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The Qualifier Connection

  1. Mar 20, 2003 #1
    Here's the idea. We can help each other study for our qualifier exams by posting qualifer-type questions along with the solutions in this thread. I think this thread will also help those studying for the physics GRE... or even those just wanting to get good at solving exam-type physics problems fast.

    eNtRopY
     
  2. jcsd
  3. Mar 20, 2003 #2

    Tom Mattson

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    Sure, I've got 5 books full of qualifer problems, plus many, many old exams from my school (Rensselaer Polytechnic Institute).
     
  4. Mar 20, 2003 #3

    chroot

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    I'm taking the Physics GRE on April 12... I've been studying my ass off for months now.

    - Warren
     
  5. Mar 23, 2003 #4
    This is a good idea, I'll post some in the next few days when I have some free time. What kinda level were you talking about pre-graduate?

    JMD
     
  6. Apr 22, 2003 #5
    I am waiting!

    Hi all,
    Well, I have waited for a month now and I had no post about this GRE Test.
    Well, as I am sure that I will find help from ppl here in this forum I have the following questions:

    1-How may I prepare for the GRE physics? and
    2-What books should I read? and
    3-Are there any useful materials on the net that I can download and use?

    hhegab
     
  7. May 6, 2003 #6
    The best way to prepare is to solve problems, lots of them. My GRE had lots of QM on it but one my friends had mostly classicaly physics, so they vary the content on each. I don't recommend buying the Green Physics GRE book, it was not helpful at all. Just solve problems, and understand the concepts behind the solution.

    JMD
     
  8. May 6, 2003 #7

    Tom Mattson

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    Jeez, I forgot all about this thread. Sorry!

    As for the GRE, I say get Princeton Review stuff. I used to work for them (taught SAT, GMAT, and GRE courses--not subject courses though).

    I will post 5 qualifier questions tomorrow, I promise.
     
  9. May 7, 2003 #8
    Pardon my ignorance, but what the hell is GRE everybody talkin' about?
     
  10. May 7, 2003 #9
    GRE

    GRE stands for Graduate Record Examinations, this is a standardized test requirred to join any american unversity to complete your graduate eduaction.
    www.gre.org is their web site.

    hhegab
     
  11. May 7, 2003 #10

    Tom Mattson

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    OK kiddies, here’s Round 1. At my school, the exam is divided into 5 sections:

    1. Classical Mechanics
    2. Classical Electrodynamics
    3. Special Relativity
    4. Quantum Mechanics
    5. Thermodynamics/Statistical Mechanics

    We’ll start off easy to warm up.

    1. (Classical Mechanics) Using Lagrangian mechanics, derive and solve the equation of motion for a simple pendulum of mass m and length L for small oscillations.

    2. (Classical Electrodynamics) Consider a spherical capacitor of outer radius a and inner radius b.
    a.) Derive the capacitance.
    b.) What is the capacitance of a single conducting sphere?

    3. (Special Relativity) Consider a reference frame in which an electric field E is observed, and B=0. Show that there is no reference frame S’ in which E’=0 and B’ is nonzero.

    4. (Quantum Mechanics)
    a.) Show that a spin-1/2 state is not invariant under 2π rotations and that a 4π rotation is required to return to the initial state.
    b.) A necessary condition for a state to be physical is that it differ by no more than a phase factor e-iθ upon 2π rotations. Consider the super position of fermion number states ψ given by

    ψ=(1/21/2)(|0>+|1>)

    Show that ψ is not physical.

    5. (Thermodynamics/Statistical Mechanics)
    In an ideal gas of molecules of mass M at temperature T, the probability for a molecule to have a velocity v is given by the Maxwell-Boltzmann distribution

    P(v)=Cexp(-Mv2/kBT)

    a.) What is the root mean square speed?
    b.) What is the most probable speed?
    c.) What is the average speed?

    You may find the following information helpful.

    (sorry, I don't know how to make matrices in this forum)
    Fμμ=0
    F0i=-Fi0=-Ei/c
    F12=-F21=-Bz
    F13=-F31=By
    F23=-F32=-Bx

    The metric tensor is g00=1, gii=-1 (i=1,2,3)
     
  12. May 7, 2003 #11

    chroot

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    Tom,

    The Physics GRE is in fact made up of 100 short questions with five multiple choice answers. Each problem should be solvable in a couple of minutes by a knowledgeable candidate. In fact, nearly a third or so of the questions can be answered immediately simply by inspection.

    In short, the kind of questions you posted, while excellent (I plan on doing them when I get home), will never appear on a GRE exam.

    - Warren
     
  13. May 7, 2003 #12

    Tom Mattson

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    Oh, I know. These are the kinds of questions that appeared on my qualifying exam. These would be among the "easy" ones (not worth many points, but embarassing if you don't get them!).

    I have 5 books full of qualifier questions, as well as 5 or 6 complete exams from my school. I'll try to post an entire exam this weekend.
     
  14. May 7, 2003 #13

    chroot

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    Terrifying. :(

    - Warren
     
  15. May 7, 2003 #14
    I would second this recommendation. I got the princeton review for mathematics and it's really helpful.
     
  16. May 7, 2003 #15

    chroot

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    I don't think there is one for the GRE physics exam.

    - Warren
     
  17. May 7, 2003 #16
    Good grief Tom. Those questions serve to show me how much I have yet to learn about all this stuff.
     
  18. May 7, 2003 #17
    I think I'll just buy basic textbooks in these fields to read through some day. There is the possibility that I may do some physics and chemistry units (maybe some robotics as well) after I complete my current degree, so it can't hurt to get a head-start.
     
  19. May 12, 2003 #18

    Tom Mattson

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    Here's the first one

    Solution:
    There is one degree of freedom, which I take to be the angular displacement θ. Let θ=0,V=0 at the bottom of the swing.

    L=T-V
    T=(1/2)m(Lθ')2
    V=mgL(1-cosθ)

    So the Lagrangian is:

    L=(1/2)m(Lθ)2-mgL(1-cosθ)

    Plug into Lagrange's equation:

    (d/dt)(∂L/∂θ')-∂L/∂θ=0
    (d/dt)(mL2θ')+mgLsinθ=0
    mL2θ''+mgLsinθ=0

    Now for the bit about small oscillations:

    sin&theta;~&theta; for &theta;<<1

    mL2&theta;''+mgL&theta;=0

    &theta;''+(g/L)&theta;=0

    let &omega;2=g/L

    &theta;''+&omega;2&theta;=0
    &omega;(t)=Acos(&omega;t)+Bsin(&omega;t)

    Now I did not give you any initial conditions (my oversight), so let's say that the pendulum is released from rest at &theta;(0)=&theta;0.

    &theta;'(0)=0 ==>B=0
    &theta;(0)=&theta;0 ==> A=&theta;0

    I know that you already know the result, and could probably do it with Newtonian mechanics (or possibly by simply writing the solution down!), but specifically asking you to solve it with Lagrangian dynamics is something that would be fair to ask. It was asked of me on my (Goldstein-based) Advanced Mechanics midterm.
     
  20. May 12, 2003 #19
    Homework

    Hello Tom,

    It is me-Ramona. Are you still in town?
    If so, can I meet you tonight?
     
  21. May 12, 2003 #20

    Tom Mattson

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    Check your Private Message box.
     
  22. May 12, 2003 #21

    Tom Mattson

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    Solution #2

    This one is worked out in Halliday, Resnick and Walker, but it is also a fair qualifier question. It appeared on a qualifier one year at my school, and it appeared on my (Jackson-based) Classical Electrodynamics midterm. It's one of those "catch you with your pants down" type questions, because you're expecting to find long, dirty problems involving roots of Bessel functions and whatnot.

    Start from Gauss' law:

    &int;E.dS=q/&epsilon;0

    The Gaussian surface here is a sphere concentric with the two plates and with radius a<r<b. Since the entire charge in enclosed, the field is that of a point charge, so:

    E(r)=(4&pi;&epsilon;0)-1q/r2

    Now for the potential difference.

    &Delta;V=-&int;CE.ds

    We use as a path C a straight line from one shell (r=a) to the other shell (r=b). Take the inner plate as positive and the outer plate as negative (this choice does not affect the result for capacitance, which depends only on geometry). Since E is everywhere antiparallel to ds, the integral reduces to:

    &Delta;V=+(4&pi;&epsilon;0)-1)&int;abdr/r2
    &Delta;V=q/(4&pi;&epsilon;0)(1/a-1/b)

    Solving for capacitance C=q/&Delta;V we get:

    C=(4&pi;&epsilon;0)/(1/a-1/b)

    This is the same as if the outer sphere were not there. We can translate this into mathematics by taking the limit of C as b-->[oo] to get:

    C=4&pi;&epsilon;0a
     
  23. May 12, 2003 #22

    Tom Mattson

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    Three's a charm!

    Note that F&mu;&nu;F&mu;&nu; is a Lorentz scalar. You can find F&mu;&nu; from F&mu;&nu; by:

    F&mu;&nu;=g&mu;&sigma;g&nu;&tau;F&sigma;&tau;

    You will find that the electric field components change sign while the magnetic field components do not.

    The inner product of the two tensors is then:

    F&mu;&nu;F&mu;&nu;=2(|B|2-|E|2/c2)

    Note that this is a Lorentz scalar, which is an invariant. That means that F&mu;&nu;F&mu;&nu;=F'&mu;&nu;F'&mu;&nu;.

    In frame S, we have nonzero E and B=0. That means that F&mu;&nu;F&mu;&nu; is less than zero. So, we cannot have nonzero B' and E'=0 because that would mean that F'&mu;&nu;F'&mu;&nu; is greater than zero.

    This is a problem from Jacskon Chapter 11.
     
    Last edited: May 14, 2003
  24. May 14, 2003 #23

    Tom Mattson

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    Numero quatro

    An arbitrary spin-1/2 state is (expanded in Sz eigenstates):

    |&alpha;>=c+|+>+c-|->

    The rotation operator is:

    D(n;&phi;)=exp(-iS.n&phi;/hbar)

    For a rotation of 2&pi; radians about the z-axis, D becomes:

    D(k;2&pi;)=exp(-2&pi;iSz/hbar)

    Apply this to |&alpha;>:

    D(k;2&pi;)|&alpha;>=c+exp(-2&pi;iSz/hbar)|+>+c-exp(-2&pi;iSz/hbar)|->
    D(k;2&pi;)|&alpha;>=c+exp(-i&pi;)+c-exp(+i&pi;)
    D(k;2&pi;)|&alpha;>=-c+|+>-c-|->
    D(k;2&pi;)|&alpha;>=-|&alpha;>

    So, we do not come back to the initial state after a 2&pi; rotation. It is easy to see how a 4&pi; rotation does return us to the initial state. (That wouldn't be accepted on a qualifier, but I have already passed mine! :p)

    First, as to the part in red, I meant to say spin-1/2 number states--sorry.

    As for the vacuum state |0>, it is unaffected by any rotation at all, so we know it will be invariant under the action of D(n;&phi;) for any n or &phi;. As for the single-particle state, we just worked out in part a.) that it will suffer a change in sign.

    Choosing n=k for definiteness, we have:

    D(k;2&pi;)|&psi;>=(1/2)1/2(D(k;2&pi;)|0>+Dk;2&pi;)|1>)
    D(k;2&pi;)|&psi;>=(1/2)1/2(|0>-|1>)

    Since the resulting state is not related to the initial state |&psi;> via a simple phase change, |&psi;> is unphysical.
     
  25. May 14, 2003 #24

    Tom Mattson

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    And 5 makes a handful!

    Sorry to anyone who tried this problem. I screwed up in writing the Boltzmann distribution: the v2 was missing.

    The problem itself is just 3 exercises in basic definitions.

    I will be making use of the following mathematical information. At my school, each examinee is given a complimentary copy of Schaum's Mathematical Handbook to keep. It helps dull the pain if you fail. Not.

    For In(a)=&int;unexp(-au2)du, where the integration is from 0 to [oo], we have:

    I3(a)=1/2a2
    I4(a)=(3/8a2)(&pi;/a)1/2

    This is exactly what the name implies: the square root of the mean of the square of the speed.

    vrms=(&int;v2F(v}dv)1/2

    where the integral is taken from 0 to [oo].

    vrms=(C&int;v4exp(-Mv2/kBT))1/2

    Recognizing that the integral above is I4(a) with a=M/kBT, we have:

    vrms=((3C(kBT)2/8M2)(kBT&pi;/M)1/2)1/2

    I am not going to simplify that any further. :p

    This is the speed at which the probability distribution is maximized.

    dP(v)/dv=(2Cv)exp(-Mv2/kBT)+(Cv2)(-2Mv/kBT)exp(-Mv2/kBT)
    dP(v)/dv=(2C)(v-2Mv3/kBT)exp(-Mv2/kBT)

    Setting dP(v)/dv=0 yields:

    vmp=(kBT/M)1/2

    The average speed is:

    vavg=&int;vP(v)dv
    vavg=C&int;v3exp(-Mv2/kBT)dv

    where the integral is taken from 0 to [oo]. Recognizing that the integral above is I3(a) with a=M/kBT, we have:

    vavg=C(kBT)2/2M2

    OK, anyone else want to put any up? If not, I'll put some more up in a few days.

    edit: fixed subscript bracket
     
    Last edited: May 14, 2003
  26. May 14, 2003 #25

    Tom Mattson

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    I just thought of something. The above solution to #5 is correct, but anyone who can see anything fishy with it gets bonus points.
     
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