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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 x_i) 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 shown^24.17 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:

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

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

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

$$p_\mu \; \equiv \; \{ {\textstyle {E \over c}}, \Vec{p} \} \; \equiv \; \{ {\textstyle {E \over c}}, p_x, p_y, p_z \}$$ (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:

$$p^\mu p_\mu \; \equiv \; {E^2 \over c^2} - \Vec{p} \cdot \Vec{p} \; = \; {E^2 \over c^2} \, - \, p^2$$ (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 out^24.18 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

$$E^2 \; = \; p^2 c^2 \; + \; m^2 c^4$$ (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.