Is the S-matrix at high energies affected by time dilation?

[geoduck]

The S-matrix can be written as the sum of the Feynman diagrams, divided by a factor of 1/sqrt[E] for each particle, where E is the particle's energy.

Does this mean at large energies, the probability amplitude to scatter is unlikely?

But how can such a statement be made when no physics is involved? There is no mention of the type of interactions, but it doesn't seem to matter: 1/sqrt[E] factor (for each external line) will make the likely-hood of scattering at high energies smaller than at a low energies.

 

[kof9595995]

I don't remember any such rule. Shouldn't it be the field strength renormalization constant that associates with each external line?

 

[torquil]

The S-matrix can be written as the sum of the Feynman diagrams, divided by a factor of 1/sqrt[E] for each particle, where E is the particle's energy.

Does this mean at large energies, the probability amplitude to scatter is unlikely?

But how can such a statement be made when no physics is involved? There is no mention of the type of interactions, but it doesn't seem to matter: 1/sqrt[E] factor (for each external line) will make the likely-hood of scattering at high energies smaller than at a low energies.

You are probably thinking about the sqrt(E) factors that arise in normalization of the field operators, e.g. as used in Landau/Lifschitz Vol.4?

Anyway, you seem to assume that the Feynman diagrams are independent of energy, which is not correct.

 

[geoduck]

As an example, the probability rate can be written as:

[tex]\frac{(2 \pi)^4 \delta(P_f-P_i)V |T|^2}{(\Pi_i 2E_iV)( \Pi_f 2E_fV)} [/tex]

where V is the volume, i stands for initial particles, and f stands for final particles.

As an example, take 2 particles colliding in phi^4 theory. Then |T|^2=\lambda^2 (to first order), where lambda is the coupling constant.

The formula for the probability rate seems to say that the greater the product of the energies of the initial colliding particles, the lower the transition rate (because of the E_i in the denominator).

But |T|^2 contains all the physics.

So I'm wondering if this division by the energy is a kinematic statement. Maybe that the more energy the initial particles have, the more possibilities that the final states can be, hence you need to lower all the probabilities so that they sum to 1.

It just seems weird to me that you can have factors of energy on the outside of |T|^2.

 

[torquil]

In Peskin/Schröder, such energy dependence is called "phase space factors", and they dictate the energy dependence of the cross-section at energies far above any particle masses in the theory, as can be seen by dimensional analysis. Check out the discussion below eq.(5.14) about QED.

But energy dependence can be much more complicated, e.g. if additional virtual particles suddenly come into play above some threshold energy, since that drastically changes T.

 

[geoduck]

In Peskin/Schröder, such energy dependence is called "phase space factors", and they dictate the energy dependence of the cross-section at energies far above any particle masses in the theory, as can be seen by dimensional analysis. Check out the discussion below eq.(5.14) about QED.

But energy dependence can be much more complicated, e.g. if additional virtual particles suddenly come into play above some threshold energy, since that drastically changes T.

Okay, I'll check it out. Thanks.

You're absolutely right about virtual particles making |T|^2 depend on energy. I set |T|^2 equal to the coupling constant, but the coupling constant itself depends on energy through the renormalization group equation.

I was hoping to try to argue the factors of E through an example: suppose you have two massless identical particles with equal and opposite momentum colliding, then they will leave the collision in equal and opposite directions. If |T|^2 is isotropic and doesn't depend on energy, then each angle that they leave is equally likely: 0-360 degrees. But the number of momentum states between 0 and 1 degrees is much larger if the two particles had a high energy. The number of momentum states between 0 and 1 degree is the arclength E*(1 degree). So if you divide by E, you lose that proportionality. I have no idea if any of this is sound, but I'll look at Peskin and Schroeder.

 

[kof9595995]

Hmm, I see, so you are talking about cross-section not S-matrix.

 

[torquil]

Another small observation: Weinberg (in QFT1) discusses these energy dependencies. E.g. in the case of a single incoming particle, we are talking about a particle decay rate. He comments that the corresponding incoming particle energy factor in the decay cross-section explains the increased particle lifetime due to time dilation when the particle is moving fast.

 

[geoduck]

Hmm, I see, so you are talking about cross-section not S-matrix.

The cross-section is interesting too, but I tried to keep it restricted to the S-matrix for simplicity. So [tex]
(f|S|i>)(<f|S|i>)^*=\frac{(2 \pi)^4 \delta(P_f-P_i)V |T|^2}{(\Pi_i 2E_iV)( \Pi_f 2E_fV)}*time [/tex]
I divided by time to get the probability rate. So it's really the S-matrix squared.

Another small observation: Weinberg (in QFT1) discusses these energy dependencies. E.g. in the case of a single incoming particle, we are talking about a particle decay rate. He comments that the corresponding incoming particle energy factor in the decay cross-section explains the increased particle lifetime due to time dilation when the particle is moving fast.

So maybe this can apply to two incoming particles? It's not so clear cut as one particle, but the faster the two particles, the slower the reactions, hence the slower the probability rate? Is this a problem for experimentalists who collide two particles, that the higher the energy, you have to wait longer because either 1) the probability rate gets lower 2) the reaction is slower via time dilation. 1) and 2) are related, but 2) is the better explanation, because it is general for all types of reactions.