B Energy production that converts Hydrogen to Iron?

[PeterDonis]

Am I misunderstanding something?

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

 

[usernamess]

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

Interesting. Do you mind sharing a source for this? I don't normally work with electronvolts, but all of my references list it exclusively as an energy. I've been unable to find a source that indicates $1 MeV == 1\frac{MeV}{c^2}$, I'd love to learn more.

 

[PeterDonis]

I've been unable to find a source that indicates $1 MeV == 1\frac{MeV}{c^2}$,

That's not what I said. Go back and read what you quoted from my post again, carefully.

 

[PeterDonis]

In those units I believe you can express it either as energy (MeV) or mass (MeV/c^2) via mass energy equivalence principle.

Mass-energy equivalence does not mean you have to measure mass and energy in the same units. For example, kilograms are not the same units as Joules.

In particle physics, eV, keV, MeV, GeV, TeV, etc. units are convenient ones, and if you implicitly assume that $c = 1$, you can put a $/ c^2$ on any of the units without changing any numerical values. That's more or less what particle physicists do, but you have to be clear about what they are doing. They are not saying that mass units and energy units must be the same because of mass-energy equivalence. They are implicitly assuming that $c = 1$ in order to be able to not have to care about whether the units they are using are "mass" units or "energy" units. But that only works if you assume $c = 1$. If you don't assume $c = 1$, then you cannot say that, for example, MeV and MeV / c^2 are the same units, because if you don't assume $c = 1$, they're not.

In your actual equation, you used MeV / c^2 everywhere, which is fine and doesn't require you to say whether you are assuming $c = 1$ or not. But it also has nothing to do with mass-energy equivalence.

 

[usernamess]

That's not what I said. Go back and read what you quoted from my post again, carefully.

Yes, you are. All of the terms in his subtraction have units MeV/c^2, so they're all nominally masses--but the key point is that the units are all the same, so the subtraction is fine.

I found the ratio of the mass per nucleon of Iron / Deuterium = 0.992, so I reasoned 1 kg of Deuterium produces 992 grams of Iron-56.

Deuterium mass per nucleon= 937.8 MeV/c^2

Iron 56 has 7.5MeV more binding energy per nucleon

Deuterium mass per nucleon 937.8 MeV/c^2 - Binding Energy 7.5MeV/c^2 = 930.3Mev/c^2 Mass Per Nucleon Iron 56

Ratio: 930.3Mev/c^2 / 937.8 MeV/c^2 = 0.992

Therefore 1 kg Deuterium Converts to 0.992 kg Iron-56 with an 8 gram energy yield.

In order for this equation to be accurate, $7.5 MeV == 7.5 \frac{MeV}{c^2}$. Every text I have thought to check has implied this relationship to be untrue. Could you please draw my attention to the error I am making?

 

[usernamess]

Mass-energy equivalence does not mean you have to measure mass and energy in the same units. For example, kilograms are not the same units as Joules.

In particle physics, eV, keV, MeV, GeV, TeV, etc. units are convenient ones, and if you implicitly assume that $c = 1$, you can put a $/ c^2$ on any of the units without changing any numerical values. That's more or less what particle physicists do, but you have to be clear about what they are doing. They are not saying that mass units and energy units must be the same because of mass-energy equivalence. They are implicitly assuming that $c = 1$ in order to be able to not have to care about whether the units they are using are "mass" units or "energy" units. But that only works if you assume $c = 1$. If you don't assume $c = 1$, then you cannot say that, for example, MeV and MeV / c^2 are the same units, because if you don't assume $c = 1$, they're not.

In your actual equation, you used MeV / c^2 everywhere, which is fine and doesn't require you to say whether you are assuming $c = 1$ or not. But it also has nothing to do with mass-energy equivalence.

Having never been taught particle physics, this is a new concept to me. Thank you for explaining. Could you elaborate on how this convenient notation avoids changing numerical values? I ran a dimensional analysis on the units and the conversion came out different.

 

[PeterDonis]

In order for this equation to be accurate, $7.5 MeV == 7.5 \frac{MeV}{c^2}$.

No, becase 7.5 MeV (as opposed to 7.5 MeV / c^2) does not appear in the actual equation. It only appears in the accompanying text. The actual equation where he does the subtraction has MeV / c^2 for every term, so it is correct.

Could you elaborate on how this convenient notation avoids changing numerical values?

It's not a "notation", it's a choice of units. If you choose units so that $c = 1$, then dividing, say, 7.5 MeV by $c^2$ gives 7.5 MeV / c^2, because $c^2 = 1$ if $c = 1$. In other words, if $c = 1$, then mass units are numerically the same as energy units. This is a common choice of units for convenience not just in particle physics but in relativity.

 

[Devin-M]

So would the craft get more delta v if it used all the energy to accelerate all the iron, or if it used the same total energy to accelerate 1/10th the iron with much higher exhaust velocity while continuously dumping 9/10ths of the produced iron overboard with low (or no) impulse?

In one scenario you’d have one exhaust stream with an exhaust velocity determined by the total iron mass and total available energy.

In the other scenario you’d have a high energy and low energy iron exhaust stream. The exhaust velocity of the high energy stream would be determined by the craft having the same total available energy as the 1st scenario. Since it’d be accelerating less propellant mass with the same amount of energy this option has higher exhaust velocity, with that high energy exhaust velocity being determined by the ratio of how much iron you dump continuously overboard as its produced at the lowest possible energy.

 

[Nugatory]

So would the craft get more delta v if it used all the energy to accelerate all the iron, or if it used the same total energy to accelerate 1/10th the iron with much higher exhaust velocity while continuously dumping 9/10ths of the produced iron overboard with low (or no) impulse?

Go back to your example of the two skaters in post #50. The math here is particularly easy because we're doing it with one shove: we have two masses stuck together, then we take some amount of add some amount of energy and the two masses are moving in opposite directions. We have two equations, one from conservation of momentum ($m_1v_1=-m_2v_2$) and one from energy. With these it's just a bit of algebra that you should do for yourself to get the basic principle of small masses at high speed or high masses at low speeds using classical ($v\ll c$) physics.
You should go through this exercise before you seriously take on the harder problem of a rocket in which the reaction mass is being released continuously instead of in one shove.

 

[Devin-M]

Go back to your example of the two skaters in post #50. The math here is particularly easy because we're doing it with one shove: we have two masses stuck together, then we take some amount of add some amount of energy and the two masses are moving in opposite directions. We have two equations, one from conservation of momentum (m1v1=−m2v2) and one from energy. With these it's just a bit of algebra that you should do for yourself to get the basic principle of small masses at high speed or high masses at low speeds using classical (v≪c) physics.
You should go through this exercise before you seriously take on the harder problem of a rocket in which the reaction mass is being released continuously instead of in one shove.

Assuming I did the math in post #50 right I think it shows the craft getting more delta v by accelerating 1/10th of the iron with the same total energy.

 

[PeterDonis]

would the craft get more delta v if it used all the energy to accelerate all the iron

What is the exhaust if it isn't the iron?

 

[Devin-M]

What is the exhaust if it isn't the iron?

The exhaust is Iron.

Suppose you have the energy in some kind of battery from converting Hydrogen into 20 iron atoms, do you use all the energy in the battery to accelerate 2 of those iron atoms or all 20? Which gives the craft more delta v?

Im imagining something like the LHC but instead of collisions you have a beam dump on both sides (beam dumps added by myself for illustration):

 

[PeterDonis]

The exhaust is Iron.

Then you can't possibly accelerate all of the iron, since you have to use at least some of it as exhaust. So I don't understand what you mean by "accelerate all the iron".

 

[Devin-M]

By "accelerate all the iron" I mean "exhaust all the iron" interchangeably.

 

[PeterDonis]

By "accelerate all the iron" I mean "exhaust all the iron" interchangeably.

You can't exhaust all the iron either since that leaves no payload. So again I don't understand what you mean. I think you need to be more precise in specifying exactly what process is going on and exactly what variable you are varying.

 

[Devin-M]

Lets talk about a falcon 9 size vehicle with a dragon capsule sized payload, but instead of kerosene and oxygen it's filled up with deuterium, which will be converted to iron with the onboard reactor. The iron will be accelerated/"exhausted" with an onboard particle accelerator.

 

[PeterDonis]

Lets talk about a falcon 9 size vehicle with a dragon capsule sized payload, but instead of kerosene and oxygen it's filled up with deuterium, which will be converted to iron with the onboard reactor.

Ok. Then there is no variable at all; you run the nuclear reaction to completion, that gives you a certain amount of energy, and how much thrust that gives you is determined by the exhaust velocity of the iron, which is determined by the mass per particle of the iron.

The iron will be accelerated/"exhausted" with an onboard particle accelerator.

Why? The nuclear reaction already gives the iron exhaust velocity. Trying to give it extra velocity with an accelerator is pointless because the energy has to come from somewhere and the only energy source is the nuclear reaction. You might as well just let the reaction heat up the iron directly.

 

[Devin-M]

Converting deuterium to (20) iron-56 atoms gives us a certain amount of energy in the battery, 8400MeV of total energy if my math is correct.

So does the Falcon 9 sized craft get more delta V using 8400MeV of energy to accelerate 2 or 20 iron atoms “out the back?”

With the option with 2 iron atoms, they would be exiting the back of the craft faster because more energy per atom. With this option the other 18 iron atoms are released non-energetically and allowed to float away from the craft.

 

[PeterDonis]

the Falcon 9 sized craft

Is way, way too big for 8400 MeV of energy to do anything useful or for the comparison you are making to be useful. You need to pick a payload that is of comparable size to the fuel.

Try this: you have a payload of 1 uranium-238 atom and your fuel is 280 deuterium atoms, which will react to form 20 iron-56 atoms giving 8400 MeV of energy. We want to exhaust N iron atoms. Any iron atoms that aren't exhausted become part of the payload. (That last item is something you left out of your previous analysis, and it makes a big difference.)

 

[Devin-M]

Try this: you have a payload of 1 uranium-238 atom and your fuel is 280 deuterium atoms, which will react to form 20 iron-56 atoms giving 8400 MeV of energy. We want to exhaust N iron atoms. Any iron atoms that aren't exhausted become part of the payload. (That last item is something you left out of your previous analysis, and it makes a big difference.)

Yes, I agree with the scenario.

So from the example on post #50, I believe the U-238 gets more delta v is N=2 than if N=20.

 

[Devin-M]

Wait, suppose the payload is 1000 U-238 atoms, not 1.

 

[Devin-M]

So we have 1000 U-238 atoms payload, we have 20 iron atoms, 8400MeV of energy and N=2 or N=20. I think the Delta V on the 1000 U-238 atoms is higher with N=2

 

[PeterDonis]

from the example on post #50

That calculation is wrong because you don't include iron atoms not in the exhaust as part of the payload.

 

[PeterDonis]

For a good discussion of the relativistic rocket equation and how it is derived, see here:

https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

Your scenarios are somewhat different because your exhaust is not light and you don't convert all of the mass of the fuel into energy, only a small fraction of it; but the analysis in that article can easily be adapted to your case.

 

[Devin-M]

That calculation is wrong because you don't include iron atoms not in the exhaust as part of the payload.

If I’m a skater, and I have a 1kg ball with a 0.1kg core, I could either throw the entire 1kg ball with 100J or I could take out the 0.1kg core, drop the rest of the ball before throwing the 0.1kg core with 100J.

 

[PeterDonis]

drop the rest of the ball

If you drop the rest of the ball, that changes the momentum and energy conservation equations. You can't just ignore what you drop; you have to include it in your analysis. You didn't.

 

[PeterDonis]

If I’m a skater, and I have a 1kg ball with a 0.1kg core, I could either throw the entire 1kg ball with 100J or I could take out the 0.1kg core, drop the rest of the ball before throwing the 0.1kg core with 100J.

Where is the 100J coming from? You have to include the energy source in your analysis as well.

 

[Devin-M]

Ok lets say a 100kg astronaut at rest has in their left hand a 0.9kg ball and their right hand 0.1kg ball… they throw the 0.1kg ball with their right hand with 100J, while opening their fingers on their left hand. The 0.9kg ball remains at rest, the astronaut goes 0.427m/s and the 0.1kg ball goes 42.6m/s in the opposite direction.

 

[PeterDonis]

Ok lets say a 100kg astronaut at rest has in their left hand a 0.9kg ball and their right hand 0.1kg ball… they throw the 0.1kg ball with their right hand with 100J, while opening their fingers on their left hand. The 0.9kg ball remains at rest, the astronaut goes 0.427m/s and the 0.1kg ball goes 42.6m/s in the opposite direction.

Again, where is the 100 J coming from? You're ignoring that. Perhaps you are relying on the non-relativistic approximation, but if so, it makes no sense to talk, as you have, about mass-energy equivalence.

Also you keep changing scenarios. But ok, fine, let's analyze this one. Here are what your numbers say:

Energy before: 100J

Energy after: (1/2) 100 (0.427)^2 + (1/2) 0.1 (42.6)^2 = 99.85 J, not quite the same but I'll assume that's rounding error.

However:

Momentum before: zero

Momentum after: 100 * 0.427 - 0.1 * 42.6 = 42.7 - 4.26 = obviously not zero

So your numbers violate momentum conservation. So however you are getting your numbers, you're doing it wrong.

 

[PeterDonis]

while opening their fingers on their left hand

The fact that an astronaut can obviously do this with an item in their hand does not mean you can obviously do this with iron that is a product of a nuclear reaction.

 

[Devin-M]

I think I found the error in my spreadsheet. After fixing it, now I get opposite results where the astronaut is better off throwing the full 1kg with 100J rather than dropping 0.9kg and then throwing 0.1kg with 100J.

 

[Devin-M]

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

0.1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.044m/s
Ball: 44.7m/s

 

[PeterDonis]

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

0.1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.044m/s
Ball: 44.7m/s

How are you determining the final speeds for these cases?

 

[Devin-M]

How are you determining the final speeds for these cases?

I used:

C=(D/B)*A

where:

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG

By slowly increasing the value of A until:

((1/2)*B*C^2)+((1/2)*D*A^2)=100J

 

[Devin-M]

Fully solving:

A=(sqrt(2)*sqrt(B)*sqrt(E))/sqrt(D*(B+D))

C=(D/B)*A

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG
E=Total_Energy_J=100J

 

[PeterDonis]

I used:

C=(D/B)*A

where:

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG

By slowly increasing the value of A until:

((1/2)*B*C^2)+((1/2)*D*A^2)=100J

Ok, good, so basically you are solving for the combined constraints of momentum and energy conservation, with the total energy being pre-set at 100 J. (You're using the non-relativistic approximation, but as I think I've already noted, that should be fine for all of the scenarios you've posed.)

The same method should work for the other scenarios you have posed, since all of the relevant quantities are known.

 

[PeterDonis]

The same method should work for the other scenarios you have posed, since all of the relevant quantities are known.

Just to expand on this some and address the question of whether it makes sense to "drop" any propellant, the relevant equations for energy and momentum conservation, with a more intuitive choice of notation, are as follows:

$$
e = \frac{1}{2} M V^2 + \frac{1}{2} \left( k - s \right) v^2
$$

$$
M V = \left( k - s \right) v
$$

where $e$ is the energy available from whatever source is being used (for example the nuclear reaction burning deuterium to iron), $M$ is the final payload mass, $V$ is the final payload velocity (in the frame where everything is initially at rest), $k$ is the total mass of propellant available, $s$ is the mass of propellant that is "dropped", i.e., not used as exhaust (so $k - s$ is the mass of propellant that is used as exhaust), and $v$ is the final velocity of the exhaust. We can use the second equation to eliminate $v$ from the first, and then rearrange to find:

$$
V^2 = \frac{2 e}{M \left( 1 + \frac{M}{k - s} \right)}
$$

All of the quantities on the RHS are fixed except $s$, so the question is what choice of $s$ will maximize $V$ (which is equivalent to maximizing $V^2$, and that means maximizing the RHS of the above). It should be obvious that any value of $s$ greater than zero will increase the factor in the denominator of the RHS above and hence will decrease $V^2$, so if we want to maximize $V^2$, we want $s = 0$, i.e., we want to use all of the propellant as exhaust and not "drop" any.

 

[Devin-M]

So from what I find online:
Falcon 9
Dragon Capsule (payload) 12000kg
Booster 1st+2nd Stage Combined Dry 26200kg
Booster 1st+2nd Stage Combined Wet 544600kg
Total Fuel Mass 518400kg

For simplicity we can call the fusion craft a single stage combined booster and payload with:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

 

[PeterDonis]

For simplicity we can call the fusion craft a single stage combined booster and payload with:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

The Falcon 9 propellant mass is optimized for chemical reactions as the energy source, not fusion reactions. If you were designing an actual fusion rocket you would probably want quite a bit less propellant.

Also, with that large a propellant mass relative to the payload mass, and fusion as the energy source, the non-relativistic approximation is no longer very accurate.

 

[Devin-M]

On a relativistic kinetic energy calculator I found 1 kg has an exhaust velocity of 0.125c with 719TJ/kg.

 

[Devin-M]

On rocket equation calculator, 37,474,057m/s exhaust velocity, initial mass 556600kg, final mass 38200kg, change in velocity 100,393,423m/s or 0.33c.

Proxima Centauri: 4.24ly
Travel Time: 12.8yrs @ 0.33c

 

[Devin-M]

It's amazing what you'd be able to do with just a tiny amount of fuel.

I tried the same dry mass with only 100kg of fuel. Initial mass 38300kg, final mass 38200kg, exhaust velocity 37474057m/s...

97km/s delta v for getting around Earth or the Solar System with only 100kg of fuel in a 38200kg craft!

 

[PeterDonis]

It's amazing what you'd be able to do with just a tiny amount of fuel.

If it's fusion fuel, sure, since the energy per unit mass is about 5 orders of magnitude larger than for chemical fuel. The hard part is actually getting the fusion reaction to go.

 

[Devin-M]

It seems possible to do even better still:

We start off the same:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 556600kg
Propellant Mass: 518400kg
Propellant Energy: 719TJ/kg

But, before we do any thrusting, we use all the energy in the Deuterium to make matter-antimatter:
518400kg deuterium * 0.008kg/kg = 4147kg matter-antimatter
518400kg-4147kg = 514253kg iron

Now we dump overboard non-energetically 510106kg of iron leaving us with:

4147kg matter-antimatter
4147kg iron

The propellant mass is now:

4147kg + 4147kg = 8294kg

This changes the pre-thrust mass of the rocket to:
Dry Mass: 38200kg (of which 12000kg is useful payload)
Wet Mass: 46494kg
Propellant Mass: 8294kg (1/2 matter-antimatter, 1/2 iron)
Propellant Total Energy: 372EJ

But this becomes difficult to put into the rocket equation because pre-thrust the propellant mass is 8294kg, but only half of that mass actually comes out of the engine, because the other half turns to pure energy.

So we're back to the scenario in which an astronaut has a 0.5kg ball in his left hand, a 0.5kg ball in his right hand, and he's letting go of one ball while throwing the other...

 

[PeterDonis]

It seems possible to do even better still

The best possible scenario in terms of propellant usage is a photon rocket, which is the scenario discussed in the article on the relativistic rocket equation that I referenced. That scenario assumes that all of the propellant mass is converted to photons that are ejected as exhaust with perfect collimation.

The downside of this method is that you can't just magically convert matter to photons. You have to find actual reactions that do it.

 

[Devin-M]

Actually, I’ll have to retract my last post. It isn’t quite clear that converting all the energy to antimatter and then dumping much of the iron would actually improve the delta-v.

 

[PeterDonis]

converting all the energy to antimatter

Doesn't make sense anyway. The energy is what it is. You can't make it more energy by converting it to antimatter. You might as well just use it directly.

 

[Devin-M]

I thought by reducing total mass before thrust begins (with the same energy on board) it would be beneficial, but isn’t clear whether that’s truly the case.

 

[PeterDonis]

I thought by reducing total mass before thrust begins (with the same energy on board) it would be beneficial

Which would work the same even if you didn't turn the energy into antimatter. And I've already shown you that it doesn't make things any better.

 

[Devin-M]

I guess it makes sense. It's not like a car so decreasing the propellant mass (and by extension total mass) with the same total energy won't make it faster.

If we increase the mass of the ball the astronaut throws for the same throw energy, the astronaut always goes faster.

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

2kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.198m/s
Ball: 9.9m/s

1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.14m/s
Ball: 14.08m/s

0.1kg Ball, 100J, 100kg Astronaut:
Astronaut: 0.044m/s
Ball: 44.7m/s

A=(sqrt(2)*sqrt(B)*sqrt(E))/sqrt(D*(B+D))

C=(D/B)*A

A=Astronaut_Final_V_m/s
B=Ball_Mass_kg
C=Ball_Final_V_m/s
D=Astronaut_Mass_KG
E=Total_Energy_J=100J