Relativistic Dynamics and Energy in R5

Roland Alfred Sprenger DOI 10.5281/zenodo.20007363 18.10.2026

Contents

The kinetic energy of the motion of a body moving in a relativistic manner in four-dimensional space is specified, and the relativistic mass increase is interpreted as the increase in its five-dimensional velocity. The existence of rest mass, the nature of mass and the nature of time are explained within a five-dimensional worldview.

Sections:

1. Introduction

2. Momentum and Mass in Four-Dimensional Space

3. The Invariance of Total Energy and Total Momentum

4. Force and Acceleration

5. Energy, Mass, and Time
6. The relativistic mass increase

7. Summary and Outlook

Attachment

1. Introduction

The hypothesis presented in [1], that relativistically moving bodies move in a four-dimensional auxiliary space, raises the question of whether the four-dimensional velocities that occur there also possess kinetic energy. Apparently, no energy has yet been measured that occurs in a fourth dimension. On the other hand, the question arises as to whether the increase in mass with velocity contains the answer to the previous question.

In [1], equations (6) and (7)

v_w = (γ / c) ∙ v_x² (1)

and v_x² + v_w² = v² (2)

are derived for the velocity component v_w in the direction of the fourth dimension w and for the motion v composed of the components v_w and v_x . (In [1], v = v´, since this is also the velocity of the inertial frame S´ of the moving body.) Furthermore, according to [1], for the angle ψ at which the body moves relative to the x-axis in the x-w-plane, the following equations (5) hold:

cos⁡ ψ = l / l´ = v_x / v , s. Fig. 1

and cos⁡ ψ = √ (1 -v_x² / c² ) = 1 / γ . (3)

Fig. 1: A rod of length l moves at the relativistic velocity v_x in the positive x-direction. In the assumed four-dimensional auxiliary space, according to [1], its rest length l´, and the velocity v (= v´) of its rest frame appear at an angle ψ, and its component v_w in the direction of the fourth dimension w.

2. Momentum and Mass in Four-Dimensional Space

The relativistic mass increase is a purely observational phenomenon without a causal explanation and is therefore actually surprising. It is an obvious interpretation of the momentum, which is mathematically too small in the observer’s frame when using the rest mass due to time dilation, because otherwise the velocity would have to be greater than the measured velocity.

However, if one rearranges the brackets in the equation p = m v_x= (m₀ γ) v_x

as follows: p = m₀ (γ v_x) ,

one obtains the greater velocity v from Eq. 2, since from Eq. 3: γ = 1 / cos ⁡ψ = v / v_x it follows that

p = m₀ [ (v / v_x ) v_x ] ,

so p = m₀ v (bold type denotes a vector) (4)

Accordingly, the four-dimensional velocity v carries the momentum; that is, the velocity is in fact greater than the observed velocity, but only in four-dimensional space. This lends the previously “auxiliary space” a semblance of reality. In three-dimensional space only the perpendicular projection of the velocity v occurs.

Mass therefore remains constant (see, however, section 6, last paragraph). In modern theoretical physics, too, the mass of a body is defined as its rest mass (and m₀ is then replaced by m). There, too, there is a four-dimensional velocity vector [2], which confirms the approach described here.

3. The Invariance of Total Energy and Total Momentum

Since the old interpretation involving variable mass also yields correct results in many cases, we may then also use its formulas

m = m₀ / √ (1 - v²/ c²) = m₀ γ , E = m c² , E₀ = m₀ c² und E_k = E - E

as already done above.

Let γ = m / m₀

it follows from (3) cos ⁡ψ = m₀ / m

and further m₀ / m = v_x/ v .

Expanding in terms of c² gives E₀ / E = v_x/ v , (5)

that is, the energies are related to one another in the same way as the corresponding velocities, and therefore form a right-angled triangle in Fig. 2 that is similar to the triangles shown in Fig. 1.

From cos⁡ ψ = v_x/ v it follows that cos ψ = E₀ / E , so

E = E₀/ cos ⁡ψ , (6)

which provides further evidence of the possible reality of the fourth dimension. The angle ψ can be constructed from the velocity v_x without calculation; see Attachment.

Fig. 2: The four-dimensional geometric relationship between rest energy E₀ , dynamic energy E and momentum p, as well as dynamic kinetic energy E_k= E - E₀

The length of the shorter leg of the red right triangle in Fig. 2 is |c p|, where p is the momentum from Eq. 4 p = m₀ v = m₀ γ v_x .

Proof:

In the blue right-angled triangle in Fig. 2, the following holds for the velocities

v² - v_w² = v_x² with v = v_x / cos ⁡ψ = v_x γ and Eq. 1

⇒ v_x² γ² - (γ² / c²) v_x⁴ = v_x² | ∙ c⁴ / v_x²

⇒ γ² c⁴ - γ² v_x² c² = c⁴

⇒ m₀² γ² c⁴ - m₀² γ² v_x² c² = m₀² c⁴

⇒ (m₀ γ c²)² - (m₀ γ v_x c)² = (m₀ c²)²

⇒ E² - (c p)² = E₀² (7)

This is the invariance of total energy and total momentum, as described in the theory of special relativity [2]. The fact that it also follows from this four-dimensional approach demonstrates that it is consistent with SRT and makes the reality of the fourth dimension seem even more likely.

From Eq. 7, we have E²= (m₀ c²)²+ (c p)²

and using Eq. 4, E = √ (m₀² c⁴ + c² m₀² v² ) . (8)

This is the total energy as a function of velocity v in four-dimensional space.

4. Force and Acceleration

In particle accelerators, a force that accelerates the charged particles—in this case, in the x-direction—acts on them, increasing their momentum. The following applies [2]

F = dp/dt = d(m₀ γ v_x)/dt = m₀ γ (dv_x/dt) + m₀ (dγ/dt) v_x = m₀ γ a_x + m₀ (γ³ v_x a_x / c²) v_x ,

so F = m₀ γ a_x (1 + (v_x γ / c)² ) . (9)

Solving for a_x, we get a_x = F / [m₀ γ (1 + (v_x γ / c)² ) ] . (10)

As v_x → 0 , γ → 1 and (v_x γ) / c → 0. Then, approximately, a_x = F / m₀ , which is Newton's second law. As the velocity v_x increases, v and γ also increase, and the acceleration decreases. If the force F acts in a linear accelerator 3.2 km long, as in Stanford, then γ → ∞ and thus a_x→ 0, so that the speed limit c cannot be exceeded, as was demonstrated there in 1974.

The speed of light was practically reached after just 1 km. However, the force that continues to act imparts further energy to the electron, which now leads only to the acceleration of v_w and thus to an increase in ψ (see Fig. 3). The particle therefore describes a curve with left-hand curvature in the x-w plane.

For v_x→ c , according to Eq. 3 , ψ → 90° and γ → ∞, and therefore v_w→ ∞ (Eq. 1) and v → ∞ (Eq. 2).

Fig. 3 illustrates the position of the velocity vectors v for increasing velocities v_x .

Fig. 3: Increase in velocity v in four-dimensional space and in the angle ψ as the velocity v_x increases in three-dimensional space, shown for v_x = 0.5c ; 0.8c ; 0.9c ; 0.99c.

From (1) it follows that v_w= v_x² / √ (c²- v_x²) . (11)

On the corresponding graph, the tips of the vectors v lie, see Fig. 3 .

In the following, the acceleration (for a || F) in four-dimensional space is derived.

For the four-dimensional acceleration vector, we have a = Δv/Δt ,

and for the three-dimensional acceleration vector, a_x= Δv_x/Δt .

For their magnitudes, we have a / a_x = Δv / Δv_x

and thus a = a_x (Δv / Δv_x) .

Fig. 4: Relativistic acceleration of a particle in four-dimensional space

As Δt → 0 and thus also Δv → 0, the secant vector a becomes a tangent vector, and Δv/(Δv_x) becomes the derivative dv/dv_x of

v = γ v_x .

a = a_x dv/dv_x

a = a_x [(dγ/dv_x) v_x + γ]

dγ/dv_x = d/dv_x [1 - (v_x²/ c²)^-1/2] = (-1/2) [1 - (v_x²/ c²)^-3/2] ∙ (-2 v_x /c² ) = γ³ v_x /c²

a = a_x (γ³ v_x²/ c² + γ) = a_x γ (γ² v_x² / c² + 1) = a_x γ (v²/ c² + 1)

Eq. 10 ⇒ a = {F / [m₀ γ (1 + (v_x γ / c)² )]} γ (v² / c² + 1) = {F / [m₀ γ (1 + (v / c)²)]} γ (v² / c² + 1)

a = F / m₀ (12)

In four-dimensional space, Newton’s second law thus also holds for relativistic velocities. This, too, is a result well-known in SRT [2], which in turn confirms the approach used here. It appears to contradict Figures 3 and 4, since in those figures, as v_x approaches c with constant Δv_x increments, the Δv-arrows become increasingly longer, so that a by no means appears to be constant. However, the times Δt must be taken into account, which increase accordingly as v_x approaches c.

5. Energy, Mass, and Time

Eq. 8 E = √ (m₀² c⁴ + m₀² c² v²) = m₀ c √ (c²+ v²) = c p_E

is now interpreted to mean that a particle with total energy E and mass m₀ has momentum p_E = m₀√ (c² + v²) and moves with the velocity v_E = √ (c² + v²) . The corresponding velocity vector v_E is the hypotenuse in the right-angled triangle with the velocities c and v.

The vector v lies in the x-w plane; the vector c must be perpendicular to it. If c were also in the x-w plane, the projection of v_E would not be v_x. In five-dimensional spacetime, therefore, the only direction left for c is the time direction. A particle with velocity v_E also moves in the time direction (Fig. 5), specifically at the speed of light. As a result, it possesses additional kinetic energy, the rest energy m₀c². The rest energy results from the motion of the universe in the time direction.

Fig. 5: Motion of a body with v_x = 0.5 c and four-dimensional velocity v, with the x-w-plane moving in the direction of time at the speed of light; resulting velocity vector v_E ; worldlines in R⁴ and R⁵ shown in blue; rest energy E₀ and total energy E as red lines, with the chosen unit of measurement m₀ c having the same magnitudes as the velocity vectors c and v_E .

Eq. 8: E = m₀ c √ (c² + v²) ⇒ E / (m₀ c) = √ (c² + v²) = v_E

E₀ = m₀ c ⇒ E₀ / (m₀ c) = c

Eq. 7 and 4: c p = c m₀ v ⇒ (c p) / (m₀ c) = v

This means that the right-angled triangle in Fig. 2, with invariance of total energy and momentum (Eq. 7), corresponds in Fig. 5 to the right-angled triangle formed by the vectors v_E , v, and c, with invariance of five-dimensional and four-dimensional velocity.

v_E²- v²= c² (13)

Hereat, v > c may apply, but v_E ≥ c always applies. Furthermore, the following applies

v_E= c γ . (14)

Proof: 1 = [1 / (1 - v_x²/ c²)] (1 - v_x²/ c²)

1 = γ²(1 - v_x²/ c²) = γ²- [(γ² v_x²) / c²]

1 + [(γ² v_x²) / c²] = γ²

γ v_x= v ⇒ √ (1 + v²/ c²) = γ (15)

√(1 + v²/ c²) ∙ c = γ c

√ (c²+ v²) = γ c

(13) ⇒ v_E= c γ

The term c p has the dimension of energy. As can be seen from the three equations above, the proportionality factor between each of the velocities and its associated energy is (m₀ c). Accordingly, the short side in Fig. 2, with the magnitude c p = c m₀ v, represents the energy E_v of the four-dimensional velocity v:

E_v= m₀ c v .

Both in the red triangle of energies in Fig. 2 and in the triangle with sides v, c, v_E and their associated energies in Fig. 5, the following holds:

E₀²+ E_v²= E² .

In R⁵, the energies are added quadratically. Accordingly, the following holds:

E_x²+ E_w²= E_v² where E_x= (1/2) m₀ v_x²

and E_w= √ (E_v²- E_x²) , see example in the attachment.

The terms for the total energy E = m₀ c √( c²+ v²) and E = m c² are equivalent.

Proof:

(15) ⇒ √ [(c²+ v²) / c²] = γ

m₀ c² √ [(c²+ v²) / c² ] = m₀ γ c²

m₀ γ = m ⇒ m₀ c √ (c²+ v²) = m c²

Furthermore, for the total energy applies

E = m₀ c √ (c²+ v²) = m₀ c² √ (1 + (v / c)²)

E = E₀ √ [1 + (v / c)²] . (16)

It follows that the kinetic energy is

E_k= E - E₀= E₀ [√ (1 + (v / c)²) - 1] (17)

See Fig. 6 for the graphs E(v) and E_k(v). As can be seen in Eq. 7, these are hyperbola branches.

The known expression for kinetic energy E_k = m₀ c² (γ - 1)

is equivalent to that in Eq. 17, since √(1 + v²/ c²) = γ , see Eq. 15.

Fig. 6: Plots of the graphs of E = E₀√ (1 + (v / c)²) ; D = [0 ; ∞ [ (red) with its asymptote E ⁄ (m₀ c² ) = v / c and E_k = m₀ c²(√ (1 + (v / c)²) - 1) ; D = [0 ; ∞ [ (black) with E₀ = 1 Nm ; Unlike v_x , v is unbounded; see Fig. 3.

From E = E₀ γ and v_E= c γ follows E = (E₀/c) v_E or E = m₀c v_E . (18)

The total energy is directly proportional to the five-dimensional velocity v_E .

6. The relativistic mass increase

The kinetic energy of a particle moving at relativistic speeds is given by

E_k= E - E₀= m c²- m₀ c²= (m - m₀) c²= Δm c²

(8) ⇒ Δm = (E - E₀) / c² = (m₀ c v_E- m₀ c²) / c²

Δm = (m₀/ c) (v_E- c) (19)

The so-called relativistic mass increase Δm is therefore a linear function of the five-dimensional velocity v_E and proportional to the velocity difference v_E - c (see Fig. 7). According to this hypothesis, it is thus the kinetic energy of this velocity difference (divided by c²). In the relativistic mass increase, the velocity v_E makes itself felt, even though it lies in five-dimensional space. And this relationship also serves as an indication of the claimed motion of our three-dimensional space at the speed of light through a five-dimensional space.

Fig. 7: Illustration of the difference between the magnitudes of v_E and c, v_E – c, to which the relativistic mass increase is proportional. For v_x → c, v → ∞ and v_E → ∞. Since the subtrahend c, the distance with magnitude c, remains constant, v_E – c also approaches infinity, just like the relativistic mass increase.

Δm can also be expressed in terms of the four-velocity v. From Eq. 17, we obtain

Δm = (E - E₀) / c² = (E₀/ c² ) [√ (1 + (v/c)² ) - 1] ,

which, together with Eq. 15, leads to the well-known formula Δm = (E₀/ c² ) (γ - 1) .

The question posed in the introduction – whether kinetic energy can be attributed to four-velocities – is now answered in the affirmative by the interpretation of relativistic mass increase set out above.

Both the so-called dynamic mass and the rest mass of bodies can be explained by their velocities v_E and c respectively, and their kinetic energies – ultimately by that of v_E alone, since at rest v_E = c. There is no fundamental difference between rest mass and dynamic mass, both are manifestations of the kinetic energy associated with a single velocity. Therefore, defining the mass of a body as its rest mass is merely a stopgap that does not do justice to the true nature of mass. Mass is a manifestation of kinetic energy, just as heat is.

7. Summary and Outlook

A momentum is associated with the body’s four-dimensional velocity, and the relativistic energy-momentum invariant is derived from the arrangement of the three-dimensional and four-dimensional velocities.

It is shown that Newton’s second law applies to the acceleration of the four-dimensional velocity.

The total energy is represented mathematically and graphically as a function of the four-dimensional velocity.

The rest energy of a body is the kinetic energy of its motion through the time dimension. It appears as rest mass.

The dynamic mass of a body moving at relativistic speeds is interpreted as the kinetic energy of its five-dimensional velocity, which, when the body is at rest, turns into its rest energy or rest mass. Mass is a manifestation of kinetic energy.

The relativistic increase in mass can be explained by the limitation of three-dimensional velocity to the speed of light and the resulting increase in four-dimensional and five-dimensional velocities and their kinetic energy.

A new world model is being established: The four-dimensional cosmos—and with it the three-dimensional universe—moves at the speed of light orthogonally through the fifth, timelike dimension. Time thereby acquires a spatial character. Accordingly, the universe is a moving three-dimensional cross-section of five-dimensional spacetime.

Time is the variable spatial coordinate of the four-dimensional cross-section of a five-dimensional Euclidean space, which consists of the three-dimensional universe plus a fourth spatial dimension, and moves at the speed of light perpendicular to all four directions of propagation in the fifth dimension.

The supplementation of the previous [1] four-dimensional relativistic kinematics by the presented statements on dynamics and energy not only closes a gap in our understanding but also explains the existence of rest mass and the essence of mass and time. This may serve as further evidence that the hypothetical fourth spatial dimension is not merely a mathematical construct but could actually exist. And this approach presumably also can explain why the speed of light sets the upper limit for all speeds of action.

References

[1] www.zenodo.org ; DOI 10.5281/zenodo.17266293

www.roland-sprenger.de: Rest length and time in five-dimensional spacetime

[2] en.wikipedia.org/Mass in special relativity

Attachment

Derivation of ψ

From Eq. 3, cos⁡ ψ = √ (1 - v_x²/ c² )

it follows that cos² ψ + v_x²/ c² = 1

and from this sin⁡ ψ = v_x/ c .

If one draws a circle of radius 1c centered at the origin and draws a line parallel to the v_x – axis at a distance of v_x/c (see Fig. eight), the free leg of ψ points through the intersection of the parallel line with the arc of the circle.

Fig. 8: Construction method for the angle ψ using only the velocity v_x

Example relating to the energies of the velocity components

Given: v_x= 0,6 c ⇒ γ = 1,25

E = m c²= m₀ γ c²= 1,25 m₀ c²

E_x= (1/2) m₀ v_x²= (1/2) m₀ (0,6 c)²= 0,18 m₀ c²

v_w= (γ / c) ∙ v_x²= 1,25 ∙ 0,6² c = 0,45 c

v = √ (v_x²+ v_w²) = √ (0,6²+ 0,45²) c = 0,75 c

E_v= m₀ c v = m₀ c ∙ 0,75 c = 0,75 m₀ c²

E_w= √ (E_v²- E_x²) = √ (0,75²- 0,18²) m₀ c²= 0,7281 m₀ c²

E = √ (E_x²+ E_w²+ E₀²) = √ (0,18²+ 0,7281²+ 1²) m₀ c²= 1,250 m₀ c²

The square root of the sum of the squared component energies gives the correct result.

v_E= c γ = 1,25 c

E = m₀ c v_E= 1,25 m₀ c²