Exploring the Strange Result of Equation 8.9 in Quantum Field Theory

[shadi_s10]

Dear all,
I am taking 'field theory' course this semster and I am reading 'quantum field theory' by mandl and shaw.
In chapter 8, equation (8.9) we see:

E_1 E_2 v_rel=〖[(p_1 p_2 )^2-m_1^2 m_2^2]〗^(1/2)

and we know that as p is a four vector:
p=(E,P)=m
so p_1 p_2 = m_1 m_2

!
Isn't it strange?!
I think the right hand side of eq(8.9)should be zero then!

what am I donig wrong?!

 

[vela]

and we know that as p is a four vector:
p=(E,P)=m

That doesn't make any sense. The mass m is a scalar. How can it be equal to p, a four-vector?

 

[bp_psy]

Dear all,
I am taking 'field theory' course this semster and I am reading 'quantum field theory' by mandl and shaw.
In chapter 8, equation (8.9) we see:

E_1 E_2 v_rel=〖[(p_1 p_2 )^2-m_1^2 m_2^2]〗^(1/2)

and we know that as p is a four vector:
p=(E,P)=m
so p_1 p_2 = m_1 m_2

!
Isn't it strange?!
I think the right hand side of eq(8.9)should be zero then!

what am I donig wrong?!

The norm of the 4-vector is equal to m not the vector itself.

 

[shadi_s10]

But if you take a look at field theory by guidry we have the exact same term!
I mean:
p= (E,P) = m
Because as you know in relativity we have:
E^2+P^2=m^2
and this is the exact result from p= m
!

 

[vela]

You must be leaving out typographical information because what you are writing simply doesn't make sense. It's akin to saying the vector (2,1,3) is equal to the number 6. It just doesn't work from a mathematical perspective.

 

[bp_psy]

But if you take a look at field theory by guidry we have the exact same term!
I mean:
p= (E,P) = m
Because as you know in relativity we have:
E^2+P^2=m^2
and this is the exact result from p= m
!

It is actually E^2-P^2=m^2

$$\vec{p}=(E,\vec{P})$$

$$\vec{p}\cdot\vec{p}=E^2-\vec{P}\cdot\vec{P}=m^2$$
|p|=m
I do not have your book but I know the notation you are using.It does not bother pointing the difference between the vector and the norm.You should be able to figure out what is he is talking about from the context.