I wasn't sure where this question should go, so I'm posting it here.

 

What's a good example of a particle that fissions to produce two particles with opposite spins?

It's not fission, but

 

[math]\pi^0 \rightarrow e^- + e^+ + \gamma[/math]

[math]\Psi^0 \rightarrow e^- + e^+ [/math]

[math]\Psi^0 \rightarrow \mu^- + \mu^+ [/math]

[math]\Upsilon^0 \rightarrow e^- + e^+ [/math]

[math]\Upsilon^0 \rightarrow \mu^- + \mu^+ [/math]

 

True fission has lots of constituent particles, so a spin 0 parent can still give you spin 0 products

It's not fission, but

 

[math]\pi^0 \rightarrow e^- + e^+ + \gamma[/math]

[math]\Psi^0 \rightarrow e^- + e^+ [/math]

[math]\Psi^0 \rightarrow \mu^- + \mu^+ [/math]

[math]\Upsilon^0 \rightarrow e^- + e^+ [/math]

[math]\Upsilon^0 \rightarrow \mu^- + \mu^+ [/math]

 

True fission has lots of constituent particles, so a spin 0 parent can still give you spin 0 products

 

Thanks swansont. What is [math]\Psi^0[/math]?

Thanks swansont. What is [math]\Psi^0[/math]?

 

It's called the J/psi particle (pause to look this up: two different discoverers, one at Brookhaven, the other Stanford so I guess they named it differently) and is made up of a charm/anticharm quark pair, so it's a meson.

I see that the right side of the equations give two opposite charge. Does it imply that the charge and the spin are corolated ?

Thanks

No, the two opposite charges simply reflect conservation of charge.

Doesn't the electron and the positron have the same spin (1/2) ?

well, electrons can have spin +1/2 or -1/2 same with the positron.

other wise orbitals would only hold 1 electon.

 

it is as athiest says, the conservation of charge.

It's called the J/psi particle (pause to look this up: two different discoverers, one at Brookhaven, the other Stanford so I guess they named it differently) and is made up of a charm/anticharm quark pair, so it's a meson.

 

That's interesting. It means that the electron and positron are never "in" the J/psi particle. The J/psi particle starts out as a pair of quarks and then becomes an electron and a positron.

Doesn't the electron and the positron have the same spin (1/2) ?

 

Same spin but not same orientation in these cases. Spin 1/2 (magnitude of the spin vector) but if the parent is spin 0, then not the same projection on the z-axis of your coordinate system: one would be spin up (+1/2) and the other spin down (-1/2)

If a particle had spin and was moving at C, does the spin have to align perpendicular to motion to avoid the surface going faster than C? In other words, if it spun like a wheel with the hub moving at C, the surface would cyclically get ahead of the hub and need to exceed C. In a perpendicular spin, the surface stays at the same translation postion of the hub.

just a quicky, but this stuff rings a bell, isn`t this what Paul Dirac worked on?

If a particle had spin and was moving at C, does the spin have to align perpendicular to motion to avoid the surface going faster than C? In other words, if it spun like a wheel with the hub moving at C, the surface would cyclically get ahead of the hub and need to exceed C. In a perpendicular spin, the surface stays at the same translation postion of the hub.

 

Spin isn't physical motion, it's also known as "intrinsic angular momentum."

If a particle had spin and was moving at C, does the spin have to align perpendicular to motion to avoid the surface going faster than C? In other words, if it spun like a wheel with the hub moving at C, the surface would cyclically get ahead of the hub and need to exceed C. In a perpendicular spin, the surface stays at the same translation postion of the hub.

 

Notwithstanding what swansont said, there really wouldn't be any difference between the speed of the center of the particle vs. the speed of its surface if it was a point particle.

 

Also, if you think about it, perpendicular spin would also result in FTL speed. Any point on the surface would be going at least as fast as the center, but you have to add to that the circular motion of that point around the center, which, when put together, would amount to diagonal motion (whose path would trace out a helix). If one of the component vectors of diagonal motion is already going at C, the diagonal motion itself would have to be going faster than C.

If a particle had spin and was moving at C, does the spin have to align perpendicular to motion to avoid the surface going faster than C? In other words, if it spun like a wheel with the hub moving at C, the surface would cyclically get ahead of the hub and need to exceed C. In a perpendicular spin, the surface stays at the same translation postion of the hub.

 

This is an important point to remember: Spin is not a classical quantity! It Take the classical electron radius, which is the Compton wavelength of the electron:

[math]r_e=\frac{1}{4 \pi \epsilon_0}\frac{e^2}{m_ec^2}[/math],

and the ``spin'' angular momentum:

[math]m_ev_0r_e = \frac{\hbar}{2}[/math].

 

No calculate [math]v_0[/math], which is the speed at the ``equator'' of an electron.

 

[math]v_0=\frac{\hbar}{2m_e r_e}[/math]

[math]\Rightarrow v_0 = \frac{4\pi\epsilon_0\hbar c^2}{2e^2}.[/math]

 

But we can define the fine structure constant [math]\alpha[/math], so that

 

[math]\Rightarrow v_0 = \frac{c}{2\alpha}[/math].

 

We know very well that [math]\alpha = \frac{1}{137}[/math], which tells us that the electron spins 260 (ish) times faster than the speed of light.

 

Here I treated the electron as a sphere of charge, but as someone already pointed out, the electron is a point particle. Because a point is zero dimensional, there is no concept of it spinning.