I was just checking on wikipedia about photon spin, and came across the
following paragraph.

"In general, a boson with spin 1 should be observable with three
distinct spin projections (-1, 0 and 1). However, the zero projection
would require a frame where the photon is at rest. Because the (rest)
mass is zero, such a frame does not exist, according to the theory of
relativity."

If the spin of a photon became zero then it would cease to exist. To
conserve spin, particles totalling spin 1 would need to be created,
such as electron-positron, positron-neutrino, or quark-antiquark.

We observe this effect in pair production and the weak interaction, so
my question is:

Is there any evidence proving that pair production is not a result of a
photons spin becoming zero?

"Golden Boar" wrote in message
oups.com...

I was just checking on wikipedia about photon spin, and came across the
following paragraph.

"In general, a boson with spin 1 should be observable with three
distinct spin projections (-1, 0 and 1). However, the zero projection
would require a frame where the photon is at rest. Because the (rest)
mass is zero, such a frame does not exist, according to the theory of
relativity."

If the spin of a photon became zero then it would cease to exist.

No. if an elementary particle has spin, S = 0, or spin
projection, S_z = 0, it's not a photon. A single photon
always has S = 1, but never has S_z = 0.

To
conserve spin, particles totalling spin 1 would need to be created,
such as electron-positron, positron-neutrino, or quark-antiquark.

No. Total angular momentum, J = j1 + j2 is conserved.
if orbital angular momentum is zero, then total spin
S = J = s1 + s2 is conserved, but a pair of photons, as
well as a pair of leptions, can have S = J = s1 + s2 = 0.

We observe this effect in pair production and the weak interaction, so
my question is:

Is there any evidence proving that pair production is not a result of a
photons spin becoming zero?

There are no proofs in physics. Assigned photon intrinsic
spin, S = 1, S_z = 0, is constant with all observations.

Actually, a photon state is described by it's total angular
momentum, J = L + S. A photon's intrinsic spin, S = 1,
and it's orbital angular momentum, L, add as vectors:
J = L + S. L and S aren't good quantum numbers, but
J is. For a given J, L = J + 1 or L = J - 1.

Because the photon possesses transverse electric and
magnetic fields, J = 0 is forbidden for a single photon,
and J = 1 is forbidden for two photons.

Thus, a two photon state, with total angular momentum,
J = 0, as in electron-positron pair creation (lepton singlet
state) is allowed and is observed.

[Old Man]

"Old Man" wrote in message
...

"Golden Boar" wrote in message
oups.com...

I was just checking on wikipedia about photon spin, and came across the
following paragraph.

"In general, a boson with spin 1 should be observable with three
distinct spin projections (-1, 0 and 1). However, the zero projection
would require a frame where the photon is at rest. Because the (rest)
mass is zero, such a frame does not exist, according to the theory of
relativity."

If the spin of a photon became zero then it would cease to exist.

No. if an elementary particle has spin, S = 0, or spin
projection, S_z = 0, it's not a photon. A single photon
always has S = 1, but never has S_z = 0.

To
conserve spin, particles totalling spin 1 would need to be created,
such as electron-positron, positron-neutrino, or quark-antiquark.

No. Total angular momentum, J = j1 + j2 is conserved.
if orbital angular momentum is zero, then total spin
S = J = s1 + s2 is conserved, but a pair of photons, as
well as a pair of leptions, can have S = J = s1 + s2 = 0.

We observe this effect in pair production and the weak interaction, so
my question is:

Is there any evidence proving that pair production is not a result of a
photons spin becoming zero?

There are no proofs in physics. Assigned photon intrinsic
spin, S = 1, S_z = 0, is constant with all observations.

Actually, a photon state is described by it's total angular
momentum, J = L + S. A photon's intrinsic spin, S = 1,
and it's orbital angular momentum, L, add as vectors:
J = L + S. L and S aren't good quantum numbers, but
J is. For a given J, L = J + 1 or L = J - 1.

Because the photon possesses transverse electric and
magnetic fields, J = 0 is forbidden for a single photon,
and J = 1 is forbidden for two photons.

Thus, a two photon state, with total angular momentum,
J = 0, as in electron-positron pair creation (lepton singlet
state) is allowed and is observed.

In addition,

a system of two photons (in electron-positron annihilation)
has invariant mass. Thus a center-of-momentum frame exists
wherein transverse spin, J_z = S_z = 0 is allowed.

[Old Man]

[Old Man]