# Paradox of energy gain

1. Nov 29, 2009

### e2m2a

The following thought experiment describes the energy gain of a system that defies a classical explanation. In this experiment we assume no friction and all velocities are at sub-relativistic speeds, hence, Newtonian mechanics applies and Galilean transformation is adequate.
Suppose we have an object, denoted as a slider with mass M, that has one degree of freedom of movement along a linear track. The slider can move in the positive y-direction along this track with respect to an x-y coordinate system that is fixed to an inertial laboratory reference frame. The slider is initially at rest. On the surface of the slider is another track which has a section that is aligned perpendicular to the y-axis. On one end of this track section is a smaller object, denoted as the mover with mass m. (We choose the ratio of M and m such that M >> m.) The mover can slide along this track in the positive x-direction until the track curves it around through a quarter turn, such that the mover emerges from the curved section in the positive y-direction. Also at the end of the track is a near-massless spring with a force constant k. We push the mover against the spring a distance ∆x, compressing it, and giving the mover an initial elastic potential energy equal to:

Elastic Energy = 1/2 k ∆x² (1).

At some point in time we release the mover. The velocity of the mover in the positive x-direction with respect to our laboratory frame as it emerges from the spring is easily derived as:

Vmover = ∆x√(k/m) (2).

Hence, the kinetic energy would be:

KEmover = 1/2 m Vmover² (3).

As the mover travels at a constant speed in the positive x-direction, we push the slider in the positive y-direction, giving the slider a velocity Vslider. Thus, the total initial kinetic energy of the system with respect to our laboratory frame would be:

Total Initial Energy = 1/2 M Vslider² + 1/2 m Vslider ² + 1/2 m Vmover² (4).

At some later point in time the mover travels counter-clockwise around the quarter section of the track and emerges traveling in the positive y-direction. With respect to the frame of the slider because we assume there is no friction, the magnitude of the velocity of the mover in the positive y-direction would be the same as the magnitude of the initial velocity in the positive x-direction expressed in equation 2. (Remember, we also assume M >> m, such that the effect the mover has on the slider as it rounds the curved section of the track due to the centrifugal reactive force acting on the track is inconsequential.) The velocity of the mover in the positive y-direction with respect to the laboratory frame is easily derived from the Galilean velocity addition theorem as:

Vmover lab = Vslider + Vmover (5).

Hence, with respect to the laboratory frame, the mover's kinetic energy would now be:

KEmover lab = 1/2 m (Vslider + Vmover)² (6).

Expanding, we have:

KEmover lab = 1/2 m Vslider² + m Vslider Vmover + 1/2 m Vmover² (7).

Combining with the kinetic energy of the slider, we have the total final energy of
the system as:

Total Final Energy = 1/2 M Vslider² +1/2 m Vslider² + m Vslider Vmover + 1/2 m Vmover² (8).

Subtracting the initial total energy, equation 4, from the above, we arrive at:

∆Energy = m Vslider Vmover (9).

The above expression represents a gain in energy of the system. The initial energy of the system can be accounted for by the work we did in compressing the spring and the work we did in moving the system in the positive y-direction. We define the system as the mass of the slider and the mass of the mover. We include the mass of the track on the slider as part of the mass of the slider. The total initial kinetic energy of the system expressed in equation 4 must by the conservation of energy be accounted for by an equal decrease in the internal energy of our body so that the net change in energy at this initial time is zero. However, when the mover emerges from the curved section of the track, the system has gained energy by the amount expressed in equation 9.
Would anyone like to venture an explanation where the source of this gain in energy
comes from? Would it be in the domain of quantum physics? Could we link it to the vacuum energy?

2. Nov 29, 2009

### wywong

When you assume M>>m and ignore something, you shouldn't be surprised to find an error of the order of m when your input is of order M.

Wai Wong

3. Nov 29, 2009

### cesiumfrog

:rofl: You waffled on and confused yourself, so now you're blaming gremlins?
Fail. Take the limit M>>m at the end, not arbitrarily. Hint: large enough mass means (since KE is nonlinear in velocity) it can lose a portion of KE without significant change of speed.

4. Nov 30, 2009

### e2m2a

Lets take a simple concrete case. Lets make the earth the slider. Lets have an inertial observer(the space observer) moving such that relative to this observer the earth moves at a constant velocity in the positive y-direciton equal to vy+. Lets have the mover be a marble that can roll on a track attached to the earth. At some point in time the marble is given an initial velocity in the positive x-direction equal to vx+. It rolls in a straight line for a time then curves around the quarter turn of the track until relative to both an observer on the earth and the space observer, the marble is now moving in the positive y-direction. I can assure you the impact the marble has on the earth's relative velocity with respect to the space oberserver would be inconsequential. Realistically, there would be a decrease in the velocity of the marble due to friction, probably by no more than 20 per cent, depending on the coefficient of friction,etc. However, there would still be an energy gain relative to the space observer as explained in the previous post.

5. Nov 30, 2009

### Staff: Mentor

Whatever mechanism you use to push the marble also pushes back on the earth. You must account for the total change in energy of both marble and earth.

6. Nov 30, 2009

### e2m2a

The change in kinetic energy of the earth compared to the change in kinetic energy of the marble would again be insignificant and would have no bearing on the gain of the marble with respect to the space observer. For example, if the marble were compressed against a spring, then released, the lion's share of the elastic potential energy would manifest in the marble, not the earth. In fact, you would not be able to measure the change in kinetic energy of the earth. But again, this would not have any bearing on this thought experiment anyway.
The gain in kinetic energy occurs after the marble makes it turn around the curved point of the track. Before it makes its turn, the total kinetic energy of the marble with respect to the space frame is found by a simple application of the Pythagoream Theorem. It would be:

KEinitial = 1/2 m vx sq + 1/2 m vy sq (1).

Where, vx = the velocity of the marble emerging from the
spring in the positive x-direction with respect to the space
frame
vy = the velocity of the earth and marble in the
positive y-direction with respect to space frame

The total kinetic energy of the marble after it emerges from the curved section, moving in the positive y-direction is found by a simple application of the binomial theorem:

KEfinal = 1/2 m vx sq + m vx vy + 1/2 m vy sq (2).

Clearly, there has been a gain in energy mathematically equal to the difference of equation 1 and 2, which is the mid-term of the expansion of equation 2.

7. Nov 30, 2009

### Staff: Mentor

Don't just say the change in the earth's KE is insignificant--do the calculation.

Rather than mess around with curved surfaces, try this simpler problem. Start off with the earth and the marble at rest, with the marble on the compressed spring. View things from a frame of reference in which the earth/marble is initially at rest. Release the spring. Say the marble ends up with speed v in the y direction. What's the speed of the earth? What's the total change in KE of everything?

Now redo the calculation from a frame in which the earth/marble is initially moving at some speed Ve in the y direction. Release the marble again. What's the total change in KE of everything viewed from this frame?

I hope you'll agree that this example contains the same issue--the difference between v2 and (Ve + v)2 and some supposed missing energy.

8. Nov 30, 2009

### cesiumfrog

Why does every crackpot have to involve centrifugal force at some point, isn't it simpler if your marble just bounces squarely off a rubbery tree-stump or something?

So, in the flying saucer frame (which has velocity $-v$ relative to our centre of mass), the marble's kinetic energy goes from $\frac 1 2 m_{marble} (v-v_{marble})^2$ to $\frac 1 2 m_{marble} (v+v_{marble})^2$, which is an increase of $2 m_{marble}\ v\ v_{marble}$. But by conservation of momentum, the earth's momentum inarguably must have decreased by a paltry $2\ m_{marble}\ v_{marble}$, and so (assuming the collision was elastic) the earth's kinetic energy went from $\frac 1 2 M_{earth} (v+\frac{m_{marble}}{M_{earth}}v_{marble})^2$ to $\frac 1 2 M_{earth} (v-\frac{m_{marble}}{M_{earth}}v_{marble})^2$, which (I trust you will verify) gives the required decrease of $2 m_{marble}\ v\ v_{marble}$, and not zero. Now, it may be true that the change in the earth's velocity is only a tiny fraction (specifically $\frac{m_{marble}}{M_{earth}}$) of the change in the marble's velocity, but nonetheless it cannot be neglected here; the total changes in kinetic energy are exactly in balance.

9. Nov 30, 2009

### e2m2a

Okay, I'm going to make a simple modification of the thought experiment to show the kinetic energy of the earth after the marble is accelerated by the spring has no bearing on the issue of energy gain. The modification is simple and is as follows. Instead of 1 marble, we have 2 marbles. Both marbles are compressed the same length in opposite directions on seperate springs that each have the same force constant and both marbles have the same mass. At the same point in time the marbles are released. One marble travels with a velocity component in the positive x-direction of 10 meters/sec. The other marble travels with a velocity component in the negative x-direction of 10 meters/sec. Each marble has a mass of .25 kg. (This would imply that each marble had an initial elastic potential energy of 12.5 joules). Hence, by symmetry equal and opposite impulses will act on the earth when the marbles are accelerated, therefore, there will be no change in the velocity of the earth with respect to the x-axis, and therefore, no change in the kinetic energy of the earth after the 2 marbles emerge from the springs. The earth relative to the space frame travels in the positive y-direction at 20 meters/sec.
Thus, with respect to the space frame the total initial kinetic energy of the two marbles combined after they emerge from the springs would be 125 joules. After both marbles round the curved sections, they emerge traveling in the positive y-direction. We can find their velocities at this point with respect to the space frame by applying the Galilean velocity transformation. Each marble would have a velocity of 30 meters per second in the positive y-direction. Hence, the final combined kinetic energy of the two marbles with respect to the space frame would be 225 joules, a net gain of 100 joules in energy with respect to the space frame.

10. Dec 1, 2009

### rcgldr

The opposing x component of forces cancel, but not the y components. In order to round the curves, the two curved sections generate a net force in positive y direction, which is the force responsible for the acceleration of the two marbles in the y direction. This force coexists with the marbles reaction force in the negative y direction equal and opposing the force from the curved sections, against the curved section and the earth that those curved sections are attached to.

The earth, curved sections, and marbles are a closed system. Momentum of this system is conserved, and if there are no losses, then kinetic energy of this system is also conserved. From an external reference frame, after exiting the curves, the marbles end up moving slightly less than 30 m/s in the +y direction and the earth ends up slightly less than 20 m/s in the +y direction. I'll leave it to the other posters here to do the math.

11. Dec 1, 2009

### Staff: Mentor

Did you bother to do the calculation I suggested or look at cesiumfrog's post?

Let's call the speed of the marbles v in the x-direction. Their total KE is mv2. When they are pushed in the y-direction, they push back on the earth. Now calculate the speed of the marbles and the earth after the collision with the curved surfaces. View things from the center of mass frame. Assuming no energy is lost, the total KE must be mv2. Total momentum is zero, so the follow holds:
(1) 2mvm = Mve (momentum conservation)
(2) mv2 = mvm2 + 1/2Mve2 = mvm2 + 2m2vm2/M (energy conservation)
Where vm and ve are the speeds of marbles and earth after the collision, viewed from the center of mass frame.

Now view things from a frame in which the earth is originally moving at speed vx in the y direction.
The initial KE of the marbles is: mvx2 + mv2 = mvx2 + mvm2 + 2m2vm2/M
The initial KE of the earth is: 1/2Mvx2

After the collision, the speeds are:
marbles: vx + vm
earth: vx - ve = vx - (2m/M)vm

Please calculate the total KE of marbles + earth and compare it to their initial KE.

12. Dec 1, 2009

### e2m2a

Your analysis is probably correct about the curved sections generating the necessary force to accelerate the marbles with respect to the space frame in the positive y-direction. And by Newton's Third Law there must be an equal and opposite force acting on the earth. However, because the mass of the earth is so immense, I predict that the decrease in the velocity of the earth would be so miniscule, that with respect to the most accurate measurement device known to man, the velocity of the earth after the marbles round the turns would still measure 20 m/s with respect to the space frame. Yes, it would be less, but it would be of the order of 19.9999999999999... m/s. Using this figure, whatever it is, to determine the final kinetic energy of the earth, then subtracting from the initial kinetic energy of the earth, I predict the decrease in the kinetic energy of the earth will no where approach the increase in kinetic energy of the marbles. I will do the calculation.

13. Dec 1, 2009

### e2m2a

I will do the calculation, but I predict the decrease in the kinetic energy of the earth would be insubstantial, far less than the increase in the kinetic energy of the marbles.

14. Dec 1, 2009

### xxChrisxx

Don't predict. Do.

The reason your are wrong is becuase you have stated you are assuming seomthing fundamentally wrong.

Just because the change in velocity of the earth is small doesnt mean the KE change is small.

No wonder you gain energy when you conveniently ignore the loss.

15. Dec 1, 2009

### sophiecentaur

Didn't the Banks have a system which was expected to work on these lines?

16. Dec 2, 2009

### e2m2a

I have done the calculations and to my surprise, there is no over-all energy gain of the system. This surprised me because of the enormous mass of the earth, I assumed its change in ke would be negligible. I admit I was wrong. There is no paradox.

17. Dec 2, 2009

### sophiecentaur

Breathes a sigh of relief!!

18. Dec 2, 2009

### xxChrisxx

Cool beans.

The main thing to learn from this, is how your assumptions affect the outcome. It's always better to check and see than to try to guess. Also it's good to try and quantify the error arisen from assumptions.