Notes on Relativity and Cosmology - Physics Department, UCSB

6.4. MORE ON MASS, ENERGY, AND MOMENTUM 155
By the way, you may notice a certain similarity between the formulas for p and
E in terms of rest mass m 0 and, say, the formulas (4.11) on page 104 for the
position x and the coordinate time t relative to the origin for a moving inertial
object in terms of it’s own proper time τ and boost parameter θ. In particular,
we have
We also have
pc
E = v = tanhθ (6.10)
c
E 2 − c 2 p 2 = m 0 c 4 ( cosh 2 θ − sinh 2 θ ) = m 2 0c 4 . (6.11)
Since m 0 does not depend on the reference frame, this is an invariant like,
say, the interval. Hmm.... The above expression even looks kind of like the
interval.... Perhaps it is a similar object?
⋆ ⋆ ⋆ Here is what is going on: a displacement (like ∆x, or the position relative
to an origin) in general defines a vector – an object that can be thought of like
an arrow. Now, an arrow that you draw on a spacetime diagram can point in
a timelike direction as much as in a spacelike direction. Furthermore, an arrow
that points in a ‘purely spatial’ direction as seen in one frame of reference points
in a direction that is not purely spatial as seen in another frame.
So, spacetime vectors have time parts (components) as well as space parts.
A displacement in spacetime involves c∆t as much as a ∆x. The interval is
actually something that computes the size of a given spacetime vector. For a
displacement, it is ∆x 2 − c∆t 2 .
Together, the momentum and the energy form a single spacetime vector. The
momentum is already a vector in space, so it forms the space part of this vector.
It turns out that the energy forms the time part of this vector. So, the size of
the energy-momentum vector is given by a formula much like the one above for
displacements. This means that the rest mass m 0 is basically a measure of the
size of the energy-momentum vector.