# B Kinetic energy: Legal to use kinetically independent frames?

1. Feb 16, 2017

### BitWiz

A rocket is drifting in gravity-free space and is observed by an external observer who is also drifting at an unchanging location using an arbitrary coordinate system.

The rocket accelerates at a fixed rate using a massless photon engine that results in a negligible change in the mass of the vehicle. The propellant has energy-mass, but for now, assume it's negligible.

An observer within the rocket sees an instant and constant acceleration (A) on his accelerometer. The rocket uses energy at a fixed rate of R. He can start and stop the engine and run it for any length of time, and he always sees the same acceleration A being used at the rate R. His total energy consumption (E) is then E = R * T. In other words, total energy usage is proportional to time.

The external observer see the rocket begin to accelerate. He notes that the rocket has mass and assigns a value of 1 to this mass. He uses his own clock (t) and sees that rocket's measured velocity (v) is t * the observed acceleration (a). Thus, v is proportional to t.

Thus, the astronaut sees total energy (E) is proportional to T.

The external observers, using the kinetic energy formula ek = mv2 / 2, sees that total ek is proportional to t2 / 2.

Since the kinetic energy has to come from the energy produced on the rocket, how can both be right?

Thanks!

2. Feb 16, 2017

### Staff: Mentor

Is this assumption viable? Make a reasonable assumption about the mass of the ship and the acceleration, see what that implies in terms of the energy, momentum, and mass of the exhaust.

3. Feb 17, 2017

### BitWiz

Hi, Nugatory, thanks for the reply.

By assigning "1 unit" to any arbitrary ship mass, I get the following:

For T,t = 2, E(T) = 2 energy units, ek(t) = 2 energy units
For T,t = 3, E(T) = 3 energy units, ek(2) = 4.5 energy units
For T,t = 10, E(T) = 10 energy units, ek(t) = 50 energy units

By special relativity, T <> t, but the difference can be made negligible at small relative velocities.

The rocket: I've simplified the motor to produce instant thrust to get rid of the calculus. Similarly, the propellant mass loss is unmeasurable over time( perhaps the energy of the photons is arbitrarily small, and they're emitted at an arbitrarily small rate).

Thus, both rocket acceleration A and the rate R that energy is expended are constants when the engine is on. It follows that the total energy expended E is proportional to T. The main thing is that E(T) is a linear function, and in fact

1) E(T) = T.

The external observer: Disregarding relativistic effects, the external observer sees the rocket's acceleration a is A, and his local time t is T. He calculates the rockets velocity v as a * t.

2) Thus v is proportional to t -- and by associations, to E.

The kinetic energy formula says ek = 1v2 / 2 (where mass = 1).

3) Thus v is proportional to SQRT( 2ek) (and ek(t) = t2 / 2 )

If the kinetic energy of a rocket ek is supposed to equal the energy E expended by the rocket, then 2) and 3) are incompatible except where t = 2. Can you please tell me why?

Thanks!

4. Feb 17, 2017

### Ibix

Maybe think about momentum? Can the momentum carried by the exhaust be negligible?

5. Feb 17, 2017

### BitWiz

Hi, Ibix. If kinetic energy is a curve over t, then another curve will have to be subtracted to get the linear E(T). Can the momentum of the exhaust from a constant output rocket vary as a power curve over t in an increasingly negative way?

6. Feb 17, 2017

### Ibix

You're getting ahead of yourself. Do you think the momentum of the exhaust can be negligible? What are the implications for the momentum of the rocket in either case?

7. Feb 17, 2017

### BitWiz

No. If the exhaust momentum is negligible, then I suppose the acceleration is negligible. ;-)

However, I'm trying to linearize acceleration on the ship without introducing mass changes that -- I think -- are an unnecessary complication. If I take the rocket's mass reduction into account, acceleration (and kinetic energy) increases over time using the same energy usage rate, making the differences between E and ek even worse.(?)

8. Feb 17, 2017

### Ibix

But if you can't treat the momentum of the exhaust as negligible, what about its energy?

9. Feb 17, 2017

### BitWiz

It's not zero. But is it really significant?

As calculated by the external observer -- the total kinetic energy of the rocket with respect to time is a power curve, yet the energy usage by the rocket is linear. What is the missing term that connects them? Doesn't have to be a curve?

Thanks!

10. Feb 17, 2017

### jbriggs444

Have you calculated how much energy winds up in the exhaust stream compared to how much winds up in the rocket?

11. Feb 17, 2017

### Ibix

Well, the power from the fuel is constant. And you're finding that the power absorbed by the rocket is not. And you're neglecting the energy in the exhaust. And something doesn't add up. I know where I'd look...

Edit: beaten to it by jbriggs, I see.

12. Feb 17, 2017

### BitWiz

Hi, jbriggs, I like your avatar. ;-)

Equal and opposite, I would think, assuming no efficiency losses. If the rocket expends 6E, then 3E goes to the rocket and -3E to the exhaust. The sum of the momenta should be zero. (?)

From the rocket's point of view, E is still proportional to V even though half of E ends up in the exhaust. We can still say that 6E produces 6V. The relationship between E and V is still linear.

Kinetic energy is not linear. 6E produces 3v2 instead of 6v. It is a power curve, and no matter what the starting conditions, a power function will always overtake a linear function of the same sign.

I have an idea why this might be, and it involves relativity, but it only works if I haven't overlooked something pretty basic. I'm still looking for that oversight.

13. Feb 17, 2017

### jbriggs444

The sum of the momenta should be zero. But the question is about the energy. Kinetic energy is never negative. The exhaust energy cannot be equal and opposite.

Don't guess. Calculate. Pick a rocket mass. Pick an exhaust velocity. Pick a mass for the exhaust. Calculate the energy of both rocket and exhaust.

14. Feb 17, 2017

### Staff: Mentor

No relativity is needed. You will get consistent results if you analyze this situation classically (that is, use the Galilean transforms and velocity addition, $p=mv$, $E_k=mv^2/2$) or relativistically (Lorentz transforms and velocity addition, $p=\gamma{m}v$, $E_k=(\gamma-1){m}c^2$).

And the thing that you are missing in both cases is the same: the kinetic energy of the reaction mass is required to balance the books so that the change in the kinetic energy of the rocket+reaction mass system is equal to the energy released by the fuel burn. However, I very very strongly recommend that you work out the classical case first - the relativistic case is a much harder way of arriving at the same insight.

Beyond that, I can only repeat jbrigg's advice: Don't guess, calculate.

15. Feb 25, 2017

### BitWiz

Hi jbriggs, Ibix, and Nugatory,

Thanks for your replies. Sorry for my tardiness (traveling).

Given: A rocket with a wet mass of 2kg at rest with respect to an outside observer floats in gravity-free space. The rocket accelerates 1 kg of propellant with a force of 1 newton for 1 second. By definition(?) the one kilogram of propellant will achieve a velocity of 1.0 m/s using an impulse of 1.0 newton-second.

Tsiolkovsky gives Δv = ve ln (m0 / m1) where Δv is delta-v, ve is exhaust velocity, m0 is initial mass of the rocket, and m1 is the dry mass. This works out to ln(2) or 0.693 m/s.

Why are the 2 one-kilogram masses moving at different velocities?

Thanks!

16. Feb 25, 2017

### Staff: Mentor

Tsiolkovsky's equation allows for the fact that the exhaust is emitted over the entire one second burn, and the speed (relative to the outside observer) of the first exhaust out the back at time zero is different than the speed of the last exhaust out the back one second later. The mass of the rocket is also different at the start of the burn and at the end. This is really important if you're designing a rocket that has to meet a particular performance specification, but when you're trying to understand the underlying physics in different frames it just gets in the way - complicates the calculations and requires you to do some non-trivial calculus without contributing any new physical insight.

Instead of using Tsiolkovsky's equation, you can make the simplifying assumption that once every second we throw a mass $m_e$ out the back with speed $v_e$ relative to the rocket. You'll end up understanding the different-frame physics just as well and the math will be much easier.

Last edited: Feb 25, 2017
17. Feb 25, 2017

### BitWiz

Thanks, Nugatory,
That's how I pictured it. So Tsiolkovsky only works with a continuous force applied to continuous *propellant* flow, but not a continuous force applied to a discrete chunk of propellant?

18. Feb 25, 2017

### Staff: Mentor

Right. But you really want to work this problem with the instantaneous force, discrete chunk of propellant assumption - imagine that the propellant is being fired out of a cannon pointing aft.

19. Feb 25, 2017

### BitWiz

Thanks, Nugatory,

By saying "instantaneous" force instead of "continuous" force, you eliminate a time dimension. Are this the ramp-up time of the force or the amount of time a chunk of propellant and the force are engaged?

Otherwise, can't "continuous propellant" still be discrete by using arbitrarily small chunks -- unless the interaction between the chunks is important(?).

Thanks!

20. Feb 25, 2017

### Staff: Mentor

Using arbitrarily small chunks is fine - you're just firing the cannon I mentioned in the previous post more often with smaller projectiles. You can work through this problem assuming that once a second we fire off a 1 kg reaction mass, or once a millisecond we fire off a 1g reaction mass. You'll get slightly different results because they're slightly different conditions, and if you're trying to answer the question you asked at the start of this thread, either approach will get you there. The advantage of doing it in discrete chunks fired off at once is that you don't need any calculus, as you would if you choose to work with a continuous fuel burn.