On Jun 14, 8:52 pm, Barrow wrote:
On 6 15 , 2 01 , Igor wrote:
On Jun 14, 11:23 am, Barrow wrote:

Dear all,
Why is the degree of freedom of a scalar field equal to 1?
Why does the massless photon(vector boson) have degrees of freedom
2? and Why is the massive photon degrees of freedom equal to 3??

I don't figure out any physical meaning of the above facts.
Actually, I read these from the introduction of higgs phonemena from
the text book by Mr. Ryder. Any instructions will be appreciated!
Sincerely Barrow

The scalar having just one degree of freedom is true by definition.
Scalars are defined as just numbers.

A vector boson, on the other hand, has at most three degrees of
freedom, corresponding to the three independent directions in which it
might be polarized. Those correspond to two independent transverse
degrees and one longitudinal degree of freedom, defined as
oscillations perpendicular or parallel to the direction of wave
motion, respectively. Massless gauge particles obey the Coulomb gauge
condition del dot A = 0. This is equivalent to saying k dot A = 0,
where k is the wave number vector. This says that k must be
perpendicular to A at all points and hence the wave is transverse, and
consists of only two degrees of freedom. The gauge condition for a
massive particle has del dot A being proportional to the particle's
mass. So the massive particle corresponds to a wave having both
transverse and longitudinal components, hence three degrees of
freedom.

Many thanks! I think I got the roughly picture of why the massless
vector boson has 2 degrees of freedom.
But allowing me to ask still one more question. When I learned the
Coulomb gauge in Electromagnetic course, I remember that the chosen of
Coulomb gauge is just to decouple the equation of scalar potential and
vector potential. It's free to choose other gauges, say, Lorentz gauge
can also be chosen. Why does the massless gauge particles have to obey
the Coulomb gauge del dot A = 0?

Nevertheless, if the massless vector boson has to choose the Coulomb
gauge, thus I can understand it can only be transverse wave.

it is not that the coulomb gauge is necessary

it is only that
in the coulomb gauge it is easy to see
that there are only two degrees of freedom

the gauge doesn't affect the physical property of degrees of freedom

a geometric illustration might help

if you have a plane in R^3 described by the constraint
x + 2y -7z = 2

the plane only has 2 degrees of freedom
even though it is embedded in 3 dimensional space

you can translate the coordinate system
though
and rotate the coordinate system
and the degrees of freedom of a figure are invariant

so when we discover that in one translation and rotation
that constraint becomes z = 0
then we see a coordinate system where only x and y
completely free
give coordinates on the figure

but there are more powerful constraints theorems
that show in these circumstances that
the mere existence of those types of constraints
reduce the degrees of freedom by one

that constraint for the polarisation is a light cone constraint
the coulomb gauge is the natural setting to explore light cone
constraints
but the same constraint exists in all the gauges

its just that the coulomb gauge always works with transverse states
so we get nice little equations like

e . q = 0
for polarisation e
and 4momentum q

and the constraint is obvious

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galathaea: prankster, fablist, magician, liar