Mandelstam variables explained

In theoretical physics, the Mandelstam variables are numerical quantities that encode the energy, momentum, and angles of particles in a scattering process in a Lorentz-invariant fashion. They are used for scattering processes of two particles to two particles. The Mandelstam variables were first introduced by physicist Stanley Mandelstam in 1958.

If the Minkowski metric is chosen to be

diag(1,-1,-1,-1)

, the Mandelstam variables

s,t,u

are then defined by

s=(p1+p

2
2)

c2=(p3+p

2
4)

c2

t=(p1-p

2
3)

c2=(p4-p

2
2)

c2

u=(p1-p

2
4)

c2=(p3-p

2
2)

c2

,where p1 and p2 are the four-momenta of the incoming particles and p3 and p4 are the four-momenta of the outgoing particles.

s

is also known as the square of the center-of-mass energy (invariant mass) and

t

as the square of the four-momentum transfer.

Feynman diagrams

The letters s,t,u are also used in the terms s-channel (timelike channel), t-channel, and u-channel (both spacelike channels). These channels represent different Feynman diagrams or different possible scattering events where the interaction involves the exchange of an intermediate particle whose squared four-momentum equals s,t,u, respectively.

For example, the s-channel corresponds to the particles 1,2 joining into an intermediate particle that eventually splits into 3,4: The t-channel represents the process in which the particle 1 emits the intermediate particle and becomes the final particle 3, while the particle 2 absorbs the intermediate particle and becomes 4. The u-channel is the t-channel with the role of the particles 3,4 interchanged.

When evaluating a Feynman amplitude one often finds scalar products of the external four momenta. One can use the Mandelstam variables to simplify these:

p1p2=

2
s/c-
2
m
1
-
2
m
2
2

p1p3=

2
m+
2
m
3
-t/c2
1
2

p1p4=

2
m+
2
m
4
-u/c2
1
2

Where

mi

is the mass of the particle with corresponding momentum

pi

.

Sum

Note that

(s+t+u)/c4=

2
m
1

+

2
m
2

+

2
m
3

+

2
m
4
where mi is the mass of particle i.[1]

To prove this, we need to use two facts:

2
p
i

=

2
m
i

(1)

p1+p2=p3+p4

p1=-p2+p3+p4(2)

So, to begin,

s/c2=(p1+p

2
2)
2
=p
1

+

2
p
2

+2p1p2

t/c2=(p1-p

2
1

+

2
p
3

-2p1p3

u/c2=(p1-p

2
1

+

2
p
4

-2p1p4

Then adding the three while inserting squared masses leads to,

2
(s+t+u)/c
1

+

2
m
2

+

2
m
3

+

2
m
4

+2

2
p
1

+2p1p2-2p1p3-2p1p4

Then note that the last four terms add up to zero using conservation of four-momentum,

2

2
p
1

+2p1p2-2p1p3-2p1p4=2p1(p1+p2-p3-p4)=0

So finally,

(s+t+u)/c2=

2
m
1

+

2
m
2

+

2
m
3

+

2
m
4
.

Relativistic limit

In the relativistic limit, the momentum (speed) is large, so using the relativistic energy-momentum equation, the energy becomes essentially the momentum norm (e.g.

E2=pp+

2
{m
0}
becomes

E2pp

). The rest mass can also be neglected.

So for example,

2=(p
s/c
1+p
2+2
2

p1p22p1p2

because
2
p
1

=

2
m
1
and
2
p
2

=

2
m
2
.

Thus,

s/c2

2p1p2

2p3p4

t/c2

-2p1p3

-2p2p4

u/c2

-2p1p4

-2p3p2

See also

References

Notes and References

  1. Book: Griffiths, David . David J. Griffiths . 2008 . Introduction to Elementary Particles . 2nd . . 978-3-527-40601-2 . 113.