(First published inside the Alice Law Version 5 physics program. November 2005)
Updated for the web.
With the topic of Energy Principles coming to the agenda, I am republishing my E=mc² work here in the form in which it was presented in the Alice Law Version 5. It has been retranslated in order to be published in a clearer language.
Animated Figure 1 – Let us consider a platform mounted on a wagon and a ball standing on this platform. If the platform is parallel to the ground, the ball will remain on it as long as the train does not move; when the train starts to move, the ball will fall.
Now imagine that when the train starts moving, a man tries to prevent the ball from falling by changing the inclination of the platform. Changing the inclination of the platform transfers the pushing force of the locomotive to the ball, and in this way the ball can stay on the platform.
Based on the examples above, I will now explain what E=mc² actually means.
Let us start by writing the forces acting on the ball:
The first force is the gravitational force in the direction of the Y–axis. What we need here is to express the potential energy arising from the gravitational force. How to calculate the magnitude (size) of the potential energy that is produced by a force was explained in the “Potential Energy” section.
We are using the Forces Principle. If we interpret the resulting potential energy with respect to the gravitational force, our arrow becomes VG (downward). If we interpret it with respect to the pushing force, it becomes VP (upward). They are equal in magnitude but opposite in direction. Here VG and VP are the magnitudes of the potential energy (acceleration) belonging to the gravitational force.
The second force acting on the ball is the pushing force of the locomotive in the direction of the X–axis. Now let us write the potential energy produced by this force.
We calculate the magnitude of the potential energy arising from the pushing force in a similar way. In accordance with the Forces Principle, we define the pushing force of the locomotive both as a and as g.
If we interpret the pushing force of the locomotive as a push, our arrow becomes VP; if we interpret it as a gravitational force, it becomes VG. Here too, VG and VP are the magnitudes of the potential energy belonging to the pushing force.
On the left, we see together the potential energies corresponding to both forces (the gravitational force and the pushing force of the locomotive).
In the figure, the inclination of the platform is balanced according to the acting forces. Therefore, even though the platform is tilted, the ball does not fall. If we change the inclination of the platform or the potential energies, the balance will be disturbed.
Using the buttons on the right, I invite you to change the forces or the inclination of the platform and make the ball move. If you cannot keep the ball on the platform, it will explode. You can press the Reset button and try again.
In our example, since the magnitude of the gravitational force does not change, the only variable force is the pushing force of the locomotive.
As long as the man can keep the direction of the resultant force perpendicular to the platform, he will be able to prevent the ball from falling off the platform. For each value of the locomotive’s pushing force, there is a specific inclination angle of the platform.
In this way we have defined the potential energies acting on the ball.
Now let us write the kinetic energies of the ball.
Animated Figure 6 – First let us write the kinetic energy in the direction of the push. As long as the pushing force continues, the kinetic energy of the ball in the X–axis direction increases in proportion to the acceleration of the train.
If the pushing stops, there will be no further increase in kinetic energy, and unless a force is applied in the opposite direction, there will be no decrease in the kinetic energy of the ball either.
As we saw in the Potential Energy section, the increase in the kinetic energy of the ball in the X–axis direction will follow the same pattern as the acceleration of the train.
Now we also need to write the kinetic energy along the Y–axis. Let us ask ourselves a question in order to understand whether the ball has kinetic energy along the Y–axis.
If the Earth we live on were to suddenly disappear, would we remain where we are, or would we be flung off into space?
Using the information coming from the Forces Principle section, we conclude that we would be thrown off into space, and that our speed would be determined by the potential energy we have at that moment along the Y–axis.
Since we know that the force value of the potential energy is equal to g, we calculate the kinetic energy on the Y–axis and the magnitude of its vector as shown in the figure. The direction of the vector is upward.
In this way, we have defined both the kinetic energies and the potential energies of the ball along both axes. In the figure, the blue arrows represent kinetic energies, and the yellow and green arrows represent potential energies.
The total energy of the ball is determined by the vector sum of the resultant vectors of its potential and kinetic energies. (Animated Figure 8)
Up to now, in our examples, the man on the wagon has been trying to prevent the ball from falling. From this point on, let us consider the opposite: the man gradually increases the inclination of the platform, and you try to keep the ball on the platform by changing the pushing force of the locomotive.
To reach the equality E=mc², we assume that the locomotive has infinite pushing power.
In the animation below, you are in control of the locomotive. While the man increases the inclination of the platform, you will try to prevent the ball from falling. You can adjust the pushing force of the locomotive by moving the slider on it. If you let the ball fall, press Reset and Play to try again.
In the animation above, as you try to keep the ball on the platform, you can see that the resultant vectors of kinetic and potential energy grow larger and larger.
For an object that has energy E=mc², the magnitudes of the resultant vectors of its potential and kinetic energies have reached the value c, and both vectors point in the same direction.
To give you an idea, I would like to show what kind of effect an object carrying energy E=mc² is under. As you know, on Earth the magnitude of the gravitational acceleration acting on us is g = 9.81 m/s².
Now imagine an enormously large gravitational force. Let us calculate the potential energy of an object that has energy E=mc². The gravitational force acting on the object will create its potential energy.
An object on the surface of a celestial body
with such a terrifying gravitational acceleration
will have energy E=mc².
Because Alice moves upward together with this celestial body with speed c. Therefore, Alice’s kinetic energy is Ek = ½mc². At the same time, Alice is also subjected to the pushing force, and thus her potential energy is Ep = ½mc².
Alice’s total energy is the sum of these two energies. Since both energies point in the same direction, Alice’s energy is E=mc².
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