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4 Energy and momentum

Writing the Minkowski 4-vector for energy-momentum

$$(m_0c)^2=(E/c)^2-p^2$$ (12)

or alternatively

$$(m_0c)^2=(\gamma m_0c)^2-(\gamma m_0v)^2$$ (13)

in Euclidean form yields:

$$(\gamma m_0c)^2=(m_0c)^2+(\gamma m_0v)^2$$ (14)

The Euclidean form becomes transparent if the identity $c=\gamma \chi$ is used:

$$(\gamma m_0c)^2=(\gamma m_0\chi)^2+(\gamma m_0v)^2$$ (15)

saying that the 4D momentum is the vector sum of spatial and proper time momentum. Equation (14) does however not yield an invariant. The factor $\gamma$ , resulting from the Minkowski 4-velocity prohibits this.

The Euclidean relativistic Lagrangian for a freely moving particle in 4D,

$$\Lambda=m_0c^2$$ (16)

is a constant of motion as a result of the universal velocity magnitude $c$ for the free particle in 4D space-time (see also the derivation of Montanus in [2]). This shows that, just as in the Euclidean 4-velocity, $\gamma$ must be left out in the Euclidean form, which again yields $m_0c$ as an invariant:

$$(m_0c)^2=(m_0\chi)^2+(m_0v)^2$$ (17)

Note that the invariance of the 4D momentum is also consistent with the invariance of the acceleration $\alpha_E$ that was discussed in the previous Section. Since the acceleration is always orthogonal to the velocity, the magnitude of the momentum vector will not change.

Figure 5: Minkowski and Euclidean components for energy-momentum.

Figure 5 again visualizes the geometries.

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