# The Stern Gerlach Experiment Revisited

## Is it possible to provide an explanation to the observations of the Stern Gerlach Experiment using the classical theories?

Some Considerations:

We consider the standard set-up for the Stern-Gerlach experiment. The predominant component of $\vec{B}$ is $B_{z}$ Again $B_{z}$ varies most strongly with changes in z

$$\vec{F}=\nabla(\vec{\mu}.\vec{B})\approx\vec{e}_{z}\mu_{z}\frac{\partial B_{z}}{\partial z}=\vec{e}_{z}F_{z}$$

The force acting on the electrons is supposed to cause the deflection.This causes acceleration in the z direction and hence an increase in the KE in the z direction. The total KE of the electron(in consideration of the three directions) cannot change since magnetic field can only curve the path of an electron. It cannot change the magnitude of speed.Increase of speed in the z direction may be compensated by decrease of speed in the x or in the y direction Changes in the value of $B_z$ due to the accelerated motion of the electrons,is accompanied by the creation of an electric field: $$Curl{\;} \vec{E}=-\frac{\partial \vec{ B}}{\partial t}$$ Decrease of magnetic energy= increase in electrical energy, if total KE remains unchanged for each particle. When the particles pass out of the region of interaction with the magnetic field the electrical energy restores the energy of the magnetic field.

Prior to this, while the interaction is going on , may write the curl B equation as: $$\int\vec{E}.\vec{dl}=-\frac{d}{dt}\int\int\vec{B}.\vec{ds}$$ The integral on LHS is a closed line integral whose plane is in the x-y direction.The electrons seem to get accelerated in the x-y direction due to the emf in action and this should tend to restore the acceleration in the z-direction. The electrical effect is just a temporary effect.

Now,the greater the amount of deflecting force , due to higher value of $\mu_{z}$ ,greater the decrease in magnetic energy and greater the amount of acc in the x-y plane.The restoring effect becomes stronger for larger values of magnetic moment in the z-direction.Incidentally for each value/magnitude of $\mu_{z}$ we have to consider the two directions the ,+ve z and the -ve z directions.

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You may imagine repeating the experiment with two sets of electrons at different times ,each set having its own constant value of $\mu_z$ for the different electrons. But the two sets have two different values for $\mu_z$.The one with a larger value of $\mu_z$ will suffer greater acceleration in the z-direction. But the restoring action will also be greater as indicated in the posting above. – Anamitra Palit Aug 9 '12 at 18:49
An important point of concern is physical origin of the retarding force in the x-y plane. Work done by total magnetic force over any infinitesimal path should be zero,for each electron individually. – Anamitra Palit Aug 10 '12 at 2:35
How is the retarding force (in the above comment) generated by spin?if $B_x=B_y\approx 0$ then acceleration in the x and the y-directions due to spin alone should be zero. – Anamitra Palit Aug 10 '12 at 2:56

The enormous power of this experiment is the excluded middle. A classical object going into analyzing region could have its magnetic moment in the X--Y plane generating no analyzing acceleration.

When you get two regions and an excluded middle you have discovered a non-classical behavior.

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Classical theory fails to predict the "excluded middle". This goes in conjunction with the "classic" exclusion or unaccountability of how spin generates retardation in the x-y plane if $B_x=B_y=0$. Work done by magnetic field for any infinitesimal path is zero. This idea is there in the classical theories and needs to be respected in any type of classical reasoning. – Anamitra Palit Aug 10 '12 at 3:06
$\vec{F}=\nabla(\vec{\mu}.\vec{B})\approx\vec{e}_{z}\mu_{z}\frac{\partial B_{z}}{\partial z}=\vec{e}_{z}F_{z}$ Force corresponding to the above formula acts along the direction of $\vec{B}=\vec{e_z}B_z$, if $B_y=B_z=0$ The above force is not identical with the force given by $F=q(\vec{v}\times \vec{B})$. Magnetic force cannot act along the magnetic line of force – Anamitra Palit Aug 10 '12 at 15:27
The point is that there is no force if there is no $\mu_z$, ie if the magnetic moment is aligned in the x-y plane. – James Aug 11 '12 at 7:13
"there is no force if there is no $\mu_z$" Correct. So the classical analysis implies a probability for atoms to occupy the center of the figure and experiment shows that they do not. The classical analysis is incomplete and insufficient to explain the observation. – dmckee Aug 11 '12 at 13:38

Let's consider classically the problem of a spinning charged sphere passing through a magnetic field given by:

$B_x=B_y=k$ where k is a small constant

$B_z$ varies strongly in the z-direction.

That is $\frac{\partial B_z}{\partial z}$ is a large constant=C

First we consider the effects of translation

Force on the charged sphere:

$\vec{F}=q(\vec{v}\times\vec{B})$

$$F_z=v_xB_y-v_yB_x$$

In consideration of thermal motion we have:

Average speed of the electron in the +ve x-direction=Time average speed in the -ve x-direction=$A$

Average speed of the electron in the +ve y-direction=Time average speed in the -ve y-direction=$A$

Mean speed (thermal) considering all directions:

$\langle v_x\rangle=\langle v_y\rangle=0$

In the x-direction motion is given by $V+\langle {v_x}\rangle= V$

$V$ represents mass motion along the x-axis.

$v_x$ in the formula for $F_z$ considers both mass motion and thermal motion

It would be interesting to consider the particle flux density in a particular direction say the + y direction.

We have $j_{+y}=\frac{1}{4} nv_{m}$ Where, $n$ is the concentration of particles

$v_m=\sqrt{\frac{8kT}{\pi m_0}}\approx 1.59\sqrt{\frac{kT}{m}}=A$

$j_{+y}$ is the particle flux density in the +ve y direction. Current Density:

$\vec{J}=-e\vec{j}$

$\vec{j}$ is the particle flux density at a point in some particular sense of direction.$\vec{J}$ is the current density

Therefore current density in +ve y direction:

$$J_{+y}=-e j_{+y}$$

And

Similarly, Current density in -ve y direction:

$${J_{-y}}=-e{j_{-y}}$$

The two corresponding currents (equal in magnitude but opposite in direction)long the positive and the negative directions of the y axis suffer equal and opposite deflections producing the two lines on the screen

Classically we are supposed to have only two lines on the screen with an "excluded middle region". This is in consideration of translation only.

Now let's come to the aspect of spin.We may consider a sphere spinning with its axis inclined at an angle $\theta$ wrt the z-axis.Net force on a rotating charged ring rotating in magnetic field is zero for all components.

$\vec{dF_R}=i(\vec{dl}\times\vec{B})$

$\vec{F_R}=i(\int \vec{dl})\times \vec{B}$ [Suffix R stands for "ring" and "i" stands for current]

$=0$

Assuming that B does not vary strongly over the ring, considering the dimensions of an electron

If variations of B over the ring are considered the two lines on the screen would perhaps become a bit broader------I suppose.

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You do realize that the analyzing direction is the one over which the field varies, right? Is there something in particular that you are trying to accomplish here? – dmckee Aug 11 '12 at 4:40
Do you find anything counter-intuitive with the above considerations(ie, the answer provided)?Classical prejudice seems to be working out excellently ,at least with the problem at hand---the Stern-Gerlach Experiment. – Anamitra Palit Aug 11 '12 at 4:47
We may take $\langle v_x \rangle=\langle v_y \rangle$. They correspond to internal motion.The averages masy be taken over time or space and they should be equal in consideration of the Ergodicity principle – Anamitra Palit Aug 11 '12 at 5:11

We examine the same problem of a charged sphere passsing through a magnetic field given by:

$B_x=B_y=k(const)$

$B_z=C_1 z+C_2$

So that

$\frac{\partial B_z}{\partial z}=C_1(large{\;\;\;} constant)$

It is assumed that $C_1>>k$

Lorentz Force:

$$\vec{F}=q(\vec{v}\times \vec{B})$$ $$F_x=q(v_y B_z-v_zB_y)$$ $$F_y=q(v_z B_x-v_xB_z)$$ $$F_z=q(v_x B_y-v_zB_x)$$ Now, $$F_x=q(C_1 z+C_2)\frac{dy}{dt}-qk\frac{dz}{dt} {\;\;\;\;\;\;\;}(1)$$
$$F_y= qk\frac{dz}{dt}- q(C_1 z+C_2)\frac{dx}{dt}{\;\;\;\;\;\;\;\;\;}(2)$$
$$F_z= qk\frac{dx}{dt}-qk\frac{dy}{dt}{\;\;\;\;\;\;\;\;\;}(3)$$

From (1) and (2): $$m \frac{d^2}{dt^2}(x+y)=C_1qz \frac{d}{dt}(y-x)+C_2q\frac{d}{dt}(y-x){\;\;\;\;\;\;\;\;\;}(4)$$

We may write (3) as:

$$m\frac{d^2 z}{dt^2}=- kq \frac{d}{dt}(y-x){\;\;\;\;\;\;\;\;\;}(5)$$

Or $$\frac{d}{dt}(y-x)= -\frac{m}{kq}\frac{d^2 z}{dt^2}{\;\;\;\;\;\;\;\;\;}(6)$$

The plus/minus sign in (6) is due to the fact that RHS of (5) may be positive or negative.

Substituting (6) in(4) we have:

$$\frac{d^2 (x+y)}{dt^2}=-(\frac{C_1}{k}z+\frac{C_2}{k})\frac{d^2 z}{dt^2}{\;\;\;\;\;\;\;\;\;}(7)$$

$$\frac{d^2}{dt^2}[(x+y)+\frac{C_1}{k}z+\frac{C_2}{k}]=0$$ Or, $$x+y+\frac{C_1}{k}z+\frac{C_2}{k}=At+B{\;\;\;\;\;\;\;\;\;}(8)$$ Initial conditions: We have to use the initial values of ,x,y,z and their derivatives it,$\frac{dx}{dt}$,$\frac{dy}{dt}$and $\frac{dz}{dt}$ for the evaluation of the constants. For $\frac{dy}{dt}$ and $\frac{dz}{dt}$ we should consider the most probable thermal speed$=\sqrt{\frac{2kT}{m_0}}\approx 1.41 \sqrt{\frac{kT}{m_0}}$

Incidentally this value is fairly close to the mean speed $=\sqrt{ \frac{8kT}{m_0 \pi}}=1.59 \sqrt{ \frac{kT}{m_0 }}$

The most probable speed and the mean speed due to thermal reasons are same for all directions. They are not influenced by the mass motion of the particles. In deciding the initial constants we have to consider both the positive and the negative values of the most probable or the mean speed which are same for all the particles. Since $\frac{C_1}{k}$ is a very large quantity the initial condition for $\frac{dz}{dt}$ will predominate. For x direction the mass motion should predominate over the mean value or the most probable value . Otherwise the gas would not move through the space between the magnet poles. So we have two solutions[due to differences in the constants] which are identical sets of particles—those having thermal motion in the positive z-direction and those having thermal motion in the negative z-direction. Time is considered zero when a particle reaches the piont x=0 on the y-z plane.The initial values of y and z have an effect on the constant B apart from x=0 condition. But The constant B would be of a much dominating nature: $B=y_0+\frac{C_1}{k}z_0$.

$A=V\pm v_m\pm v_m\pm\frac{C_1}{k}v_m$

Or, $A=\approx V\pm v_m \frac{C_1}{k}$

Since $\frac{C_1}{k}$ a large quantity.

Finally we have: $$x+y+\frac{C_1}{k}z+\frac{C_2}{k}=(V\pm v_m \frac{C_1}{k})t$$

The effect of the constant $B$ gets swamped off by the thage value of $\frac{C_1}{k}$ and by the multiplicative action of time.

There are two possible solutions for z for particular instant of time

It would be interesting to consider neutral particles(example:neutral atoms)which have some effective spin.The relative distances and the orientation of the charges do not change with time.

Due to the effect of spin the values of velocity $\vec{v_i}$ are different for the different charges.

Magnetic force: on the positive charges:$\vec{F_m}=+q(\Sigma\vec{v_i}\times \vec{B}) Magnetic force: on the negative charges:$\vec{F_m}=-q(\Sigma\vec{v'_i}\times \vec{B})

Net Magnetic force$=q((\Sigma\vec{v_i}-\Sigma\vec{v’_i})\times \vec{B})$

Using the previous concepts(average speed , most probable speed their proximity to each other etc we may derive the same conclusion as before.Here B has been assumed to be constant over the dimensions of the atom.But even if B is assumed to vary over the dimensions of the neutral atoms we will have some net Lorentz force to work out the problem in a more or less similar manner

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In so far as the last equation in the answer is concerned, the value of x is the same for all particles passing though the region of interaction of the magnetic field. The value of y is confined to the dimemsions of the region of interaction. Its becomes negligible due to the large vale of $\frac{C_1}{k}$. We have in effect:$\frac{C_1}{k}z+\frac{c_2}{k}=(V\pm v_m \frac{C_1}{k})t$ – Anamitra Palit Aug 12 '12 at 11:18