25-11-2001 11:21									#151
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GP....hmmm.

Let's see.

Again assuming non-relativistic non-zero motion (v/c < 0.01, mv >> hf/c) and to first order:

We have the impetus delivered by the photon as 2hf/c, so original momentum is mv and the final momentum is mv + 2hf/c. Energy transfer is elastic (or so) so the change in energy of the probe is (p_f^2)/2m - (p_i^2)/2m

To first order again, the energy increment is 4hfv/c, therefore this is how much of the original energy the photon loses. Note that we can see why the constraints are necessary, since for v/c > 0.25 we have that the photon descends into a negative energy state with this approximation, and for v=0 we get "something for nothing" as the photon does not lose energy in the process.

	25-11-2001 22:19									#152
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Ok so how much does the photon change wavelength? (What percent of the energy is transferred in each bounce? How much does the wavelenght change. Let's say we start with an MBT sized body (75 tons) at rest and the initial light wavelength is 500nm.

I guess I should be able to crunch this myself but I get confused with nu (funny looking v) and f and all that.

	25-11-2001 22:26									#153
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The mirror bounce system onbiously seems difficult (optical alignment is tough enough on an optical table...but you're gonna do it over millions of miles?) but let's say we can do it. The less energy is transferred by a bounce, the more dependant you are on precision alignment...) Or looking at it another way, given a set ability in precision alignment...(let's say an average of 4 bounces before the photon is scattered) than the lower the energy transfer the more significant the wasted energy becomes.

*GP as always...bowing to the high poobas of physics...but attempting to approach engineering problems with an engineering mindset*

	25-11-2001 23:14									#154
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Actually, I just realised something. Even with the mirror-bounce system with no light absorbtion, the energy trasfer could not be 100% efficient, because energy is also transfered to the other mirror.

	25-11-2001 23:50									#155
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yeah. I was thinking about this too. I think that the lighter the "struck body" the more energy is transferred. So that you would expect more energy to go to the spacecraft. But I'm not sure if you would need a very rigid mounting on the Moon mirror for instance to stop loss of energy through vibrations. Oh and if it were on the moon and the spaceship was being sent to very far away from the moon, you'd obviously have frequent lost contact because of the rotation of the moon around the Earth. (The Moon has a fixed face, right?)

	26-11-2001 00:35									#156
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Well, yes, if the mirror was properly fixed on the Moon the energy loss would be negligible.

Tracking the probe would indeed present a problem. Maybe it could be boosted intermittently - only when the beam can reach it.

	26-11-2001 00:44									#157
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If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth.

	26-11-2001 03:46									#158
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Why did we do something so stupid?

	26-11-2001 04:15									#159
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quote: . Oh and if it were on the moon and the spaceship was being sent to very far away from the moon, you'd obviously have frequent lost contact because of the rotation of the moon around the Earth. (The Moon has a fixed face, right?)

The Moon rotates once every 28 days, however the solution to this problem is to place the system at either of the the lunar poles. These will maintain a fixed orientation all year round.

	26-11-2001 04:43									#160
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BC, that would work ok for flights out of the plane of the ecliptic. (I said that.)

	26-11-2001 04:54									#161
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quote: If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth.

The Earth does not block the path between the Moon very often. What's the problem?

The moon follows a plane that is inclined ~5 degrees to the Earths rotation plane around the sun. Which is why we don't get eclipses every month.

	26-11-2001 06:00									#162
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quote: Originally posted by Big Crunch quote: If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth. The Earth does not block the path between the Moon very often. What's the problem? The moon follows a plane that is inclined ~5 degrees to the Earths rotation plane around the sun. Which is why we don't get eclipses every month.

I'm wasn't talking about the "blocking problem" but about the moving around the earth issue. Imagine a dime on a phonograph.

Having a polar laser/mirror takes car of this problem excepts for louanch in the ecliptic. (unless you assume that the Moon is a perfect sphere...you need to have some elavation above 0 degrees or else the laser will be occluded by hills and stuff.)

	26-11-2001 06:22									#163
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So anyway...enough of that, let's say you've got an unoccluded path. I grant you that for most geometries/times that will be so.

How many bounces do I need to get 25% of enerrgy expended in making the light converted into spacecraft movement.

Rough idea: 1? 10? 10^6?

	26-11-2001 06:27									#164
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OK, so you mean because you are firing towards the horizon.

I don't believe that to be a serious problem. As I say the Moon is inclined about 5 degrees to the plane of rotation of the planets, so only if the poles are located in a valley, and surrounded by mountains are you going to have a problem.

Any planetary craft would be 5 degrees above the horizon, which means the line of sight would only be blocked by hills above 80m high within a 1km radius, 800m within 10km.

	26-11-2001 06:31									#165
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quote: Originally posted by Big Crunch OK, so you mean because you are firing towards the horizon. I don't believe that to be a serious problem. As I say the Moon is inclined about 5 degrees to the plane of rotation of the planets, so only if the poles are located in a valley, and surrounded by mountains are you going to have a problem. Any planetary craft would be 5 degrees above the horizon, which means the line of sight would only be blocked by hills above 80m high within a 1km radius, 800m within 10km.

OK, cool. You're right. Geometery will be fine almost all the time.

Could you address some of the other questions?

1. How many bounces ar required to transfer energy.

2. What impact does the bounces at the fixed mirror have. (Is any significant energy expended there?)

	26-11-2001 06:42									#166
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quote: Originally posted by GP 1. How many bounces ar required to transfer energy. 2. What impact does the bounces at the fixed mirror have. (Is any significant energy expended there?)

2) Practically none.

Momentum transfer (making "classical" assumptions) is 2p. Energy transfer per hit will be the increase in kinetic energy ~p²/m_ref where m_ref is the mass of the reflecting object. Therefore the allocation of energy to the fixed item (the moon) and to the spacecraft is the ratio of masses.

Mass of moon = 10²² kg.
Mass of craft = 10³ kg at most.

With regards to 1) I will have to check. I'll get back to you, but it depends on craft velocity and mass. The faster and lighter the less it requires.

Edit - Getting masses of the Earth and Moon mixed up.

	26-11-2001 07:20									#167
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OK. I know it's an it dpends answer. I just want a feel it. 4 or 4 million.

Pick some reasonable numbers. If you can't come up with any how about. An MBT at rest with visible light.

MBT = 75tons =80,000 kg. Visible light...say 500 nm.

	26-11-2001 08:46									#168
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I guess Bush is afraid of terrorists sabataging his trip to pluto or something.

	26-11-2001 16:28									#169
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quote: Originally posted by GP OK. I know it's an it dpends answer. I just want a feel it. 4 or 4 million. Pick some reasonable numbers. If you can't come up with any how about. An MBT at rest with visible light. MBT = 75tons =80,000 kg. Visible light...say 500 nm.

I did a quick calculation assuming those figures and two opposable mirrors and came up with 10⁴⁰.

I have serious doubts over my calculation.

	26-11-2001 18:09									#170
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For launches in the Ecliptic, just mount the laser on a circumlunar railroad. It wouldn't have to move very fast to counter the Moon's one-month day. And you could do the same with Mercury, using huge fixed or rail-mounted solar cell arrays for power. With solar-powered self-replicating robots, the whole setup could build itself from local materials.

	26-11-2001 21:56									#171
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quote: Originally posted by Big Crunch I did a quick calculation assuming those figures and two opposable mirrors and came up with 1040. I have serious doubts over my calculation.

1. AHHH...who cares about a few tens of orders of magnitude....let the engineers fix it!

2. (more humor) That's one tired little photon...he spends decades shuttling back and forth to make all those bounces...sure hope there isn't an atom out of place on the mirror...or a particle of dust in space anywhere...or our tired friend gets scattered! Poor little guy. Cute little tired photon.

3. (serious question) To check your calcs just see how much energy is transferred in a single bounce. (for instance if the mirror system was impractical...perhaps because the moving craft has an imperfect mirror/optic.) Do we only get ~1/(10^40)* of our laser energy transferred to the craft?


* I realize that the amount of energy will probably change as the photon changes to a lower wavelength...but at this stage orders of ,magnitude seem to be enough to talk about.

	26-11-2001 21:57									#172
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quote: Originally posted by Jack the Bodiless For launches in the Ecliptic, just mount the laser on a circumlunar railroad. It wouldn't have to move very fast to counter the Moon's one-month day. And you could do the same with Mercury, using huge fixed or rail-mounted solar cell arrays for power. With solar-powered self-replicating robots, the whole setup could build itself from local materials.

Or you could just have a few seperate sites and hand off periodically. I agree that geography is not a big issue.

	26-11-2001 22:04									#173
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just a check on your calculation, BC. That was for transferring 25% of the energy to the craft? I assume that there are diminishing returns as the photon becomes lower in energy. That's why I picked 25%.

	26-11-2001 23:05									#174
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quote: I assume that there are diminishing returns as the photon becomes lower in energy

The final equation I got for wavelength shift per hit was:

l - l' = h/m_craftc.

Which is identical to the equation for Compton scattering, which raises alarm bells.

So anyway I plugged in the figures:

h = 6.6 x 10^-34
m = 10⁵
c = 3 x 10⁸

Therefore Dl = 2.2 x 10 ^-47, per hit. We want the wavelength to change by 1500nm so divide 1.5 x 10 ⁶ by 2.2 x 10 ^-47 and we get ~ 10 ⁴⁰

Using classical assumptions the energy transfer per hit should be p²_photon/m_craft = (h/l)²/10⁵

Plug in the figures and the energy transfer is ~10^-40 of the photons total energy per hit.

Looks like my calc wasn't out by any serious order of magnitude.

	26-11-2001 23:11									#175
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quote: Cute little tired photon

Don't worry. Given the relativistic time dilation, photons (from their POV) are destroyed as soon as they are created.

	26-11-2001 23:36									#176
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quote: Originally posted by GP So anyway...enough of that, let's say you've got an unoccluded path. I grant you that for most geometries/times that will be so. How many bounces do I need to get 25% of enerrgy expended in making the light converted into spacecraft movement. Rough idea: 1? 10? 10^6?

GP, I'm disappointed in you. If you'd read my conclusion, you'd see that the energy loss is dependent on "v", the veloctiy of the probe. For classically-dominated regions, the energy transferral is directly proportional to v.

If the probe is moing at, say, 3000 m/s (a reasonable "starting speed" to be achieved using chemical propellants) then 4% of the photon's energy is used up for each bounce. Solving for log_0.960.75, we get ~7 bounces.

	26-11-2001 23:40									#177
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BC, I'm disappointed in you, too.

You forgot to take the probe's motion into acount.

Your calculations are only good for when the probe is "starting out" with 0 velocity wrt the mirror/laser.

	27-11-2001 00:12									#178
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quote: Originally posted by KrazyHorse BC, I'm disappointed in you, too. You forgot to take the probe's motion into acount. Your calculations are only good for when the probe is "starting out" with 0 velocity wrt the mirror/laser.

Just testing you.

I was too busy looking for a calculation error to spot that assumption error. I overlooked it by saying to myself "from the crafts frame of reference".

	27-11-2001 00:15									#179
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Yeah, I was doing the same thing too, at first, but I was trying to work out the Doppler shift...

Instead, I just ignored it and put in a caveat about non-relativistic speeds.

I love approximations. They make my life much easier.

	27-11-2001 00:16									#180
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Why did I not just use the Doppler shift equation?

quote: a reasonable "starting speed" to be achieved using chemical propellants

Or my other favourite, the lunar rail gun, another energy saving device.

As you may have guessed, I am bit of an energy conservationist.