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Derivation of the Laue Equation -- Problem with path difference

  1. Sep 1, 2014 #1
    Hi everyone

    1. The problem statement, all variables and given/known data
    I'm trying to understand the Laue equation but I have problems unterstanding the derivation, especially one part in my text book about the path difference (see attachment)



    3. The attempt at a solution

    My text book says that the path difference here is:
    [tex] \Delta s = \vec T \hat{s_0} - \vec T \hat{s} [/tex]

    But why is there a minus? I thought there has to be a plus (like in the derivation of the Bragg equation) because when I look at the picture in the attachment, the ray which is on bottom has to travel dS1 and dS2 in addition. Where is my mistake?

    Thanks for your help everyone. If anything is unclear just ask, I know my English is not the best.
     

    Attached Files:

  2. jcsd
  3. Sep 1, 2014 #2

    TSny

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    Hello.

    Note the signs of the dot products ##\vec{T} \cdot \hat{s}_0## and ##\vec{T} \cdot \hat{s}'##.

    In particular, for your diagram, is ##\Delta s_1 = \vec{T} \cdot \hat{s}_0## a positive quantity or a negative quantity?
     
  4. Sep 1, 2014 #3
    I think it's positive. I didn't create the sketch, I took it out of my textbook. I still don't understand why dS2 is negative.
     
  5. Sep 1, 2014 #4

    nrqed

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    What can you say about the angle between ##\vec{T}## and ##\vec{s}_0##? (Not a precise value, just a range). What can you say about the dot product between vectors with an angle in that range?
     
  6. Sep 2, 2014 #5
    I know when the angle is between 0 and 90 degrees the dot product is positive. for 90 to 180 degrees it's negative. This may sound stupid, but I have a problem understanding this when I take a look at the sketch because the tip of the S0 vector touches the T vector. How do I cope with this? It's quite embarrassing that I don't know how I calculate a dot product this way, but I've never stumpled upon such a problem. I add an attachment so you know what I mean by "when the tip of one vector touches the other".

    Thanks for your help
     

    Attached Files:

    • vec.png
      vec.png
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  7. Sep 2, 2014 #6

    vanhees71

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    The angle between vectors as it occurs in the cross product, [itex]\vec{x} \cdot \vec{y}=|\vec{x}| |\vec{y}| \cos \theta[/itex] with [itex]\theta \in [0,\pi][/itex] is defined to be the angle between the vectors with their starting points coinciding, i.e., in your figure you have to parallel shift one of the vectors such that it's starting point coincides with the starting point of the other vector. Then the angle at this pivot point is the one relevant in the scalar product! In your drawing this angle is obviously between [itex]\pi/2[/itex] and [itex]\pi[/itex] and thus the scalar product id negative (because the cosine is negative in this range).
     
  8. Sep 2, 2014 #7
    So the dot product of T and S0 in my first attachment is negative and the dot product of T and S' is positive? If that's the case I think my textbook is wrong :eek:
     
  9. Sep 2, 2014 #8

    TSny

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    Yes.

    You might be right. In the diagram, Δs = Δs1 - Δs2 would be a negative number, whereas the diagram seems to be deriving an expression for the positive extra distance that the bottom ray travels relative to the top ray.

    But note that the magnitude of Δs = Δs1 - Δs2 does give the correct magnitude of the path difference of the two rays. So, it depends on how the text makes use of the expression for Δs as to whether the overall derivation of the Laue equation is wrong.
     
  10. Sep 2, 2014 #9
    Alright, I got it now. Thank your very much for your help
     
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