Science > Physics > Electrodynamics questions, 4: radiation field of point particle
| Topic: |
Science > Physics |
| User: |
"Erland Gadde" |
| Date: |
09 May 2004 09:56:37 AM |
| Object: |
Electrodynamics questions, 4: radiation field of point particle |
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
with "x" as cross product and "." as dot product, and:
E = electric field
q = charge of particle
n = unit vector from particle to field point
v = velocity of particle
c = light speed
s = 1 - n.v/c
R = distance from particle to field point
t = time
and all vectors and quantities should not be calculated at the present
time but at the _retarded time_ (if the particle's distance at the
time t' to the field point is R, and if t = t' + R/c, where t is the
present time, then this time t' is the retarded time).
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
For example, assume that |dv/dt| = 10^6 m/s^2. An acceleration that to
mee seems enormous. If the above condition should hold, then we must
have R >> 9*10^10 m, and the latter distance is 60% of the distance
Earth-Sun!
As another example, consider a laboratory experiment, with, say, R = 1
m. If the condition is to hold in this case, then we should have
|dv/dt| >> 9*10^16 m/s^2,
meaning that the particle should accelerate from rest to near light
speed in few nanoseconds! I don't know what the typical accelerations
are for subatomic particles, but such an acceleation doesn't seem
realistic to me.
So, to say that the radiation field dominates at large distances is
misleading.
Yet, at least Eyges is primarily interested in the radiation field and
ignores the velocity field.
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy? If not, mustn't it be
accounted for when calculating the energy loss?
All help and comments would be appreciated.
Erland Gadde
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| User: "Old Man" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
09 May 2004 03:01:12 PM |
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"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405090656.2a34f7ff@posting.google.com...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
with "x" as cross product and "." as dot product, and:
E = electric field
q = charge of particle
n = unit vector from particle to field point
v = velocity of particle
c = light speed
s = 1 - n.v/c
R = distance from particle to field point
t = time
and all vectors and quantities should not be calculated at the present
time but at the _retarded time_ (if the particle's distance at the
time t' to the field point is R, and if t = t' + R/c, where t is the
present time, then this time t' is the retarded time).
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
Could this be a slight exaggeration? For circular orbits in
a magnetic field, |dv / dt| = v^2 / R. Consider an electron
synchrotron of radius 10 km for which v / c = 0.9. Then,
for radial distances greater than 10 + 12.1 = 22.1 km, the
field strength from the acceleration term exceeds that of the
velocity term.
For example, assume that |dv/dt| = 10^6 m/s^2. An acceleration that to
mee seems enormous. If the above condition should hold, then we must
have R >> 9*10^10 m, and the latter distance is 60% of the distance
Earth-Sun!
As another example, consider a laboratory experiment, with, say, R = 1
m. If the condition is to hold in this case, then we should have
|dv/dt| >> 9*10^16 m/s^2,
meaning that the particle should accelerate from rest to near light
speed in few nanoseconds! I don't know what the typical accelerations
are for subatomic particles, but such an acceleation doesn't seem
realistic to me.
So, to say that the radiation field dominates at large distances is
misleading.
Yet, at least Eyges is primarily interested in the radiation field and
ignores the velocity field.
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy?
Yes. The velocity term doesn't contribute to charged particle
energy loss in free space. [Old Man]
If not, mustn't it be
accounted for when calculating the energy loss?
All help and comments would be appreciated.
Erland Gadde
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| User: "Erland Gadde" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
13 May 2004 07:34:48 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<DeCdnUMiM4iWEwPdRVn-gQ@prairiewave.com>...
"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405090656.2a34f7ff@posting.google.com...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
with "x" as cross product and "." as dot product, and:
E = electric field
q = charge of particle
n = unit vector from particle to field point
v = velocity of particle
c = light speed
s = 1 - n.v/c
R = distance from particle to field point
t = time
and all vectors and quantities should not be calculated at the present
time but at the _retarded time_ (if the particle's distance at the
time t' to the field point is R, and if t = t' + R/c, where t is the
present time, then this time t' is the retarded time).
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
Could this be a slight exaggeration? For circular orbits in
a magnetic field, |dv / dt| = v^2 / R. Consider an electron
synchrotron of radius 10 km for which v / c = 0.9. Then,
for radial distances greater than 10 + 12.1 = 22.1 km, the
field strength from the acceleration term exceeds that of the
velocity term.
OK, but still, if Ea >> Ev, (Ea = acc. field, Ev = vel. field) we must
be several tenths of kilometers from the synchroton, while the
interesting things happends _inside_ the synchroton (or...?), where Ev
still dominates.
Erland Gadde
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| User: "Old Man" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
13 May 2004 06:14:20 PM |
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"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405130434.7d60b95a@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<DeCdnUMiM4iWEwPdRVn-gQ@prairiewave.com>...
"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405090656.2a34f7ff@posting.google.com...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
with "x" as cross product and "." as dot product, and:
E = electric field
q = charge of particle
n = unit vector from particle to field point
v = velocity of particle
c = light speed
s = 1 - n.v/c
R = distance from particle to field point
t = time
and all vectors and quantities should not be calculated at the present
time but at the _retarded time_ (if the particle's distance at the
time t' to the field point is R, and if t = t' + R/c, where t is the
present time, then this time t' is the retarded time).
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
Could this be a slight exaggeration? For circular orbits in
a magnetic field, |dv / dt| = v^2 / R. Consider an electron
synchrotron of radius 10 km for which v / c = 0.9. Then,
for radial distances greater than 10 + 12.1 = 22.1 km, the
field strength from the acceleration term exceeds that of the
velocity term.
OK, but still, if Ea >> Ev, (Ea = acc. field, Ev = vel. field) we must
be several tenths of kilometers from the synchroton, while the
interesting things happends _inside_ the synchroton (or...?), where Ev
still dominates.
Erland Gadde
For v -> c, the velocity field is peaked at right angles to v, but
the acceleration field is peaked in the forward v direction (dipolar
+ headlight effect). There is no acceleration field at the center
of a synchrotron. It's emitted along a tangent to the periphery.
[Old Man]
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| User: "Old Man" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
09 May 2004 01:42:31 PM |
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"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405090656.2a34f7ff@posting.google.com...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
with "x" as cross product and "." as dot product, and:
E = electric field
q = charge of particle
n = unit vector from particle to field point
v = velocity of particle
c = light speed
s = 1 - n.v/c
R = distance from particle to field point
t = time
and all vectors and quantities should not be calculated at the present
time but at the _retarded time_ (if the particle's distance at the
time t' to the field point is R, and if t = t' + R/c, where t is the
present time, then this time t' is the retarded time).
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
The velocity field declines as 1 / R^2 and the acceleration
field ought to decline as 1 / R, but Erland shows a 1 / r^2
dependence whereof r is undefined. [Old Man]
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
For example, assume that |dv/dt| = 10^6 m/s^2. An acceleration that to
mee seems enormous. If the above condition should hold, then we must
have R >> 9*10^10 m, and the latter distance is 60% of the distance
Earth-Sun!
As another example, consider a laboratory experiment, with, say, R = 1
m. If the condition is to hold in this case, then we should have
|dv/dt| >> 9*10^16 m/s^2,
meaning that the particle should accelerate from rest to near light
speed in few nanoseconds! I don't know what the typical accelerations
are for subatomic particles, but such an acceleation doesn't seem
realistic to me.
So, to say that the radiation field dominates at large distances is
misleading.
Yet, at least Eyges is primarily interested in the radiation field and
ignores the velocity field.
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy? If not, mustn't it be
accounted for when calculating the energy loss?
All help and comments would be appreciated.
Erland Gadde
.
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| User: "Erland Gadde" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
09 May 2004 04:47:22 PM |
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(Erland Gadde) wrote in message news:<c114637b.0405090656.2a34f7ff@posting.google.com>...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
Correction: the denominator in the second term should be
c*s^3*R
and nothing else.
Sorry!
Erland Gadde
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| User: "Edward Green" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
23 May 2004 07:13:22 AM |
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(Erland Gadde) wrote in message news:<c114637b.0405090656.2a34f7ff@posting.google.com>...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
I have Eyges. What equation number is that?
<...>
And while we are waiting, something cute just occured to me. Given
conservation of energy, and given that intensity ~ E^2, then we expect
to the product r^2 E^2 to be constant around a point radiating source.
Hence your 1/r dependence of E in the radiation field.
Now what about the near field? Through long familiarity, it had begun
to seem "obvious" to me that it fell off as 1/r^2 by a similar
argument; something like "the amount of field is conserved", aka
Guass's law, which is really a restatement of the observation that the
field falls off as 1/r^2 .
It seems like an obvious argument for the radial dependence of the
electrostatic field is merely suggestive, whereas that of the
radiation field makes contact with other physical concepts, like
energy.
Comments?
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| User: "Keith Stein" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
24 May 2004 09:06:59 AM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0405230413.7d494fa3@posting.google.com...
<...>
And while we are waiting, something cute just occured to me. Given
conservation of energy, and given that intensity ~ E^2, then we expect
to the product r^2 E^2 to be constant around a point radiating source.
Hence your 1/r dependence of E in the radiation field.
While this would be true for a spherically symetric radiating source, it
would not be true in general, i think. Note that while a stationary 'point'
is spherically symetrical, a moving point is a line, which is of course
NOT spherically symetrical eh!
Now what about the near field? Through long familiarity, it had begun
to seem "obvious" to me that it fell off as 1/r^2 by a similar
argument; something like "the amount of field is conserved", aka
Guass's law, which is really a restatement of the observation that the
field falls off as 1/r^2 .
Same point again, Mr. Green. If electrons were little 'darts' then we
wouldn't expect the near field to fall off as 1/r^2, otoh if they were
little 'spheres' then we might, although to be honest i can't the field
couldn't fall off as 1/r^3, but maybe it would violate something or
other. Who knows?
It seems like an obvious argument for the radial dependence of the
electrostatic field is merely suggestive, whereas that of the
radiation field makes contact with other physical concepts, like
energy.
I see what you mean, Mr.Green. Bottom line is:
Comments?
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| User: "Erland Gadde" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
24 May 2004 07:11:16 PM |
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(Edward Green) wrote in message news:<eca320d0.0405230413.7d494fa3@posting.google.com>...
erland@bredband.net (Erland Gadde) wrote in message news:<c114637b.0405090656.2a34f7ff@posting.google.com>...
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
I have Eyges. What equation number is that?
Eq. 14.14, p. 282, in my edition (Dover, 1980), with some changed
notation, and a typo, the last denominator should be c*s^3*R.
Regards,
Erland Gadde
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| User: "Timo Nieminen" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
09 May 2004 08:17:04 PM |
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On Mon, 9 May 2004, Erland Gadde wrote:
The electric field of a moving point particle (point current) is given
by (Lienard-Wiechert form, gaussian units):
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*r^2))
Should be:
E = q*((n-v/c)*(1-(|v|/c)^2)/(s^3*R^2) + n x ((n-v/c) x
(dv/dt)/c)/(c*s^3*R))
The two terms in the expression for E above are called the _velocity
field_ and the _radiation field_.
"The radiation field dominates at large distances and it is this that
must be used to calculate energy loss by radiation." writes Leonard
Eyges i "The Classical Electromagnetic Field, p. 282 (Dover, 1980).
But, as Eyges also points out, the condition that the radiation field
dominates is:
R*|dv/dt| >> c^2.
And that is my problem. As far as I can see, this condition can only
be satisfied at very large distances: light years or so.
For example, assume that |dv/dt| = 10^6 m/s^2. An acceleration that to
mee seems enormous.
Consider a cyclotron with say, charged particles at v = 10^8 m/s.
|dv/dt| = 10^16 / radius m/s^2, so 10^6 doesn't seem so ernormous.
So, to say that the radiation field dominates at large distances is
misleading.
Yet, at least Eyges is primarily interested in the radiation field and
ignores the velocity field.
If the energy in the velocity field is constant (eg cyclotron with beam at
constant speed), then the radiation field is the important thing, since
this is energy lost by the charge(s).
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy? If not, mustn't it be
accounted for when calculating the energy loss?
If the energy content of the velocity field changes, then yes, one had
better consider this. Still true to say that it doesn't contribute to the
_radiated_ energy, but it can contribute to the energy loss. Note that for
a particle in periodic motion, the average energy loss due to the velocity
field over one period must be zero.
I was recently directed to some interesting reading (thanks, Bill):
http://www.mathpages.com/home/kmath528/kmath528.htm
which considers this issue.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Erland Gadde" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
13 May 2004 11:16:31 AM |
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Timo Nieminen wrote:
Consider a cyclotron with say, charged particles at v = 10^8 m/s.
|dv/dt| = 10^16 / radius m/s^2, so 10^6 doesn't seem so ernormous.
Old Man had a similar example. But still, you must be quite far away
from the cyclotron for the acceleration field to be dominant.
If the energy in the velocity field is constant (eg cyclotron with beam at
constant speed), then the radiation field is the important thing, since
this is energy lost by the charge(s).
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy? If not, mustn't it be
accounted for when calculating the energy loss?
If the energy content of the velocity field changes, then yes, one had
better consider this. Still true to say that it doesn't contribute to the
_radiated_ energy, but it can contribute to the energy loss. Note that for
a particle in periodic motion, the average energy loss due to the velocity
field over one period must be zero.
Where can I find a proof of this?
Thinking about all this, I realize that the concepts are not clearly
defined (as far as I have seen). What do we mean by "radiated energy"
and "energy loss"?
Do we only consider radiated energy through spherical surfaces, and if
so, centered where? For example, is it still true that no energy is
radiated through a non-spherical surface in the case of constant
velocity of the particle? Could this perhaps be proved by vector
analysis?
Let Ev = velocity field, Ea = acceleration field,
Mv = n x Ev, Ma = n x Ea (where n = unit vector to field point from
particle's position at retarded time.)
The total magnetic field is then M = Mv + Ma, and the Poynting vector
is
S = (c/(4*pi))E x M. But the Poynting vector then consists of four
terms
S = c/(4*pi)(Ev x Mv + Ev x Ma + Ea x Bv + Ea x Ba)
Now, when we talk about the energy loss or radiation due to "velocity
field" or "acceleration field", it is not clear to me which terms we
refer to (since two of them are mixed). And of these terms, which of
them are zero or don't contribute to radiation, and through which
surfaces?
I think these matters need to be clarified, but that is not done in
any of the books I have consulted...
Regards,
Erland Gadde
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| User: "Old Man" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
13 May 2004 07:29:51 PM |
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"Erland Gadde" <erland@bredband.net> wrote in message
news:c114637b.0405130816.714d07e7@posting.google.com...
Timo Nieminen wrote:
Consider a cyclotron with say, charged particles at v = 10^8 m/s.
|dv/dt| = 10^16 / radius m/s^2, so 10^6 doesn't seem so ernormous.
Old Man had a similar example. But still, you must be quite far away
from the cyclotron for the acceleration field to be dominant.
If the energy in the velocity field is constant (eg cyclotron with beam
at
constant speed), then the radiation field is the important thing, since
this is energy lost by the charge(s).
I know that if dv/dt = 0, then the particle will radiate no energy,
but what if dv/dt =/= 0? Is it then still true that the velocity field
doesn't contribute to the radiated energy? If not, mustn't it be
accounted for when calculating the energy loss?
If the energy content of the velocity field changes, then yes, one had
better consider this. Still true to say that it doesn't contribute to
the
_radiated_ energy, but it can contribute to the energy loss. Note that
for
a particle in periodic motion, the average energy loss due to the
velocity
field over one period must be zero.
Where can I find a proof of this?
Thinking about all this, I realize that the concepts are not clearly
defined (as far as I have seen). What do we mean by "radiated energy"
and "energy loss"?
Do we only consider radiated energy through spherical surfaces, and if
so, centered where? For example, is it still true that no energy is
radiated through a non-spherical surface in the case of constant
velocity of the particle? Could this perhaps be proved by vector
analysis?
Let Ev = velocity field, Ea = acceleration field,
Mv = n x Ev, Ma = n x Ea (where n = unit vector to field point from
particle's position at retarded time.)
The total magnetic field is then M = Mv + Ma, and the Poynting vector
is
S = (c/(4*pi))E x M. But the Poynting vector then consists of four
terms
S = c/(4*pi)(Ev x Mv + Ev x Ma + Ea x Bv + Ea x Ba)
Now, when we talk about the energy loss or radiation due to "velocity
field" or "acceleration field", it is not clear to me which terms we
refer to (since two of them are mixed). And of these terms, which of
them are zero or don't contribute to radiation, and through which
surfaces?
I think these matters need to be clarified, but that is not done in
any of the books I have consulted...
Regards,
Erland Gadde
I think that the total radiated power is proportional to the
square of the acceleration and, for v << c, is not dependent
upon the velocity. Jackson says that total radiated power
is Lorentz invariant and that for linear acceleration total
radiated power is proportional to (dp / dt)^2 where p is the
charged particle's relativistic momentum,.
If total radiated power is independent of velocity, can one infer
that the "velocity term" doesn't contribute to energy loss in
free space? [Old Man]
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| User: "Edward Green" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
22 May 2004 08:50:45 AM |
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"Old Man" <nomail@nomail.net> wrote in message news:<g6KdnbW7wI6ejjndRVn-iQ@prairiewave.com>...
If total radiated power is independent of velocity, can one infer
that the "velocity term" doesn't contribute to energy loss in
free space? [Old Man]
I don't know. But can one infer from the names that the velocity
field/term is simply the electrostatic field as seen by a relatively
moving observer?
Proceding as if this is correct, can we perhaps guess that even if we
jiggle the particle, the velocity field amounts to the electrostatic
field as seen by an observer who was, at different epochs, moving at
different relative velocities (related to use of retarded time): the
electrostatic field, now chopped up into sections appropriate to
various v's.
Proceding as if this meant something, perhaps we can infer that the
acceleration/radiation field/term is some component over and above
that which can be explained by this first construction.
Hmm... Ok. Say all of this meant something. The velocity field of an
unaccelerated particle has a different total energy (modulo some
discussion about cutoff radius) or energy density for different
observers, although the field obviously doesn't give a figo how we
observe it. So for an unaccelerated, say relativistic, electron,
some/all of the kinetic energy resides in the field, whereas from the
field's point of view, that's our problem -- it's just the plain old
field of a static particle, shucks.
Now, say we disturb the electron, and guess that the observed field
is just the same as before, except that at a given location, we
observe the field appropriate to an earler motion of the electron, at
a time dependent on the distance from said some earlier motion, and
the speed of light.
And it was good.
But it was not entirely correct, since we found that there was an
additional component of the energy not attributable to this simple
construction, but instead attributable to the resulting kinkiness of
the field itself -- a kinked EM field carrying energy and momentum --
aka "light" -- aka "the radiation field".
And it was better.
And now one need to take just perhaps a peek at the equations, to see
if this fanstasy is supportable, and if we can tease out these
components pointwise or only globally, by observing the fields to be
the sum of two components, which locally, just result in "the field".
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| User: "Keith Stein" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
23 May 2004 06:18:17 AM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0405220550.362972@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<g6KdnbW7wI6ejjndRVn-iQ@prairiewave.com>...
If total radiated power is independent of velocity, can one infer
that the "velocity term" doesn't contribute to energy loss in
free space? [Old Man]
I don't know.
I think the Old Man is suggesting that the total radiated power from
a star, or anything, does not change according to who is observing it,
and if Old Man means what I think Old Man means then
the answer is "Yes" imho, Mr. Green, but i did enjoy your digression eh!
But can one infer from the names that the velocity
field/term is simply the electrostatic field as seen by a relatively
moving observer?
Proceding as if this is correct, can we perhaps guess that even if we
jiggle the particle, the velocity field amounts to the electrostatic
field as seen by an observer who was, at different epochs, moving at
different relative velocities (related to use of retarded time): the
electrostatic field, now chopped up into sections appropriate to
various v's.
Proceding as if this meant something, perhaps we can infer that the
acceleration/radiation field/term is some component over and above
that which can be explained by this first construction.
Hmm... Ok. Say all of this meant something. The velocity field of an
unaccelerated particle has a different total energy (modulo some
discussion about cutoff radius) or energy density for different
observers, although the field obviously doesn't give a figo how we
observe it. So for an unaccelerated, say relativistic, electron,
some/all of the kinetic energy resides in the field, whereas from the
field's point of view, that's our problem -- it's just the plain old
field of a static particle, shucks.
Now, say we disturb the electron, and guess that the observed field
is just the same as before, except that at a given location, we
observe the field appropriate to an earler motion of the electron, at
a time dependent on the distance from said some earlier motion, and
the speed of light.
And it was good.
But it was not entirely correct, since we found that there was an
additional component of the energy not attributable to this simple
construction, but instead attributable to the resulting kinkiness of
the field itself -- a kinked EM field carrying energy and momentum --
aka "light" -- aka "the radiation field".
And it was better.
And now one need to take just perhaps a peek at the equations, to see
if this fanstasy is supportable, and if we can tease out these
components pointwise or only globally, by observing the fields to be
the sum of two components, which locally, just result in "the field".
And better still
Would be to appreciate that "the field" (be it electric or magnetic) is
actually a distortion in the medium, and these "distortions" in the
medium travel in the medium at the velocity of light in the medium,
and this velocity is RELATIVE TO THE MEDIUM,which is why
Maxwell insisted "A MEDIUM NECESSARY " eh!
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| User: "Cecil Moore" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
23 May 2004 08:43:09 AM |
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Keith Stein wrote:
Would be to appreciate that "the field" (be it electric or magnetic) is
actually a distortion in the medium, and these "distortions" in the
medium travel in the medium at the velocity of light in the medium,
and this velocity is RELATIVE TO THE MEDIUM,which is why
Maxwell insisted "A MEDIUM NECESSARY " eh!
Since those fields are probably prohibited from traveling outside
of our universe, Maxwell was probably right. Empty space is not
empty. Didn't relativity solve the speed of propagation problem
through the original ill-conceived "Ęther"?
--
cheers, Cecil
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| User: "Timo Nieminen" |
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| Title: Re: Electrodynamics questions, 4: radiation field of point particle |
20 May 2004 08:48:14 PM |
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Didn't see original reply, so piggybacking:
"Erland Gadde" <erland@bredband.net> wrote:
Timo Nieminen wrote:
If the energy content of the velocity field changes, then yes, one had
better consider this. Still true to say that it doesn't contribute to
the
_radiated_ energy, but it can contribute to the energy loss. Note that
for
a particle in periodic motion, the average energy loss due to the
velocity
field over one period must be zero.
Where can I find a proof of this?
Rate of work done against the radiation reaction force is
dW/dt = const v da/dt = const a^2 + const d(v da/dt)/dt
The integral over one period of the second term must be zero, since
v(t) = v(t+T), da(t)/dt = da(t+T)/dt.
For this to be a proof of the above, one would also need to show that the
Larmor formula term (ie the first term on the RHS) doesn't involve the
velocity field.
Thinking about all this, I realize that the concepts are not clearly
defined (as far as I have seen). What do we mean by "radiated energy"
and "energy loss"?
Do we only consider radiated energy through spherical surfaces, and if
so, centered where? For example, is it still true that no energy is
radiated through a non-spherical surface in the case of constant
velocity of the particle? Could this perhaps be proved by vector
analysis?
We can consider radiated energy to be energy that is lost, and cannot
return to the particle regardless of its future motion. I'd consider
energy temporarily stored in the velocity field as a loss, but this loss
is negative (so a gain) when this stored energy is returned to the
particle.
But the Poynting vector then consists of four
terms
S = c/(4*pi)(Ev x Mv + Ev x Ma + Ea x Bv + Ea x Ba)
Now, when we talk about the energy loss or radiation due to "velocity
field" or "acceleration field", it is not clear to me which terms we
refer to (since two of them are mixed). And of these terms, which of
them are zero or don't contribute to radiation, and through which
surfaces?
It's a good question. I haven't looked at this in terms of fields yet,
just from conservation of energy, which is a nice simple approach, but
doesn't say anything about the fields.
I think these matters need to be clarified, but that is not done in
any of the books I have consulted...
Rohrlich "Classical charged particles" might be worth reading. A copy sits
here next to my desk, but I'm only reading it very slowly.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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