Free Energy Puzzle

Thanks for the reply Jorrie,

You stated: "Note that E= KE_f - KE_i has no meaning unless you specify the frame of reference;"

My reply: You appear to be attempting to associate relative velocity to kinetic potential.

Since kinetic potential is measured in Joules, which I believe is a scalar quantity; I can't wrap my mind around "frame of reference" as applied to scalar quantities.

Please give us some insight here.

Thanks

Gavilan

Hi Gavilan

Waiting for Jorrie to reply to your post I would like to stress that when we talk about "energy potential" we actually imply the capability to make something out of it (just using common language). For example, if we move along with a train, maybe in a helicopter above it, it doesn't matter energy-wise if we both run at walking pace or at almost the speed of light. Maybe Jorrie will enlighten us about the cosmological significance of a train running at this or another speed, but in the limited system we are considering for our calculations, the train has no kinetic energy. We cannot make any use of the train's speed whatsoever!

Now, lets make a more earthly example, where the train exchanges some of its kinetic energy to heat. For example consider a massive object, maybe another train which lies stopped on the rails. The first train is running in a collision route with the second one having a speed v, while we are watching the scene from above (again traveling at speed v). By the way, one can extend this thought experiment to consider friction.

Now, the system of the two trains (with masses m₁,m₂ respectively) has for us kinetic energy equal to 1/2m₂v² , while an observer sitting by the tracks, sees 1/2m₁v²_.

If we assume that after the collision the two trains continue to move on as one mass, we can calculate the part of the kinetic energy that will be transformed to heat: By using along the momentum preservation theory and doing some school math, this equals to

1/2.m ₁m₂/(m₁+m₂).v²,

whether you do the calculations in the helicopter or down on the earth. It should be so, as heat is supposed to be independent of the frame of reference. If you want to consider the case we add a Δv to the first train's speed, then the math takes a little bit longer, but still you get the same result (that is, by substituting v to v+Δv in the above formula) for both observers. What differs between the two observers is the persentage of the initial kinetic energy transformed to heat, but I cannot see why this should bother us.

Now that I'm thinking of it, even heat can be considered relative in a sense. Not to a moving frame but to the temperature of the entities it is going to interact with. One may seek to find what's the energy potential of a cup of hot tea. You can always get a figure in Joules, but if the environment has the same temperature with that cup of tea, there is no energy potential you can make use of. Of course, heat can have an absolute reference point which is 0K, unlike speed, which can not be measured against an absolute zero-speed frame of the universe, because there is none (or am I wrong?)

Maybe Jorrie can tell us more about energy as an absolute quantity, but I feel this lies outside the context of this thread!

Hi tkot:

tkot states:"Waiting for Jorrie to reply to your post I would like to stress that when we talk about "energy potential" we actually imply the capability to make something out of it (just using common language). For example, if we move along with a train, maybe in a helicopter above it, it doesn't matter energy-wise if we both run at walking pace or at almost the speed of light. Maybe Jorrie will enlighten us about the cosmological significance of a train running at this or another speed, but in the limited system we are considering for our calculations, the train has no kinetic energy. We cannot make any use of the train's speed whatsoever!"

Gavilan replies: Dynamic and regenerative braking is a very mature technology. It has been used in the railroad industry since the beginning of electric traction and is utilized in all regenerative capable transportation systems. These same principles are now moving to space applications. KE is indeed recoverable.

http://www.bestsyndication.com/Articles/2006/c/carter_mark/031206_hybrid_cars.htm

http://www.bestsyndication.com/?q=072407_solar-power-sailing-in-outer-space-extend-long-distance-travel.htm

Gavilan

Hi Gavilan

When I said:

"We cannot make any use of the train's speed whatsoever!"

I meant that we on the helicopter traveling at the same speed over the train cannot take advantage of the train's kinetic energy, because there is no kinetic energy in this helicopter-train system!

If we, on the other hand, widen our system by introducing another train, like in my example, or maybe the air and the rails which cause friction, then we have a system whose total kinetic energy is not 0 of course. It may differ from reference frame to reference frame, but it is not 0, because we have relative movement. Certainly we can have work production out of this. In the example I put forward, I showed how we may have some of this energy converted to heat.

Hi again Gavilan, you wrote: "You appear to be attempting to associate relative velocity to kinetic potential."

I don't know what you mean by "kinetic potential";^[1] perhaps you meant "kinetic energy"? It is a scalar measured in joules.

You wrote: "I can't wrap my mind around "frame of reference" as applied to scalar quantities."

Think of gravitational potential energy (PE = GMm/r). It is a scalar, but totally dependent upon the reference point you choose, e.g. Earth's center, Earth's surface, barycenter, etc. Kinetic energy (KE=½mv²⁾ is not dependent on the reference point, but rather on the reference frame you choose, more particularly upon the speed of the object in the reference frame.

Consider a stationary car being hit by a 1 kg mass moving at 10 m/s, expending a kinetic energy of 50 joule on the car. Now let the car move at 10 m/s in the opposite direction and choose the car as your reference frame. You will agree that the resulting collision energy is now 200 joule and the damage on the car will show it. KE and PE are scalars, but totally variant under a change of frame of reference. Rest mass-energy (E=mc²) of an object under constant temperature and pressure is one scalar that is invariant under a change of coordinates/frame of reference.

Jorrie

[1] I've seen the term potential function being used, which is the potential energy per unit mass, i.e. Pf = GM/r and is measured in joule/kg. It is sometimes easier to also leave the mass out of kinetic energy and just work with KE = ½v² joule/kg and I suppose one can call this the kinetic energy function or kinetic energy parameter.

Hi Jorrie;

Jorrie points out: "[1] I've seen the term potential function being used, which is the potential energy per unit mass, i.e. Pf = GM/r and is measured in joule/kg. It is sometimes easier to also leave the mass out of kinetic energy and just work with KE = ½v² joule/kg and I suppose one can call this the kinetic energy function or kinetic energy parameter."

If I may add: In "Fundamentals of Astrodynamics" - I think the term used is Specific Mechanical Energy which is the "sum of the kinetic energy per unit mass and potential energy per unit mass remains constant along its orbit."

Gavilan

Jorrie,

Again, your time and thought provoking posts are appreciated.

In Jorrie's post #61 he states: "The amount of energy (or fuel) needed to give an object a small velocity boost (fortunately) depends only on its rest energy, i.e. on it's mass and speed in its own frame of reference. Its speed is obviously zero in its own frame of reference; hence it has zero Newtonian kinetic energy in its own frame."

My reply: A train begins moving away from the station.

The superintendent, standing on the platform is at reference frame rf1.

I am on the engineer of the locomotive, my reference frame is rf2.

My wife, who is a big fan of the engineer, is pacing the train alongside the tangent track at a constant speed of V. This frame of reference is rf3. Her speed relative to the rf1 is Vfr1 the same as the train.

As the train passes the yard limit sign it begins to accelerate by increasing the rate of doing work ΔKE/t = power of the governor controlled constant output prime mover at full throttle.

The train increases its speed to 2Vfr1, the wife, who is being followed by a police car, maintains the speed of Vfr1. The differential velocity of the car and the train is now equal to the differential velocity between the car and superintendent.

The energy required to increase the velocity of the train is independent of the frame of reference point. The amount of energy required to achieve the final velocity of the train is the same regardless if you are the lady in the car, the superintendent at the station, or the engineer.

Gavilan

I havent the energy to read this mass of posts, suffice to say:

What's all this frame of reference and 'relative to' Kodswhallop? Kinetic energy is not a vector quantity, it's absolute and relates to nothing except a velocity of nil, zero, zilch. Kinetic energy, as it is defined using velocity squared, cannot be divided up into incremental chunks at a whim.

Hi Jorrie,

Truly an excellent question. I've seen this sort of point several times on the Physics Forum and most people get it totally screwed up. Here, I count on you and Fyz to give us good challenges.

Codemaster and Randall really answered the question (or at least got within whispering distance). The "missing" 100,000 J is hidden in the booster or whatever you use to bump up the speed.

You can switch back and forth between inertial frames without any difficulty provided you remember to use the correct bookkeeping. We're all used to transforming x and v; the same can be done with KE using something like the work-energy theorem, provided we keep track of everything, i.e., keep it conservative. From the earth observer's frame, you measure a different KE than the observer in the distant frame and that holds for both the final result and for the booster; from the earth's frame, you will measure that booster as having and providing 105,000 J.

The potential lack of clarity arises from thinking about the booster in terms of fuel or chemical energy. If we go back to Maxwell and always equate all energies to the motion of a mass, we get an easier calculation. In the aboslutely simplest example I can think up, put a two ball Newton's cradle on a train travelling at 100 m/s on a straight, level track and you will see the same effect and all the energies and momenta can be easily calulated in either frame and the two frames can be equated.

Again, thank you for the question. It caused me to let go of my facile explanation and really think about this.

Tom

Hi Tom. Your comment: "The "missing" 100,000 J is hidden in the booster ..." intrigues me.

Do you mean that because the booster had to be accelerated to 100 m/s relative to Earth in the first place, giving it a larger total energy relative to Earth, is the reason why it could deliver a whopping 100,000 joule of ΔE to the spacecraft in Earth's reference frame?

I'm not sure how one would go about to theoretically 'prove' that, but it is an interesting thought!

Jorrie

Jorrie,

If I configure this correctly, I will find that an observer, outside the space craft, in the moving frame of reference, prior to the boost, sees the booster having 5000J and the spacecraft having 0J. After the boost, he will see the booster having 0J and the spacecraft having 5000J. He thus sees energy conserved (as well as momentum though I don't show that).

From the point of view of an observer on the earth, the booster will have 605,000J before the boost and 500,000J afterwards. The spacecraft will have 500,000J before the boost and 605,000J afterwards. He sees energy (and, again, momentum) conserved.

I believe there are other numbers that satisfy the equations, but I'm lazy and didn't work them out. The point is that you can't compare the delta in one frame withe delta in another frame unless you are careful to use something like

dKE = Fdx

and

dKE' = F'dx'

where ' refers to the moving frame, KE is kinetic energy, F is the force from the booster, and x is the distance over which it is applied. Actually, it is probably easier to just say dx = vdt where v is the velocity measured in that frame.

So, you see, I'm both agreeing with you that the missing 100,000J is an artifact of the jumping back and forth and I'm also saying we could have found it by looking at the booster. How's that for a fence-straddling answer; perhaps I could be a politician if nothing else works out.

Tom

Why,why,why are intelligent hooman beeings discussing such unproductive themes?- don't they have anything better to do?- such as is there a GOD?- you see pilgrims, all is modelled on THEORY- & quite likely is flawed- I will give you all a fact - in the long run, we are all dead- why don't you all try to do something productive in these relevant facts- not your monetary stupid bickering about something that matters not to an originating principle/s- could it be that you, the highest intelligent monkeys, don,t know any better?. And in fact, the stupid bickering about mathematical/ algebra/ etc means nothing - as far as I can see , you are all just animals(as am I)- but I would love to see a live & let live attitude as against I am right- so you are wrong- get stuffed!.

Hi Neil - I would not quite call energy an "unproductive theme"...

I also think that in all these discussions, we give everyone a shot to say their say - isn't that a "live & let live attitude"?

Jorrie

Besides, we did do something about immortality, but then we all decided not to tell you about it.

Thank you for thinking me an intelligent human; according to my sisters, both descriptors are in serious question.

Anyway, I'm not bickering with Jorrie, I'm arguing. And I do that in order to learn from him; I have figured out more about reference frames these past two or three days than I did in the last forty-some years.

And, I close with the old chestnut:

Arguing with an engineer is a lot like rassling a pig in the mud; after a coupla hours, you realize the pig likes it.

Tom

And we're still agreed on not telling him about the immortality thing, right?

Dear Gang of Cerebes:

Jorrie indeed knows how to stir the pot. Have you seen the OTHER post dealing with the ball being dropped? That is my next visit.

Yes indeed, a constant force on a mass constant system will indeed impart a constant acceleration regardless of velocity BUT - in a mass constant system, at constant power (rate of doing work) the acceleration decreases as velocity increases. In the case of chemical propellants the mass of the system decreases, which maintains acceleration. But in the end, in all Newtonian Propulsion Systems the sum energy imparted is equal to 1/2 Δm V²

Where m is the fuel mass and V is the rocket or ion exhaust velocity.

Perhaps Jorrie or someone else of greater skill can show us that exact relationship. It is given in the basic rocket equation; but I believe in that equation the initial velocity is always 0.

Can we agree that Power is the rate of doing work? Power = ΔW/Δt ?

Can we agree that Force = mass X acceleration?

Can we agree that Power = Force x Velocity?

Where V is much less that C and if the mass and power remain constant, as the velocity continues to increase, what happens to the force? And if the force decreases what happens to the acceleration?

As you can see, you can not maintain constant acceleration at constant power.

As far as the "frame of reference" arguments being made.

In a stable orbit, the specific mechanical energy will be -u/2a whether the observer is on the surface of the prime focus body, in the satellite, on a nearby planet, or in my mother inlaws car parked on I-5 near LA on a bad traffic day. The energy required to alter a will also be the same, regardless of the reference frame. Frame of reference is a wonderful thing, and I am sure the experts use in very convincing and cleaver ways, but it really is quite meaningless in the context of Kinetic or orbital energies where the "frame of reference" is inherently defined.

Now, for the fellar who just doesn't understand why we could be interested in such things.

Let us now consider Hyperbolic Excess Speed. As the PHA crosses earths orbital path at r _heliocentric its defined orbital velocity at r _heliocentric must be greater than √2u_earth /r_{earth at closest approach} or earth gravitational capture will occur. It doesn't make a diddly squat ( Squat _diddly ) what the frame of reference is. HOWEVER the flight path angle WILL determine whether the capture will result in capture orbit or capture impact. The acceleration of earths gravitational field will = Squat _diddly in terms of its Hyperbolic Excess Speed even thought it will have very significant affect on the impact energy.

Gavilan

Hi Gavilan. I agree with you up to "As you can see, you can not maintain constant acceleration at constant power ...", but I think none of it is really relevant to this puzzle, which uses constant impulse (mΔv), not constant power.

You also wrote: "The energy required to alter a will also be the same, regardless of the reference frame."

Can I then take it that you agree with what I stated in the opening post: "However, the spacecraft's speed has only changed by 10 m/s, so its actual kinetic energy gain is: ΔKE = ½x100x10² = 5,000 joule."?

I realize that you were talking orbital energy and parameters, which is different to this puzzle, set in a free-space scenario. The OTHER puzzle that you referred to is more the gravity/orbital thing.

Jorrie