Natural Science Forum / Physics / Relativity / July 2006

what is a 4 vector, a function a 4 parameters?

draw a 4 vector

Yes, in SR. Note that your "p" is shorthand for 3 components, and there
is no gamma anywhere.

The 4-velocity V has components relative to coordinates {x^i} defined by:
    V^i = (d/d\tau) x^i    where \tau is the proper time of
                           the object.

The sometimes-seen equation p = m*gamma*v, for 3-vectors p and v, is not
an invariant equation and must be written in a specific inertial frame.
In particular, both p and v use _coordinate_time_  in their definition,
not the object's proper time (the change from coordinate to proper time
involves a factor of gamma in this case).

Tom Roberts

I've read a couple of responses and most are very good.
However, I thought I'd add a couple of my thoughts for
completeness.

Mostly when m/=0. For the photon, since m=0, there is a
little bit of a different approach. We know the 4-wave
vector is

k^i = [2*Pi/lambda; k (3-vector)]

Then p^i = h/(2*Pi) * k^i. This is then consistent with
the observed relation E=p*c with E=h*c/lambda and p=h/lambda.

Firstly it is only true in flat space, because how it is
derived. Secondly, you have to be careful how you use
P=m*V, since m=0 and p/=0 for photons.

Bingo! Note that since v=c for photons, gamma is undefined
for photons in free space (vacuum) as a quantity.

I usually define the above as Vproper = d/dtau * X
then Vproper = gamma * V = dt/dtau * d/dt * X
thus P = m * Vproper = m * gamma * V
This way things tend to remind me that I need to be more
careful since m*gamma is like zero*infinity for photons.
This is where a lot of older books introduce m*gamma as
relativistic mass. However, today it is more common to
think of Vproper=gamma*V and P=m*Vproper instead of
mrel=gamma*m and P=mrel*V. The modern way tends to not
introduce all the logical errors that the older way did.