Transforming a 4-vector integration measure

1. Nov 27, 2012

Sekonda

Hello,

I have a particular derivation of a four-vector integration measure, basically changing the measure to some related more useful measure - but I'd like to do this in 3-vector notation. Here it is, from the integral:

$$-i\lambda\int_{-\infty}^{\infty}\frac{d^4k}{(2\pi)^4}\frac{i}{k^2-m^2}$$

where k is the four vector: $$k=(E,\mathbf{p})$$

We transform the integration measure like so:

$$d^3k=4\pi k^{2}dk\: ,\: \frac{dk^2}{dk}=2k\: ,\: d^3k=2\pi kdk^2$$

Where the first step is done due to spherical symmetry (can anyone explain why this is spherically symmetric? I'm guessing the 'k' is isotropic in some sense?)

Anyway we attain that the integration measure is:

$$\frac{d^4k}{(2\pi)^4}=\frac{k^2dk^2}{16\pi^2}$$

My main question is how to change this derivation to a 3-vector one? I'm guess I pretty much subsitute 'k' for 'p' but I'm not sure.

Thanks,
SK

2. Nov 27, 2012

tom.stoer

too many identical k's w/o index, vector notation

3. Nov 27, 2012

cosmic dust

Sorry, but I did not quite get what you want to do. You want to turn the given integral in 4-space to an integral in 3-space? Well, you cannot do it in general, unless you have a constrain (like the "mass shell": k2 = m2) . If you do, please let me know so I can tell you what to do in this case. About the spherical symmetry question, the integrand is a function of k2 = k02 - k2 i.e. it depents only of the magnitude of k and not it's direction.

4. Nov 28, 2012

andrien

d4k=dk0d3k,also k2=k02-k2

5. Nov 28, 2012

tom.stoer

again too many k's w/o index ;-)

k2=k02-k2 → 2k2=k02

but this is definitly NOT what you mean ...

6. Nov 28, 2012

andrien

ahh yes,so
k2=k02-$k^2$

7. Nov 28, 2012

tom.stoer

but this is what cosmic dust says: you cannot do that in general w/o assuming that there is a constraint; and in the above mentioned integral there isn't such a constraint, so the mass-shell condition does not apply.

8. Nov 28, 2012

andrien

one can use it by simply giving a negative imaginary part to mass,if you wish.

9. Nov 28, 2012

Sekonda

What I wanted to do was just take this out of 4 vector notation, how would I express this with a scalar part and a vector part.

I think if k is a four vector then:

$$d^4k=dEd^3\mathbf{p} \: ,\: k=(E,\mathbf{p})$$,

is this right? If so I just want to change the derivation I stated intially but using the integration measure on the RHS above this line.

Thanks,
SK

10. Nov 28, 2012

tom.stoer

yes, that's correct