Why S, T and U Must Be Positive in Mandelstam Variables

[LAHLH]

Hi,

Srednicki says in Ch20, we must remember that the mandelstam variable s is positive. However s is defined as $$s=-(k_1+k_2)^2=(E_1+E_2)^2-(\vec{k_1+k_2}).(\vec{k_1+k_2})$$. Using metric (-+++). I can't quite see why this must be always positive? or why for that matter t and u are negative?

Also he says we get $$V_3(s)$$, from the general $$V_3(k_1,k_2, k_3)$$ by setting two of the three k's to -m^2, and the remaining one to -s. Does anyone know why these values? I would have imagined one would be $$sqrt(-s)=k_1+k_2$$ and the other two were just left as $$k_1$$ and $$k_2$$ for scattering in s channel kind of way.

Thanks again

 

[ansgar]

just work it out,

$$s=-(k_1+k_2)^2=(E_1+E_2)^2-(\vec{k_1+k_2}).(\vec{k_1+k_2})$$

write k_1 dot k_2 etc as functions of E and mass m.

$$s = m_1^2 + m_2^2 + 2( E_1E_2 - cos \theta \sqrt{E_1^2-m_1^2} \sqrt{E_2^2-m_2^2} \:\: )$$

then you'll see that s >= 0

 

[LAHLH]

Thanks for the reply ansgar. Is the expression you posted postivie just becase $$\sqrt{E_1^2-m_1^2} < \sqrt{E_1^2}=E_1$$ and sim for second factor. So even if $$cos(\theta)=1$$, the $$cos \theta \sqrt{E_1^2-m_1^2} \sqrt{E_2^2-m_2^2}$$ term is less than $$E_1E_2$$. Therefore the whole of s must be positive?

The way I convinced myself s was postivie before seeing you post, was looking at it in the CM frame, and using the fact s is a Lorentz scalar. In the CM frame $$s=(E_1+E_2)^2$$ which is obviously positive.

I still have been unable to convince myself t and u are negative however, either by switching to a conveinient frame or otherwise.

Thanks again for the help

 

[ansgar]

Thanks for the reply ansgar. Is the expression you posted postivie just becase $$\sqrt{E_1^2-m_1^2} < \sqrt{E_1^2}=E_1$$ and sim for second factor. So even if $$cos(\theta)=1$$, the $$cos \theta \sqrt{E_1^2-m_1^2} \sqrt{E_2^2-m_2^2}$$ term is less than $$E_1E_2$$. Therefore the whole of s must be positive?

The way I convinced myself s was postivie before seeing you post, was looking at it in the CM frame, and using the fact s is a Lorentz scalar. In the CM frame $$s=(E_1+E_2)^2$$ which is obviously positive.

I still have been unable to convince myself t and u are negative however, either by switching to a conveinient frame or otherwise.

Thanks again for the help

You should not struggle with this if u want to learn QFT ;)

Yes that is also a good way to see it :)

 

[LAHLH]

You should not struggle with this if u want to learn QFT ;)

Yes that is also a good way to see it :)

Yes, yes, I know it's trivial, just tripping myself up trying to see it exactly, but think I got there from multiple directions in the end. Having said that is there an analagous way to show t and u are negative? I've tried various frames I thought might be useful without anything that has convinced me.

 

[ansgar]

Yes, yes, I know it's trivial, just tripping myself up trying to see it exactly, but think I got there from multiple directions in the end. Having said that is there an analagous way to show t and u are negative? I've tried various frames I thought might be useful without anything that has convinced me.

show us what u have tried

 

[LAHLH]

Well for example $$t=-(k_1-k_3)^2$$

I naivley tried CM frame again at first:

$$t=-[(E_1-E_3, 2\vec{k_1})]^2=(E_1-E_3)^2-4\vec{k^{2}_1}$$

Then subbing in the usual Einstein energy momentum relation:

$$t=-[(E_1-E_3, 2\vec{k_1})]^2=(E_1-E_3)^2-4(E^{2}_1-m^{2}_1)$$

$$t=E^{2}_3+4m^{2}_1-2E_3E_1-3E^{2}_1$$

I can't think of a clever way to argue why this must always be negative right now

-----------------------------

I also tried working in the rest frame of one of the particles with a similar result.

 

[ansgar]

have u used what E_3 will be in the CM frame?

have u tried going to e.g. 3's rest frame?

also you can try with all m -> 0 and then add a "small" mass?

 

[Avodyne]

t and u can be positive if the mass is not zero.