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What does the |-sign mean in this text?

  1. Nov 25, 2014 #1
    http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Exam_tfy4240_Dec_2013.pdf
    solutions: http://web.phys.ntnu.no/~ingves/Teaching/TFY4240/Exam/Solution_tfy4240_Dec_2013.pdf

    In problem 3 a) I'm supposed to reform the integrand from ##\vec{J}(r',t_r)## to something like ##\vec{J}(r'|\omega)##. In practice, one just extracts the ##t_r## dependent part of the function and writes what remains as ##\vec{J}(r'|\omega)##. But what does my professor mean with that ##"|"## bar? Why not just write the remaining part of the function as ##\vec{J}(r')##?

    And excuse me if this is physics notation and not math notation.. In that case pls move this thread to the physics section!
     
    Last edited: Nov 25, 2014
  2. jcsd
  3. Nov 25, 2014 #2

    Doug Huffman

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    Gold Member

    The most general meaning that I know is "evaluated at" or, with super and subscripts, evaluated between. I know it best as indicating conditional probability.
     
  4. Nov 25, 2014 #3
    I don't think it's evaluated at as the ##\omega## dependent part was taken out I think. oh well i guess I'll have to bother my prof and send him an email
     
  5. Nov 25, 2014 #4

    DrClaude

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    Staff: Mentor

    I would interpret that to mean that ##\vec{J}## depends parametrically on ##\omega##.
     
  6. Nov 27, 2014 #5
    That cannot be correct in my specific example. I think..
     
  7. Nov 27, 2014 #6

    DrClaude

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    Why?
     
  8. Nov 27, 2014 #7
    Because in my example the only dependence ##\vec{J}(\vec{r},t)## has on ##\omega## is thru ##e^{i \omega t}##, and ##\vec{J}(\vec{r},t) = \vec{J}(\vec{r}|\omega) ~e^{i \omega t}## according to my professor. This is physics stuff though, not math.
     
  9. Nov 27, 2014 #8

    DrClaude

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    That is not correct. You have that "the only dependence ##\vec{J}(\vec{r},t)## has on ##t## is thru ##e^{i \omega t}##." This form for ##\vec{J}(\vec{r},t)## is to give it an explicit dependence on ##t##, but it certainly doesn't mean that the prefactor is independent of ##\omega##.
     
  10. Nov 27, 2014 #9
    How can that be the case? ##\vec{J}(\vec{r},t)## is given by ##I(t) = I_0 e^{i \omega t}## in the problem.
     
  11. Nov 28, 2014 #10
    He responded to my mail; it means the same as conditional probability, "r given omega " in this case.
     
  12. Dec 1, 2014 #11

    DrClaude

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    Staff: Mentor

    That's very unconventional. But it basically amounts to what I was saying: it means that ##\vec J## depends parametrically on ##\omega##.
     
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