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I Wu Experiment and Parity Violation

  1. Apr 27, 2017 #1
    I have a question concerning the nature of Ms. Wu's experiment confirming parity violation. I'm very familiar with this experiment and its outcomes, but the setup of the experiment itself, alludes me.
    Wu found that the electron's emitted from the Cobalt-60 atom always went in the direction opposite to their spin. My misunderstanding; I do not understand what was the independent variable of the experiment. What did she change to observe that parity was violated? http://www.doublexscience.org/wp-content/uploads/2014/01/Wu_experiment.png This picture accurately describes my dilemma. The "parity transformation" is what I don't get, it seems theoretical and not practical or even experimental. Yes, we expect that the electrons in the mirrored world would fly in the direction described by the top right of the image. How do we know the electrons don't do that? Without something changing in the experiment, could you know? The image also shows the magnetic field pointed in the same direction so she couldn't have changed that?
    I understand parity operations and why this result would violate, I just don't see how it was done experimentally. However, perhaps I don't see how they did it experimentally because I'm misunderstanding parity operations.
     
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  3. Apr 28, 2017 #2

    vanhees71

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    The point is that you can't do the space reflection and time-reversal transformations in the "active sense". It's available only in the theory. Rather, you have to use observables to check whether parity is conserved or not. Wikipedia gives a quite good description:

    https://en.wikipedia.org/wiki/Wu_experiment#The_experiment
     
  4. Apr 28, 2017 #3

    PeterDonis

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    I'm not sure if that's correct. The diagram at the right of the Wikipedia article that @vanhees71 linked to shows the solenoid coils with opposite handedness in the "mirror image" version; that seems to me to indicate that the field was reversed.
     
  5. Apr 29, 2017 #4

    vanhees71

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    Let's see. It's always a bit complicated to define the correct transformations for all quantities. I look at parity in the usual sense, i.e., the space-time coordinates transform under parity as
    $$t \rightarrow t'=t, \quad \vec{x} \rightarrow \vec{x}'=-\vec{x}.$$
    That's different from the mirror drawn in the Wikipedia, which is the parity operation followed by a 180-degree rotation.

    Now let's see, how the various quantities transform. Let's start with kinematics of point particles. It's sufficient to argue within Newtonian physics. Obviously ##\vec{v}=\dot{\vec{x}}## and ##\vec{a}=\ddot{\vec{x}}## transform as vectors, i.e.,
    $$\vec{v} \rightarrow \vec{v}'=-\vec{v}, \quad \vec{a} \rightarrow \vec{a}'=-\vec{a}.$$
    Now to make the Lagrangian of a free particle,
    $$L=\frac{m}{2} \vec{v}^2$$
    invariant under parity trafos, you have to set
    $$m \rightarrow m'=m,$$
    i.e., mass is a scalar under parity.

    Now according to Lex Sec. for an object with constant mass you have
    $$\vec{F}=m \vec{a} \; \Rightarrow \; \vec{F} \rightarrow \vec{F}'=-\vec{F}.$$
    Now let's go about electromagnetic quantities (I'm using the Heaviside-Lorentz units). To make contact with the mechanical quantities of point particles we think about a many-body system of charged particles and treat them in the usual continuum-mechanics way. So let ##\vec{v}(t,\vec{x})## denote the velocity field of the charged fluid (plasma) and ##n(t,\vec{x})## it's particle-number density. We've already seen that ##\vec{v}## is a vector field, i.e.,
    $$\vec{v}(t,\vec{x}) \rightarrow \vec{v}'(t',\vec{x}')=-\vec{v}(t,\vec{x})=-\vec{v}(t',-\vec{x}').$$
    The particle-number density is obviously a scalar field,
    $$n(t,\vec{x}) \rightarrow n'(t',\vec{x}')=n(t,\vec{x})=n(t',-\vec{x}').$$
    Now we assume the electric charge to be a scalar too. We'll see that then electrodynamics allows for parity transformations and is parity symmetric. To see this we note that the charge density and current density are ##\rho=q \vec{n}## and ##\vec{j}=q \vec{n} \vec{v}##, i.e.,
    $$\rho(t,\vec{x}) \rightarrow \rho'(t',\vec{x}')=\rho(t,\vec{x})=-\rho(t',-\vec{x}'), \quad \vec{j}(t,\vec{x}) \rightarrow \vec{j}'(t',\vec{x}')=-\vec{j}(t,\vec{x})=-\vec{j}(t',-\vec{x}').$$
    Now we can look at the Maxwell equations. The nabla operator is a vector
    $$\vec{\nabla} \rightarrow \vec{\nabla}'=-\vec{\nabla}.$$
    So from
    $$\vec{\nabla} \cdot \vec{E}=\rho$$
    we see that in order to make this equation parity invariant we have to assume that ##\vec{E}## is a vector field, i.e.,
    $$\vec{E}(t,\vec{x}) \rightarrow \vec{E}'(t',\vec{x}')=-\vec{E}(t,\vec{x})=-\vec{E}(t',-\vec{x}').$$
    For the magnetic field we have the Ampere-Maxwell Law,
    $$\vec{\nabla} \times \vec{B}+\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}.$$
    The right-hand side is a vector. On the left-hand side the term involving ##\vec{E}## is a vector too. Thus, also the term involving the magnetic field must be a vector and thus ##\vec{B}## is a pseudo vector
    $$\vec{B}(t,\vec{x}) \rightarrow \vec{B}'(t',\vec{x}')=+\vec{B}(t,\vec{x})=\vec{B}(t',-\vec{x}').$$
     
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