| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
11 Mar 2006 09:43:14 AM |
| Object: |
Energy conservation in moving frames |
I have a thought experiment that is puzzling to me.
Suppose there is a spacecraft headed towards earth. It has a laser on
board that emits a sequence of short pulses, aimed towards a receiver
on earth. These pulses have a duration T when measured by an on-board
clock, and a wavelength lambda. The repitition rate is fixed according
to the clock on the spacecraft.
Assume the spacecraft is accelerating twoards earth, so the pulses
received on earth are no longer evenly spaced in local earth time, and
the measured pulse duration T is also variable (and always shorter,
given that the spacecraft is accelerating). Also, the wavelength lambda
will be, in general, shorter.
Now assume that after some time, the laser stops emitting pulses, and
the total energy of the pulses (since the generator was turned on) is
measured both on the spacecraft and on earth. Will the answers be the
same? Does wavelength come into the energy calculation somehow, in a
way that compensates for the shorter pulse duration received on earth?
Thanks for any insight.
Bob Adams
.
|
|
| User: "Hexenmeister" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 10:42:56 AM |
|
|
<robert.w.adams@verizon.net> wrote in message
news:1142091794.069543.162290@j33g2000cwa.googlegroups.com...
I have a thought experiment that is puzzling to me.
Suppose there is a spacecraft headed towards earth. It has a laser on
board that emits a sequence of short pulses, aimed towards a receiver
on earth. These pulses have a duration T when measured by an on-board
clock, and a wavelength lambda. The repitition rate is fixed according
to the clock on the spacecraft.
Assume the spacecraft is accelerating twoards earth, so the pulses
received on earth are no longer evenly spaced in local earth time, and
the measured pulse duration T is also variable (and always shorter,
given that the spacecraft is accelerating). Also, the wavelength lambda
will be, in general, shorter.
Now assume that after some time, the laser stops emitting pulses, and
the total energy of the pulses (since the generator was turned on) is
measured both on the spacecraft and on earth. Will the answers be the
same? Does wavelength come into the energy calculation somehow, in a
way that compensates for the shorter pulse duration received on earth?
Thanks for any insight.
Bob Adams
Insight is free with hindsight.
Wavelength:
http://www.androcles01.pwp.blueyonder.co.uk/Catalina/Drive.htm
Doppler: http://www.androcles01.pwp.blueyonder.co.uk/Doppler/Doppler.htm
Einstein: http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
Androcles.
.
|
|
|
|
| User: "Timo Nieminen" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 02:04:42 PM |
|
|
On Sun, 11 Mar 2006 wrote:
I have a thought experiment that is puzzling to me.
Suppose there is a spacecraft headed towards earth. It has a laser on
board that emits a sequence of short pulses, aimed towards a receiver
on earth. These pulses have a duration T when measured by an on-board
clock, and a wavelength lambda. The repitition rate is fixed according
to the clock on the spacecraft.
Assume the spacecraft is accelerating twoards earth, so the pulses
received on earth are no longer evenly spaced in local earth time, and
the measured pulse duration T is also variable (and always shorter,
given that the spacecraft is accelerating). Also, the wavelength lambda
will be, in general, shorter.
Now assume that after some time, the laser stops emitting pulses, and
the total energy of the pulses (since the generator was turned on) is
measured both on the spacecraft and on earth. Will the answers be the
same? Does wavelength come into the energy calculation somehow, in a
way that compensates for the shorter pulse duration received on earth?
Don't bother making it more complicated by having a series of pulses, and
acceleration. Just stick to a single pulse, from a spaceship moving at
constant velocity towards the detector.
If the pulse is at a frequency f, as determined on board the spaceship,
and consists of N photons, the spaceship-measured energy in Nhf, where h
is Planck's constant.
For slow approach velocity v<<c, the detector-measured energy will be
Nh(f+df), where df = vf/c. This energy is different.
Note that the same thing will apply if a rock is thrown. A proper
relativistic calculation for the Doppler shift of the pulse doesn't change
the conclusion.
OK, so what does this mean? Firstly, while energy is conserved in an
inertial reference frame, there is no required for energy to be the same
as measured in different reference frames (something which has nothing to
do with conservation of energy).
It also means that, as measured from the _accelerating_ frame of your
shaceship, the energy of each already emitted pulse keeps getting less and
less as the ship accelerates; this looks like a nice simple
thought-demonstration that energy is _not_ conserved in accelerating
reference frames. This would be the case under either Galilei transforms
or Lorentz transforms.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
|
|
|
| User: "Hexenmeister" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 05:21:27 PM |
|
|
"Timo Nieminen" <uqtniemi@mailbox.uq.edu.au> wrote in message
news:20060312055535.L70045@emu.uq.edu.au...
| Don't bother making it more complicated by having a series of pulses, and
| acceleration. Just stick to a single pulse, from a spaceship moving at
| constant velocity towards the detector.
LOL! Doppler shift with a single cycle.
Here, debug this single pulse.
http://www.androcles01.pwp.blueyonder.co.uk/Doppler/Doppler.htm
Do you know have any idea how stupid you are, Nieminen?
Androcles.
.
|
|
|
| User: "Timo Nieminen" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 06:01:48 PM |
|
|
On Sat, 11 Mar 2006, Hexenmeister wrote:
"Timo Nieminen" <uqtniemi@mailbox.uq.edu.au> wrote:
| Don't bother making it more complicated by having a series of pulses, and
| acceleration. Just stick to a single pulse, from a spaceship moving at
| constant velocity towards the detector.
LOL! Doppler shift with a single cycle.
"Single pulse", not "single-cycle pulse". There is a difference.
Btw, you posted earlier asking about Lorentz transformations for velocity
in an arbitrary direction. I posted a derivation of the homogeneous
Lorentz transformations for arbitrary relative velocity and orientation of
the two coordinate systems.
You did reply to that post, but without comment on the derivation. Since
you've repeated your original request about Lorentz transformations for
velocity in an arbitrary direction, perhaps you didn't understand it. Do
you want to go over the derivation I provided in steps so I can explain it
to you?
--
Timo
.
|
|
|
| User: "Hexenmeister" |
|
| Title: Re: Energy conservation in moving frames |
12 Mar 2006 07:08:48 AM |
|
|
"Timo Nieminen" <uqtniemi@mailbox.uq.edu.au> wrote in message
news:20060312095712.F70045@emu.uq.edu.au...
| On Sat, 11 Mar 2006, Hexenmeister wrote:
|
| > "Timo Nieminen" <uqtniemi@mailbox.uq.edu.au> wrote:
| > | Don't bother making it more complicated by having a series of pulses,
and
| > | acceleration. Just stick to a single pulse, from a spaceship moving at
| > | constant velocity towards the detector.
| >
| > LOL! Doppler shift with a single cycle.
|
| "Single pulse", not "single-cycle pulse". There is a difference.
Idiot:
http://www.falstad.com/fourier/
Oh, you snipped.
Refusal to discuss physics noted.
"Timo Nieminen" <uqtniemi@mailbox.uq.edu.au> wrote in message
news:20060312055535.L70045@emu.uq.edu.au...
| Don't bother making it more complicated by having a series of pulses, and
| acceleration. Just stick to a single pulse, from a spaceship moving at
| constant velocity towards the detector.
LOL! Doppler shift with a single cycle.
Here, debug this single pulse.
http://www.androcles01.pwp.blueyonder.co.uk/Doppler/Doppler.htm
Do you know have any idea how stupid you are, Nieminen?
Androcles.
.
|
|
|
|
|
|
| User: "Henning Makholm" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 04:48:35 PM |
|
|
Scripsit Timo Nieminen <uqtniemi@mailbox.uq.edu.au>
It also means that, as measured from the _accelerating_ frame of your
shaceship, the energy of each already emitted pulse keeps getting less
and less as the ship accelerates; this looks like a nice simple
thought-demonstration that energy is _not_ conserved in accelerating
reference frames.
Simple? I would think it simpler to just take a standard-issue
accelerating rocketship and let a massive object drop to the floor. If
you release it from velocity zero (with respect to the accelerating
frame), its kinetic energy starts increasing all by itself, so energy
conservation is broken. This does not presuppose any relation between
the wavelength and energy contents of light.
(We can invent a potential to explain away the energy increase, but
that works only until we _change_ the acceleration of our rocketship,
at which point the potential energies of all objects would jump
wildly, definitely violating energy conservation again).
--
Henning Makholm "Also, the letters are printed. That makes the task
of identifying the handwriting much more difficult."
.
|
|
|
| User: "Timo Nieminen" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 05:15:42 PM |
|
|
On Sat, 11 Mar 2006, Henning Makholm wrote:
Scripsit Timo Nieminen <uqtniemi@mailbox.uq.edu.au>
It also means that, as measured from the _accelerating_ frame of your
shaceship, the energy of each already emitted pulse keeps getting less
and less as the ship accelerates; this looks like a nice simple
thought-demonstration that energy is _not_ conserved in accelerating
reference frames.
Simple? I would think it simpler to just take a standard-issue
accelerating rocketship and let a massive object drop to the floor. If
you release it from velocity zero (with respect to the accelerating
frame), its kinetic energy starts increasing all by itself, so energy
conservation is broken. This does not presuppose any relation between
the wavelength and energy contents of light.
Yes, that certainly is simpler, and perhaps the simplest.
The light-pulse version does have the nice touch that the rocket never
catches up to any of the pulses, whereas it will, given the ability to
maintain the acceleration for long enough, catch up to any dropped or
thrown rock.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
|
|
|
|
|
|
| User: "Henning Makholm" |
|
| Title: Re: Energy conservation in moving frames |
11 Mar 2006 10:45:23 AM |
|
|
Scripsit
Now assume that after some time, the laser stops emitting pulses, and
the total energy of the pulses (since the generator was turned on) is
measured both on the spacecraft and on earth. Will the answers be the
same?
No, because energy is not relativistically invariant.
For light pulses, when you move from the spaceship frame to the earth
frame (and thereby start moving _towards_ the light), the total
measured energy will increase by the factor sqrt((c+v)/(c-v)).
Does wavelength come into the energy calculation somehow,
Yes: shorter wavelength means higher energy content.
The easiest way to calculate the energy increase is to apply a Lorentz
transformation to the energy-momentum four-vector of the light
pulses. This takes care of the blueshift and the shorter pulse length
in one operation.
--
Henning Makholm "So? We're adaptable. We'll *switch missions*!"
.
|
|
|
|
|
Related Articles |
|
|
