- #1
Chenkb
- 41
- 1
For two-body decay, in the center of mass frame, final particle distribution is,
$$
W^*(\cos\theta^*,\phi^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)
$$
We have the normalization relation , ##\int W^*(\cos\theta^*,\phi^*)d\cos\theta^* d\phi^*=1##.
And we also know that in CM frame ##p^*## is a constant, say, ##p^*=C^*##.
So, the final particle momentum distribution can be write as(I'm not sure),
$$
W^*(\cos\theta^*,\phi^*,p^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)\delta(p^*-C^*)
$$
If the above momentum distribution in CM frame is right,
then what does it look like in the lab frame,
$$W(\cos\theta,\phi,p)=?$$
assume that the mother particle moves with velocity ##\beta## along ##z## axis.
I know it is just a Lorentz transformation, but how to handle this, especially the ##\delta##-function.
Best Regards!
$$
W^*(\cos\theta^*,\phi^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)
$$
We have the normalization relation , ##\int W^*(\cos\theta^*,\phi^*)d\cos\theta^* d\phi^*=1##.
And we also know that in CM frame ##p^*## is a constant, say, ##p^*=C^*##.
So, the final particle momentum distribution can be write as(I'm not sure),
$$
W^*(\cos\theta^*,\phi^*,p^*) = \frac{1}{4\pi}(1+\alpha\cos\theta^*)\delta(p^*-C^*)
$$
If the above momentum distribution in CM frame is right,
then what does it look like in the lab frame,
$$W(\cos\theta,\phi,p)=?$$
assume that the mother particle moves with velocity ##\beta## along ##z## axis.
I know it is just a Lorentz transformation, but how to handle this, especially the ##\delta##-function.
Best Regards!