Science > Physics > coherent and incoherent optical pulse combining ...
| Topic: |
Science > Physics |
| User: |
"Q-man" |
| Date: |
21 Jul 2004 10:54:09 PM |
| Object: |
coherent and incoherent optical pulse combining ... |
I'd like to ask a conceptual question.
Here is two optical pulses.
Somehow without energy loss, two optical pusles are combined
incoherently.
Energy will be twice.
If two pulses are combined coherently, energy will be four times
because amplitude of electric field become twice.
But if energy conservation is right, then it should be twice.
How can it satisfy the conservation of energy ?
.
|
|
| User: "Timo Nieminen" |
|
| Title: Re: coherent and incoherent optical pulse combining ... |
22 Jul 2004 12:42:19 AM |
|
|
On Thu, 21 Jul 2004, Q-man wrote:
I'd like to ask a conceptual question.
Here is two optical pulses.
Somehow without energy loss, two optical pusles are combined
incoherently.
Energy will be twice.
If two pulses are combined coherently, energy will be four times
because amplitude of electric field become twice.
But if energy conservation is right, then it should be twice.
How can it satisfy the conservation of energy ?
Because it isn't possible to combine two optical pulses coherently without
adding extra energy. It can't be done with just simple passive components
like beamsplitters.
If you want to combine two electromagnetic waves coherently, it's easiest
to do in at RF. This gives rise to exactly the same problem.
Consider, for example, two adjacent antennas. Drive one with a current I,
get a field E at some point. Drive the other with current I, also get a
field E at that point. Drive both with current I, and get a field of 2E,
so 4 times the power is radiated.
Where does the extra power come from?
Answer 1: the fields from both antennas act against the current in each
antenna, so to maintain the current I, twice the driving voltage is need.
Thus, each antenna uses twice as much power to drive.
Answer 2: consider the two antennas to be effectively a single antenna of
the same length, with current 2I. The radiation resistance R is the same,
depending only on the length for a thin antenna, so I^2 R is now 4 times
higher.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
|
|
|
| User: "Q-man" |
|
| Title: Re: coherent and incoherent optical pulse combining ... |
22 Jul 2004 10:37:38 AM |
|
|
Timo Nieminen <timo@physics.uq.edu.au> wrote in message news:<Pine.LNX.4.50.0407221531060.4547-100000@localhost>...
On Thu, 21 Jul 2004, Q-man wrote:
I'd like to ask a conceptual question.
Here is two optical pulses.
Somehow without energy loss, two optical pusles are combined
incoherently.
Energy will be twice.
If two pulses are combined coherently, energy will be four times
because amplitude of electric field become twice.
But if energy conservation is right, then it should be twice.
How can it satisfy the conservation of energy ?
Because it isn't possible to combine two optical pulses coherently without
adding extra energy. It can't be done with just simple passive components
like beamsplitters.
If you want to combine two electromagnetic waves coherently, it's easiest
to do in at RF. This gives rise to exactly the same problem.
Consider, for example, two adjacent antennas. Drive one with a current I,
get a field E at some point. Drive the other with current I, also get a
field E at that point. Drive both with current I, and get a field of 2E,
so 4 times the power is radiated.
Where does the extra power come from?
Answer 1: the fields from both antennas act against the current in each
antenna, so to maintain the current I, twice the driving voltage is need.
Thus, each antenna uses twice as much power to drive.
Answer 2: consider the two antennas to be effectively a single antenna of
the same length, with current 2I. The radiation resistance R is the same,
depending only on the length for a thin antenna, so I^2 R is now 4 times
higher.
Would you please check if I understand correctly ?
Energy becomes 4 times but to get this amount of energy we need more power.
That means that the consevation of energy is still valid.
But you mean optically it is impossible.
.
|
|
|
| User: "Timo Nieminen" |
|
| Title: Re: coherent and incoherent optical pulse combining ... |
22 Jul 2004 06:03:10 PM |
|
|
On Fri, 22 Jul 2004, Q-man wrote:
Would you please check if I understand correctly ?
Energy becomes 4 times but to get this amount of energy we need more power.
That means that the consevation of energy is still valid.
Yes.
But you mean optically it is impossible.
Impossible with passive components. With active components (ie components
that require an input of energy to operate), such as a medium with gain
used as a laser amplifier, you can do it.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
|
|
|
|
|
| User: "Mark Fergerson" |
|
| Title: Re: coherent and incoherent optical pulse combining ... |
22 Jul 2004 03:59:39 AM |
|
|
Timo Nieminen wrote:
On Thu, 21 Jul 2004, Q-man wrote:
I'd like to ask a conceptual question.
Here is two optical pulses.
Somehow without energy loss, two optical pusles are combined
incoherently.
Energy will be twice.
If two pulses are combined coherently, energy will be four times
because amplitude of electric field become twice.
But if energy conservation is right, then it should be twice.
How can it satisfy the conservation of energy ?
Because it isn't possible to combine two optical pulses coherently without
adding extra energy. It can't be done with just simple passive components
like beamsplitters.
If you want to combine two electromagnetic waves coherently, it's easiest
to do in at RF. This gives rise to exactly the same problem.
Consider, for example, two adjacent antennas. Drive one with a current I,
get a field E at some point. Drive the other with current I, also get a
field E at that point. Drive both with current I, and get a field of 2E,
so 4 times the power is radiated.
Where does the extra power come from?
Answer 1: the fields from both antennas act against the current in each
antenna, so to maintain the current I, twice the driving voltage is need.
Thus, each antenna uses twice as much power to drive.
Answer 2: consider the two antennas to be effectively a single antenna of
the same length, with current 2I. The radiation resistance R is the same,
depending only on the length for a thin antenna, so I^2 R is now 4 times
higher.
IIRC the same exact problem causes the SHIVA-NOVA
amplifiers to be so damn big.
Mark L. Fergerson
.
|
|
|
|
|
|
Related Articles |
|
|
