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Should I correct a professor?

  1. Sep 18, 2017 #1
    I sat in an introductory physics course at my university and the professor was explaining Gauss's Law.
    While I was in there I noticed he was incorrectly teaching the mathematics of surface integrals.
    For example:
    The professor stated that for a sphere centered at the origin, the area element dA was found as follows.
    Since, for a sphere the surface area is A=4πr2 it follows that dA=8πrdr. He gave similar arguments for cylindrical and other symmetries. So far, this has not affected the examples since most of them have symmetries which have the E field constant on the surfaces in question so that it reduces to just an integral over dA.

    Normally, I don't mind correcting a professor if there is a simple error, but this shows there is a severe lack of fundamental understanding of the math required. Never was there mention of parameterizing the surface and obtaining the correct area element by means of the vector product of partial derivatives. Even worse, the fact that he is integrating over r on the surface of the sphere is bothersome. I feel awkward correcting him because this is such a fundamental requirement for surface integrals. I don't know what I should do.
     
  2. jcsd
  3. Sep 18, 2017 #2

    Orodruin

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    You should definitely say something. Then it is a matter of how you phrase it. A reasonable way is non-confrontative. Do not blurt out what you might think (he has no clue), but give him a chance to correct himself, possibly by phrasing your correction as a question:
    "Sorry, professor, but how can there be a dr when r is fixed on the sphere?"
    "How does that relate to the surface element in terms of ##d\theta## and ##d\phi## that we learned in xxxx?"
    If he does not correct himself after that, take it up in private after the lecture. It depends on the person, but many people do not like being made to look bad and professors generally are in a position of relative power.

    That being said, when I took vector analysis 17 years ago my teacher was struggling with obtaining the correct surface element for the proof of the divergence theorem (it should be mentioned that ability to think as a teacher drops significantly in front of a group of students and a black board). I stood up, went to the black board, and completed the proof. That teacher became my PhD supervisor 4 years later so not everyone will take correction the wrong way.
     
  4. Sep 19, 2017 #3

    Andy Resnick

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    I would approach the professor during office hours and discuss this.
     
  5. Sep 19, 2017 #4
    What is so wrong about simply asking for clarification?

    "Ummmm....Dr. Jones? I'm confused. Why wouldn't that term be 'r-squared' in the equation?"
     
  6. Sep 26, 2017 #5

    bhobba

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    That's exactly what you do.

    My analysis teacher - and I took 3 classes with him - on the first day had a challenge to any student - find a fault with what he says or in his notes and you immediately pass. A few, including me, did just that. Of course we passed. But as he told us privately anyone capable of doing that was gong to pass anyway - it's a bit of an empty challenge. It was really meant for the good students to make them think. I certainly did that - far too much for some of my lecturers who said things like - I knew you would say that - just knew it - forget it for now or see me after class. One was why is the Heaviside function undefined at the discontinuity. I saw him later - he smiled - and said - I will leave that one for you to investigate. I did - but it wasn't until much later I found the answer - it's because if you take the inverse Fourier transform its value is 1/2 there - but explaining that much more advanced area would have taken him too far from the main topic which was at the time differential equations.

    Thanks
    Bill
     
  7. Sep 26, 2017 #6
    By all means tell him about it, but do it face to face in private. Be prepared to explain what you think is correct lest you may have missed the whole point of what he was about.
     
  8. Sep 26, 2017 #7
    Hey, can I talk with you privately, just for a minute, about ...
    Mostly that's OK.
     
  9. Sep 27, 2017 #8

    ISamson

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    Yes, talking privately is the best decision, because then the professor will not be embarrassed in class in front of all the students, but just discuss it with you alone. You will also keep your relationship with him/her with positive.
     
  10. Sep 27, 2017 #9

    FactChecker

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    Taken on its own, the statement that A=4πr2 implies that dA=8πrdr seems correct. And you say that it works in the examples he has done so far. Could you be more specific about what he is doing wrong?
     
  11. Sep 27, 2017 #10

    Orodruin

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    According to the OP, he is claiming that it is the area element involved in a surface integral in connection to Gauss’ theorem. It is not. It is expressing how area changes when the radius changes.
     
  12. Sep 28, 2017 #11

    FactChecker

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    Doesn't that depend on how the charge enclosed is integrated? If he integrates the charge in shells of thickness dr, then that definition of dA seems appropriate in determining the volume containing charge. He is explaining Gauss's law.

    PS. I haven't thought this out in detail, but it seems plausible that the teacher is right.
     
  13. Sep 28, 2017 #12

    Orodruin

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    No. He is referring to a surface element, not the vokume integration. In addition, the volume of the thin shell would be ##4\pi r^2 dr##.
     
  14. Sep 28, 2017 #13

    FactChecker

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    I'm not so sure. Gauss's law involves both and the OP does not give details.
    It's not clear to me what he does with it, but it is true that if A(r) = 4πr2, then dA = 8πrdr. I hesitate to assume what he does with that while explaining Gauss's law.
     
    Last edited: Sep 28, 2017
  15. Sep 28, 2017 #14

    Charles Link

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    The one place that I have seen ## dA=8 \pi r \, dr ## is in writing the equations of a spherical droplet for surface tension, where the increase in surface area requires a certain amount of work ## dW=\gamma \, dA ##. That application shows up in Adkins' book "Equilibrium Thermodynamics" (second edition) on p.39.
     
    Last edited: Sep 28, 2017
  16. Sep 30, 2017 #15

    haushofer

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    But seriously, I'd ask during or after class how in a surface integral you integrate in the r-direction while a surface is described by r=constant. That shows conceptual understanding, I'd say.

    (If I understood the problem correctly)
     
  17. Dec 13, 2017 at 11:11 AM #16

    Mister T

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    That is the starting point for a private discussion in the professor's office. Ask him, don't tell him, if his expression for ##dA## will work in the more general cases you're thinking about, and why you're being taught that particular expression if indeed it won't work there.

    I don't recommend going in there with that attitude. Instead, assume that there's something about it that you don't understand properly and give him the chance to explain his side of it.

    You may not get satisfaction if he is indeed wrong. He may be stubborn. My E&M professor was horrible in that regard. It was obvious that he didn't understand what he was teaching and eventually all the students knew it. For example we had a problem where the Gaussian surface was a cube and a point charge was centered in that cube. He told us that you can't use Gauss's Law to find the total flux of ##\vec{E}## by adding up the flux "coming through the six surfaces because you'll miss the flux coming through the corners". It confused the heck out of me until I figured out he was wrong.

    Edit: Oops! I just noticed that this thread is a few months old. Sorry. :sorry:
     
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