Property of photon

[rAgAv]

Okay guys.This is a serious question.

What is the property of a photon which makes it to move in a constant
velocity?

Thanx in advance.
rAgAv.

[Sean Case]

In article <1147616828.294066.163490@v46g2000cwv.googlegroups. com>,
"rAgAv" <ragav.payne@googlemail.com> wrote:

> What is the property of a photon which makes it to move in a constant
> velocity?

Zero mass.

Sean Case

[Arkadiusz Jadczyk]

On Sun, 14 May 2006 20:27:16 +0000 (UTC), "rAgAv"
<ragav.payne@googlemail.com> wrote:

>Okay guys.This is a serious question.
>
>What is the property of a photon which makes it to move in a constant
>velocity?
>
>Thanx in advance.
>rAgAv.

Photons (if massless) do not necessarily move with a "constant
velocity". They move along "null geodesics". But "light" is not the same
"photons". It is related to, but not the same. According to QED we are
dealing with quantized electromagnetic field, and "photon states" are
only "asymptotic states". So in order to answer your question, one must
know what context is your question to be put into.
--

Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
--

[Stolen_Humanity@Yahoo.com]

rAgAv wrote:
> Okay guys.This is a serious question.
>
> What is the property of a photon which makes it to move in a constant
> velocity?
>
> Thanx in advance.
> rAgAv.

Photons do NOT move at a constant velocity. They move at a constant
speed. Their direction can change.

Anyhow, the reason for this is because they are massless. All massless
particles/waves move at the speed of light in a vacuum.

-Matthew Paul Finnigan-

[Uncle Al]

rAgAv wrote:
>
> Okay guys.This is a serious question.
>
> What is the property of a photon which makes it to move in a constant
> velocity?

For all massless particles, Lorentz Invariance. Or, the permeability
and permittivity of free space (vacuum).

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz3.pdf

[rAgAv]

If a photon is really a particle of zero mass then why does a light
beam bend due to gravity?

Remember the formula F=(GMm)/r^2

[Oh No]

Thus spake rAgAv <ragav.payne@googlemail.com>
>Okay guys.This is a serious question.
>
>What is the property of a photon which makes it to move in a constant
>velocity?
>

Before we can talk about the velocity of anything at all we have to know
how we determine the time and distance of an event remote from the clock
which we use as a reference for coordinates. The fact that a photon has
zero mass makes it suitable for passing information between one event
and another at the maximum possible speed, and we define space and time
coordinates with reference to the speed of light, not the other way
about. This makes it necessarily true that the speed of light is
constant.

Regards

--
Charles Francis
substitute charles for NotI to email

[Oh No]

Thus spake rAgAv <ragav.payne@googlemail.com>
>Okay guys.This is a serious question.
>
>What is the property of a photon which makes it to move in a constant
>velocity?
>

Before we can talk about the velocity of anything at all we have to know
how we determine the time and distance of an event remote from the clock
which we use as a reference for coordinates. The fact that a photon has
zero mass makes it suitable for passing information between one event
and another at the maximum possible speed, and we define space and time
coordinates with reference to the speed of light, not the other way
about. This makes it necessarily true that the speed of light is
constant.

Regards

--
Charles Francis
substitute charles for NotI to email

[Norm Dresner]

"rAgAv" <ragav.payne@googlemail.com> wrote in message
news:1147774093.945376.126610@j55g2000cwa.googlegr oups.com...
| If a photon is really a particle of zero mass then why does a light
| beam bend due to gravity?
|
| Remember the formula F=(GMm)/r^2

Remember the formula m = E /c^2

[Jon Bell]

In article <1147774093.945376.126610@j55g2000cwa.googlegroups. com>,
rAgAv <ragav.payne@googlemail.com> wrote:
>If a photon is really a particle of zero mass then why does a light
>beam bend due to gravity?
>
>Remember the formula F=(GMm)/r^2

Remember that even in Newtonian gravity, an object's gravitational
acceleration does not depend on its mass:

ma = (GMm)/r^2
a = (GM)/r^2

So we can view a zero-mass particle as a limiting case of this.

But a better description is given by general relativity, in which
gravitation is no longer a force, but instead the visible effect of an
object's motion in curved spacetime. A massless object travels along a
geodesic in spacetime, and such geodesics have spatial curvature near
massive ("gravitating") objects.

--
Jon Bell <jtbell@presby.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA

[Igor]

rAgAv wrote:
> If a photon is really a particle of zero mass then why does a light
> beam bend due to gravity?
>
> Remember the formula F=(GMm)/r^2

It's traveling on a null geodesic, where the geodesic acceleration
vanishes, maintaining the same speed c throughout its trajectory.

[rAgAv]

Igor wrote:
> rAgAv wrote:
> > If a photon is really a particle of zero mass then why does a light
> > beam bend due to gravity?
> >
> > Remember the formula F=(GMm)/r^2
>
> It's traveling on a null geodesic, where the geodesic acceleration
> vanishes, maintaining the same speed c throughout its trajectory.

If you don't mind, can you explain on what a geodesic is? Its a very
familiar term for me, but its meaning is kind of vague in my mind.

Thanx.
rAgAv.

[Filipi Nascimento]

rAgAv wrote:
> If a photon is really a particle of zero mass then why does a light
> beam bend due to gravity?
> Remember the formula F=(GMm)/r^2

In general relativity the concept of gravity is related to a time-space
deformation, so fields dependent on time and space are deformed too.

Remember too, that without mass, a eletromagnectic wave transports
momentum(sorry, i dont remember the correct term in english, we use
momentum in brazil) that can be interpreted not only by a modern
quantized photon analysis as by a classical field wave aproach.

[Igor]

rAgAv wrote:
> Igor wrote:
> > rAgAv wrote:
> > > If a photon is really a particle of zero mass then why does a light
> > > beam bend due to gravity?
> > >
> > > Remember the formula F=(GMm)/r^2
> >
> > It's traveling on a null geodesic, where the geodesic acceleration
> > vanishes, maintaining the same speed c throughout its trajectory.
>
> If you don't mind, can you explain on what a geodesic is? Its a very
> familiar term for me, but its meaning is kind of vague in my mind.
>
> Thanx.
> rAgAv.

A geodesic is described as the path of least action, least amount of
effort in other words. In Euclidean space, the natural path is a
straight line. In non-Euclidean spaces, it can be curved and doesn't
even need to be closed. A null geodesic is simply a geodesic where the
test particle never changes it's speed. It may change direction, but
not speed.

[John Bell]

rAgAv wrote:
> Igor wrote:
> > rAgAv wrote:
> > > If a photon is really a particle of zero mass then why does a light
> > > beam bend due to gravity?
> > >
> > > Remember the formula F=(GMm)/r^2
> >
> > It's traveling on a null geodesic, where the geodesic acceleration
> > vanishes, maintaining the same speed c throughout its trajectory.
>
> If you don't mind, can you explain on what a geodesic is? Its a very
> familiar term for me, but its meaning is kind of vague in my mind.
>
Simply put, a null geodesic is the trajectory you have to travel on to
avoid "experiencing" gravity.( i.e. it is a free fall trajectory)

Bell.

[Henning Makholm]

Scripsit "John Bell" <john.bell@accelerators.co.uk>
> rAgAv wrote:

>> If you don't mind, can you explain on what a geodesic is? Its a very
>> familiar term for me, but its meaning is kind of vague in my mind.

> Simply put, a null geodesic is the trajectory you have to travel on to
> avoid "experiencing" gravity.( i.e. it is a free fall trajectory)

I don't think the word 'null' should have been in your sentence.

A _null_ geodesic is a geodesic trajectory that moves at lightspeed
(as measured by a sublight observer whose worldline momentarily
intersects the null geodesic).

--
Henning Makholm "I, madam, am the Archchancellor!
And I happen to run this University!"

[Tom Roberts]

We've had two incorrect answers here, so I thought I'd set the record
straight. 'Most everybody around here knows this, but some might not....

John Bell wrote:
> Simply put, a null geodesic is the trajectory you have to travel on to
> avoid "experiencing" gravity.( i.e. it is a free fall trajectory)

This is not at all the distinguishing characteristic of null geodesics.

In GR, all timelike geodesics are paths such that an observer following
the path would not feel the effects of gravity (to first order). That
is, all non-spacelike geodesics are free-fall trajectories. A human
observer cannot possibly follow a null geodesic.


Igor wrote:
> A null geodesic is simply a geodesic where the
> test particle never changes it's speed. It may change direction, but
> not speed.

This, too, is not the distinguishing characteristic of null geodesics.

It is indeed true that a null geodesic never changes speed relative to
any locally inertial frame. But in general there are timelike geodesics
that do not change speed relative to some coordinate system, and for
certain manifolds constant speed relative to locally inertial frames can
be achieved. Note that a test particle of nonzero mass cannot follow a
null geodesic. Note also that "speed" is coordinate dependent, and is
thus not a very general characteristic of anything.

The distinguishing characteristic of a null geodesic is that it is a
geodesic with a null tangent vector. That implies that only an
electromagnetic wave or a massless particle can follow such a
trajectory. In GR we often use the geometrical optics approximation for
light, and light rays follow null geodesics.


Tom Roberts

[rAgAv]

Tom Roberts wrote:
> We've had two incorrect answers here, so I thought I'd set the record
> straight. 'Most everybody around here knows this, but some might not....
>
> John Bell wrote:
> > Simply put, a null geodesic is the trajectory you have to travel on to
> > avoid "experiencing" gravity.( i.e. it is a free fall trajectory)
>
> This is not at all the distinguishing characteristic of null geodesics.
>
> In GR, all timelike geodesics are paths such that an observer following
> the path would not feel the effects of gravity (to first order). That
> is, all non-spacelike geodesics are free-fall trajectories. A human
> observer cannot possibly follow a null geodesic.
>
>
> Igor wrote:
> > A null geodesic is simply a geodesic where the
> > test particle never changes it's speed. It may change direction, but
> > not speed.
>
> This, too, is not the distinguishing characteristic of null geodesics.
>
> It is indeed true that a null geodesic never changes speed relative to
> any locally inertial frame. But in general there are timelike geodesics
> that do not change speed relative to some coordinate system, and for
> certain manifolds constant speed relative to locally inertial frames can
> be achieved. Note that a test particle of nonzero mass cannot follow a
> null geodesic. Note also that "speed" is coordinate dependent, and is
> thus not a very general characteristic of anything.
>
> The distinguishing characteristic of a null geodesic is that it is a
> geodesic with a null tangent vector. That implies that only an
> electromagnetic wave or a massless particle can follow such a
> trajectory. In GR we often use the geometrical optics approximation for
> light, and light rays follow null geodesics.



So, when light gets trapped by the gravitational force of a black hole,
does the light get neatly spread over the surace like butter on bread,
or does it still have a constant speed but just keeps circulating the
black hole like a satellite?


Thanx,
rAGAv

[John (Liberty) Bell]

Doubly corrected response
Tom Roberts wrote:
> We've had two incorrect answers here, so I thought I'd set the record
> straight.

> The distinguishing characteristic of a null geodesic is that it is a
> geodesic with a null tangent vector.

The enquirer's question was "what is a geodesic?"
Henning Makholm wrote, in reply to my original response:
> I don't think the word 'null' should have been in your sentence.

That is correct. What I described was actually a timelike geodesic. A
spacelike geodesic would represent a trajectory of forbidden Einstein
causality. Once a geodesic has been defined, you surely then need to
define the
meaning of "null", if you want to clarify what a null geodesic is. This
you have
not done.

In the absence of gravity (i.e.in SR) any light trajectory in a vacuum
has zero
squared length. I suggest that something very similar is true in GR. I
quote from http://en.wikipedia.org/wiki/Pp-wave_spacetime :

"Penrose also pointed out that in a pp-wave spacetime, all the
polynomial scalar invariants of the Riemann tensor vanish identically,
yet the curvature is almost never zero. If you view the Riemann tensor
as a second rank tensor acting on bivectors, this phenomenon is
analogous to the fact that a nonzero null vector has vanishing squared
length."

>From this I would say that the sum of the ds^2 elements is zero over
all parts of a null geodesic, at the infinitesimal calculus limit where
ds tends towards zero.

John (Liberty) Bell
http://global.accelerators.co.uk
(Change John to Liberty to respond by email)

[John Bell]

Henning Makholm wrote:
> Scripsit "John Bell" <john.bell@accelerators.co.uk>
> > rAgAv wrote:
>
> >> If you don't mind, can you explain on what a geodesic is? Its a very
> >> familiar term for me, but its meaning is kind of vague in my mind.
>
> > Simply put, a null geodesic is the trajectory you have to travel on to
> > avoid "experiencing" gravity.( i.e. it is a free fall trajectory)
>
> I don't think the word 'null' should have been in your sentence.
>
You are quite right. The question was what is a geodesic. The answer
should have read:
Simply put, a geodesic is the trajectory you have to travel on to
avoid "experiencing" gravity.( i.e. it is a free fall trajectory).

This does appear to be the simplest explanation of a geodesic.

The answer to the unasked question what is a null geodesic? could then
be any of the following, according to taste:
a) the geodesic of anything with zero rest mass.
b) the geodesic of anything which experiences zero elapsed time.
c) the geodesic of anything which travels at the speed of light.

I am not so sure how I would answer the alternative question: what are
a geodesic and a null geodesic in a refractive medium such as water?
or even if that question makes any sense.
Any suggestions?

Bell

[John Bell]

Tom Roberts wrote:
> We've had two incorrect answers here, so I thought I'd set the record
> straight. 'Most everybody around here knows this, but some might not....
>
> The distinguishing characteristic of a null geodesic is that it is a
> geodesic with a null tangent vector.

Once a geodesic has been defined, you surely then need to define the
meaning of "null", to clarify what a null geodesic is. This you have
not done.

In the absense of gravity (i.e. SR) any light trajectory has zero
squared length. I suggest that something very similar is true in GR. I
quote from http://en.wikipedia.org/wiki/Pp-wave_spacetime :

"Penrose also pointed out that in a pp-wave spacetime, all the
polynomial scalar invariants of the Riemann tensor vanish identically,
yet the curvature is almost never zero. If you view the Riemann tensor
as a second rank tensor acting on bivectors, this phenomenon is
analogous to the fact that a nonzero null vector has vanishing squared
length."

>From this I would infer that the integral of the ds^2 elements of a
null geodesic also vanish.

Is that what you meant by it having a null tangent vector?

Bell

[John Bell]

Corrected response.
Tom Roberts wrote:
> We've had two incorrect answers here, so I thought I'd set the record
> straight.

> The distinguishing characteristic of a null geodesic is that it is a
> geodesic with a null tangent vector.

The enquirer's question was "what is a geodesic?"

> rAgAv wrote:
>> If you don't mind, can you explain on what a geodesic is? Its a very
>> familiar term for me, but its meaning is kind of vague in my mind.

Henning Makholm wrote, in reply to my original response:
> I don't think the word 'null' should have been in your sentence.

That is correct. What I described was actually a timelike geodesic. A
spacelike geodesic would represent a trajectory of forbidden Einstein
causality.

Once a geodesic has been defined, you surely then need to define the
meaning of "null", if you want to clarify what a null geodesic is. This
you have not done.

In the absence of gravity (i.e.in SR) any light trajectory in vacuum
has zero squared length. I suggest that something very similar is true
in GR. I quote from http://en.wikipedia.org/wiki/Pp-wave_spacetime :

"Penrose also pointed out that in a pp-wave spacetime, all the
polynomial scalar invariants of the Riemann tensor vanish identically,
yet the curvature is almost never zero. If you view the Riemann tensor
as a second rank tensor acting on bivectors, this phenomenon is
analogous to the fact that a nonzero null vector has vanishing squared
length."

>From this I would infer that the sum of the ds^2 elements is zero over
all parts of the geodesic, at the infinitesimal calculus limit when ds
tends towards zero.

John (Liberty) Bell
http://global.accelerators.co.uk
(Change John to Liberty to respond by email)