On Thu, 4 May 2006 05:42:49 +0000 (UTC), Arkadiusz Jadczyk
<arkiREMOVETHIS@ANDTHIScassiopaea.org> wrote:
>If you want just the result, not necessarily the way it is derived in
>Landau, then there is a simple method through the formalism of
>differential forms.
>The 4-form d^4 p is invariant because Lorentz transformations
>are of determinant one. Then d^3p is the result of the application of
>the vector d/dp^0=d/dE to d^4 p . Thus follows the transformation law as
>in Landau
>
>ark
In fact, my answer above is wrong. What I wrote concerns a different
question. To use the above method to answer your question, let us
suppose, to get calculation shorter, that we have only p_0,p_1,p_2 (I
skip p_3). The vector is supposed to be on the mass hyperboloid
p0^2 -p_1^2-p_2^2=m^2 (1)
p0 is also denoted as E
p0=E.
We parametrize the hyperboloid by p_1,p_2,p_3 alone.
In these coordinates on the hyperboloid the tangent vectors (to the
hyperboloid) are
d/dp_1, d/dp_2
(For simplicity I am using d/dp for partial derivatives)
But in the 4D momentum space these tangent vectors have also
0-component.
You can check that these tangent vectors in 4D are
d/dp1+(p1/p0)d/dp0, d/dp2+(p2/p0)d/dp0
(I am writing p1 instead of p^2 etc)
Now, differentiating the hyperboloid equation
p_mu p^mu = m^2
you see that the vector orthogonal to the hyperboloid at (p1,p2), and of
"length" m, is (p0,p1,p2) - it is like on the sphere where the radius
vector is orthogonal to the surface.
The 3-form dx0^dx1^dx2 is Lorentz invariant in 4D space. To get from the
invariant 2-form on the mass shell we insert the radius vector p_mu.
So, let's calculate: we want to compute our invariant 3-form on two
tangent vectors, so we calculate the exterior product of the two tangent
vectors and the normal vector
V=(p0d/dp0+p1d/dp1+p2d/dp2) /\ (d/dp1+(p1/p0)d/dp_0) /\
(d/dp2+(p2/p_0)d/dp_0)
Simple calculation using wedge product gives (*)
(1 - (p1^2+p2^2)/E) d/dp0 /\ d/dp1 /\ d/dp2 /\ d/dp3 =(E^2-p1^2-p2^2) /E
d/dp0 /\ d/dp1 /\ d/dp2 /\ d/dp3 = (m^2/E) d/dp0 /\ d/dp1 /\ d/dp2 /\
d/dp3
Applying the invariant 3-form dx0^dx1^dx2 to this 3-vector (and skipping
the constant m^2) we get 1/E. This is the same as applying the form
(1/E) dx1/\Dix^2 on the hyperboloid to the vector d/dp1 /\ d/dp2 along
the coordinate lines p1,p2 on the hyperboloid.
Thus answer:
The invariant form on the hyperboloid, coordinatized by p1,p2 is
(1/E) dx1/\Dix^2
Adding one more dimension does not change anything. The only funny part
is the calculation of the wedge product where one has to pay attention
to the order of factors. That is how the sign "minus" gets into the
formula (*)
ark
--
Arkadiusz Jadczyk
http://quantumfuture.net/quantum_future/jadpub.htm
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