Momentum Four Vector & Four Momentum Vector
Chalky
scienceworld.wolfram.com/physics/MomentumFour-Vector.html)
and Four Momentum Vector (http://en.wikipedia.org/wiki/Four-momentum)
are the same thing or not? If they are the same thing, why are
virtually all the symbols and definitions different.
If they are different things, why do they both have the same norm?
Oh No
They are the same. p is generally used for 4-momentum, and carries over
into relativistic quantum theory, where p, rather than v (or yta) is
considered fundamental - so I am not so keen on the Wolfram definition.
If you move down to Relation to four velocity in the Wiki article, and
study it carefully you will find that the notational differences are not
so great, but the Wiki article has lowered the index on p, which means
multiplying by yta_alpha_beta. The components of yta_alpha_beta are the
reciprocals of yta^alpha^beta. This is done because the summation
convention sums indices when we have a repeated index, one up, one down.
For more on the maths behind this, see http://www.teleconnection.info/rq
g/IntroductionToVectorSpace.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
Hendrik van Hees
There is nothing different. In god-given units (c=1, \hbar=1) the
energy-momentum vector is
(p^{\mu})=(E,p1,p2,p3)
For a (asymptotically) free quantum particle it fullfils
the "on-shell" condition, i.e., the dispersion relation of the
corresponding momentum eigenmodes:
p_{\mu} p^{\mu}=E^2-p1^2-p2^2-p3^2=m^2,
where m is the (invariant) mass of the particle. BTW: Never use the
socalled "relativistic mass" which in the modern (i.e., 100 years old
Minkowskian analysis in terms of space-time) is the be identified
with the energy of the particle. I use the socalled west-coast
metric, where (g_{\mu \nu})=diag(1,-1,-1,-1) which is more familiar
to me than the east-coast metric used in Wikipedia.
The relation to the four-velocity vector is
p^{\mu}=m u^{\mu},
i.e.
u^{\mu}=(E/m,p1/m,p2/m,p3/m)
Obviously
u_{\mu} u^{\mu}=1.
The three-velocity in the used inertial frame of reference is defined
by
\vec{v}=\vec{p}/E
and the Lorentz factor
gamma=1/sqrt(1-\vec{v}^2)=E/m
Thus, in terms of the three-velocity the energy-momentum four-vector
reads
(p^{\mu})=m u^{\mu}=m gamma (1,\vec{v})
One must be aware that \vec{v} are not the spatial components of a
four-vector. That's only the case for the spatial components of the
four-velocity vector \vec{u}=gamma \vec{v}.
--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/