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8 Euclidean Four-Vectors

The traditional Minkowski line element with metric $(+1,-1,-1,-1)$ is:

$\displaystyle ds^2=c^2dt^2-dx^2-dy^2-dz^2$ (34)

where $ds=cd\tau$ . Four-vectors with the Euclidean metric $(+1,+1,+1,+1)$ as used in the previous Sections use the 4D velocity of the moving object and 4D Euclidean distances as invariants, which is in fact the essence of Eq. (2):

$\displaystyle c^2=v_1^2 +v_2^2 +v_3^2 +\chi^2$ (35)

Multiplication with $dt^2=dx_5^2$ yields (recall that $\chi=cd\tau/dt$ ):

$\displaystyle c^2dt^2=dx_1^2 +dx_2^2 +dx_3^2 +c^2d\tau^2$ (36)

where the factors $c^2d\tau^2$ and $c^2dt^2$ from Eq. (34) have switched roles.

The Euclidean metric thus gives rise to four-vectors for position, velocity and energy/momentum:

Euclidean Minkowskian
$( {x_1 ,x_2 ,x_3 ,c\tau } )$ $(x_1,x_2,x_3,ct)$
$( {v_1 ,v_2 ,v_3 ,\chi} )$ $ \gamma (v_1,v_2,v_3,c)$
$( {m_0 v_1 ,m_0 v_2 ,m_0 v_3 ,m_0 \chi} )$ $(p_1,p_2,p_3,E/c)$


Equation (36) is not really new. It is merely Eq. (34) written in a different form, with as a main input the definition of $\chi$ , being the time-speed of an object as measured by an observer at rest, which has three effects:

In general, the new Euclidean four-vectors can be derived from the Minkowski four-vectors by using the time component in the Minkowski four-vector as the invariant (the vector sum) for the new four-vector. It is essentially doing Pythagoras ``the other way around'', i.e., calculating the hypotenuse from the rectangular sides, instead of calculating a rectangular side from the hypotenuse and the other rectangular side (refer to [9] for a detailed treatment of Minkowski and Euclidean four-vectors).


next up previous 14 16 Note: this is Euclidean relativity, not Minkowski.
Next: Bibliography Up: Dimensions in Special Relativity Previous: 7 Transformation of Energy
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