Mass-Beam Behaviour: An Analysis of Efficiency for Particle-Beam Pushed Sails

[qraal]

Particle beams to push starships have been proposed by different researchers and I've been contemplating using a particle beam to push a Magnetic-Sail. So in pondering the particle-beam pushed sails I realized there hadn't been an analysis of the efficiency of mass-beams: just how efficiently is the energy of the beam transferred to the sail?

I went with the classic approximation first. Initially I considered an inelastic collision in which almost all the relative momentum between beam and sail is transferred from beam to sail. If you imagine the beam particles after impact washing around the sail with just enough energy to get out of the way of the incoming beam, then that's what I considered. Here's the interesting thing I found - the lowest beam energy is achieved when its relative velocity is twice the sail's velocity and thus the beam speed is 3 times the sail's speed in the rest frame.

One point I had to consider was the need to account for the mass-flow of the beam itself, which should change with the relative speed of the beam to the sail. Imagine the mass-beam is composed of a series of pulses which occur with a frequency, fo, in the rest frame. In the sail's reference frame that pulse frequency must be lower in a fashion physically analogous to the red-shift for light. I parameterised it as dm.fo.(Vr/V), where dm is the mass in each pulse, fo the rest frame frequency, Vr the relative velocity and V the velocity of the beam.

After finding the minimum beam-energy for a given thrust was at a beam velocity of 3 times the sail velocity, it meant the energy transfer efficiency was ~8/27 - just 29.63% - for inelastic momentum transfer.

What about elastic energy transfer? Firstly I am not altogether sure that's realistically possible, but I entertained the idea. Turns out the efficiency is 16/27 - just 59.26%, twice the first case.

Now I have to work out the relativistic version, though I am not altogether sure how to go about it. Any thoughts?

 

[bcrowell]

Cool idea! It's certainly going to be more energy efficient than propelling a solar sail with a laser, which is the idea I'd seen before. There's also the possibility of using an electromagnetic beam to provide energy, which is then used to run an ion drive.

With a material beam, you're going to have practical problems with beam optics; I don't know of any way to make an extremely well collimated beam of material particles. Also, when you accelerate the beam, presumably you're doing it using electromagnetic fields, but then your beam has to be charged. As a charged beam travels through space, it's going to get deflected and defocused by the ambient fields, and if it's dense enough you'll have issues with self-repulsion.

Re the relativistic version of the material beam, if the beam is relativistic in the frame of the receiver, then you're going to have a big problem with radiation damage and thermal damage to the sail. The sail also needs to be thick enough to stop the beam. It might make more sense for the receiver to use electromagnetic forces to decelerate the beam and then throw it backward. If the beam is neutral, conceivably the receiver could ionize it with microwaves first?

 

[qraal]

Hi Lawrence

So any pointers on the maths?

 

[pervect]

There are a couple of mathematical methods the OP can use to describe the relativistic material beam. The most standard would be to use the stress-energy tensor - I suspect that's what he's asking about.

The beam can probably be considered to be accelerated dust - i.e. we can assume it doesn't have any internal pressure (which would tend to cause the beam to disperse).

In that case we can express the beam in terms of its energy density T_00 per unit volume, and it's momentum density T_01 per unit volume. The needed tensor transformation laws for a boost in using geometric units should then be

$$T_{0'0'} = \gamma^2 T_{00} - 2 \beta \gamma T_{01}$$

$$T_{0'1'} = -\beta \gamma T_{00} + (\gamma^2 + \beta^2) T_{01}$$

unless I've made an error. The long form of the tensor transformation laws is

$$T_{0'0'} = \Lambda^{0}{}_{0} \Lambda^{0}{}_{0} T_{00} + \Lambda^{0}{}_{0} \Lambda^{1}{}_{0} T_{01} + \Lambda^{1}{}_{0}\Lambda^{0}{}_{0} T_{10}$$

$$T_{0'1'} = \Lambda^{0}{}_{0} \Lambda^{0}{}_{1} T_{00} + \Lambda^{0}{}_{0} \Lambda^{1}{}_{1} T_{01} + \Lambda^{1}{}_{0} \Lambda^{0}{}_{1} T_{10}$$

with
$$\Lambda^{0}{}_{0} = \Lambda^{1}{}_{1} = \gamma$$
$$\Lambda^{0}{}_{1} = \Lambda^{1}{}_{0} = -\beta$$

being the usual coefficients of the Lorent'z transform , $\beta$ = v/c and $\gamma$ = 1/sqrt(1-beta^2)

T_{10} in the long form is numerically equal to T_{10} because the stress_energy tensor is symmetric, in the "short form" I've taken this into account already.

 

[Bob S]

With a material beam, you're going to have practical problems with beam optics; I don't know of any way to make an extremely well collimated beam of material particles. Also, when you accelerate the beam, presumably you're doing it using electromagnetic fields, but then your beam has to be charged. As a charged beam travels through space, it's going to get deflected and defocused by the ambient fields, and if it's dense enough you'll have issues with self-repulsion.

In 1989, the United States built a negative hydrogen ion (one proton plus two electrons) accelerator, launched it to an elevation of 195 km (above the atmosphere), and produced a neutral hydrogen beam by stripping off the "outer" electron after acceleration. See

http://www.google.com/url?sa=t&sour...3ZHP-wb9nLbkn2D-A&sig2=PZ3IziXq-8Gr7_ap6IrtAw

neutral hydrogen beams are not deflected by the Earth's magnetic fields, and there is no intrabeam Coulombic self repulsion.

Bob S

 

[yuiop]

Now I have to work out the relativistic version, though I am not altogether sure how to go about it. Any thoughts?

Hi, Wikipedia has done some of the hard work for you here: http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

I will just clarify and expand on the Wiki article a little.

Just as Newtonian physics the total momentum before and after the collision is conserved but we use the relativistic expressions for momentum so that:

$$\frac{m_{1}\;u_{1}}{\sqrt{1-u_{1}^{2}/c^{2}}} +<br /> \frac{m_{2}\;u_{2}}{\sqrt{1-u_{2}^{2}/c^{2}}} = <br /> \frac{m_{1}\;v_{1}}{\sqrt{1-v_{1}^{2}/c^{2}}} +<br /> \frac{m_{2}\;v_{2}}{\sqrt{1-v_{2}^{2}/c^{2}}} \qquad (1)$$

where u1 and u2 are the initial velocities of m1 and m2 respectively before the collision and v1 and v2 are the final velocities respectively after the collision. Similarly the energy in an elastic collision is also still conserved in the relativistic case but now we use the relativistic expression for energy so that:

$$\frac{m_{1}c^{2}}{\sqrt{1-u_1^2/c^2}} +<br /> \frac{m_{2}c^{2}}{\sqrt{1-u_2^2/c^2}} =<br /> \frac{m_{1}c^{2}}{\sqrt{1-v_1^2/c^2}} +<br /> \frac{m_{2}c^{2}}{\sqrt{1-v_2^2/c^2}} \qquad (2)$$

The final velocity of m1 after the collision is:

$$v_1 = \frac{v_c(2c^2-u_1 v_c) - u_1c^2}{v_c(v_c - 2u_1) +c^2} \qquad (3)$$

where $$v_c$$ is the velocity of the centre of mass of the total system (which remains the same before and after the collision). $$v_c$$ is found by dividing the lhs of equation (1) by the lhs of equation (2) above and multiply by c^2 and the result is:

$$v_c = \frac{ m_1 u_1 \sqrt{1-u_2^2/c^2} + m_2 u_2 \sqrt{1-u_1^2/c^2}}{ m_1 \sqrt{1-u_2^2/c^2} + m_1 \sqrt{1-u_1^2/c^2}} \qquad (4)$$

Similarly, the final velocity of m2 is:

$$v_2 = \frac{v_c(2c^2-u_2v_c) - u_2c^2}{v_c(v_c-2u_2) +c^2} \qquad (5)$$

 

[qraal]

Thanks yuiop! Very helpful. Now for the rest of the hard-work ;-)

Hi, Wikipedia has done some of the hard work for you here: http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_relativistic

I will just clarify and expand on the Wiki article a little.

Just as Newtonian physics the total momentum before and after the collision is conserved but we use the relativistic expressions for momentum so that:

$$\frac{m_{1}\;u_{1}}{\sqrt{1-u_{1}^{2}/c^{2}}} +<br /> \frac{m_{2}\;u_{2}}{\sqrt{1-u_{2}^{2}/c^{2}}} = <br /> \frac{m_{1}\;v_{1}}{\sqrt{1-v_{1}^{2}/c^{2}}} +<br /> \frac{m_{2}\;v_{2}}{\sqrt{1-v_{2}^{2}/c^{2}}} \qquad (1)$$

where u1 and u2 are the initial velocities of m1 and m2 respectively before the collision and v1 and v2 are the final velocities respectively after the collision. Similarly the energy in an elastic collision is also still conserved in the relativistic case but now we use the relativistic expression for energy so that:

$$\frac{m_{1}c^{2}}{\sqrt{1-u_1^2/c^2}} +<br /> \frac{m_{2}c^{2}}{\sqrt{1-u_2^2/c^2}} =<br /> \frac{m_{1}c^{2}}{\sqrt{1-v_1^2/c^2}} +<br /> \frac{m_{2}c^{2}}{\sqrt{1-v_2^2/c^2}} \qquad (2)$$

The final velocity of m1 after the collision is:

$$v_1 = \frac{v_c(2c^2-u_1 v_c) - u_1c^2}{v_c(v_c - 2u_1) +c^2} \qquad (3)$$

where $$v_c$$ is the velocity of the centre of mass of the total system (which remains the same before and after the collision). $$v_c$$ is found by dividing the lhs of equation (1) by the lhs of equation (2) above and multiply by c^2 and the result is:

$$v_c = \frac{ m_1 u_1 \sqrt{1-u_2^2/c^2} + m_2 u_2 \sqrt{1-u_1^2/c^2}}{ m_1 \sqrt{1-u_2^2/c^2} + m_1 \sqrt{1-u_1^2/c^2}} \qquad (4)$$

Similarly, the final velocity of m2 is:

$$v_2 = \frac{v_c(2c^2-u_2v_c) - u_2c^2}{v_c(v_c-2u_2) +c^2} \qquad (5)$$

 

[bcrowell]

In 1989, the United States built a negative hydrogen ion (one proton plus two electrons) accelerator, launched it to an elevation of 195 km (above the atmosphere), and produced a neutral hydrogen beam by stripping off the "outer" electron after acceleration. See

http://www.google.com/url?sa=t&sour...3ZHP-wb9nLbkn2D-A&sig2=PZ3IziXq-8Gr7_ap6IrtAw

neutral hydrogen beams are not deflected by the Earth's magnetic fields, and there is no intrabeam Coulombic self repulsion.

Bob S

Interesting! But the thing is, the OP is talking about using this for interstellar probes. I really don't believe that there's any known or foreseeable technology that would keep a material beam (neutral or otherwise) collimated over distances of light-years. There's also the problem of scattering by the interstellar medium. With roughly 1 hydrogen molecule per cubic centimeter, a back-of-the-envelope calculation shows that an atom can only go about a tenth of a light-year without being scattered.

 

[qraal]

Interesting! But the thing is, the OP is talking about using this for interstellar probes. I really don't believe that there's any known or foreseeable technology that would keep a material beam (neutral or otherwise) collimated over distances of light-years. There's also the problem of scattering by the interstellar medium. With roughly 1 hydrogen molecule per cubic centimeter, a back-of-the-envelope calculation shows that an atom can only go about a tenth of a light-year without being scattered.

My idea requires an acceleration distance of just 0.1-1.0 AU. To avoid those issues.

 

[bcrowell]

My idea requires an acceleration distance of just 0.1-1.0 AU. To avoid those issues.

I see. It would be interesting to see some order-of-magnitude figures for the engineering of the thing, for comparison with, e.g., the starwisp concept http://en.wikipedia.org/wiki/Starwisp For starwisp the cruising velocity is supposed to be about 0.1c. To get that in a distance of 1 AU, you'd need an acceleration of 3000 m/s2.