Question on photons inertia ?
David Smyth
I would like to define inertia as having the property of requiring a force to
undergo a change in momentum (if that's OK with you).
Does a photon then have inertia?
As a specific example, when light is bent around a massive object (such as
the sun) it has its momentum changed. Does this mean the sun will undergo a
(very small) change in momentum in order to satisfy the law of conservation
of momentum?
If the sun does change its momentum slightly (and more than likely its
kinetic energy) does this indicate the light will be red shifted (or perhaps
blue shifted) in order to conserve energy?
If you are wondering why I am asking this question it is because I have read
the odd posting about the place describing the photon as accelerating to c
instantly upon formation and this implying a photon has no inertia. As far
as I can see if a photon has no inertia then it would be impossible for both
the law of conservation momentum and the law of equal and opposite reactive
forces to be satisfied when light is bent around a massive object.
Please keep in mind I am using the definition of inertia I have stated above.
Thanks in advance
David Smyth
CPL
University of Queensland
Jonathan Scott
on 24 Oct 1995 12:02:14 GMT,
David Smyth <dav...@psych.psy.uq.oz.au> writes:
>Hi Guys,
>
>I would like to define inertia as having the property of requiring a force to
>undergo a change in momentum (if that's OK with you).
>
>Does a photon then have inertia?
Yes.
Inertia in this sense is equivalent to the total energy expressed in
mass units, also known rather misleadingly as the "relativistic mass".
Note that gravitational effects are proportional to energy rather than
to rest mass, so not only does the sun exert a gravitational force on
photons, but photons also exert an equal and opposite (and totally
negligible) force on the sun.
Jonathan Scott
jonatha...@vnet.ibm.com or jsc...@winvmc.vnet.ibm.com
john baez
>I would like to define inertia as having the property of requiring a force to
>undergo a change in momentum (if that's OK with you).
>Does a photon then have inertia?
Using your definition, yes. Well, I'm also using the standard
definition of force, which means "that which is required to change the
momentum"... so it's pretty darn tautologous. Nonetheless, it's a fact
that a photon has a nonzero momentum that depends on which way the
photon is going, so making it go a different direction (e.g. by a
mirror) requires applying a force to it.
The concept of "force" becomes rather misleading when we go to general
relativity, so it's not really kosher to say that a gravitational
"force" is required to bend the path of a photon, or planet, or
whatever. But if we work in a semiclassical approximation allowing us
to speak of "forces" in this context, we may say that the Sun needs to
exert a force on the photon to bend its path, and that the photon exerts
an opposing force on the Sun.
>As a specific example, when light is bent around a massive object (such as
>the sun) it has its momentum changed. Does this mean the sun will undergo a
>(very small) change in momentum in order to satisfy the law of conservation
>of momentum?
Yup.
>If the sun does change its momentum slightly (and more than likely its
>kinetic energy) does this indicate the light will be red shifted (or perhaps
>blue shifted) in order to conserve energy?
Yup, though of course the effect is ridiculously teeny, since the sun
picks up just a teeny bit of momentum, and p = mv, E = mv^2/2 so it picks
up an even teenier bit of kinetic energy. Nobody will ever measure this
effect... well, not in the next few centuries.
gl...@gnn.com
In article <46me4e$t...@guitar.ucr.edu> john baez wrote:
> Nonetheless, it's a fact
>that a photon has a nonzero momentum
Are you therefore asserting that it has nonzero mass? If not, why
not?
Oz
> Nonetheless, it's a fact
>that a photon has a nonzero momentum that depends on which way the
>photon is going, so making it go a different direction (e.g. by a
>mirror) requires applying a force to it.
A slightly off-thread question here.
Does a photon have to 'travel' in the direction of it's
momentum? Now this might sound a stupid question (and it
probably is) but 'light as a wave' would have it propagating
in all directions from it's source, whilst it interacts with
matter with a well-defined momentum (and hence direction).
This is no problem if momentum and propagation direction are
not 'connected' and I cannot think of any a priori reason
why this should not be the case.
For example a photon with momentum in the x-direction
starting at the origin could travel to a place on the y-axis
and impart momentum to an electron there in the x-direction.
This only looks stupid if you keep thinking of a photon as a
massive body with momentum and velocity related.
This sort of question is one that tends to get ignored when
examinations are of great importance.
-------------------------------
'Oz "When I knew little, all was certain. The more I learnt,
the less sure I was. Is this the uncertainty principle?"
Gordon Long
Oz <O...@upthorpe.demon.co.uk> wrote:
>
>A slightly off-thread question here.
>
>Does a photon have to 'travel' in the direction of it's
The only sensible way to define a photon's direction of 'travel'
is to have it be along the direction of momentum.
If I understand where you are coming from, I think your question
can be expressed more clearly in term of Young's double slit
experiment, where a single photon can, in a sense, interfere with
itself. This experiment highlights the difficulties of thinking of
something as both a wave and a particle. In other words, in order
to interfere with itself, it must travel (as you said) "in all
directions at once"; however, as a particle, it only goes through
one slit.
This experiment is very often used as a motivation for quantum
mechanics, when you substitute something like an electron (which
also has wave-like behavior) for a photon. The reason you make the
substitution is that you can then use non-relativistic quantum
mechanics, and resolve the paradox by applying the Heisenberg
uncertainty principle.
In elementary QM, if a particle has a well defined momentum (i.e. it
is in a momentum eigenstate), then it's position is undefined (i.e. it
exists everywhere with equal probability. Mathematically, this is
essentially the same as taking the Fourier transform of a plane wave -
you get a constant). In this sense, a particle
>with momentum in the x-direction
>starting at the origin could travel to a place on the y-axis
>and impart momentum to an electron there in the x-direction.
The reason is that, if the particle has a well defined momentum (in
the x direction), it can't really start at the origin and travel to a
place on the y-axis. Instead, it exists at both places simultaneously.
Of course, this isn't very physical, and so real particles are usually
described with wave packets, where both position and momentum have some
uncertainty.
If you don't make the substitution photon->electron, then your task
would be much more difficult, since you can't describe a photon using
non-relativistic QM. A proper description of photons comes from QED.
Interestingly, according to QED, there is no such thing as a "real"
photon. The photon only exists as a carrier of the electromagnetic
force between two fermions (supersymmetry theories notwithstanding).
This means that all photons are in fact "virtual", so that the magnitude
of the photon's 4-momentum (which equals the mass for real particles)
is not quite zero. When we say that a photon is massless, we really
mean that p^2 can come arbitrarily close to zero, or (equivalently)
there is no limit to the range of the electromagnetic interaction.
As an aside, I have always been waiting for the "photons have mass"
crackpots^H^H^H^H^H^H^H^H^H advocates to bring this up, but so far no
one ever has.
--
#include <std.disclaimer>
Gordon Long | Grad. Student, High Energy Physics
gl...@vsophd.cern.ch | University of Maryland
john baez
>
>In article <46me4e$t...@guitar.ucr.edu> john baez wrote:
>> Nonetheless, it's a fact
>>that a photon has a nonzero momentum
>Are you therefore asserting that it has nonzero mass? If not, why
>not?
You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring.
By the way, could you please post articles in a way that says who they
are from? Right now it just says:
Article 79518 (90 more) in sci.physics:
From:
Subject: Re: Question on photons inertia ?
Date: Wed, 25 Oct 1995 20:06:45
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X-Mailer: GNNmessenger 1.2
This is sort of impolite.
john baez
>ba...@guitar.ucr.edu (john baez) wrote:
>> Nonetheless, it's a fact
>>that a photon has a nonzero momentum that depends on which way the
>>photon is going, so making it go a different direction (e.g. by a
>>mirror) requires applying a force to it.
>A slightly off-thread question here.
Fine.
>Does a photon have to 'travel' in the direction of it's
>momentum? Now this might sound a stupid question (and it
>probably is) but 'light as a wave' would have it propagating
>in all directions from it's source, whilst it interacts with
>matter with a well-defined momentum (and hence direction).
Well, above I was treating the photon as a particle, which is a fine
approximation for the problem I was talking about... the so-called
geometrical optics approximation, where you think of the front of a wave
as a bunch of particles all moving the way the wavefront at that point
is going. Now you're getting more serious on me and treating light as a
wave satisfying Maxwell's equations! This is more accurate, at least
unless you want to get even MORE serious and treat light using quantum
electrodynamics, in which case it's best understood as a wave-particle
thingie!
In other words, we have got to agree what we're talking about before we
talk about it: quantum electrodynamics in all its glory, the classical
approximation thereto known as Maxwell's equations, or the geometrical
optics approximation to Maxwell's equations. The tricky thing about
physics (which completely befuddles the layman or crackpot) is that
different levels of approximation bring with them very different ways of
talking and very different pictures of reality, and one must learn to
nimbly dart between them depending on the level of accuracy needed.
Let me assume you are talking about the classical Maxwell's equations in
empty space... with no gravity around to confuse us. Then there ain't
no photon, there is just a superposition of plane waves, each one moving
at the speed of light in some direction. Each plane wave is described
by a vector called its momentum. The direction of this vector says
which way the plane wave moves. The magnitude of this vector says the
spatial frequency (2pi over the wavelength) of the wave. Since
Maxwell's equations are linear these plane waves each march along
independently, heedless of the rest; we just add them up to get the
solution. The momentum of the solution can be calculated by totalling up
the momenta of the plane waves in a suitable way. So if for example we
have a spherically symmetric wave of light radiating outwards from a
central source, the total momentum is zero, even though it's made of
lots of plane waves with nonzero momentum.
>For example a photon with momentum in the x-direction
>starting at the origin could travel to a place on the y-axis
>and impart momentum to an electron there in the x-direction.
>This only looks stupid if you keep thinking of a photon as a
>massive body with momentum and velocity related.
Hmm, well, now we have our light interacting with some matter, so you
had better tell me if you are thinking about this in terms of the
classical Maxwell's equations (in which case there ain't no photon), or
in terms of quantum electrodynamics (in which case both the electron and
the photon are wave-particle thingies), or in terms of the semiclassical
approximation (in which case the electron is a wave-particle thingie but
the light is treated classically), or using some other approximation.
john baez
>>>As a specific example, when light is bent around a massive object (such as
>>>the sun) it has its momentum changed. Does this mean the sun will
>>>undergo a (very small) change in momentum in order to satisfy the law
>>>of conservation of momentum?
>>Yup.
>>>If the sun does change its momentum slightly (and more than likely its
>>>kinetic energy) does this indicate the light will be red shifted (or
>>>perhaps blue shifted) in order to conserve energy?
>>Yup, though of course the effect is ridiculously teeny, since the sun
>>picks up just a teeny bit of momentum, and p = mv, E = mv^2/2 so it picks
>>up an even teenier bit of kinetic energy. Nobody will ever measure this
>>effect... well, not in the next few centuries.
>Ok so the energy/momentum of the photon plays a part in the interaction when
>light is bent by a gravitational field.
Yup. In general relativity, two things can only interact gravitationally if
both of them have energy/momentum. Of course, everything we know of
(except possibly empty space) has energy/momentum, so this isn't a big
limitation!
>Next question:
>How would the bending of light be perceived when viewed from the frame of
>reference fixed to the centre of the source of the gravitational field
>(which is accelerating ever so slightly). Specifically, would the light
>still appear to bend and undergo Doppler shift from this reference frame?
First let me hassle you (for your own benefit, of course), and then let
me tell you what you want to know. For everything I teach someone that
they want to know, I must teach them 5 things they didn't want to know.
Or didn't KNOW they wanted to know, anyway.
Ok:
You have to be careful with "frames of reference" in general relativity.
A "frame of reference" in special relativity is a way of
assigning coordinates -- lists of 4 numbers -- to every point in
spacetime. But in fact the frames of reference people really care
about in special relativity are the "inertial frames", and often people
implicitly mean "inertial frame" when they say "frame of reference".
What's an inertial frame? It's one such that
1) anything "at rest" in this frame --- i.e., having unchanging x,y, and
z coordinates as t varies --- feels no acceleration, and
2) given any TWO things "at rest" in this frame, the light emitted by one
will not be redshifted or blueshifted when it is received by the other.
(This is a precise way of saying that they are "at rest relative to one
another" --- a phrase that can be dangerously ambiguous if we don't make
it precise like this.)
Now, not everyone uses this definition of "inertial frame" in special
relativity, but it's equivalent to the usual one, and it has the
advantage of being completely operational. That is, we can experimentally
test whether a coordinate system is an inertial frame in this sense.
Here's the problem, then: there are typically NO inertial frames in this
sense, in general relativity. Spacetime is simply too wiggly... or in
other words, gravity screws things up.
So if one uses the words "frame of reference" in general relativity, and
secretly expects this to be sort of like an inertial frame in special
relativity, one is headed for trouble. One can use the term to mean
simply "coordinate system" --- but why not save trouble and say
"coordinate system", then? Anyway, there are zillions of coordinate
systems available in general relativity, and there is no sense in which
each observer has some "best" coordinate system, except in special
cases... so one can't go around talking about the "frame of reference
fixed to the center of the source of the gravitational field" without
receiving a long grumpy lecture, because that doesn't actually specify
one particular coordinate system!
Often I just nod my head, with a slight grimace of pain, when people say
this kind of thing, and then I try to figure out what they probably
really mean. But it's best, when starting out, to confine oneself to
questions that have operational significance. That is, to ask questions
that basically amount to specifying an experiment you could do, at least
in principle.
So, for example, when you talk of how the light would be bent, you can't
really say how it would be perceived "from this reference frame", unless
you specify a coordinate system in detail; but you can say "suppose we
did this and that, then what would we actually see with our own very
eyes?"
Now let me try to answer the intent of your question. Basically, I
suspect you are wondering if the fact that the passing photon affects
the sun winds up changing the story that you've heard, namely that the
photon's path is bent by the sun. And basically, the answer is no, it
doesn't; the effect of the photon on the sun is too small to matter, but
the photon's path is measurably bent. The redshift of the photon due to
the fact that it spent some energy on pulling the sun a bit is also too
small to detect, by the way... we never see any effects due to that.
Erik Max Francis
> Are you therefore asserting that it has nonzero mass? If not, why
> not?
Mass and momentum are not linearly related in special relativity; that
is, the equation
p = m v
is not valid in relativity. It's only approximately correct in the
case where m is nonzero and v << c.
Erik Max Francis, &tSftDotIotE || uuwest!alcyone!max, m...@alcyone.darkside.com
San Jose, California, U.S.A. || 37 20 07 N 121 53 38 W || the 4th R is respect
H.3`S,3,P,3$S,#$Q,C`Q,3,P,3$S,#$Q,3`Q,3,P,C$Q,#(Q.#`-"C`- || 1love || folasade
_Omnia quia sunt, lumina sunt._ || GIGO Omega Psi || http://www.spies.com/max/
"Hands that once picked cotton can now pick Presidents." -- Jesse Jackson
David Smyth
> I don't understand what you're asking. Let's say someone in a inertial
> frame far, far away (zero acceleration) watched this photon bend and
undergo
> Doppler shift, etc.
> Since everyone agrees that the acceleration of the center of the star due
> to the photon is so small as to be unmeasureable (practically zero), why
> wouldn't someone in this frame see the same thing (unmeasureable
> difference) as the first person ?
>
> That was a rhetorical question.
>
The thing I am interested in is something I read about in and old text on
GR. It was described as a relativistic inertial frame of reference. This
is one where not only the speed of light is constant but also its velocity.
I believe it arises if the frame has no external constraints i.e. free to
move with any gravitational accelerations. I guess the principal is that as
light is bent (or accelerated) by a gravitational field the reference frame
will move along with the bending.
So the questions I have:
a) Does such a thing exist?
b) If it does then how does differing energies/momentums of light come into
play?
It seems to me that light having a gravitational field, the intensity
of which is a function of its energy, would prevent such a thing existing
as each frequency of light would be accelerated by slightly different
amounts. However this may be exactly offset by some other factor (such
as the momentum?).
This leads to yet another question:
When white light is bent around say the sun, does it split into its
constituent frequencies?
I was once given a response to a question about relativistic inertial frames
of reference on another thread. But the mathematics was beyond me.
So as much hand waving as possible would be appreciated.
Jonathan Scott
on 29 Oct 1995 00:55:09 GMT,
David Smyth <dav...@psych.psy.uq.oz.au> writes:
>On Sat, 28 Oct 1995, David Cary wrote:
>> I don't understand what you're asking. Let's say someone in a inertial
>> frame far, far away (zero acceleration) watched this photon bend and
>> undergo Doppler shift, etc.
>> Since everyone agrees that the acceleration of the center of the star due
>> to the photon is so small as to be unmeasureable (practically zero), why
>> wouldn't someone in this frame see the same thing (unmeasureable
>> difference) as the first person ?
I'm not quite sure exactly what you mean, and I know you said this
question was rhetorical, but I'd like to point out there there are
significant differences in viewpoint between a local observer in a
gravitational field and a remote observer using a Euclidean background
set of coordinates.
In particular, the rulers and spatial coordinate system of an observer
in a gravitational field are effectively curved downwards by an amount
that corresponds to the curvature of a light beam at that acceleration
relative to flat space. If the observer is standing still in a
gravitational field, everything (including objects or waves travelling
at or near the speed of light) appears to accelerate downwards locally
at the same rate relative to local rulers. This means that light
appears to accelerate at TWICE that rate relative to the flat background
coordinates used by the distant observer.
>The thing I am interested in is something I read about in and old text on
>GR. It was described as a relativistic inertial frame of reference. This
>is one where not only the speed of light is constant but also its velocity.
>I believe it arises if the frame has no external constraints i.e. free to
>move with any gravitational accelerations. I guess the principal is that as
>light is bent (or accelerated) by a gravitational field the reference frame
>will move along with the bending.
Yes, that is correct. However, it can only be established over a
frame which is small enough that gravitational tidal effects do not
apply. Note for example that both tides and internal gravitational
forces are sufficient to cause detectable deviations from inertial
behaviour at different locations within the Space Shuttle, although
these are only of the order of a millionth of a g of acceleration,
hence the term "microgravity" for that environment.
>It seems to me that light having a gravitational field, the intensity
>of which is a function of its energy, would prevent such a thing existing
>as each frequency of light would be accelerated by slightly different
>amounts. However this may be exactly offset by some other factor (such
>as the momentum?).
This is wrong; the acceleration is the same for everything. The force
is proportional to the energy, just as for ordinary massive objects.
>This leads to yet another question:
>
>When white light is bent around say the sun, does it split into its
>constituent frequencies?
No, for the same reason.
>David Smyth
>CPL
>University of Queensland
Jonathan Scott
jonatha...@vnet.ibm.com or jsc...@winvmc.vnet.ibm.com
D. Y. Kadoshima
> In article <8146984...@upthorpe.demon.co.uk>,
(snip)
> A proper description of photons comes from QED.
> Interestingly, according to QED, there is no such thing as a "real"
> photon. The photon only exists as a carrier of the electromagnetic
> force between two fermions (supersymmetry theories notwithstanding).
(snip)
>>>>
I agree with what you posted, but wouldn't it be 'cleaner' to think of the
particular frequency (or range of frequencies) within the electromagnetic
spectrum as the 'carrier', and that which fixes the dynamic properties of
the interacting body upon which this particular electromagnetic force is
applied, with a 'photon' as a measure of this particular EM force that
impresses the energy of a quantum upon this particular interacting mass?
Thus the frequency (which is time related) of the EM force excites this
interacting mass and causes its momentum to vary from the general
direction of the EM force at any particular point in time, but the
summation of the varying momenta over a full cycle (which may be more
or less than that due to a 'photon') is in the direction of the applied EM
force.
Regards;
D. Y. K.
john baez
>What I am really thinking about (as a concrete example) is
>whether a photon emitted from the sun, can transfer momentum
>to a distant star with some of its momentum perpendicular to
>the line joining the two stars.
>Or am I still being vague?
No, that's quite precise. The answer is, for all practical purposes,
no: the photon can't go from one star to another and transfer momentum
perpendicular to the line joining the two stars.
But to expand a bit on it: as I pointed out in my other post, this
effect *can* happen with virtual particles. In your example above, the
reaction going on can be drawn in terms of the following Feynman
diagram:
>> e- e-
>> \ /
>> \ /
>> \__photon____/
>> / \
>> / \
>> / \
(Even though you didn't mean it as a Feynman diagram, we can think of it
as that if we want!) Since the photon line is an "internal line", the
photon can be thought of as a virtual particle. As I pointed out, the
Heisenberg uncertainty principle puts a limit on how accurately we can
measure the position of the two vertices in the diagram, simultaneously
measuring the momentum of the photon. So it *is* possible in principle,
for the photon to transfer momentum in a direction other than that of
its measured line of flight, as long as the difference is less than that
allowed by the uncertainty principle (uncertainty in momentum x
uncertainty in position is less than or equal to hbar/2). However, in
the case of two distant stars, this difference is ridiculously small,
since the stars are very far apart.
The uncertainty principle matters much more when we slam two electrons
at each other, so they are very close together. In such situations, the
fact that the photon is "virtual" rather than "real" can make a
tremendous difference.
Gordon Long
>gl...@hpplus02.cern.ch (Gordon Long) wrote:
>
>> The reason is that, if the particle has a well defined momentum (in
>>the x direction), it can't really start at the origin and travel to a
>>place on the y-axis. Instead, it exists at both places simultaneously.
>
>>Of course, this isn't very physical, and so real particles are usually
>>described with wave packets, where both position and momentum have some
>>uncertainty.
>
>when has QM been 'physical'! :-) Does it happen? As I say,
>somewhat to my surprise I can't quite see why not.
>
Well, believe it or not, QM really does try to describe the physical
world :-).
Anyway, when you ask 'does it happen?', the answer is that it depends
on exactly what you mean. If you did manage to give a particle a
precise value for momentum, then yes, it would happen. In other words,
let's say the particle starts at the origin. While it is there, you
give it this precise momentum, in the x-direction (say). Then the
particle can show up anywhere, not just along the x-axis, and interact
with another particle. The reason is that as soon as you give it the
momentum, it is no longer at the origin, but is (mathematically)
everywhere at the same time. So it can show up anywhere.
In practice, it is impossible to give a particle a precise momentum
(or position, for that matter). The position and momentum of particles
we see in the lab are usually described by wave packets, which means
that there is a peaked, symmetric distribution for the probability that
the particle has a given position or momentum. The position probability
density and the momentum probability density are Fourier transforms of
each other. This, btw, is how you derive the uncertainty principle.
>[...] I was
>specifically using the photon exactly for the reason you
>indicated above, for a massive particle momentum and
>velocity will correlate closely over quite small distances
>(or should I say angle), which is boring.
>
Actually, the behavior of an electron is just as wave-like as the
behavior of light. This fact is crucial to the understanding of quantum
mechanics. Also, the correlation between momentum and velocity is the
same for massive and massless particles. In fact, the direction of
velocity and the direction of momentum are the same *by definition*.
>What you seem to be saying is that for small distances an
>appreciable amount of 'force' is mediated by 'massive' (in
>some sense) photons, but at very long distances this drops
>to a negligible level and the photon appears to be massless.
>I have likely got this screwed up, but I expect you will
>correct me.
When I wrote my original post, I was afraid I'd get into a discussion
about the mass of photons. Let me try to be a little more precise, at
the probable cost of completely confusing the issue. First, the force
is the same in the short-range and long-range cases. What changes is
the magnitude of the 4-momentum of the photon. As the distance that the
photon travels gets longer and longer, the magnitude of a photon's
4-momentum (which is sqrt(E^2 - p^2)) comes closer and closer to zero.
Now for the confusing part, which centers around the question: what do
you want to call this magnitude? Some people call it 'mass'. Why? Well,
for real particles, special relativity tells you that a particle's mass
is equal to the magnitude of its 4-momentum. This is the well-known
equation E^2 = p^2 + m^2 (in units where c=1). You can go ahead and
think of the mass of virtual particles using this definition, but then
'mass' is no longer an intrinsic property of the (virtual) particle.
Alternatively, you can define mass such that it remains an intrinsic
property (this invariant mass is what you use when you write down the
propagator in a Feynman diagram), but then the energy (or momentum) of
the virtual particle is not what energy (or momentum) conservation tells
you it should be. These are both interpretations of the fact that the
magnitude of the 4-momentum is not equal to the invariant mass of a
virtual particle.
The range (or lifetime) of a virtual particle drops off exponentially,
where the exponent is proportional to the invariant mass of the virtual
particle. For a photon, this invariant mass is 0. This means that there
is no exponential drop off, and the photon's range is essentially infinite.
> So for interactions at very short distances (say
>in an accelerator target) the momentum and 'direction' of a
>photon would be quite strongly correlated. In a very long
>distance interaction, say between a distant galaxy and
>ourselves, momentum and 'direction' might not be. In some
>sense that is.
>
Well, as I said before, the 'direction' is given by the momentum.
Again, I think you are really referring to the uncertainty principle,
where the momentum of a photon can turn out to have a whole range of
directions. I'm not sure that the distance has anything to do with
it, except in the sense that a large distance gives you a more precise
measurement of direction of momentum.
Now, you might ask "Wait a minute, I thought you said that there is no
such thing as a 'real' photon. How can you apply the uncertainty
principle?" The answer is that you can, but indirectly. A photon is
the carrier of the EM force, and its position and momentum are
determined by the particles that exchange the photon. So, when you talk
about "where" the photon is, you are really talking about the position
of the particles with which the photon interacts. Similarly, the only
way to figure out the momentum of the photon is to figure out the
momenta of the particles exchanging the photon. This is how the
uncertainty principle sneaks in; it applies to the particles that
exchange the photon.
>> As an aside, I have always been waiting for the "photons have mass"
>>crackpots^H^H^H^H^H^H^H^H^H advocates to bring this up, but so far no
>>one ever has.
>
>OhMyGod, me a crackpot? I really try hard not to be, but you
>must realise that crackpottery come from ignorance and I
>have plenty of that. OhMyGod, a large glass of Whisky is
>seriously needed here. OhMyGod!
>
Were you a "photons have mass" advocate? I was not aware. Well,
(with appropriate ceremonial music in the background) subject to the
condition that you henceforth cease to insist that photons have mass,
and furthermore no longer disregard all experiments to the contrary,
I hereby remove you from the Gordon List of Crackpots.
Something I'm sure you have been spending many a sleepless night
worrying about.
Oz
<A lot. Snipped for conservation of bandwidth. :-)>
<Generally about uncertainty.>
Hmm, this is getting a mite philosophical. More so than I
originally intended. Also I think it is perhaps being more
generalised than I intended. My real question was a simple
one, or so I thought.
"Can a photon emitted from the sun transmit some momentum to
the earth in a direction perpendicular to the line joining
the sun and the earth." In other words does 'direction of
travel' have to be exactly the same as it's direction of
momentum over long distances. I *think* you said no.
> Now, you might ask "Wait a minute, I thought you said that there is no
>such thing as a 'real' photon. How can you apply the uncertainty
>principle?" The answer is that you can, but indirectly. A photon is
>the carrier of the EM force, and its position and momentum are
>determined by the particles that exchange the photon. So, when you talk
>about "where" the photon is, you are really talking about the position
>of the particles with which the photon interacts. Similarly, the only
>way to figure out the momentum of the photon is to figure out the
>momenta of the particles exchanging the photon. This is how the
>uncertainty principle sneaks in; it applies to the particles that
>exchange the photon.
Ahh, this is a better way of looking at it. I much prefer
this. Are you saying that we can in effect ignore the photon
since the transferred momentum is (nontechspeak) smeared out
by the uncertainty principle of the particles taking part in
the exchange. So in this case the answer to the above would
be yes (rather than no), and would be rather independent of
distance. (Caveats in all directions too).
> Were you a "photons have mass" advocate? I was not aware. Well,
>(with appropriate ceremonial music in the background) subject to the
>condition that you henceforth cease to insist that photons have mass,
>and furthermore no longer disregard all experiments to the contrary,
>I hereby remove you from the Gordon List of Crackpots.
<Long roll of timpani>
Oz hereby declares that photons do not have mass <cymbal>
and is humbly grateful to be struck off the infamous and
godless list of crackpots in the high and mighty book of the
Gordon Long <cymbal, timpani crescendo)
<in small voice> "But I never did believe they had mass."
Skies open and a momentous bolt of photons fries Oz to a
cinder where he stands, horrorstruck. AP comes out to wipe
up the mess. <Funereal music>
Bronis Vidugiris
Oz <O...@upthorpe.demon.co.uk> wrote:
)Hmm, this is getting a mite philosophical. More so than I
)originally intended. Also I think it is perhaps being more
)generalised than I intended. My real question was a simple
)one, or so I thought.
)
)"Can a photon emitted from the sun transmit some momentum to
)the earth in a direction perpendicular to the line joining
)the sun and the earth." In other words does 'direction of
)travel' have to be exactly the same as it's direction of
)momentum over long distances. I *think* you said no.
I think it depends on what one means by a photon. If you call
everything that mediates an electromagnetic force a photon, it's clear
that photons don't always have momentum in their direction of travel,
where direction of travel is defined by the vector from the starting
particle to the ending particle.
This usually gets called a "virtual" photon, as I understand it.
Unfortunately I don't know enough QFT to really be more clear on this
stuff.
Gordon Long
>
>
Sorry about the confusing answer. My problem was that I see these two
questions as being qualitatively different from each other, so it's a
little hard to answer them as one single question. My answer to the
first was both-yes-and-no (hence the long discussion of the uncertainty
principle); my answer to the second was yes, direction of travel is
exactly the same as direction of momentum, by definition.
Maybe I should give more one-word answers :-)
>>[...] This is how the
>>uncertainty principle sneaks in; it applies to the particles that
>>exchange the photon.
>
>Ahh, this is a better way of looking at it. I much prefer
>this. Are you saying that we can in effect ignore the photon
>since the transferred momentum is (nontechspeak) smeared out
>by the uncertainty principle of the particles taking part in
>the exchange. So in this case the answer to the above would
>be yes (rather than no), and would be rather independent of
>distance. (Caveats in all directions too).
>
Exactly. If you're interested in the position or momentum of the
photon, you should really apply the uncertainty principle to the
particles that exchange the photon, rather than to the photon itself.
In this sense (and only in this sense) the answer to the first question
above would be yes. Also, as you said, this answer would be rather
independent of distance, with appropriate caveats about a longer
distance enabling a more precise measurement of direction.
>
><Long roll of timpani>
>Oz hereby declares that photons do not have mass <cymbal> ...
>
><in small voice> "But I never did believe they had mass."
>
Somehow, I always figured that. (But I can't undo the ceremony,
since that means playing the music backwards, and any music played
backwards always reveals recordings of satanic rituals and incurs
the wrath of Tipper Gore.)
john baez
>Mark Krumholz <krum...@princeton.edu> wrote:
>> There is a simple answer to this question: Yes, a photon travelling
>>from the sun to the earth has momentum and gives that momentum to the
>>earth when it impacts. The momentum vector of the photon is always
>>parallel to its direction of travel. Consequently, photons from
>>travelling from the Sun to the Earth cannot impart to the Earth a
>>momentum perpendicular to the line joining their point of departure and
>>their point of impact.
>OK, that's fine then. Not even a little bit due uncertainty.
>So we know pretty accurately the energy, positions and
>momentum of this photon then. After the event of course.
>Somehow this doesn't quite feel right, but if you as an
>expert say so, then who am I to disagree?
You are Oz the Magnificent, and thus always have the right to disagree.
>So we should distinguish between a completed interaction,
>where the positions, energy and momentum of a single photon
>(at least at each end) can be known (retrospectively) rather
>exactly, and a prediction where they cannot be accurately
>known, and thus predicted.
Quantum mechanics can't be wiggled around that easily. Of course you
can measure both position and momentum of a photon; the uncertainty
principle however says you might be WRONG.
The photons coming from the sun to the earth, being virtual photons, CAN
push the earth in a direction perpendicular to their motion, due to the
uncertainty principle. But as I explained in an earlier post, because
the sun is so far away, this effect is ridiculously, insanely tiny.
For a back-of-the-envelope type calculation, I suspect that the main
number that matters is the wavelength of light divided by the distance
of the sun from the earth! Small. That's why Krumholz was perfectly
justified in ignoring it. Only someone insanely interested in matters
of principle, such as myself, would ever worry about it. As someone
once said when I noted that a spinning brick alone in outer space would
eventually slow down due to gravitational radiation, "only a
mathematician would think of that."
Bronis Vidugiris
Oz <O...@upthorpe.demon.co.uk> wrote:
)Mark Krumholz <krum...@princeton.edu> wrote:
(hasn't made it over here directly yet)
)
)> There is a simple answer to this question: Yes, a photon travelling
)>from the sun to the earth has momentum and gives that momentum to the
)>earth when it impacts. The momentum vector of the photon is always
)>parallel to its direction of travel. Consequently, photons from
)>travelling from the Sun to the Earth cannot impart to the Earth a
)>momentum perpendicular to the line joining their point of departure and
)>their point of impact
Exception #1 - unlike charges attracting are, by some descriptions,
exchanging "virtual photons" - which implies that a photon can carry
momentum opposite to the way it's moving. Or consider the "photons"
that carry the force from a magnetic field. (This is what I alluded
to in my previous post).
)OK, that's fine then. Not even a little bit due uncertainty.
)So we know pretty accurately the energy, positions and
)momentum of this photon then. After the event of course.
Nah.
)Somehow this doesn't quite feel right, but if you as an
)expert say so, then who am I to disagree?
The following may shed some "light" :-) on the topic of the
uncertanity principle:
ps - the reference for this was a popularization "Quantum Reality"
by Nick Herbert, Herbet also has more technical references on
the "stellar interferometer" as I recall.
Start: old post
-------
If you look at the light from a distant start, the 'wave function' of
the photon is pretty big. Why is it big? In order for the photons to
stay within the beam during their long transit time, they must have a
very tiny spread in sideways momentum. This means that the sideways
position must be very large by Heisenberg's uncertanity principle.
The *smallest* size of a photon wave function from a distant star is from
Beelgeuse (according to Herbert) - and is about 10ft! Others are much larger
- in fact, some were larger than the size of the experimental apparatus
used to measure them (400ft).
But when we 'look' at these photons, we do *not* see 10ft fuzzy disks,
we see points of light just like every other photon. Somehow, this spread
out wavefunction collapses when we 'look' at it. They don't "look" any
different than any other photons. (A photon is a photon is a photon.)
Of course, after we "look" at them, they are no longer in the
same eigenstate, in fact the purer the measurement approaches an
eigenstate of position, the less well defined the momentum is.
We can confirm the size of the photon wave function, however, by a neat
gadget called a 'stellar interferometer', built by physicist Robert Hanbury
Brown in 1965. It consists of two movable mirrors which focus light from
the same star on a photon detector. If we separate the mirrors of the
interferometer by more than the width of the wave function of the photon, the
coherence and interference pattern disappears.
Bronis Vidugiris
Mark Krumholz <krum...@princeton.edu> wrote:
) As for the virtual photon point: virtual photons do mediate the
)electromagnetic force, and they can impart a negative momentum
)(anti-parallel to their direction of travel). However, they _do
)not_ have momenta perpendicular to their direction of travel
)(except for that due to quantum uncertainty in that direction).
Dunno about that. Consider a particle moving in a magnetic field.
The magnetic field generates a transverse force.
)However,
)this is academic because the Coulomb potential between the Earth and the
)Sun is basically zero, since both are essentially neturally charged
)objects. Over the distance between the Earth and the Sun, the small
)irregularities in charge distribution in the Earth and the Sun have no
)measurable effect. Thus, while individual virtual photons may impart
)parallel or anti-parallel momentum to the Earth, the net effect of
)virtual photons is 0.
Yes, it's academic - OTOH the sun has a magnetic field, and while
v is almost exactly parallel to b, I suspect offhand that it's not
exactly parallel, implying some vxb force for any charge on the earth!
Now all we need to know is if the earth has a net charge or not
(it may, consider the solar wind).
john baez
>QED seems to me to be primarily a prediction tool.
Just like the rest of physics, I guess...
>Any given
>Feynman diagram accurately models a completed individual
>interaction (correct?).
Hmm, to my mind, Feynman diagrams are things whose amplitudes you *add
up* to figure out the probability that something or other happens. As
for the individual diagrams, you can think of them as cute pictures of the
various individual processes whose amplitudes you are summing up. All
Feynman diagrams with the same external legs (same particles coming
in and out, with same momenta) get summed over when doing any
calculation of an *experimentally measurable* scattering amplitude.
If you ask how "real" these individual processes are that you are
summing up, or how "real" are the internal edges (virtual particles) in
an individual Feynman diagram, all I can say is that it's a matter of
taste! Do you like to think of all mathematical constructs like the
electric field, the vector potential, or the wavefunction as "real", or
only some of them, or none of them? It's up to you; just get the
calculations of observable phenomena right!
>So (and neatly contradicting himself in one posting) are
>virtual particles to be considered a construct reflecting
>something underlying, or real (but invisible) beasties?
Worrying about what's "real", eh? I can't help you there; I don't even
know if *I'm* real. I could easily just be an abstraction devised by
the universe to summarize the results of the calculations my atoms are
performing.
Bronis Vidugiris
Oz <O...@upthorpe.demon.co.uk> wrote:
)Now I'm slightly unclear (an improvement). Perhaps if we
)took an actual example to clear it up. An excited atom emits
)a photon which is absorbed by another. Lets take two cases,
)one where the distance between the atoms is say 10mm, and
)one where it is 10^7m. Hmmm, perhaps we should take a large
)number of interactions like this to get a distribution of
)momentum for simplicity. If I took the distribution of the
)transferred momentum directions for each of these cases I
)would naively expect it to be the same for both distances
)since the uncertainty of the interaction is the same for
)both (ie for each atom pair). On the other hand the
)direction of the interacting photon(s) in each case is far
)more 'exact' over the longer distance than the shorter
)distance so one might expect a difference.
)
)Of course you might say that 'it all cancels out', my
)reaction would be to ask if this is calculated, or just (OK,
)I know) and opinion/experience. :-)
OK, here's my understanding of the situation.
Momentum will be conserved in total for the system when/if the
momentum of each atom is measured before and/or after the photon
emission, to any desired degree of accuracy that one cares
to measure, assuming there are no unaccounted for outside
interactions/influences, of course.
To maintain the conservation of momentum at all times, one
needs to introduce the idea of the momentum of the photon.
Otherwise one sees a a "lag" where the momentum appears to
temporarily dissappear.
The uncertainity principle enters here in that if you know the
momenta of the source and destination atoms, you have to have
some uncertanity about their position, so you never have
an exact "path". You're always uncertain either about the
momentum of an atom, or it's position. In the ideal case when
you know the momentum of the source and destionation atoms
exactly, you have *no* idea of where they are at all! When
you start to know something about where the source and
destination atoms are, you loose a bit of certanity about what
their momentum is. So when you start to get *too* accurate about
measuring the momentum of the atoms, you are constrained to not
know where they are.