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Next: The Mass of Light Up: Relativistic Kinematics Previous: Conversion of Energy into Mass

Lorentz Invariants   [Advanced Topic]

In the previous Chapter we encountered the notion of 4-vectors, the prototype of which is the SPACE-TIME vector, $x_\mu \equiv \{ct, \Vec{x} \} \equiv \{x_0, x_1, x_2, x_3 \}$, where the ``zeroth component'' x0 is time multiplied by the speed of light ( $x_0 \equiv c t$) and the remaining three components are the three ordinary spatial coordinates. [The notation is new but the idea is the same.] In general a vector with Greek indices (like $x_\mu$) represents a 4-vector, while a vector with Roman indices (like xi) is an ordinary spatial 3-vector. We could make up any old combination of a 3-vector and an arbitrary zeroth component in the same units, but it would not be a genuine 4-vector unless it transforms like spacetime under LORENTZ TRANSFORMATIONS. That is, if we ``boost'' a 4-vector $a_\mu$ by a velocity $u = \beta c$ along the x1 axis, we must get (just like for $x_\mu = \{ ct, x, y, z \}$)

\begin{eqnarray*}a_0' &=& \gamma (a_0 - \beta a_1) \\
a_1' &=& \gamma (a_1 - \beta a_0) \\
a_2' &=& a_2 \\
a_3' &=& a_3 \; .
\end{eqnarray*}


It can be shown24.17 that the INNER or SCALAR PRODUCT of any two 4-vectors has the agreeable property of being a LORENTZ INVARIANT - i.e., it is unchanged by a LORENTZ TRANSFORMATION - i.e., it has the same value for all observers. This comes in very handy in the confusing world of Relativity! We write the SCALAR PRODUCT of two 4-vectors as follows:

\begin{displaymath}a_\mu b^\mu \; \equiv \; \sum_{\mu=0}^3 \; a_\mu b^\mu
\; = . . . 
 . . . ec{b}
\; = \; a_0 b_0 \; - \; (a_1 b_1 + a_2 b_2 + a_3 b_3)
\end{displaymath} (24.13)

where the first equivalence expresses the EINSTEIN SUMMATION CONVENTION - we automatically sum over repeated indices. Note the - sign! It is part of the definition of the ``metric'' of space and time, just like the PYTHAGOREAN THEOREM defines the ``metric'' of flat 3-space in Euclidean geometry.

Our first LORENTZ INVARIANT was the PROPER TIME $\tau$ of an event, which is just the square root of the scalar product of the space-time 4-vector with itself:

\begin{displaymath}c \tau \; = \; \sqrt{x_\mu x^\mu}
\; = \; \sqrt{c^2t^2 - \Vec{x} \cdot \Vec{x}}
\end{displaymath} (24.14)

We now encounter our second 4-vector, the ENERGY-MOMENTUM 4-vector:

\begin{displaymath}p_\mu \; \equiv \; \{ {\textstyle {E \over c}}, \Vec{p} \}
\; \equiv \; \{ {\textstyle {E \over c}}, p_x, p_y, p_z \}
\end{displaymath} (24.15)

where $c p_0 \equiv E = \gamma m c^2$ is the TOTAL RELATIVISTIC ENERGY and $\Vec{p}$ is the usual MOMENTUM 3-vector of some object in whose kinematics we are interested. [Check for yourself that all the components of this vector have the same units, as required.] If we take the scalar product of $p_\mu$ with itself, we get a new LORENTZ INVARIANT:

\begin{displaymath}p^\mu p_\mu \; \equiv \; {E^2 \over c^2} - \Vec{p} \cdot \Vec{p}
\; = \; {E^2 \over c^2} \, - \, p^2
\end{displaymath} (24.16)

where $p^2 \equiv \Vec{p} \cdot \Vec{p}$ is the square of the magnitude of the ordinary 3-vector momentum.

It turns out24.18 that the constant value of this particular LORENTZ INVARIANT is just the c4 times the square of the REST MASS of the object whose momentum we are scrutinizing: ${E^2 \over c^2} - p^2 = m^2 c^2$ or E2 - p2 c2 = m2 c4. As a result, we can write

 \begin{displaymath}E^2 \; = \; p^2 c^2 \; + \; m^2 c^4
\end{displaymath} (24.17)

which is a very useful formula relating the ENERGY E, the REST MASS m  and the MOMENTUM p of a relativistic body.

Although there are lots of other LORENTZ INVARIANTS we can define by taking the scalar products of 4-vectors, these two will suffice for my purposes; you may forget this derivation entirely if you so choose, but I will need Eq. (17) for future reference.



 
next up previous
Next: The Mass of Light Up: Relativistic Kinematics Previous: Conversion of Energy into Mass
Jess H. Brewer
2001-03-26