1. Mar 22, 2009

### Punkanzee

Hello Physics Forums, I am new here and I thank the community for any input I might receive on my physics paradox. To some of you this may not be a paradox and I hope that you may be able to explain to me how this makes sense from all reference points.

It has long been held that kinetic energy for objects is completely frame-dependent, that the energy content is relative to other observers. But for this I present a thought experiment, please excuse any miscalculations.

From our perspective on earth, earth is stationary. In this thought experiment we have a rocket, rocket one, weighing 1,000,000 kg that is traveling in a straight line directly away from the earth at 40,000 km/hour. The kinetic energy of this rocket given by the equation KE = 1/2MV^2 is 6.172 x 10^13 joules. Keep in mind the rocket is moving at 40,000 km/hour only when compared to an observer on earth. Traveling directly beside rocket one is rocket two that is traveling in the same direction as rocket one and is identical in mass, velocity, and kinetic energy. If rocket one is going to accelerate from 40,000 km/hour to 50,000 km/hour the amount of energy this would require is the difference in kinetic energy of rocket one moving 50,000 km/hour and 40,000 km/hour. The kinetic energy of a rocket one moving 50,000 km/hour is 9.645 x 10^14 joules. So the amount of energy in order to accelerate rocket one from 40,000 km/hour to 50,000 km/hour is 9.465 x 10^14 joules – 6.172 x 10^13 joules = 8.848 x 10^14 joules. As we can see the amount of energy required to accelerate the rocket is enormous compared to the initial kinetic energy of the rocket moving 40,000 km/hour. This is caused of course by the exponential factor in the equation for kinetic energy. However 50,000 km/hour from an observer on earth is only 10,000 km/hour from an observer on rocket two since to begin with rocket one had a velocity of 0 km/hour because both rockets were moving beside one another at the same velocity. And because this is the case from rocket two's observation, only 3.858 x 10^12 joules of kinetic energy will be required from rocket two's perspective because rocket one is accelerating from 0 km/hour to 10,000 km/hour. So clearly kinetic energy depends on the perspective of the observer.

However, this doesn't make sense once you take into account amounts of energy that will stay universally constant regardless of observation point such as light. In order to take a deeper look into the problem two scenarios are presented. In both scenarios it is irrelevant how much energy both rockets required to accelerate from the earth to their current velocity and position in space. Both rockets are powered by Variable Specific Impulse Magnetoplasma Rocket (VASIMR) thrusters because it is a highly efficient means of propulsion and requires no explosive chemicals. VASIMR is an is an electro-magnetic thruster that uses radio waves to ionize a propellant and magnetic fields to accelerate the resulting plasma to generate thrust. This system of propulsion is important because it can be powered electrically. In the next two scenarios microwaves will be emitted by either the earth or rocket two and absorbed by rocket one and then converted into electricity and thus power the VASIMR to accelerate rocket one. In both scenarios the energy to produce the microwaves comes from fission nuclear reactions. For this thought experiment assume all transfers of energy are 100% efficient.

In scenario one microwaves will be emitted from the earth only to rocket one in order to accelerate rocket one from 40,000 km/hour to 50,000 km/hour. The energy required in the form of microwaves can be calculated and would be equal to the energy calculated earlier to accelerate rocket one to 50,000 km/hour so we would need 8.848 x 10^14 joules of energy in the form of microwaves.

In scenario two, instead of the earth emitting microwaves to rocket one, rocket two will be emitting microwaves to rocket one. As stated earlier from rocket two's perspective rocket one will be accelerated from 0 km/hour to 10,000 km/hour and will require 3.858 x 10^12 joules of energy. This amount of energy will be transferred between rockets also in the form of microwaves.

Take an alternate scenario, where the microwaves are transferred first from the earth to rocket two. The microwaves are stored by rocket two in a battery of some sort and then transferred to rocket one. Now which energy is rocket two going to send to rocket one? The smaller amount which makes sense for rocket two or the larger amount that makes sense for earth?

Light energy is given by the equation: Planck's constant x frequency of the light. There is no such exponential factor and light energy remains constant regardless of where the observer is or how fast the observer is moving. The Doppler effect, responsible for red shift is not going to affect the overall energy if the Doppler effect can even be said to be making an impact in this thought experiment. Therein lies the paradox.

2. Mar 22, 2009

### Mapes

Don't forget to include the kinetic energy acquired by the plasma.

3. Mar 22, 2009

### Staff: Mentor

The Doppler effect changes the frequency and as you mentioned the energy is proportional to the frequency, therefore the Doppler effect must have an impact. When you properly account for the Doppler effect and the energy of the plasma then you will find that energy is conserved in both frames. The easiest way to do this, IMO, is to use the 4-momentum.

4. Mar 22, 2009

### Punkanzee

Hmm, yes I see where you are coming from but let me try to rephrase the alternate scenario.

Well first of all I suppose the sure Doppler effect must always be taken into account. However does the Doppler effect the actual overall energy when including energy over time of the microwave 'pulse' laser or what have you from earth to rocket one or from rocket two to rocket one? That's another question entirely I suppose.

And sure the plasma jetting out the back of rocket two must also be taken into account. The rockets are so far from earth that the gravity from earth is negligible and thus neither rocket is required to run the VASIMR hardly ever. In order for rocket one to accelerate, rocket one only has to thrust more plasma, not make the plasma go any faster.

If you were going to send the microwave energy from the earth to rocket two, the rocket that doesn't use any energy for VASIMR, how much energy would you send to it with the expectation of resending it to rocket one? Lets say you send it the full sale 8.848 x 10^14 joules. Still from rocket two's perspective, rocket one only needs to rush ahead at 10,000 km/hour so you will only be needing 3.858 x 10^12 joules. So how is it not possible that you could make rocket one go 10,000 km/hour(rocket two)/50,000 km/hour(earth) only using 3.858 x 10^12 joules and thus have 8.809 x 10^14 joules left over provided you use rocket two as an intermediary? You could then just send left over energy back to earth and thus you accelerated rocket one for cheap because of the intermediary rocket two.

5. Mar 22, 2009

### Mapes

Wrong. Again, you haven't included the kinetic energy of the plasma that is ejected from rocket one. This same plasma has a different kinetic energy from the point of view of rocket two than from Earth.

EDIT: Different kinetic energy, not necessarily greater kinetic energy.

Last edited: Mar 22, 2009
6. Mar 22, 2009

### Punkanzee

So can you please tell me how much energy it will actually need?

7. Mar 22, 2009

### Mapes

You can determine this by calculating the first rocket's change in momentum; the plasma needs to acquire this momentum in the opposite direction (to satisfy momentum conservation). Then calculate the plasma's kinetic energy; you'll first need to assume a certain mass or velocity.

8. Mar 22, 2009

### Punkanzee

I thank you for your feed back.

9. Mar 22, 2009

### Punkanzee

I see that momentum is the confusing element. On another note, mostly to simplify things, say we have two objects, object A and object B traveling in front of another object, object C moving perpendicular to Object C's point of view (please ignore resistances, efficiencies, etc again). To start with, object A and object B have masses of 20 kg and each are moving with respect to object C at 10 m/s. (Please excuse the very crude text drawing)

AB ---->

C

As you can see A and B are so close that they are practically touching. Now B pushes off A so that From C's pov B is now moving 15 m/s and A is moving 5 m/s. From B's pov, A slowed down 5 m/s and B sped up 5 m/s so there is a 10 m/s difference between the two.

Now from either pov, A, B, or C's momentum is conserved, adds up, whatever you want to call it. This was the whole point of what I mentioned earlier, correct?

Now then if kinetic energy is taken into account:
From C's pov
A has 250 joules and B has 2250 joules.
From B's pov
C has 2250 joules and A has 1000 joules.
From A's pov
B has 1000 joules and C has 250 joules.

Lets again use Microwave energy, this time to initiate the jump of B off of A. Some mechanism is used to push A off of B and it is powered by microwaves, imagine what ever you'd like to.

Originally Both A and B had 1000 joules a piece when moving 10 m/s from C for a combined 2000 joules. Now when B pushes off of A because of the exponential factor existent in the equation for kinetic energy both A and B will now have a combined kinetic energy of 2500 joules from C's pov. This whole thing is boiling down to momentum versus kinetic energy I suppose, please stay with me on this. Another good question should really just be the relationship between momentum and kinetic energy but continuing...

Does the amount of microwave energy required obey momentum or kinetic energy? For things to add up it is momentum that should be obeyed but how can we measure the momentum of light? Why does there exist a difference between momentum and kinetic energy?

Thanks again in advance for any of your input on the matter.

10. Mar 23, 2009

### Mapes

The laws of conservation of energy and conservation of momentum include electromagnetic waves (e.g., microwaves). The momentum of an EM wave carrying energy E is E/c.