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4 - MASS IN OUR RELATIVITY

( 4.1 ) In this section we shall derive the measured mass of the moving train relative to the static observer. Now, suppose a static observer on the earth surface, also a static train which inside it a static rider and clock. Then, both the rider and the observer are agreeed at the time = 0, where at this time the velocity of the train was equal to zero.

Now, at 0 the train moved with constant velocity . after the train had travelled the distance relative to the static observer, it stopped. In this case, the static observer registered the time according to his clock to the train to travel the distance . Then, he secured the velocity of the train was , where

Then, the measured momentum of the train relative to the observer was

( 4.1.1 )

Where

is the measured momentum for the static observer when the train was moving.

is the measured mass for the static observer when the train was moving.

For the rider of the train, when the train stopped, the measured distance that the train is travelled for the rider would be [ as we have seen in ( 3.4 ) ], but he would secure, this distance was travelled at the time according to his clock, where

Where, is the time measured by the static observer.

Thus, the rider would secure, the velocity of his train was , where

( 4.1.2 )

We see equation ( 4.1.2 ) is in contrast with equation ( 3.1.1 ), where in equation ( 3.1.1 ) , but in equation ( 4.1.2 )

that is because, equation ( 3.1.1 ) is just applied during the motion of the train of the rider, where in this case the measured distance that the train travels during the motion for the rider is , where is the measured distance that the train travels for the static observer. when the train stops, then , [ as we have seen in figure ( 3.4.1 ) ].

Thus, the measured momentum of the train relative to the rider ( according to his clock ) was , where

( 4.1.3 )

Where, is the rest mass of the train, where we proposed the measured mass of the moving train for the rider is the rest mass. If we proposed both, the rider and the observer will be agreed at the measured momentum, then by equalizing equations ( 4.1.1 ) and ( 4.1.3 ) we get

Then

( 4.1.4 )

Equation ( 4.1.4 ) represents the relativistic mass of the moving train for the static observer.

( 4.2 ) Now, suppose the rider of the moving train B desire measuring the mass of the static train A during his train motion. As we have seen, the rider will secure, the time that is measured by his clock is equal to the time that is measured by the clock of the static train, where , where is the time that is measured by the clock of the rider for himself, and is the time that is measured by the clock of the static train A for the rider. Now, if train B travelled the distance for the rider, then, it is also considered for the rider that train A is travelled the distance , where it is considered for the rider, train A is moving with velocity , where is the velocity of train B. Thus, the measured momentum of train A for the rider ( according to his clock ) is where

( 4.2.1 )

The measured momentum of train A for itself relative to the reference frame of the moving train B ( according to the clock of train A for the rider ) is , where

( 4.2.2 )

is the rest mass of the train.

If we equalized equations ( 4.2.1 ) and ( 4.2.2 ), we get

From that we get

( 4.2.3 )

Thus we get, the rider of the moving train B will measure the mass of the static train A equals to the rest mass of it as for a static observer on the earth surface.

( 4.3 ) Now, suppose train A started at rest to move with constant velocity , and after it had travelled the distance for the rider of the moving train B, it stopped. In this case, the rider secured, train A travelled the distance in a time separation according to his clock, then its measured velocity for him was , where

And then the measured momentum is , where

( 4.3.1 )

Where is the measured momentum of the moving train A for the rider of the moving train B.

According to the clock of the moving train A for the rider, this distance was travelled at a time where

Then the velocity of train A for itself relative to the reference frame of the moving train B is , where

Thus its momentum for itself for the reference frame of the rider of the moving train B is , where

( 4.3.2 )

If we equalize equations ( 4.3.1 ) and ( 4.3.2 ), we get

Thus we get

( 4.3.3 )

Equation ( 4.3.3 ) represents the measured mass of the moving train A for the rider of the moving train B ( relativistic mass ), where according to the equation, the mass is increased during the motion. For an observer static on the earth surface, the relativistic mass of the moving train A is given also as in equation ( 4.3.3 ), where both the rider and the observer will measure the velocity and rest mass , and from that we see, the motion of the train of the rider does not affect to his calculations comparing to the calculations of the static observer on the earth surface, where both will get the same calculations, but, the motion of the train of the rider makes him to get these calculations slower than the observer.

5 - EQUIVELANCE OF MASS AND ENERGY

( 5.1 ) The kinetic energy of train A for a static observer on the earth surface or a rider of the moving train B is given by calculating the work done in increasing its speed from zero to . Suppose the force acts parallel to the distance of train A, then the work done on it is given by

Newton’s second law is expressed by

Thus

Since , then we get

As we have seen previously, the measured mass of train A for the observer and the rider is , then we get

By integrating the equation above we get

Since the work is done to increase the velocity of train A from zero to , then it produced a changing in its kinetic energy from zero to , thus

( 5.1.1 )

Equation ( 5.1.1 ) represents the kinetic energy of the moving train A for the observer or the rider. We can write equation ( 5.1.2 ) as

( 5.1.2 )

is the total energy of train A.

is the rest mass energy of train A.

Thus, we can express the total energy of any particle for any frame of reference as

( 5.1.3 )

Equation ( 5.1.3 ) leads to the equivalence of mass and energy, where according to this equation we conclude the mass and energy are two faces for one currency.

6 - ENERGY IN OUR RELATIVITY

( 6.1 ) As we know from the Quantum theory, the energy is considered as photons, have a rest mass equals to zero, and we can express the energy that is held by any photon is given by the equation

( 6.1.1 )

Where, is the energy that is held by the photon, is Plank’s constant, and is the frequancy of the wave of the photon.

According to the concept of equivalence of mass and energy, the equivalent mass of any photon is given by

( 6.1.2 )

Now, suppose a train moving with constant velocity and a static observer on the earth surface. As we know, the motion of the clock inside the train is slower than on the the earth surface for the static observer, where, the time that is measured by the observer is via his clock, and via the clock of the moving train. Since the frequency of a wave is defined as the number of cycles at the unit of time, thus, the frequency of any wave on the earth surface for any observer static on the earth surface is given by

( 6.1.3 )

Now, suppose this wave entered inside the moving train, then the frequency of the wave will be as

Thus

( 6.1.4 )

Where, from equation ( 6.1.4 ), the frequency of the wave will be less inside the moving train than outside for the static observer. Thus, the energy that is held by any photon inside the moving train is given by

Thus from equation ( 6.1.1 ) we get

( 6.1.5 )

Equation ( 6.1.5 ) represents the endured energy inside the moving train for the reference frame of the static observer on the earth surface. The difference of energy of the reference frame of the earth surface and the moving train is given by

( 6.1.6 )

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