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

On Jun 14, 10:23 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

This a bit like asking why an insect has six legs and why a quadruped
has four legs.
Particles are *classified* according to their observed properties.
It is observed that some particles are vector bosons, which by
definition means that they have three degrees of freedom. It is right
to ask how being massless (like the photon) removes one of those
degrees of freedom.

If you are reading Ryder's textbook and have gotten stuck on this
point, then you are reading above your level of preparation and need
to read more introductor materials. I suggest the Feynman Lectures on
Physics for a start. There will be about four other books you will
need to read before you get to Ryder's book.

PD

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!

I know almost nothing about this, which would seem to place us on
equal footing. I read PD's reply, and I don't think your question is
like asking why a mouse has four legs, a starfish five, and an insect
six: it's more like asking, given some creatures with _no_ obvious
legs, why we say that this one has three, this one four, and so forth.

I think it's clear why a scalar field has one degree of freedom --
there is exactly one number to specify at each point (actually, an
infinite number of degrees of freedom, if we are talking about the
complete field).

It's not so clear why light would have two degrees of freedom at a
point. I presume this involves polarization state an intensity,
speaking macroscopically. What I'm not sure is why we don't get yet
more degrees of freedom for the wavevector -- which direction the
light is going.

I take it this is the kind of insight you were looking for. Three
degrees? I don't know. Polarization state and angular momentum, which
is possibly fixed by other considerations for a zero-mass particle.

[We actually have no two-legged animals -- if you want to count limbs
instead of "legs", leaving us with 4,5,6,8 and 2n. 2n covers
creatures like the king crab, which have an obscene number of
appendages, all waving and groping and carrying on in seemingly
autonomous purpose -- but presumably in pairs. Can we ask "why" 5 is
the only odd number? Why not 3, or 7?]

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.

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.

Sincerely Barrow

On Jun 14, 11:52 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?

It doesn't have to. Any gauge may be chosen. The Coulomb gauge just
provides a convenient way of looking at it. And the Coulomb gauge is
simply the Lorentz gauge where the scalar potential is set equal to
zero.

On Jun 14, 2:01 pm, 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.

You know, I concede that one is indeed a tautology: asking why a
scalar field has one degree of freedom is like asking why a monopod
has one foot.

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.

Hmm... so I was right that it had something to do with polarization
state, but wrong in guessing how the third degree of freedom came into
play (maybe). I notice that the direction of the wave vector, which
we might think gives us several more degrees of freedom, doesn't count
for some reason.

In this scheme is there implicitly a local propagation vector --
something like a fluid velocity -- built into the one dimensional
case, which also doens't count?

On Jun 17, 10:50 am, Edward Green wrote:
On Jun 14, 2:01 pm, 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.

You know, I concede that one is indeed a tautology: asking why a
scalar field has one degree of freedom is like asking why a monopod
has one foot.

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.

Hmm... so I was right that it had something to do with polarization
state, but wrong in guessing how the third degree of freedom came into
play (maybe). I notice that the direction of the wave vector, which
we might think gives us several more degrees of freedom, doesn't count
for some reason.

In this scheme is there implicitly a local propagation vector --
something like a fluid velocity -- built into the one dimensional
case, which also doens't count?

Particles in one dimension would have no choice but to be polarized
longitudinally, if that's what you asking. So massless gauge
particles in one dimension would probably make no sense, but then
restricting things to just one dimension is usually begging for
problems to begin with. Interesting physics doesn't usually result
until you talk about two of them.

On Jun 17, 1:33 pm, Igor wrote:
On Jun 17, 10:50 am, Edward Green wrote:

In this scheme is there implicitly a local propagation vector --
something like a fluid velocity -- built into the one dimensional
case, which also doens't count?

Particles in one dimension would have no choice but to be polarized
longitudinally, if that's what you asking. So massless gauge
particles in one dimension would probably make no sense, but then
restricting things to just one dimension is usually begging for
problems to begin with. Interesting physics doesn't usually result
until you talk about two of them.

Poor phrasing. I should have said "degrees of freedom".

I was noting that if we roughly said the electromagnetic field had two
degrees of freedom, then we are apparently regarding k as a given. I
was wondering is something similar were implicit in a "scalar field".

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

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar