проф. др Миодраг Матељевић,
Универзитет у Београду, Математички факултет

„Једначине елементарних кретања“


Лоренцове трансформације »

In relativity theory we often want to calculate the position of a point from a different reference point.
Suppose we have a second reference frame $S'$, whose spatial axes and clock exactly coincide with that of $S$ at time zero, but it is moving at a constant velocity v with respect to S along the $x$-axis.
Since there is no absolute reference frame in relativity theory, a concept of "moving" does not strictly exist, as everything is always moving with respect to some other reference frame. Instead, any two frames that move at the same speed in the same direction are said to be comoving. Therefore $S$ and $S'$ are not comoving.
The linear momentum of a system of particles can also be defined as the product of the total mass, $m$, of the system times the velocity, $v_{cm}$, of the center of mass. $$\sum{\mathbf{F}} = {\mathrm{d}\mathbf{p} \over \mathrm{d}t}= m \frac{\mathrm{d}\mathbf{v}_{cm}}{\mathrm{d}t}=m\mathbf{a}_{cm}\, $$ This is a special case of Newton's second law (if mass is constant).

Modern definitions of momentum, Momentum in relativistic mechanics

In relativistic mechanics, in order to be conserved, the momentum of an object must be defined as $$\mathbf{p} = \gamma m_0\mathbf{v}\,, $$ where $m_0$ is the invariant mass of the object and $\gamma $ is the Lorentz factor, given by $$\gamma = \frac{1}{\sqrt{1 - (v/c)^2}}\,, $$ where $v$ is the speed of the object and $c$ is the speed of light.

Rotation

In geometry, Euler's rotation theorem states that, in three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point. It also means that the composition of two rotations is also a rotation. Therefore the set of rotations has a structure known as a rotation group.
The theorem is named after Leonhard Euler, who proved it in 1775 by an elementary geometric argument. The axis of rotation is known as an Euler pole. The extension of the theorem to kinematics yields the concept of Instant axis of rotation.
In linear algebra terms, the theorem states that, in $3$-dimensional ($3D$) space, any two Cartesian coordinate systems with a common origin are related by a rotation about some fixed axis. This also means that the product of two rotation matrices is again a rotation matrix and that for a non-identity rotation matrix it must happen that: one of its eigenvalues is $1$ and the other two are $-1$, or it has only one real eigenvalue which is equal to unity. The eigenvector corresponding to this eigenvalue is the axis of rotation connecting the two systems.
The center of mass $\mathbf{R}$ of a system of particles of total mass $M$ is defined as the average of their positions, $\mathbf{r}_i$, weighted by their masses, $m_i$ :[3] $$\mathbf{R} = \frac{1}{M} \sum m_i \mathbf{r}_i.$$ For a continuous distribution with mass density $\rho(\mathbf{r})$, the sum becomes an integral:[4] $$\mathbf R =\frac 1M \int \mathbf{r} \; dm = \frac 1M \int\rho(\mathbf{r})\, \mathbf{r} \ dV.$$ If an object has uniform density then its center of mass is the same as the centroid of its shape.[5]
Examples

- The center of mass of a two-particle system lies on the line connecting the particles (or, more precisely, their individual centers of mass). The center of mass is closer to the more massive object; for details, see below.
- The center of mass of a uniform ring is at the center of the ring; outside the material that makes up the ring.
- The center of mass of a uniform solid triangle lies on all three medians and therefore at the centroid, which is also the average of the three vertices.
- The center of mass of a uniform rectangle is at the intersection of the two diagonals.
- In a spherically symmetric body, the center of mass is at the geometric center.

This approximately applies to the Earth: the density varies considerably, but it mainly depends on depth and less on the latitude and longitude coordinates.
More generally, for any symmetry of a body, its center of mass will be a fixed point of that symmetry.

In general, when a rigid body moves, both its position and orientation vary with time. In the kinematic sense, these changes are referred to as translation and rotation, respectively. Indeed, the position of a rigid body can be viewed as a hypothetic translation and rotation (roto-translation) of the body starting from a hypothetic reference position (not necessarily coinciding with a position actually taken by the body during its motion).
Denote by $D(t)$ rigid-body in moment $t$; set $D=D(0)$.
When a rigid body moves, mapping $A(t,r)$ is defined on $I\times D$, where $D$ subset in $R^3$ (the rigid body in moment for example $t=0$).
For each fixed $t$, $A(t,\cdot)$ is a Euclidean isometry and continuous in $t$. %is a rigid body motion
For $t_1 < t_2$, there is a composition of translation and rotation $B$ such that $B(D(t_1))=D(t_2)$.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Reduction the Minkowski isometry to the Euclidean isometry

Since $ds=ds'$, $y=y'$, $z=z'$, we find the mapping $(x,c t) \mapsto (x',ct')$ is the Minkowski isometry. Hence $$x^2 -c^2 t^2= x'^2 -c^2 t'^2,$$ and therefore $$ x^2 + c^2 t'^2 = x'^2 +c^2 t^2 .$$ Thus $(x,c t') \mapsto (x',ct)$ is a Euclidean isometry and therefore by complex notation we find $$ x'+ i ct=e^{-i \phi}(x + i ct' ) \mbox{ and } x + i ct' =e^{i \phi} (x'+ i ct).$$ Hence $$\begin{equation}\label{loren1} \left.\begin{array} {ll} x & = x' \cos \phi - ct \sin \phi\\ c t'&= x' \sin \phi + c t \cos \phi \end{array}\right \} \end{equation}$$ Hence, from the second equation we find $$ct= \frac{c t'- x' \sin \phi }{\cos \phi}= \frac{c t'}{\cos \phi}- x' \tan \phi$$ and therefore $$ct \sin \phi = (c t'- x' \sin \phi )\tan \phi .$$ Substituting this in the first equation we find $$x=\frac{ x'}{\cos \phi}- c t' \tan \phi $$ From the first equation for $x'=0$, we get $x= - ct \sin \phi$; (and from the second equation $c t'= c t \cos \phi$). Since $ x=Vt$, $ \sin \phi= - V/c$, $\cos \phi= \sqrt{1- \sin^2 \phi}= \sqrt{1-v^2/c^2}= \Omega$, $\gamma=\frac{1}{\cos \phi}$ and $\tan \phi= - \gamma V/c$.
Hence $$t= \gamma (t' + V x'/c^2) \mbox{ and } x= \gamma (x' + V t')$$ Draw the orthogonal axis $(X,c T')$ and rotate them for the angle $\phi$ to obtain $(X',c T)$ and give geometric explanation of time dilatation and length contraction.
The $y$ and $z$ coordinates are unaffected; only the $x$ and $t$ axes transformed. These Lorentz transformations form a one-parameter group of linear mappings, that parameter being called rapidity.
A quantity invariant under Lorentz transformations is known as a Lorentz scalar.