for gravity and as the inertial resistance in Newton s equations is know as the principle of equivalence and is under precise experimental study at the UW We don t have precision tests at length scales below about a fraction of a millimeter The other side of the similarity between accelerating frames and gravity is that a frame of reference that is freely falling in a gravitational field and is thus accelerating yields the same physics as an inertial reference frame but with no gravitational interaction Physics 505 Lecture 3 3 Autumn 2005 It is important to note that if we are comparing experiments in 2 different laboratories in two different reference frames moving with respect to each other we should use both identical apparatus and identical initial conditions in the two frames Thus the initial conditions are identical and not related by the transformation of Eq 34 So it is reasonable to ask what is the complete set of coordinate transformations the Galilean transformations that preserve the inertial properties of the frame of reference To answer this question we want consider a general functional form for the transformations ie the transformations are allowed to depend on the coordinates themselves The mathematical background here includes the study of vector spaces our Euclidean configuration space E is such a space linear algebra and group theory see Lectures 4 and 5 on the web page for my Phys 557 course We can express the most general transformation of this nature including the possibility that it is nonlinear as using component notation Wtgkltnltrgtr2ltrgtr3lttt rk 701hk7139tr239tr339tyt 39 The second equation expresses the inverse transformation which we require to exist The only other required features are that gk and hk are differentiable singlevalued and invertible Note that in the first line we are treating 1739 as a function of f t and I while 7 t is a function of only I ie dFdt arat The converse is true for the inverse transformation in the second line Considering the time derivatives of these expressions and using the chain rule we can obtain the corresponding expressions for the transformations of the velocity and the acceleration as usual in such an expression repeated indices are summed over J dt ark quot at ark quot at g 310 szkrk 6 Physics 505 Lecture 3 4 Autumn 2005 and u dzr 62 62 3r quot 62 J 2 2 rkr 2 rk rk 7 d t 3rk6r arkar ark at u 2 39 azr 62W 311 Rjkrk Jrkr 2 Jrk 2 arkar arkat a t where we have introduced the 3x3 matrix 31639 jk 2 312 The inverse transformation for the velocity for example has the corresponding form 6131 6th ah j drzark k 67quot E 1 313 7 v J Rjkrk E with the inverse matrix 61 R7 3 14 J ark We can verify that this is the inverse by considering 6739 67quot 6739 71 RijH quot 5 315 r J ark a a where 51 is the Kronecker delta function the unit matrix that is l for j l and zero otherwise Physics 505 Lecture 3 5 Autumn 2005 Now we want to consider the other side of Newton s second law the forces For now we focus on conservative forces Since the potential energy is described by a scalar function we require that it transforms as U39f tUftU f tt 316 Now we can again use the chain rule of partial derivatives to derive the transformation of the Cartesian components of the corresponding force F jrfrt aUaiE T 6 1 6r ark at It ark 317 T mm 5 1k 1k We note that Tik R or in the more efficient notation of linear algebra R 1 T 318 where the right hand side is the transpose of T Our goal is to ensure that the two sides of Newton s equation transform in the same way ie we want M2 dt k dt d miquot kF 319 Thus in order to be certain that we are transforming to another inertial frame assuming we started in one it is necessary that the acceleration transform just like the force Comparing Eqs 311 and 317 we first notice that the transformation of the force exhibits no eXplicit velocity or time dependence Thus we require that the final 3 terms in Eq 3 l l vanish for any value of 4 the velocity in the original frame This requires that each term separately vanish which is accomplished by transformations of the form Physics 505 Lecture 3 6 Autumn 2005 gqr2r3tajbj I3r2r3cjt 320 where aj and of are constants and bj is a linear function of the 3 components of r The first and last terms are just the constant translation and transformation to a reference frame with constant relative velocity mentioned earlier The fact that the middle term is a linear function ensures that the matriX R is constant ie not a function of t or the 3 components of r The final step in defining the Galilean transformation is to require that jk RJk 321 In this case both the acceleration and the force transform in the same way and we have dmf39 d mi 71 RjkFc0 322 From Eqs 318 and 321 it follows that Rjk 2R gtR 1 2R 323 Thus the linear transformation included in the Galilean transformation is described by a 3X3 constant matriX for which the inverse is equal to the transpose These matrices are the orthogonal matrices and the associated transformations are the global or rigid rotations in 3dimensions The name derives from the fact that right angles are preserved by the transformations The associated group is 03 although we typically focus on the group SO3 where the S means we do not include re ections Examples of re ections are x x or xi y Zr 2 x J 2 Mathematically this means that we include only matrices with determinant 1 More generally the above property for orthogonal matrices requires only that detRl detR2 det1gtdetRi1 324 ASIDE The orthogonal 3X3 matrices with determinant l constitute the fundamental representation of the abstract group SO3 ie the smallest matrices Physics 505 Lecture 3 7 Autumn 2005 whose multiplication properties provide a faithful representation of the multiplication properties of the elements of the group These matrices and the group elements are parameterized by three continuous parameters the Euler angles which we can think of as the angles by which we rotate in the three independent planes xy yz and zx or in the special case of 3dimensions as the angles of rotation about the three unique axes perpendicular to these planes This parameterization is differentiable so that SO3 is a socalled Lie group and we can define derivatives with respect to each of the angles infinitesimally close to the origin in parameter space ie close to the identity operator in the group space The derivatives serve to define operators that generate infinitesimal transformations and are called simply the generators of the group For SO3 there are 3 generators which with appropriate normalizations are the angular momentum operators Lx Ly Lz familiar from quantum mechanics The generators serve to define a vector space called the algebra of the group in which the vector product is provided by the commutator for example LwLy iL 44 isjlel The relevant structure of the algebra and thus the structure of the group near the identity is specified by the 3index tensor on the righthand side of the commutator equation which is called the structure constant of the group the factor of 139 is a result of the choice to define the Lk as Hermitian The transformations studied by Lie will play an important role in our discussion of flows in phase space In summary we have seen that the form of Newton s second law is covariant under the Galilean transformations which include translation of the origin by a constant vector 3 parameters transformation to a reference frame moving with constant relative velocity 3 parameters and constant rigid rotations of the Cartesian coordinate axes 3 parameters Although we have not discussed it explicitly it should be clear that there is also covariance with respect to a constant translation in time I39IT 325 ie physics is the same in different time zones Overall the Galilean transfomiations are described by 10 parameters Note that if the form of the force exhibits invariance with respect to time translations no explicit time dependence then we expect the energy to be conserved Physics 505 Lecture 3 8 Autumn 2005


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Allison Fischer University of Alabama

"I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.