E=mc2

AliceInPhysics Publications

The Principles Of Energy

E=mc²

Han Erim

August 30 , 2011

(First release is in the Alice Law Version 5 physics program, November 2005)

Reprinted for web

As the issue of the principles of energy has come to the fore, I hereby republish my work on E=mc², in the form I published it in Alice Law Version 5. It has been retranslated with the aim of achieving a better language.

E=mc²

Animated Figure 1 – Let’s think of a plate mounted on a wagon and a ball lying on it. If the plate is in parallel with the ground, the ball lies on the ball as long as the train stays inert, whereas it falls down if the train moves.

Let’s assume that a man tries to prevent the ball from falling down by changing the slope of the plate when the train starts moving. The change in the slope of the plate transfers the pushing force of the locomotive to the ball and thus the ball will stay on the plate.

Here, with reference to the examples above, I will explain you what E=mc² means.

Let’s start by writing the forces affecting the ball:

The first force is gravitational force in the direction of Y axis. What we need is to state the potential energy caused by gravitational force. The way to calculate the size of the potential energy born by the force has been described in the chapter named “Potential Energy.”

We are employing the Principle of Forces.

If we interpret the potential energy incurred by adhering to gravitational force, our arrow will be VG (pointing downwards), whereas if we interpret it by adhering to repulsive force, it will be VP (pointing upwards). They have the same size, but opposite directions.

Here, VP or VG is the the size of the potential energy which belong to gravitational force.

The second force affecting the ball is the repulsive force of the locomotive on X axis direction. Now, let’s write down the potential energy born by this force. The size of the potential energy stemming from the repulsive force is calculated in a similar way. In compliance with the Principle of Forces, we define the repulsive force of the locomotive both as a and as g. If we interpret the pushing force of the locomotive as repulsive force, our arrow will be VP, whereas if we interpret it as gravitational force, it becomes VG.

Similar to above, VP or VG is the size of the potential energy which belong to repulsive force of the locomotive.

On the left, we can see the potential energies belonging to both forces (gravitational force and the repulsive force of the locomotive).

In the figure, the slope of the plate is balanced according to the forces in effect. Therefore, the ball does not fall down, even though the plate is slanted. If we change the slope of the plate or the potential energies, the balance will be disturbed.

I kindly request that you use the buttons on the right and move the ball by altering the forces or the slope of the plate. The ball will explode if you cannot keep it on the plate. You can retry by clicking on reset button.

In our sample, as the size of the gravitational force does not change, the only variable force is the repulsive force of the locomotive. The man will be able to prevent the ball from falling off the plate, as long as he manages to keep the direction of the resultant force perpendicular to the plate. For each value of the repulsive force of the locomotive, there is a certain standing angle of the plate.

Thus, we have defined the potential energies affecting the ball.

Let’s write the kinetic energies of the ball.

Animated Figure 6 – First of all, let’s write the kinetic force in the direction of repulsion. As long as the repulsive force is prolonged, the kinetic energy of the ball in X axis direction increases, depending on the acceleration of the train. If repulsion does not continue, the kinetic energy does not increase, and it does not decrease unless a negative force is applied. The increase in the kinetic energy of the ball in X axis direction will be in the way that we have seen in the chapter named “Potential Energy”, in parallel with the acceleration of the train.

Now, we need to write down the kinetic energy on Y axis. Let’s see if the ball has kinetic energy on Y axis or not by asking a question to ourselves.

If the Earth, on which we live, disappeared in a moment, would we stay in our places, or would we be scattered in space?

If we make use of the information we have from the Principle of Forces chapter, we deduce the conclusion that we would be scattered in space, and our speed would be determined by the potential energy we have on Y axis at that moment. As we know that the force value of potential energy equals to g, we can calculate the kinetic energy on Y axis and the size of the vector belonging to it as seen in the figure on the right. The direction of the vector is upwards.

This way, we have defined the kinetic energies and the potential energies of the ball in both directions. The blue arrows in the figure demonstrate the kinetic energies, while yellow and green arrows show the potential energies. The total energy of the ball is determined by the vector sum of the resultant vectors belonging to the potential and kinetic energies. (Animated Figure 8).

In the samples we have seen until now, the man on the wagon was trying not to drop the ball. From this point on, let’s think in the opposite way. Let the man slowly increase the slope of the plate and let it be you who tries to keep the ball on the plate by changing the repulsive force of the locomotive. In order to achieve E=mc² equation, we assume that the locomotive has an unlimited repulsive force.

In the animation below, you have the control of the locomotive. While the man increases the slope of the plate, you will try not to drop the ball. You can adjust the repulsive force of the locomotive by using the slider on it. If you drop the ball, please retry by using Reset and Play buttons (Increase the power gradually while the train is setting off, the hardest parts of the animation are the first moments when the train starts moving.)

In the animation above, we have seen that the resultant vectors belonging to the kinetic and potential energies increase as we try to keep the ball on the plate. The sizes of the resultant vectors belonging to the potential and kinetic energies of an object having E=mc² energy have reached c value and both vectors point at the same direction.

With the aim of making a suggestion to you, I would like to show you under what kind of an effect an object carrying E=mc² energy is. As you already know, the size of the gravitational acceleration affecting us on the Earth is g=9,81 m/sn².

Now, please think of a gravitational force with an enormous size. Let’s calculate the potential energy of an object having E=mc² energy. The gravitational force affecting the object will constitute the potential energy of that object.

The energy of an object on a celestial body having such an enormous gravitational force is E=mc².

The reason for that is Alice moves upwards at c speed together with the celestial body. Therefore, the kinetic energy of Alice is Ek=½mc². At the same time, Alice is exposed to repulsive force as well. Therefore, her potential energy is Ep=½mc². Alice’s total energy is the sum of both energies. As both energies are in the same direction, Alice’s energy is E=mc².

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