Minkowski Versus Euclidean 4-Vectors

Minkowski versus Euclidean 4-vectors

R.F.J. van Linden e-mail [email protected], web http://www.euclideanrelativity.com February 27, 2006

Abstract Minkowski 4-vectors are written in Euclidean form. It is shown that in this way the physical meaning of their components can be made more intuitive and directly associated with geometric properties in Euclidean space-time.

1 1 Displacement

ds2=c2dt2-dA2 c2dt2=ds2+dA2 If the Minkowski 4-vector components for displace-

ment in 4D space-time t) n ia r a 2 2 2 2 2 v in ds = d(ct) − dx − dy − dz (1) (

(invariant) t d cdt c ds ds are alternatively written in Euclidean form (see also [1]) they read:

dA dA 2 2 2 2 2 2 2 2 2 d(ct) = ds +dx +dy +dz = d(cτ) +dx +dy +dz Minkowskian Euclidean (2) Equations 1 and 2 both contain the same informa- tion but the essential dierence is that the roles of Figure 1: Geometric visualization of Minkowski the variables have changed. In Eq. 1, 2 is the ds and Euclidean displacement 4-vectors. Spatial dis- Lorentz invariant. In Eq. 2, 2 is the Lorentz dct placement is represented as . invariant. The mathematics of this switch remain dA consistent with special relativity. Equation 2 shows that the speed in the Euclidean time dimension, 2 Velocity which is now formed by the proper time τ is: We will henceforth use, conform [1], the notation 2 2 2 (3) for the speed in the proper vτ = c − vspace χ = ds/dt = cdτ/dt time dimension and v for spatial speed. An at- The displacement in the proper time dimension for tractive benet in the Euclidean components of the a moving object in an interval dt (according to an 4-vector for velocity is that they are now the reg- observer at rest) equals: ular derivatives of the displacement 4-vector components with respect to t: 2 2 2 2 2 2 2 2 2 2 vτ dt = c dt − vxdt − vydt − vz dt d(ct)2 ds2 dA2 = c2dt2 − dx2 − dy2 − dz2 = + (7) dt dt dt = ds2 (4) or So ds is now no longer the invariant Minkowski dis- c2 = χ2 + v2 (8) placement but the displacement in the proper time In Fig. 2 this is again illustrated. dimension. The factor that played an equiva- dct The corresponding Minkowski 4-vector compo- lent role in the Minkowski 4-vector has become the nents for velocity are derived by multiplying the invariant 4D displacement in Euclidean space-time. time derivative of the Minkowski displacement 4- This is visualized in Fig. 1. Here the spatial dis- vector with the factor γ = 1/p1 − v2/c2, placement pdx2 + dy2 + dz2 is written as a single variable . The 'Minkowski triangle' (left) then dA c2 = γ2(c2 − v2) (9) is: ds2 = dct2 − dA2 (5) which is the same as taking the derivative with respect to the proper time τ = t/γ instead of t. and the 'Euclidean triangle' (right) is: Without this factor the 4-vector does not yield a Lorentz invariant value. Figure 3 shows the back- dct2 = ds2 + dA2 (6) ground of this multiplication and here it shows that, from an Euclidean perspective, the Minkowski The dotted lines in Fig. 1 represent the values for 4-vector components seem to have a pure mathe- dA and dct that would result from a Lorentz trans- matical function only. The components of the Eu- formation. clidean 4-vector on the other hand have an intuitive

2 physical meaning: they represent the actual speeds in Euclidean space and proper time. The vector- sum of these has universal magnitude c and rotates in 4D if the spatial velocity accelerates in 3D.

3 Acceleration c2= Ȗ2(c2-v2) c2=Ȥ2+v2 Figure 4 shows the geometry for the components

) t of, respectively, the Minkowski 4-vector for accel- n a i r a eration α and the Euclidean one, α . Here, a v M E n i ( dierent geometric representation is chosen for the

Ȗc (invariant) c Ȥ

c factors γc and γv to remain consistent with the di- rection of spatial velocity vector v (note that the scalar values of these factors have been used in the Ȗv v previous examples). The components of the Eu- Minkowskian Euclidean

Ȗv dv /dt Į Figure 2: Geometric visualization of Minkowski U E and Euclidean velocity 4-vectors. E dȤ /dt ĮE Euclidean

) t n a i proper time r c a v ĮM n J d(Jc)/dt i ( c J d(Jv)/dt F U Ȗc M Minkowskian

v space

Figure 4: Geometry of Minkowski and Euclidean Ȗv acceleration 4-vectors. Ȗv Euclidean clidean acceleration 4-vector are

Ȗc ) t n a i r dUE c a µ Ȥ Ȗ time proper v (10) αEµ = = (dχ/dt, dv/dt)

(invariant) in ( dt c c Ȥ Any acceleration in 3D space corresponds to a ro- Minkowskian v tation in SO(4) of the Euclidean 4D velocity vector v space UE with invariant magnitude c, implying that the acceleration 4-vector must always be orthogonal to Figure 3: Geometry of Minkowski and Euclidean it, or αE · UE = 0. This is in agreement with the velocity 4-vectors. Minkowski acceleration 4-vector αM with components

αMν = dUMν /dτ = [γd(γc)/dt, γd(γv)/dt] = γ(cdγ/dt, vdγ/dt + γdv/dt) (11)

3 that is also orthogonal to the Minkowski velocity 4- Note that the invariance of the 4D momentum is vector. The components for the Minkowski acceler- also consistent with the invariance of the accelera-

ation 4-vector have their origin in the change of the tion αE that was discussed in the previous Section. line elements γc and γv as is shown in the Figure. Since the acceleration is always orthogonal to the As these line elements were shown to have a math- velocity, the magnitude of the momentum vector ematical function only in Euclidean space-time, the will not change. derived acceleration components will either. In the Euclidean 4-vector on the other hand, dv/dt and dχ/dt form the orthogonal vector components and Ȗm0v these maintain their intuitive physical interpreta-

tion, while the magnitude of αE is invariant under rotations in SO(4) (see also the next Section for the physical signicance of this invariance). Note that, although and in Fig. 4 are parallel, m c time proper ) αM αE Ȗ 0 t n a in general their magnitudes are not equal. i r a v n m Ȥ i 0 ( c

0 4 Energy and momentum m

Writing the Minkowski 4-vector for energy- m0v space momentum 2 2 2 (12) (m0c) = (E/c) − p Figure 5: Minkowski and Euclidean components for or alternatively energy-momentum.

2 2 2 (m0c) = (γm0c) − (γm0v) (13) Figure 5 again visualizes the geometries. in Euclidean form yields:

2 2 2 5 Current density (γm0c) = (m0c) + (γm0v) (14) The Euclidean form becomes transparent if the The relativistic density ρ of a collection of moving identity is used: c = γχ charges is dened as γρ0, where ρ0 is the charge density in the rest frame of the moving charges, also 2 2 2 (15) (γm0c) = (γm0χ) + (γm0v) referred to as proper density. A derived quantity is saying that the 4D momentum is the vector sum of the current density 4-vector Jµ that is dened as: spatial and proper time momentum. Equation (14) dx does however not yield an invariant. The factor γ, J = ρ µ (18) resulting from the Minkowski 4-velocity prohibits µ 0 dτ this. and is constructed quite similar to the energy- The Euclidean relativistic Lagrangian for a freely momentum 4-vector: moving particle in 4D, dxµ 2 (16) p = m . (19) Λ = m0c µ 0 dτ is a constant of motion as a result of the univer- The current density 4-vector can be rewritten into sal velocity magnitude c for the free particle in 4D Euclidean form in the same way as with the energy- space-time (see also the derivation of Montanus in momentum 4-vector: [2]). This shows that, just as in the Euclidean 4- velocity, must be left out in the Euclidean form, 2 2 2 γ (ρ0c) = (γρ0c) − (γρ0v) which again yields m0c as an invariant: 2 2 2 (γρ0c) = (ρ0c) + (γρ0v) 2 2 2 2 2 2 (m0c) = (m0χ) + (m0v) (17) (γρ0c) = (γρ0χ) + (γρ0v) (20)

4 saying that the 4-dimensional current density is the the vector potential A is vector sum of the current density in space and the q v current density in the proper time dimension. A A2 + A2 + A2 = φ (23) x y z c2 similar eect of the factor γ is seen in the current density as in the energy-momentum 4-vector and while φ for a moving charge can be written in terms this suggests that there will also be a justication of the retarded potential φr: that allows to leave out this factor in the equa- φ = γφ . (24) tion for current density to reach an invariant result, r equivalent to the relativistic Lagrangian for energy- Using these identities, K can be determined as: momentum. That relativistic Lagrangian more or 2 less represents the view on energy that a 'Hyper- 2 2 2 v 2 (25) K = γ φr 1 − = φr spacelander' with full 4D observational skills would c2 have. Such an upgrade from 3D to 4D observational The Euclidean form thus reads: skills eliminates length contraction eects (length is 2 2 2 2 2 2 (26) invariant in 4D) and for the 4D observer this turns φ = φr + c Ax + Ay + Az the relativistic charge density into the proper den- ρ The variable φ is however not invariant. In order sity ρ0. It is therefor justied to leave out γ in Eq. to make the 4-vector invariant under rotations in (20). SO(4) (the Euclidean equivalent of Lorentz trans- With a single unit of charge the charge density formations, see also [3]) the expression would have will be the same for all inertial frames (charge is to be multiplied with 1/γ but this is inconsistent invariant). The equation for a single charge reads: with the approach in the other 4-vectors where the aim was to get rid of the gamma's in the Euclidean 2 2 2 (ρ0c) = (ρ0χ) + (ρ0v) (21) expressions. This raises the question whether the classical potential 4-vector could be the Euclidean and is the only equation that is on equal footing form already, although at rst sight this seems in with the Euclidean 4-vector for momentum, Eq. conict with the traditional + − −− pattern in its (14), when applied to a single elementary mass components, necessary to yield an invariant. particle. It strikes that the factor γ is eliminated Various operations on the potential 4-vector, like automatically here. There is no need for an up- e.g. the derivation of the elds of E and B in F = grade from 3D to 4D observational skills to jus- µν ∇µAν − ∇ν Aµ, involve the operator ∇µ which is tify any omission of γ like was done for the energy- dened as: momentum 4-vector. This markedly distinguishes ∂ ∇ = , −∇ (27) the properties charge and current in the electro- µ ∂t magnetic eld from the properties mass and energy- The derivative with respect to in is incon- momentum in the gravity eld and seems to sug- t ∂/∂t sistent with the derivation with respect to in gests a dimensional hierarchy, i.e., the electromag- τ Minkowski 4-vectors but consistent with the Eu- netic eld seems to have one less dimension than clidean 4-vectors. Furthermore, the components gravity. φ and A both already have intuitive physical mean- ings, showing in particular from operations like 6 Electromagnetic potential E = −∇φ − ∂A/∂t and B = ∇ × A. The pattern + − −− therefor seems an intrinsic property of electromagnetic potentials rather than being re- µ Finally, the potential 4-vector A = (φ, cA) for a lated to Minkowski geometry. uniformly moving charge could be rewritten in Eu- clidean form using the same method as used so far: References 2 2 2 2 2 2 (22) φ = K + c Ax + Ay + Az [1] R.F.J. van Linden, "Euclidean Special relativ- with K still to be determined as the temporal com- ity", available at www.euclideanrelativity.com ponent in the Euclidean form. The magnitude of (Sep 2017)

5 [2] H. Montanus, "Proper time formulation of relativistic dynamics", Foundations of Physics 31 (9) 1357-1400 (September 2001)

[3] A. Gersten, Euclidean special relativity, Foun- dations of Physics 33 (8) 1237-1251 (August 2003).