 # Derivation method of numerous dynamics in the Special Theory of Relativity

From the journal Open Physics

## Abstract

The article presents innovative method of deriving dynamics in the Special Theory of Relativity. This method enables to derive infinitely many dynamics in relativistic mechanics. The authors have shown five examples of these derivations. In this way,It is presented that the dynamics known today as the dynamics of Special Theory of Relativity is only one of infinitely of theoretically possible.

## 1 Introduction

The currently accepted dynamics of STR have different experimental confirmations. Present article does not deal with these experiments. They are not analyzed, nor evaluated. The purpose of this article is to show that the concrete dynamics of STR does not only result from the kinematics of STR only. Formally and correctly mathematically, many STR dynamics, which have different properties, can be derived within the kinematics STR. Relativistic dynamics is derived based on the relativistic kinematics and one additional assumption, which allows the concept of mass, momentum and kinetic energy to be introduced into the theory. This paper is a discussion about possible assumptions and the dynamics of STR resulting from these assumptions. It presents the author’s method of deriving numerous dynamics for this theory.

The decision, in which dynamic STR is correct, can only result from experiments. Available publications show that the dynamics indicated by Albert Einstein is correct, that is for x = 3/2 (see Section 5). However, because each experiment is fraught with errors, it is possible that more accurate experiments carried out in the future will show that the optimal model of dynamics for x = 3/2 ± Δx, where Δx is a noticeable correction.

Kinematics deals with the movement of bodies without taking their physical characteristics into account. The basic concepts of kinematics are: time, location, transformation, speed and acceleration.

Dynamics deals with the movement of material bodies under the action of forces. The basic concepts of dynamics are: inertial mass, force, momentum and kinetic energy.

Kinematics and dynamics are resulting in mechanics. This study deals with relativistic mechanics, i.e. the Special Theory of Relativity, which unlike classical mechanics, also applies to high-speed.

## 2 Kinematic assumptions of the Special Theory of Relativity

The kinematics of the Special Theory of Relativity is based on the following assumptions:

### I. All inertial systems are equivalent

This assumption means that there is no physical phenomenon, which distinguishes the inertial system. In a particular case, it means that there is no such phenomenon for which the absolute rest is needed to explain. Mathematically, it results from this assumption that each coordinate and time transformation have coefficients with exactly the same numerical values as inverse transformation (with the accuracy to the sign resulting from the velocity direction between the systems).

### III. Transformation of time and position coordinates between the inertial systems is linear

These assumptions are often written in other equivalent forms.

Based on mentioned assumptions, it is possible to derive Lorentz transformation on which the Special Theory of Relativity is based. There are many different derivation ways of this transformation. Two derivations are presented in monograph .

Markings adopted in Figure 1 will be convenient for our needs. Inertial systems move along their x-axis. Symbol v2/1 stands for a velocity of U2 system measured by the observer from U1 system, while v1/2 is a velocity of U1 system measured by the observer from U2 system. In the Special Theory of Relativity occurs that v2/1 = −v1/2. Figure 1

Relative movement of inertial systems U1 and U2 (v2/1 = −v1/2)

Lorentz transformation from U2 to U1 system has a form of:

(1) t 1 = 1 1 ( v 2 / 1 / c ) 2 ( t 2 + v 2 / 1 c 2 x 2 )
(2) x 1 = 1 1 ( v 2 / 1 / c ) 2 ( v 2 / 1 t 2 + x 2 )
(3) y 1 = y 2 , z 1 = z 2

Lorentz transformation from U1 to U2 system has a form of:

(4) t 2 = 1 1 ( v 1 / 2 / c ) 2 ( t 1 + v 1 / 2 c 2 x 1 )
(5) x 2 = 1 1 ( v 1 / 2 / c ) 2 ( v 1 / 2 t 1 + x 1 )
(6) y 2 = y 1 , z 2 = z 1

Transformation (1) – (3) and (4) – (6) includes complete information on the relativistic kinematics.

## 3 Selected properties of relativistic kinematics

In order to derive dynamics two formulas from kinematics, i.e. (20) and (23) from kinematics will be needed. They will be derived out of transformation (1) – (3).

### 3.1 Transformation of velocity

Determine the differentials from transformation (1) – (3)

(7) d t 1 = 1 1 ( v 2 / 1 / c ) 2 ( d t 2 + v 2 / 1 c 2 d x 2 )
(8) d x 1 = 1 1 ( v 2 / 1 / c ) 2 ( v 2 / 1 d t 2 + d x 2 )
(9) d y 1 = d y 2 , d z 1 = d z 2

From the inertial system U1 and U2, the moving body U3 is observed. In U1 system, it has a velocity of v3/1,while in U2 system it has a velocity of v3/2. The components of these velocities are presented in Figure 2. Figure 2

Movement of the body from two inertial systems U1 and U2

The coordinates of body U3 position in U1 system are x1, y1, z1. At the same time in U2 system these coordinates are x2, y2, z2. Since the body U3 moves, these coordinates change in time. When time dt1 elapses in U1 system then time dt2 elapses in U2 system. For such indications the changes of coordinates of body U3 position in U1 system in the time interval dt1 are dx1, dy1, dz1. Changes of coordinates of body U3 position in U2 system in the time interval dt2 are dx2, dy2, dz2.

The body velocity U3 in inertial system U2 has the following components:

(10) v 3 / 2 x = d x 2 d t 2 , v 3 / 2 y = d y 2 d t 2 , v 3 / 2 z = d z 2 d t 2

The body velocity U3 in inertial system U1 has the following components:

(11) v 3 / 1 x = d x 1 d t 1 , v 3 / 1 y = d y 1 d t 1 , v 3 / 1 z = d z 1 d t 1

When differentials (7) – (9) are put into Eqs. (11), one will receive

(12) v3/1x=11(v2/1/c)2(v2/1dt2+dx2)11(v2/1/c)2(dt2+v2/1c2dx2)v3/1y=dy211(v2/1/c)2(dt2+v2/1c2dx2)v3/1z=dz211(v2/1/c)2(dt2+v2/1c2dx2)

i.e.

(13) v3/1x=v2/1+dx2/dt21+v2/1c2(dx2/dt2)v3/1y=1(v2/1/c)2dy2/dt21+v2/1c2(dx2/dt2)v3/1z=1(v2/1/c)2dz2/dt21+v2/1c2(dx2/dt2)

On the basis of (10) the desired velocity transformation from U2 to U1 system is obtained

(14) v3/1x=v3/2x+v2/11+v3/2xv2/1c2v3/1y=1(v2/1/c)2v3/2y1+v3/2xv2/1c2v3/1z=1(v2/1/c)2v3/2z1+v3/2xv2/1c2

In special case,when U3 body moves parallel to x-axis then occurs

(15) v 3 / 1 x = v 3 / 1 , v 3 / 2 x = v 3 / 2 , v 3 / 1 y = v 3 / 2 y = 0 , v 3 / 1 z = v 3 / 2 z = 0

Then velocity transformation (14) takes the form of formula to sum-up parallel velocities

(16) v 3 / 1 = v 3 / 2 + v 2 / 1 1 + v 3 / 2 v 2 / 1 c 2

### 3.2 Change of velocity seen from different inertial systems

The body at rest in U3 system has momentary acceleration to U3 system. The body movement is observed from U1 and U2 systems. The velocities of inertial systems are parallel to each other. Markings shown in Figure 3 are adopted. Figure 3

Increases in the velocity seen in inertial systems U1 and U2

The differentials from formula (16) will be determined:

(17) d v 3 / 1 = d v 3 / 2 + v 2 / 1 1 + ( v 3 / 2 v 2 / 1 ) / c 2 d v 3 / 2 d v 3 / 2 = 1 + v 3 / 2 v 2 / 1 c 2 ( v 3 / 2 + v 2 / 1 ) v 2 / 1 c 2 1 + v 3 / 2 v 2 / 1 c 2 2 d v 3 / 2
(18) d v 3 / 1 = 1 v 2 / 1 2 c 2 1 + v 3 / 2 v 2 / 1 c 2 2 d v 3 / 2

If U3 system is U2 system then it is necessary to replace index 3 with 2. Then,

(19) d v 3 / 1 = d v 2 / 1 , v 3 / 2 = v 2 / 2 = 0 , d v 3 / 2 = d v 2 / 2

On this basis, the formula (18) takes a form of

(20) d v 2 / 2 = d v 2 / 1 1 ( v 2 / 1 / c ) 2

Relation (20) is related to the change of body velocity seen in the inertial system U2, in which the body is located (dv2/2), and the change of velocity seen from another inertial system U1 (dv2/1).

### 3.3 Time dilatation

If motionless body is in U2 system, then for its coordinates occurs

(21) d x 2 d t 2 = 0

Based on time transformation (7) one receives

(22) d t 1 d t 2 = 1 1 ( v 2 / 1 / c ) 2 ( 1 + v 2 / 1 c 2 d x 2 d t 2 ) d x 2 d t 2 = 0 d t 1 d t 2 = 1 1 ( v 2 / 1 / c ) 2

On this basis we receive the formula for time dilatation of motionless body with regard to U2 system

(23) d x 2 d t 2 = 0 d t 2 = 1 ( v 2 / 1 / c ) 2 d t 1

Recording of time dilatation in a form of (23) is more precise than the commonly used recording, as it has a form of implications. Such a record makes it clear that such dilatation applies only to motionless bodies in relation to U2 system (or for events occurring in the same position in relation to U2 system).

## 4 Dynamics in the Special Theory of Relativity

All dissertations will be conducted only for one-dimensional model, i.e. all analyzed vector values will be parallel to x-axis. Each derived dynamic can easily be generalized into three-dimensional cases.

In order to derive dynamics in the Special Theory of Relativity (STR), it is necessary to adopt an additional assumption, which allows the concept of mass, momentum and kinetic energy to be introduced into the theory. Depending on the assumption, different dynamics of bodies are received.

The inertial mass body resting in inertial frame of reference is determined by m0 (rest mass). The rest mass is determined on the base unit of mass and the method of comparing any mass with this base unit. The inertial mass body at rest in U2, as seen from U1 system, is determined by m2/1 (relativistic mass). It is worth to note that the relativistic mass in this case is an inertial mass that occurs in the Newton’s second law, rather than mass occurring in the formula for momentum, as assumed in the STR. In this way, a different definition of relativistic mass has been adopted, than one in the STR. Such a definition of the relativistic mass is more convenient in deriving dynamics.

The body of m0 inertial mass is in U2 system. It is affected by force F2/2 that causes acceleration of dv2/2/dt2. Therefore, for the observer from U2 system, the Newton’s second law takes a form of

(24) F 2 / 2 := m 0 a 2 / 2 = m 0 d v 2 / 2 d t 2

For the observer from U1 system, inertial mass of the same body is m2/1. For this observer, the force F2/1 acts on the body, causing acceleration of dv2/1/dt1. Therefore, for the observer from U1 the Newton’s second law takes the form of

(25) F 2 / 1 := f ( v 2 / 1 ) m 0 a 2 / 1 = m 2 / 1 ( v 2 / 1 ) a 2 / 1 = m 2 / 1 a 2 / 1 = m 2 / 1 d v 2 / 1 d t 1

Equation (25) means that a generalized form of the Newton’s second law is postulated. This generalized form contains an additional parameter f(v). From the formula (24) shows that always f (0) = 1. In classical mechanics f(v) = 1, while in the current dynamics STR f(v) = 3 (formula (32)). Determining another form of parameter f(v) leads to other dynamics for STR. The inertial relativistic mass m2/1 is the product of this additional parameter f(v) and the inertial mass body at rest m0. In this article, the parameter f(v) will not be used, only the inertial relativistic mass m2/1.

Definitions identical as in classical mechanics apply for momentum and kinetic energy.

For the observer from U2 system, the change of this body momentum can be recorded in the following forms

(26) d p 2 / 2 := F 2 / 2 d t 2 = m 0 a 2 / 2 d t 2 = m 0 d v 2 / 2 d t 2 d t 2 = m 0 d v 2 / 2

For the observer from U1 system, the change of this body momentum can be recorded in the following forms

(27) d p 2 / 1 := F 2 / 1 d t 1 = m 2 / 1 a 2 / 1 d t 1 = m 2 / 1 d v 2 / 1 d t 1 d t 1 = m 2 / 1 d v 2 / 1

where:

1. dp2/2 is a change of body momentum with rest mass m0 in the inertial system U2, measured by the observer from the same inertial system U2,

2. dp2/1 is a change of body momentum in the inertial system U2, measured by the observer from the same inertial system U1.

Kinetic energy of the body is equal of the work into its acceleration. For the observer from U1 system, the change of kinetic energy of this body is as follows

(28) d E 2 / 1 := F 2 / 1 d x 2 / 1 = m 2 / 1 a 2 / 1 d x 2 / 1 = m 2 / 1 d v 2 / 1 d t 1 d x 2 / 1 = m 2 / 1 d x 2 / 1 d t 1 d v 2 / 1 = m 2 / 1 v 2 / 1 d v 2 / 1

where:

– dE2/1 is a change of kinetic energy of the body in inertial system U2, measured by the observer from the inertial system U1.

### 4.1 STR dynamics with constant force (STR/F)

In this section, a model of dynamics of bodies based on the assumption that the force accelerating of the body (parallel to x-axis) is the same for an observer from every inertial system will be derived (hence indication F).

#### 4.1.1 The relativistic mass in STR/F

In the model STR/F it is assumed, that

(29) F 2 / 1 F := F 2 / 2

Having introduced (24) and (25), one obtains

(30) m 2 / 1 F d v 2 / 1 d t 1 = m 0 d v 2 / 2 d t 2

On the base (20) and (23), one has

(31) m 2 / 1 F d v 2 / 1 d t 1 = m 0 d v 2 / 1 1 ( v 2 / 1 / c ) 2 1 ( v 2 / 1 / c ) 2 d t 1

Hence, a formula for relativistic mass of the body that is located in the system U2 and is seen from the system U1 is obtained, when assumption (29) is satisfied, as below

(32) m 2 / 1 F = m 0 1 1 ( v 2 / 1 / c ) 2 3 / 2

#### 4.1.2 The momentum in STR/F

The body of rest mass m0 is associated with the system U2. To determine the momentum of the body relative to the system U1 a substitution of (32) to (27)

(33) d p 2 / 1 F = m 2 / 1 F d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 3 / 2 d v 2 / 1 = m 0 c 3 1 ( c 2 v 2 / 1 2 ) 3 / 2 d v 2 / 1

The body momentum is a sum of increases in its momentum, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(34) p 2 / 1 F = m 0 c 3 0 v 2 / 1 1 ( c 2 v 2 / 1 2 ) 3 / 2 d v 2 / 1

From the work  (formula 72, p. F167) it is possible to read out, that

(35) d x ( a 2 x 2 ) 3 / 2 = x a 2 a 2 x 2 , a 0

After applying the integral (35) to (34) the formula for the body momentum in U2 system is received and measured by the observer from U1 system in a form of

(36) p 2 / 1 F = m 0 c 3 v 2 / 1 c 2 c 2 v 2 / 1 2 = m 0 1 ( v 2 / 1 / c ) 2 v 2 / 1

This formula is identical to the formula for momentum known from the STR, for the same reasons as in the case of momentum. This is because the dynamics known from the STR is derived from the assumption (29). It was adopted unconsciously, because it was considered as necessary. The awareness of this assumption allows to its change and derives other dynamics.

As already mentioned above, the definition of relativistic mass adopted is different from the definition adopted in the STR. In this case, the relativistic mass is the one, which occurs in the Newton’s second law (25). In this particular case, it is expressed in terms of dependency (32). In the STR, the relativistic mass is the one, which occurs in the formula (36) per momentum.

#### 4.1.3 The momentum in STR/F for small velocities

For small velocity v2/1 << c momentum (36) comes down to the momentum from classical mechanics, because

(37) v 2 / 1 << c p 2 / 1 F m 0 v 2 / 1

#### 4.1.4 The kinetic energy in STR/F

A determination of the formula for kinetic energy will be given. The dependence for the relativistic mass (32) is introduced to the formula (28)

(38) d E 2 / 1 F = m 2 / 1 F v 2 / 1 d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 3 / 2 v 2 / 1 d v 2 / 1 = m 0 c 3 v 2 / 1 ( c 2 v 2 / 1 2 ) 3 / 2 d v 2 / 1

The kinetic energy of body is a sum of increases in its kinetic energy, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(39) E 2 / 1 F = m 0 c 3 0 v 2 / 1 v 2 / 1 ( c 2 v 2 / 1 2 ) 3 / 2 d v 2 / 1

From the work  (formula 74, p. 167) it is possible to read out, that

(40) x d x ( a 2 x 2 ) 3 / 2 = 1 a 2 x 2

After applying the integral (40) to (39) the formula for the kinetic energy of the body in U2 system and measured by the observer from U1 system in a form of

(41) E 2 / 1 F = m 0 c 3 1 c 2 x 2 0 v 2 / 1 = m 0 c 3 1 c 2 v 2 / 1 2 1 c = m 0 c 2 1 1 ( v 2 / 1 / c ) 2 m 0 c 2

This formula is identical to the formula for kinetic energy known from the STR, for the same reasons as in the case of momentum (36).

#### 4.1.5 The kinetic energy in STR/F for small velocities

Formula (41) can be written in the form

(42) E 2 / 1 F = m 0 c 2 1 1 ( v 2 / 1 / c ) 2 1 ( v 2 / 1 / c ) 2 1 + 1 ( v 2 / 1 / c ) 2 1 + 1 ( v 2 / 1 / c ) 2
(43) E 2 / 1 F = m 0 v 2 / 1 2 2 2 1 v 2 / 1 2 c 2 + 1 v 2 / 1 2 c 2

On this basis, for small values v2/1 ≪c one receives

(44) v 2 / 1 c E 2 / 1 F m 0 v 2 / 1 2 2 2 1 + 1 = m 0 v 2 / 1 2 2

#### 4.1.6 The force in STR/F

Due to the assumption (29) value measurement of the same force by two different observers is identical.

### 4.2 STR dynamics with constant momentum change (STR/Δp)

In this section, a model of dynamics of bodies based on the assumption that the change in momentum of the body (parallel to x-axis) is the same for an observer from every inertial system will be derived (hence indication Δp).

These dynamics seem particularly interesting, because the conservation law of momentum is a fundamental law. Assumption that the change of body momentum is the same for every observer seems to be a natural extension of this law.

#### 4.2.1 The relativistic mass in STR/Δp

In the model STR/Δp it is assumed, that

(45) d p 2 / 1 Δ p := d p 2 / 2

Having introduced (26) and (27), one obtains

(46) m 2 / 1 Δ p d v 2 / 1 = m 0 d v 2 / 2

On the base (20), one has

(47) m 2 / 1 Δ p d v 2 / 1 = m 0 d v 2 / 1 1 ( v 2 / 1 / c ) 2

Hence, a formula for relativistic mass of the body that is located in the system U2 and is seen from the system U1 is obtained, when assumption (45) is satisfied, as below

(48) m 2 / 1 Δ p = m 0 1 1 ( v 2 / 1 / c ) 2

#### 4.2.2 The momentum in STR/Δp

The body of rest mass m0 is associated with the system U2. To determine the momentum of the body relative to the system U1 a substitution of (48) to (27) is made

(49) d p 2 / 1 Δ p = m 2 / 1 Δ p d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 d v 2 / 1 = m 0 c 2 1 c 2 v 2 / 1 2 d v 2 / 1

The body momentum is a sum of increases in its momentum, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(50) p 2 / 1 Δ p = m 0 c 2 0 v 2 / 1 1 c 2 v 2 / 1 2 d v 2 / 1

From the work  (formula 52, p. 160) it is possible to read out, that

(51) d x a 2 x 2 = 1 2 a ln a + x a x , a 0

After applying the integral (51) to (50) the formula for the body momentum in U2 system and measured by the observer from U1 system is received in a form of

(52) p 2 / 1 Δ p = m 0 c 2 1 2 c ln c + x c x 0 v 2 / 1 = m 0 c 2 ln c + v 2 / 1 c v 2 / 1

#### 4.2.3 The momentum in STR/Δp for small velocities

Formula (52) can be written in the form

(53) p 2 / 1 Δ p = m 0 v 2 / 1 2 c v 2 / 1 ln c + v 2 / 1 c v 2 / 1 = m 0 v 2 / 1 2 ln ( 1 + v 2 / 1 / c ) c / v 2 / 1 ( 1 v 2 / 1 / c ) c / v 2 / 1
(54) p 2 / 1 Δ p = m 0 v 2 / 1 2 ln 1 + 1 c / v 2 / 1 c / v 2 / 1 1 1 c / v 2 / 1 c / v 2 / 1

On this basis, for small values v2/1 << c one receives

(55) v 2 / 1 c p 2 / 1 Δ p m 0 v 2 / 1 2 ln e 1 / e = m 0 v 2 / 1 2 ln ( e 2 ) = m 0 v 2 / 1

#### 4.2.4 The kinetic energy in STR/Δp

A determination of the formula for kinetic energy will be given. The dependence for the relativistic mass (48) is introduced to the formula (28)

(56) d E 2 / 1 Δ p = m 2 / 1 Δ p v 2 / 1 d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 v 2 / 1 d v 2 / 1 = m 0 c 2 v 2 / 1 c 2 v 2 / 1 2 d v 2 / 1

The kinetic energy of body is a sum of increases in its kinetic energy, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(57) E 2 / 1 Δ p = m 0 c 2 0 v 2 / 1 v 2 / 1 c 2 v 2 / 1 2 d v 2 / 1

From the work  (formula 56, p. 160) it is possible to read out, that

(58) x a 2 x 2 d x = 1 2 ln a 2 x 2

After applying the integral (58) to (57) the formula for the kinetic energy of the body in U2 system and measured by the observer from U1 system in a form of

(59) E 2 / 1 Δ p = m 0 c 2 1 2 ln c 2 x 2 0 v 2 / 1 = m 0 c 2 2 ln ( c 2 v 2 / 1 2 ) + m 0 c 2 2 ln ( c 2 )
(60) E 2 / 1 Δ p = m 0 c 2 2 ln c 2 c 2 v 2 / 1 2 = m 0 c 2 2 ln 1 1 ( v 2 / 1 / c ) 2

#### 4.2.5 The kinetic energy in STR/Δp for small velocities

Formula (60) can be written in the form

(61) E 2 / 1 Δ p = m 0 v 2 / 1 2 2 c 2 v 2 / 1 2 ln 1 1 ( v 2 / 1 / c ) 2 = m 0 v 2 / 1 2 2 ln 1 [ 1 ( v 2 / 1 / c ) 2 ] ( c / v 2 / 1 ) 2
(62) E 2 / 1 Δ p = m 0 v 2 / 1 2 2 ln 1 1 1 ( c / v 2 / 1 ) 2 ( c / v 2 / 1 ) 2

On this basis, for small values v2/1 << c one receives

(63) v 2 / 1 c E 2 / 1 Δ p m 0 v 2 / 1 2 2 ln 1 1 / e = m 0 v 2 / 1 2 2

#### 4.2.6 The force in STR/Δp

Body with rest mass m0 is related to U2 system. It is affected by force that causes acceleration. For the observer from this system, the acceleration force has in accordance with (24) the following value

(64) F 2 / 2 = m 0 d v 2 / 2 d t 2

For the observer from U1 system, acceleration force has in accordance with (25) the following value

(65) F 2 / 1 Δ p = m 2 / 1 Δ p d v 2 / 1 d t 1

If to divide parties’ equation (65) by (64), then on the basis of (20) and (23) one will receive

(66) F 2 / 1 Δ p F 2 / 2 = m 2 / 1 Δ p m 0 d t 2 d t 1 d v 2 / 1 d v 2 / 2 = m 2 / 1 Δ p m 0 ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2

On the basis of (48) a relation between measurements of the same force by two different observers is obtained

(67) F 2 / 1 Δ p = 1 ( v 2 / 1 / c ) 2 F 2 / 2

The highest value of force is measured by the observer from the inertial system in which the body is located.

### 4.3 STR dynamics with constant mass (STR/m)

In this section, a model of dynamics of bodies, based on the assumption that body weight is the same for an observer from each inertial reference system, will be derived (hence indication m).

#### 4.3.1 The relativistic mass in STR/m

In the model STR/m it is assumed, that

(68) m 2 / 1 m := m 0

Therefore, for the observer from inertial system U1, the body mass in U2 system is the same as the rest mass.

#### 4.3.2 The momentum in STR/m

The body of rest mass m0 is associated with the system U2. To determine the momentum of the body relative to the system U1 a substitution of (68) to (27)

(69) d p 2 / 1 m = m 2 / 1 m d v 2 / 1 = m 0 d v 2 / 1

The body momentum is a sum of increases in its momentum, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(70) p 2 / 1 m = m 0 0 v 2 / 1 d v 2 / 1 = m 0 v 2 / 1

In this relativistic dynamics the momentum is expressed with the same equation as in classical mechanics.

#### 4.3.3 The kinetic energy in STR/m

A determination of the formula for kinetic energy will be given. The dependence for the relativistic mass (68) is introduced to the formula (28)

(71) d E 2 / 1 m = m 2 / 1 m v 2 / 1 d v 2 / 1 = m 0 v 2 / 1 d v 2 / 1

The kinetic energy of body is a sum of increases in its kinetic energy, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(72) E 2 / 1 m = m 0 0 v 2 / 1 v 2 / 1 d v 2 / 1 = m 0 v 2 / 1 2 2

In this relativistic dynamics the kinetic energy is expressed with the same equation as in classical mechanics.

#### 4.3.4 The force in STR/m

Body with rest mass m0 is related to U2 system. It is affected by force that causes acceleration. For the observer from this system, the acceleration force has in accordance with (24) the following value

(73) F 2 / 2 = m 0 d v 2 / 2 d t 2

For the observer from U1 system, acceleration force has in accordance with (25) the following value

(74) F 2 / 1 m = m 2 / 1 m d v 2 / 1 d t 1 = m 0 d v 2 / 1 d t 1

If to divide parties’ equation (74) by (73), then on the basis of (20) and (23) one will receive

(75) F 2 / 1 m F 2 / 2 = d t 2 d t 1 d v 2 / 1 d v 2 / 2 = ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2

i.e.

(76) F 2 / 1 m = ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2 F 2 / 2

The highest value of force is measured by the observer from the inertial system in which the body is located.

#### 4.3.5 Discussion on the STR/m dynamics

Obtaining a relativistic dynamics, in which there is no relativistic mass, and equations for kinetic energy and momentum are identical as in classical mechanics can be surprising, because in relativistic mechanics it is believed that the accelerated body can achieve maximum speed c. However, this dynamics is formally correct.

If the body velocity v2/1 reaches c value, then according to (76)

(77) F 2 / 1 m = ( 1 1 ) 3 / 2 F 2 / 2 0

In the inertial system U2, in which the body is located, can be affected by acceleration force F2/2 of any, but finite value. However, from a perspective of the inertial system U1, towards which the body has c velocity, the same force is zero. This means that from a perspective of U1 system, it is not possible to perform work on the body, which will increase its kinetic energy indefinitely. From the relation (72) it results that the kinetic energy, that a body with mass m0 and velocity c has, a value has

(78) E max m = m 0 c 2 2

### 4.4 STR dynamics with constant force to its operation time (STR/F /Δt)

In this section, a model of dynamics of bodies based on the assumption that the force that accelerates of the body (parallel to x-axis) divided by the time of operation of this force is the same for an observer from every inertial system will be derived (hence indication F/Δt).

#### 4.4.1 The relativistic mass in STR/F/Δt

In the model STR/F/Δt it is assumed, that

(79) F 2 / 1 F / Δ t d t 1 := F 2 / 2 d t 2

Having introduced (24) and (25), one obtains

(80) m 2 / 1 F / Δ t d v 2 / 1 d t 1 1 d t 1 = m 0 d v 2 / 2 d t 2 1 d t 2

On the base (20) and (23), one has

(81) m 2 / 1 F / Δ t d v 2 / 1 d t 1 2 = m 0 d v 2 / 1 1 ( v 2 / 1 / c ) 2 ( 1 ( v 2 / 1 / c ) 2 ) d t 1 2

Hence, a formula for relativistic mass of the body that is located in the system U2 and is seen from the system U1 is obtained, when assumption (79) is satisfied, as below

(82) m 2 / 1 F / Δ t = m 0 1 1 ( v 2 / 1 / c ) 2 2

#### 4.4.2 The momentum in STR/F/Δt

The body of rest mass m0 is associated with the system U2. To determine the momentum of the body relative to the system U1 a substitution of (82) to (27)

(83) d p 2 / 1 F / Δ t = m 2 / 1 F / Δ t d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 2 d v 2 / 1 = m 0 c 4 1 ( c 2 v 2 / 1 2 ) 2 d v 2 / 1

The body momentum is a sum of increases in its momentum, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(84) p 2 / 1 F / Δ t = m 0 c 4 0 v 2 / 1 1 ( c 2 v 2 / 1 2 ) 2 d v 2 / 1

From the work  (formula 54, p. 160) it is possible to read out, that

(85) d x ( a 2 x 2 ) 2 = x 2 a 2 ( a 2 x 2 ) + 1 4 a 3 ln a + x a x , a 0

After applying the integral (85) to (84) the formula for the body momentum in U2 system and measured by the observer from U1 system in a form of

(86) p 2 / 1 F / Δ t = m 0 c 4 x 2 c 2 ( c 2 x 2 ) + 1 4 c 3 ln ( c + x ) ( c x ) 0 v 2 / 1 = m 0 c c v 2 / 1 2 ( c 2 v 2 / 1 2 ) + 1 4 ln ( c + v 2 / 1 ) ( c v 2 / 1 )
(87) p 2 / 1 F / Δ t = m 0 v 2 / 1 1 2 1 1 ( v 2 / 1 / c ) 2 + ln c + v 2 / 1 c v 2 / 1 c 2 v 2 / 1

#### 4.4.3 The momentum in STR/F/Δt for small velocities

Formula (87) can be written in the form

(88) p 2 / 1 F / Δ t = m 0 v 2 / 1 1 2 ( 1 ( v 2 / 1 / c ) 2 ) + 1 4 ln ( 1 + v 2 / 1 / c ) c / v 2 / 1 ( 1 v 2 / 1 / c ) c / v 2 / 1
(89) p 2 / 1 F / Δ t = m 0 v 2 / 1 1 2 ( 1 ( v 2 / 1 / c ) 2 ) + 1 4 ln 1 + 1 c / v 2 / 1 c / v 2 / 1 1 1 c / v 2 / 1 c / v 2 / 1

On this basis, for small values v2/1 << c one receives

(90) v 2 / 1 c p 2 / 1 F / Δ t m 0 v 2 / 1 1 2 + 1 4 ln e 1 / e = m 0 v 2 / 1 1 2 + 1 4 ln ( e 2 ) = m 0 v 2 / 1

#### 4.4.4 The kinetic energy in STR/F/Δt

A determination of the formula for kinetic energy will be given. The dependence for the relativistic mass (82) is introduced to the formula (28)

(91) d E 2 / 1 F / Δ t = m 2 / 1 F / Δ t v 2 / 1 d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 2 v 2 / 1 d v 2 / 1 = m 0 c 4 v 2 / 1 ( c 2 v 2 / 1 2 ) 2 d v 2 / 1

The kinetic energy of body is a sum of increases in its kinetic energy, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(92) E 2 / 1 F / Δ t = m 0 c 4 0 v 2 / 1 v 2 / 1 ( c 2 v 2 / 1 2 ) 2 d v 2 / 1

From the work  (formula 58, p. 160) it is possible to read out, that

(93) x d x ( a 2 x 2 ) 2 = 1 2 ( a 2 x 2 )

After applying the integral (93) do (92) the formula for the kinetic energy of the body in U2 system and measured by the observer from U1 system in a form of

(94) E 2 / 1 F / Δ t = m 0 c 4 1 2 ( c 2 x 2 ) 0 v 2 / 1 = m 0 c 4 2 1 ( c 2 v 2 / 1 2 ) m 0 c 4 2 1 c 2
(95) E 2 / 1 F / Δ t = m 0 c 2 2 1 1 ( v 2 / 1 / c ) 2 m 0 c 2 2 = m 0 v 2 / 1 2 2 1 1 ( v 2 / 1 / c ) 2

The formula for kinetic energy (95) was derived from the work , due to the fact that the author adopted a different assumption than the one on which the dynamics known from the STR was based.

#### 4.4.5 The kinetic energy in STR/F/Δt for small velocities

For small velocity v2/1≪c kinetic energy (95) comes down to the kinetic energy from classical mechanics, because

(96) v 2 / 1 c E 2 / 1 F / Δ t m 0 v 2 / 1 2 2 1 1 = m 0 v 2 / 1 2 2

#### 4.4.6 The force in STR/F/Δt

Body with rest mass m0 is related to U2 system. It is affected by force that causes acceleration. For the observer from this system, the acceleration force has in accordance with (24) the following value

(97) F 2 / 2 = m 0 d v 2 / 2 d t 2

For the observer from U1 system, acceleration force has in accordance with (25) the following value

(98) F 2 / 1 F / Δ t = m 2 / 1 F / Δ t d v 2 / 1 d t 1

If to divide parties’ equation (98) by (97), then on the basis of (20) and (23) one will receive

(99) F 2 / 1 F / Δ t F 2 / 2 = m 2 / 1 F / Δ t m 0 d t 2 d t 1 d v 2 / 1 d v 2 / 2 = m 2 / 1 F / Δ t m 0 ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2

On the basis of (82) relation between measurements of the same force by two different observers is obtained

(100) F 2 / 1 F / Δ t = 1 1 ( v 2 / 1 / c ) 2 F 2 / 2

The lowest value of force is measured by the observer from the inertial system in which the body is located.

### 4.5 STR dynamics with constant mass to elapse of observer’s time (STR/m/Δt)

In this subchapter a model of body dynamics will be derived based on the assumption that the body mass divided by the elapse of time in observer system is the same for the observer from each inertial frame of reference (hence indication m/Δt).

#### 4.5.1 The relativistic mass in STR/m/Δt

In the model STR/m/Δt it is assumed, that

(101) m 2 / 1 m / Δ t d t 1 := m 0 d t 2

On the base (23), one obtains

(102) m 2 / 1 m / Δ t d t 1 = m 0 1 ( v 2 / 1 / c ) 2 d t 1

Hence, a formula for relativistic mass of the body that is located in the system U2 and is seen from the system U1 is obtained, when assumption (101) is satisfied, as below

(103) m 2 / 1 m / Δ t = m 0 1 1 ( v 2 / 1 / c ) 2

#### 4.5.2 The momentum in STR/m/Δt

The body of rest mass m0 is associated with the system U2. To determine the momentum of the body relative to the system U1 a substitution of (103) to (27)

(104) d p 2 / 1 m / Δ t = m 2 / 1 m / Δ t d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 d v 2 / 1 = m 0 c 1 c 2 v 2 / 1 2 d v 2 / 1

The body momentum is a sum of increases in its momentum, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(105) p 2 / 1 m / Δ t = m 0 c 2 0 v 2 / 1 1 c 2 v 2 / 1 2 d v 2 / 1

From the work  (formula 71, p. 167) it is possible to read out, that

(106) d x a 2 x 2 = arcsin x a , a > 0

After applying the integral (106) to (105) the formula for the body momentum in U2 system and measured by the observer from U1 system in a form of

(107) p 2 / 1 m / Δ t = m 0 c arcsin v 2 / 1 c 0 v 2 / 1 = m 0 c arcsin v 2 / 1 c

#### 4.5.3 The momentum in STR/m/Δt for small velocities

Formula (107) can be written in the form

(108) p 2 / 1 m / Δ t = m 0 v 2 / 1 arcsin v 2 / 1 c v 2 / 1 c

On this basis, for small values v2/1 << c one receives

(109) v 2 / 1 c p 2 / 1 m / Δ t m 0 v 2 / 1

#### 4.5.4 The kinetic energy in STR/m/Δt

A determination of the formula for kinetic energy will be given. The dependence for the relativistic mass (103) is introduced to the formula (28)

(110) d E 2 / 1 m / Δ t = m 2 / 1 m / Δ t v 2 / 1 d v 2 / 1 = m 0 1 1 ( v 2 / 1 / c ) 2 v 2 / 1 d v 2 / 1 = m 0 c v 2 / 1 c 2 v 2 / 1 2 d v 2 / 1

The kinetic energy of body is a sum of increases in its kinetic energy, when the body is accelerated from the inertial system U1 (the body has velocity 0) to the inertial system U2 (the body has velocity v2/1), i.e.

(111) E 2 / 1 m / Δ t = m 0 c 0 v 2 / 1 v 2 / 1 c 2 v 2 / 1 2 d v 2 / 1

From the work  (formula 73, p. 167) it is possible to read out, that

(112) x a 2 x 2 d x = a 2 x 2

After applying the integral (112) do (111) the formula for the kinetic energy of the body in U2 system and measured by the observer from U1 system in a form of

(113) E 2 / 1 m / Δ t = m 0 c c 2 v 2 / 1 2 0 v 2 / 1 = m 0 c c 2 v 2 / 1 2 + m 0 c c 2
(114) E 2 / 1 m / Δ t = m 0 c 2 m 0 c c 2 v 2 / 1 2 = m 0 c 2 ( 1 1 ( v 2 / 1 / c ) 2 )

#### 4.5.5 The kinetic energy in STR/m/Δt for small velocities

Formula (114) can be written in the form

(115) E 2 / 1 m / Δ t = m 0 v 2 / 1 2 2 2 c 2 v 2 / 1 2 ( 1 1 ( v 2 / 1 / c ) 2 ) ( 1 + 1 ( v 2 / 1 / c ) 2 ) 1 + 1 ( v 2 / 1 / c ) 2
(116) E 2 / 1 m / Δ t = m 0 v 2 / 1 2 2 2 c 2 v 2 / 1 2 1 ( 1 ( v 2 / 1 / c ) 2 ) 1 + 1 ( v 2 / 1 / c ) 2 = m 0 v 2 / 1 2 2 2 1 + 1 ( v 2 / 1 / c ) 2

On this basis, for small values v2/1 ≪c one receives

(117) v 2 / 1 << c E 2 / 1 m / Δ t m 0 v 2 / 1 2 2 2 2 = m 0 v 2 / 1 2 2

#### 4.5.6 The force in STR/m/Δt

Body with rest mass m0 is related to U2 system. It is affected by force that causes acceleration. For the observer from this system, the acceleration force has in accordance with (24) the following value

(118) F 2 / 2 = m 0 d v 2 / 2 d t 2

For the observer from U1 system, acceleration force has in accordance with (25) the following value

(119) F 2 / 1 m / Δ t = m 2 / 1 m / Δ t d v 2 / 1 d t 1

If to divide parties’ equation (119) by (118), then on the basis of (20) and (23) one will receive

(120) F 2 / 1 m / Δ t F 2 / 2 = m 2 / 1 m / Δ t m 0 d t 2 d t 1 d v 2 / 1 d v 2 / 2 = m 2 / 1 m / Δ t m 0 ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2

On the basis of (103) relation between measurements of the same force by two different observers is obtained

(121) F 2 / 1 m / Δ t = ( 1 ( v 2 / 1 / c ) 2 ) F 2 / 2

The highest value of force is measured by the observer from the inertial system in which the body is located.

## 5 The general form of dynamics

In presented examples, assumptions have been adopted which can be written in forms (30), (46), (68), (80) and (101). On this basis, it can be seen that the assumption for relativistic dynamics is as follows

(122) m 2 / 1 { a , b } d v 2 / 1 a d t 1 b = m 0 d v 2 / 2 a d t 2 b , a , b R

The physical meaning of the formula (122) depends on the value of the parameters a and b to be determined. For example, if a = b = 1, then this formula takes the form (29), equivalent to the form (30), from the first example.

On the basis of (20) and (23) one receives

(123) m 2 / 1 { a , b } d v 2 / 1 a d t 1 b = m 0 d v 2 / 1 a ( 1 ( v 2 / 1 / c ) 2 ) a ( 1 ( v 2 / 1 / c ) 2 ) b / 2 d t 1 b = m 0 1 1 ( v 2 / 1 / c ) 2 a + b / 2 d v 2 / 1 a d t 1 b

(124) { x } { a , b } x = a + b 2 R

Now on the basis of (123) the relativistic inertial mass of body in U2 system, seen from U1 system, when an assumption is fulfilled (122), is expressed in dynamics {x} by the following formula

(125) m 2 / 1 { x } = m 0 1 1 ( v 2 / 1 / c ) 2 x

Each such relativistic mass defines a different relativistic dynamics.

According to presented examples, based on formulas (27) and (125), the momentum in dynamicsalign {x} is expressed by the following formula

(126) p 2 / 1 { x } = 0 v 2 / 1 d p 2 / 1 { x } = 0 v 2 / 1 m 2 / 1 { x } d v 2 / 1 = m 0 0 v 2 / 1 1 1 ( v 2 / 1 / c ) 2 x d v 2 / 1
(127) p 2 / 1 { x } = m 0 c 2 x 0 v 2 / 1 1 ( c 2 v 2 / 1 2 ) x d v 2 / 1

According to presented examples, based on formulas (28) and (125), the kinetic energy in dynamics {x} is expressed by the following formula

(128) E 2 / 1 { x } = 0 v 2 / 1 d E 2 / 1 { x } = 0 v 2 / 1 m 2 / 1 { x } v 2 / 1 d v 2 / 1 = m 0 0 v 2 / 1 1 1 ( v 2 / 1 / c ) 2 x v 2 / 1 d v 2 / 1
(129) E 2 / 1 { x } = m 0 c 2 x 0 v 2 / 1 v 2 / 1 ( c 2 v 2 / 1 2 ) x d v 2 / 1

According to presented examples, based on formulas (24), (25) and (20), (23), the relation between forces in dynamics {x} is expressed by the following formula

(130) F 2 / 1 { x } F 2 / 2 = m 2 / 1 { x } d v 2 / 1 d t 1 m 0 d v 2 / 2 d t 2 = m 2 / 1 { x } d v 2 / 1 d t 1 m 0 d v 2 / 1 1 ( v 2 / 1 / c ) 2 1 1 ( v 2 / 1 / c ) 2 d t 1 = m 2 / 1 { x } m 0 ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2

On the basis of (125) one receives

(131) F 2 / 1 { x } F 2 / 2 = 1 1 ( v 2 / 1 / c ) 2 x ( 1 ( v 2 / 1 / c ) 2 ) 3 / 2 = 1 1 ( v 2 / 1 / c ) 2 x 3 2

On the basis of (25) and (125) the Newton’s second law for dynamics {x} is obtained

(132) F 2 / 1 { x } = m 0 1 1 ( v 2 / 1 / c ) 2 x a 2 / 1

## 6 Summary of dynamics

Summary derived formulas for momentum and kinetic energy:

Dynamics x = 0

(133) p 2 / 1 m = m 0 v 2 / 1
(134) E 2 / 1 m = m 0 v 2 / 1 2 2

Dynamics x = 1/2

(135) p 2 / 1 m / Δ t = m 0 c arcsin v 2 / 1 c = m 0 v 2 / 1 arcsin ( v 2 / 1 / c ) v 2 / 1 / c
(136) E 2 / 1 m / Δ t = m 0 c 2 ( 1 1 ( v 2 / 1 / c ) 2 ) = m 0 v 2 / 1 2 2 2 1 + 1 ( v 2 / 1 / c ) 2

Dynamics x = 1

(137) p 2 / 1 Δ p = m 0 c 2 ln c + v 2 / 1 c v 2 / 1 = m 0 v 2 / 1 ln c + v 2 / 1 c v 2 / 1 c 2 v 2 / 1
(138) E 2 / 1 Δ p = m 0 c 2 2 ln 1 1 ( v 2 / 1 / c ) 2 = m 0 v 2 / 1 2 2 ln 1 [ 1 ( v 2 / 1 / c ) 2 ] ( c / v 2 / 1 ) 2

Dynamics x = 3/2 (recognized STR dynamics)

(139) p 2 / 1 F = m 0 v 2 / 1 1 1 ( v 2 / 1 / c ) 2
(140) E 2 / 1 F = m 0 c 2 1 1 ( v 2 / 1 / c ) 2 m 0 c 2 = m 0 v 2 / 1 2 2 2 1 v 2 / 1 2 c 2 1 + 1 v 2 / 1 2 c 2

Dynamics x = 2

(141) p 2 / 1 F / Δ t = m 0 v 2 / 1 1 2 1 1 ( v 2 / 1 / c ) 2 + ln c + v 2 / 1 c v 2 / 1 c 2 v 2 / 1
(142) E 2 / 1 F / Δ t = m 0 c 2 2 1 1 ( v 2 / 1 / c ) 2 m 0 c 2 2 = m 0 v 2 / 1 2 2 1 1 ( v 2 / 1 / c ) 2

Figure 4 shows compared momentums from derived relativistic dynamics. Figure 4

Module of the momentum in dynamics: STR/m (x=0), STR/m/Δt (x=1/2), STR/Δp (x=1), STR/F (x=3/2) and STR/F /Δt (x=2)

Figure 5 shows compared kinetic energies from derived relativistic dynamics. Figure 5

Kinetic energies in dynamics: STR/m (x=0), STR/m/Δt (x=1/2), STR/Δp (x=1), STR/F (x=3/2) and STR/F /Δt (x=2)

Figure 6 shows relation between measurements of the same force from derived relativistic dynamics. Figure 6

Relation between measurements of the same force by two different observers in dynamics: STR/m (x=0), STR/m/Δt (x=1/2), STR/Δp (x=1), STR/F (x=3/2) and STR/F /Δt (x=2)

## 7 Even more general form of dynamics

Relation (125) to the relativistic mass can be even more generalized. In the general case, it is possible to assume that the relativistic mass is expressed by the following formula

(143) m 2 / 1 { f } = m 0 f ( v 2 / 1 )

Where f(v2/1) is any continuous function with the following properties

(144) f ( v 2 / 1 ) 0
(145) f ( 0 ) = 1
(146) f ( v 2 / 1 ) = f ( v 2 / 1 )

Each function f(v2/1) defines a different dynamics of the STR.

## 8 Final conclusions

This study presents the author’s method of deriving dynamics in the Special Theory of Relativity (STR). Five examples of such deriving were shown.

Derivation of dynamics is based on two formulas applicable in the kinematics of STR, i.e. (20) and (23). In order to derive the dynamics of STR, it is necessary to adopt an additional assumption in kinematics, which allows the concept of mass, kinetic energy and momentum to be introduced into the theory.

The dynamics of STR/F (x = 3/2) is nowadays recognized as the dynamics of the STR. It is based on the assumption that each force parallel to x-axis has the same value for the observer from each inertial frame of reference. Formally, however, other dynamics are possible in accordance with the kinematics of the STR. In order to derive them, it is necessary to base on a different assumption.

The currently accepted dynamics of STR has numerous experimental confirmations. However, it is not excluded that more accurate experiments designed specifically for this purpose will show that the optimal model is the dynamics for x = 3/2 ± Δx, where Δx is a noticeable correction. A calorimeter can be useful for verification of different dynamics. This device can measure the amount of heat released when stopping particles to high speed. On this basis, it is possible to determine graphs of the kinetic energy of accelerated particles (for example in accelerators of elementary particles) as a function of their velocity, analogous to those presented in Figure 5. On this basis, it is possible to indicate the dynamics in which the kinetic energy of particles is compatible with experiments.

The presented method of dynamism derivation can also be used in other theory of body kinematics. In the monograph  this method was used to derive four dynamics in the Special Theory of Ether, which are allowed for kinematics derived in the articles [4, 5]. The presented method is analogous to that used in another area, in the article .

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