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    Relativistic Compton Scattering Calculation

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    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please see attached picture. Note that the formula for Compton scattering is wrong.

    © BrainMass Inc. brainmass.com October 2, 2022, 6:33 pm ad1c9bdddf
    https://brainmass.com/physics/nuclear-forces/relativistic-compton-scattering-calculation-592135

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    A condensed version of the solution is provided below. See attachment for equations.

    The current density vector J is defined as the change in charge per unit area per unit time.
    So, if we integrate the current density over a surface S, we get the change in charge per unit time on that surface - the current flowing through the surface.
    We write it as:
    (1.1)
    Now assume we have a surface that encloses a volume V and we have charges flowing through this volume.

    If the number of charges that go into the volume (a cube in this case) is equal to the number of charges that leave the cube at the same time, then there is no change in the charges per unit time on the entire surface.
    In this case, and (1.1) becomes
    (1.2)
    Now, let's say that some charges do not leave the volume at the same rate of charges that enter the volume.
    This means that we have a charge build-up inside the volume (for example, in the drawing above, per unit time 5 charges enter the enclosed surface but only three leave through the surface, having a volume charge density of 2 charges per volume per unit time).
    So we have:
    (1.3)
    Now, anytime we have free charges floating around enclosed surface, we immediately think of Gauss law - the electric flux through a closed surface is proportional to the enclosed charges:
    (1.4)
    Taking the derivative of both sides with respect to time, we get:
    (1.5)
    But this change in enclosed charge is exactly the change in charge due to the difference in the current density over the surface, hence we can add (1.3) and (1.5) to write:
    (1.6)
    If we assume that the enclosed surface does not change with time, we can bring the time derivative into the integral (The electric field is the only time dependent quantity there):
    (1.7)
    Since we integrate over the same closed surface, the two integrals can be combined

    (1.8)
    If we now "redefine" the concept of current density and define it as:
    (1.9)
    Now, lets "squash" our cube, so it looks like this:

    The contribution to integral (1.8) from the squashed sides disappear, and the integral becomes a simple surface integral. We integrate the "new" current density over the surface area and we get the current through the surface:
    (1.10)
    And now we can apply this current to Ampere's law:
    (1.11).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 2, 2022, 6:33 pm ad1c9bdddf>
    https://brainmass.com/physics/nuclear-forces/relativistic-compton-scattering-calculation-592135

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