Transforming a 4-vector integration measure

[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

 

[tom.stoer]

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

 

[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.

 

[andrien]

d⁴k=dk⁰d³k,also k²=k₀²-k²

 

[tom.stoer]

also k²=k₀²-k²

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

k²=k₀²-k² → 2k²=k₀²

but this is definitely NOT what you mean ...

 

[andrien]

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

k²=k₀²-k² → 2k²=k₀²

but this is definitely NOT what you mean ...

ahh yes,so
k²=k₀²-$k^2$

 

[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.

 

[andrien]

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

 

[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

 

[tom.stoer]

yes, that's correct