# Why can't I do this to get infinite energy?

I know that I cannot do this because of conservation of energy, so I am looking for an answer as to why this will not work.

So by my understanding of Einstein's whole famous $E=mc^2$ thing it is possible to turn matter into energy, and energy into matter. So what if you were to create a space station way up high and run a wire down to Earth, then pump some energy up it.

The space station would then turn that energy into matter and let it drop down to Earth. People down at Earth would harvest the kinetic energy somehow, then turn the matter back into energy, pump it back up, then repeat the process. The result would be the energy they started with plus the kinetic energy. Where is the flaw in this reasoning?

(I'm guessing it might take energy for the electrons to fight gravity up to the space station).

• "it is possible to turn matter into energy, and energy into matter." that's a common misunderstanding. The corresponding Einstein principle is not that matter can change into energy and vice versa, but that energy is to be redefined to include contribution due to mass. So 1kg of uranium, steel, stone or anything else already has energy 9E16 J, there is no turning of matter to energy involved. What's changing in nuclear reactions is not the amount of energy, but its form - from potential energy in the nuclei to thermal energy of all the particles. Apr 26, 2015 at 11:00
• @JánLalinský That's an answer, not a comment.
– rob
Apr 26, 2015 at 15:13
• @Ján Lalinský :When we say matter-energy conversion,we generally mean converting light to matter(electrons,protons etc) and vice versa. actually mass is equivalent to energy not matter.
– Paul
Apr 26, 2015 at 15:25
• @Paul, the question and my comment relates to Einstein's relation $E=mc^2$, where $m$ is mass, $E$ is energy. The concepts of matter and light or their transformation are not subject of my comment. Apr 26, 2015 at 19:10
• Or in other words, mass is a reflection of or manifestation of energy content, as I understand. To regard either as an addition to the other would be double-counting since they are the same (morally/really at least, if not technically so) in the way that, say, ice and water are. Mar 19, 2016 at 3:06

[5/3 - Extended the answer, made some corrections, and responded to John Duffield's comment]

This is actually the paradox that led Einstein to General Relativity. Consider a special case: An electron and positron are at the Earth's surface. Bring them together and they annihilate, creating gamma rays (which is very energetic light). The gamma rays travel up to the Space Station, where they are converted back to a positron and electron. These are dropped back to Earth where the kinetic energy is harvested.

Einstein found a way out of the paradox: Gravity must affect energy just like matter. Light must lose energy if it climbs against gravity. Likewise, it must gain energy if it drops from the Space Station to Earth.

Minor point - the space station has a large horizontal velocity. Let us pretend that it is stationary at the top of a very tall tower. Let us also ignore the rotation of the Earth.

This leads to another of Einstein's thought experiments. Unlike other forces, gravity attracts all particles and therefore must affect all energy. Einstein could not think of an experiment that could distinguish between a freely falling elevator in a uniform gravitational field and an elevator floating in space where there is no gravity. (There are some caveats. The elevator must be small or the effects of a non-uniform field will be noticeable. Looking outside the elevator is cheating. See this for more details.) He concluded the laws of physics in a freely falling elevator are the same as those in an inertial frame of reference without gravity.

Furthermore, the laws in a rocket in empty space accelerating at 1 g are the same as the laws in a uniform 1 g gravitational field. On Earth the ground pushes you upward with enough force to accelerate you at 1 g. This is completely equivalent to a rocket. This is called the equivalence principle.

In a rocket, you can use an inertial frame of reference. You take into account for your acceleration, and then you see that particles with no forces move in straight lines, particles at rest remain at rest.

You can also work in a frame of reference where the rocket is at rest. To do this, you say that a force toward the tail of the rocket acts on everything in the universe. This kind of force is an accounting trick you must use if you want to work in a non-inertial frame of reference. It is called a pseudo force.

The equivalence principal is a statement that gravity is a pseudo force, and that a freely falling frame of reference is inertial. Unlike a rocket, the pseudo force of gravity is not always uniform.

We can use the Equivalence Principal to calculate the effect of gravity on light.

Suppose you are in a rocket at rest a distance L from a space station far from Earth. You start accelerating toward it at 1 g just as a beam of light leaves the station toward you. You would see a blue Doppler shift, a higher frequency. At low speeds, the Doppler shift is related to velocity as below.

$$\nu = \nu_0 \left[1 + \dfrac{v}{c}\right]$$

Neglecting the distance you travel, light takes time $$L/c$$ to reach you. During this time you would accelerate to velocity $$v= gL/c$$. So

$$\nu = \nu_0 \left[1 + \dfrac{gL}{c^2}\right]$$

Since gravity and the acceleration of a rocket are completely equivalent, you see the same Doppler shift on Earth. (See this for more info.)

This seems very odd when you think about it. The laws of physics are the same when you don't look out of the elevator or rocket. When you do look out, you see an obvious difference. The acceleration of the rocket gives the universe a velocity, the velocity causes time dilation and length contraction, and these generate the Doppler sift. On Earth, the upward acceleration keeps you from falling. The universe stays still.

If the frequency of light increases, then its period decreases. If you measure a shorter period on Earth than the space station finds, it is because your clock is slower. The acceleration of gravity causes the same time dilation as the acceleration of the rocket.

$$t = \dfrac{t_0} {\left[1 + \dfrac{gL}{c^2}\right]}$$

This is valid for points separated by a small L. To compare time intervals on Earth to time intervals not influenced by gravity far from Earth, we would have to use $$g = GM/r^2$$, take into account length changes, and integrate. The result of a more exact calculation is

$$t = \dfrac{t_0}{\sqrt{1 - \dfrac{2GM}{rc^2}}}$$

From quantum mechanics, the energy of a photon is related to its wavelength. $$E = h \nu = \dfrac{h c}{\lambda}$$. So light that climbs against gravity is red shifted and has a drop in energy.

I need to be careful about statements like "Light loses energy as it climbs against gravity."

Light doesn't slow down as it climbs. Light travels at the speed of light.

You also cannot follow a photon and watch its energy change. There is no frame of reference that follows a photon. You get singularities if you try to look at the universe from the point of view of a photon.

All you can do is measure the frequency or energy of a photon as it leaves and as it arrives. You can measure a frequency shift or energy difference between two points.

If you aim a laser upwards, many photons in identical states climb against gravity. You can measure the frequency at various altitudes with a diffraction grating or measure the recoil when a particle is struck.

If the diffraction grating or particle is at rest with respect to earth, you will find that there is a frequency and energy drop as the beam climbs.

If the diffraction grating or particle is fired from a cannon on earth with enough speed to coast up to the space station and stop there, you would find no frequency or energy drop. The values of the entire beam would be the as same measured at the space station.

In both cases, the energy is not some "stuff" inside the photon that is lost as the photon rises. If is a result of an interaction between the photon and the diffraction grating or particle.

In the stationary with respect to Earth case, the rate of the clock of the diffraction grating or particle changes with altitude, and so do the measured values.

In the coasting case, the diffraction grating or particle are at rest in an inertial frame of reference. Their clocks run at the same rate as the space station's.

• Wonderful answer. Just one question: "If the frequency of light increases, then its period decreases. If you measure a longer period on Earth than the space station finds, it is because your clock is slower" -- If your clock is slower, wouldn't you measure a shorter period (as less time would pass)? I noticed a previous edit of the question said shorter as well, maybe it's a mistake? If not, please kindly help me understand? Thanks.
– user137288
Dec 30, 2020 at 1:34
• @RobertoValente - Thanks for pointing that out. The frequency increases on Earth. The period is shorter on Earth. Dec 30, 2020 at 3:45
• Do we measure a shorter period on Earth (increased frequency) due to time dilation or due to the effect gravitational attraction exerts on incoming light?
– user137288
Jan 2, 2021 at 23:17

Your guess at the solution to this paradox is correct. "Pumping energy up" to the space station, regardless of the method you choose, would require an input of at least the amount of energy you would gain in kinetic energy on the way down.

This is just a variation on the impossible perpetual motion machine concept. In practice, you would not only not gain energy, you wouldn't even break even - there would inevitably be energy losses to heat, stray radiation, etc, at every step.

Note that you have a misconception here that energy can be "converted" into mass. In fact, energy is mass - they are actually equivalent.

You cannot get infinite energy this way. When sending back energy to the space station electromagnetic wave is also affected by gravity. Photons lose momentum trying to counteract this force (decrease frequency - increase wave length).