Solving the Callan-Symanzik Equation

  • Thread starter latentcorpse
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In summary, the person is trying to solve for the massless case in a paper from 2009 and is stuck on two parts. They are unsure of what the Taylor expansion for the integral is and are also struggling to get Feynman rules from it.
  • #36
fzero said:
The energy of a massive particle is never zero!

Nevertheless, you can use conservation of energy and momentum to show that [tex]E_1 = E_{1'}[/tex], so the rest of your calculation works out.



Yex.

So I have been trying another CS question. I am asked at the end of q3 in this paper:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2006/Paper49.pdf [Broken]
to explain why [itex]\mu \frac{dF}{d \mu}=0[/itex]

For this I wrote that we can substitute [itex]h=\mu^{\frac{1}{2}\epsilon} Z^{\frac{1}{2}}h_0[/itex] (where a subscript 0 denotes a bare quantity) to rewrite the definition of F all in terms of bare quantities

i.e. [itex]e^{iF(\lambda,m^2,h;\mu)}=\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}[/itex]
[itex]\Rightarrow {iF(\lambda,m^2,h;\mu)}=\ln{\frac{1}{N_0} \int d[\phi] e^{iS_0[\phi]+h_0 \int d^dx \phi_0}}[/itex]
Now since bare quantities are by construction independent of the RG scale [itex]\mu[/itex], the RHS of this equality will be [itex]\mu[/itex] independent and therefore by equality so will the LHS.
Therefore [itex]\frac{dF}{d \mu}=0 \Rightarrow \mu \frac{dF}{d \mu}=0[/itex]

Does that look ok?

What about the next part though? We need to derive the CS equation. At first glance it looks like you just bash out the same chain rule procedure as we used previously but it turns out to be a bit more complicated. Previously, we defined the [itex]\gamma_\phi[/itex] term to go with the [itex]\frac{\partial}{\partial \sqrt{Z}}[/itex] term so that when we acted it on [itex]G_n^B=Z^{n/2}G_n[/itex] we pulled down a factor of n by the chain rule. However, this was because we were able to write the bare green's function in terms of the renormalised greens function with that factor of [itex]Z^{n/2}[/itex] to operate on.

But now we have the bare quantity F (which we aren't able to write in terms of any renormalised quantity) so I don't understand how we will get the h next to the [itex]\gamma_h[/itex] term?

Do we just define [itex]\gamma_h[/itex] differently from [itex]\gamma_\phi[/itex]? I guess I could just write [itex]\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}[/itex]. Does this work ok?

Thanks again!
 
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  • #37
latentcorpse said:
Do we just define [itex]\gamma_h[/itex] differently from [itex]\gamma_\phi[/itex]? I guess I could just write [itex]\gamma_h=\frac{\mu}{h} \frac{dh}{d \mu}[/itex]. Does this work ok?

That's the definition they're using. It's very natural for anomalous dimensions to be related to logarithmic derivatives, since they naturally appear as factors like [tex]\mu^\gamma[/tex] in scaling relations.
 
<h2> 1. What is the Callan-Symanzik equation? </h2><p> The Callan-Symanzik equation is a mathematical equation used in quantum field theory to describe the evolution of a physical quantity as a function of a scale parameter. It is often used to study the behavior of physical systems at different energy scales. </p><h2> 2. Why is solving the Callan-Symanzik equation important? </h2><p> Solving the Callan-Symanzik equation allows us to understand how physical quantities change as a function of energy scale, which is crucial in studying the behavior of physical systems. It also helps us make predictions about the behavior of these systems at different scales. </p><h2> 3. How is the Callan-Symanzik equation solved? </h2><p> The Callan-Symanzik equation is typically solved using perturbation theory, which involves expanding the equation in a series of terms and solving for each term. Other methods, such as renormalization group techniques, can also be used to solve the equation. </p><h2> 4. What are some applications of the Callan-Symanzik equation? </h2><p> The Callan-Symanzik equation has many applications in theoretical physics, including in the study of phase transitions, critical phenomena, and the behavior of quantum field theories. It is also used in particle physics to study the behavior of fundamental particles and their interactions. </p><h2> 5. Are there any limitations to the Callan-Symanzik equation? </h2><p> While the Callan-Symanzik equation is a powerful tool in theoretical physics, it does have some limitations. It is most accurate for systems with weak interactions and small energy scales, and it becomes less accurate at higher energy scales. Additionally, the equation can be difficult to solve for complex systems with many interacting particles. </p>

1. What is the Callan-Symanzik equation?

The Callan-Symanzik equation is a mathematical equation used in quantum field theory to describe the evolution of a physical quantity as a function of a scale parameter. It is often used to study the behavior of physical systems at different energy scales.

2. Why is solving the Callan-Symanzik equation important?

Solving the Callan-Symanzik equation allows us to understand how physical quantities change as a function of energy scale, which is crucial in studying the behavior of physical systems. It also helps us make predictions about the behavior of these systems at different scales.

3. How is the Callan-Symanzik equation solved?

The Callan-Symanzik equation is typically solved using perturbation theory, which involves expanding the equation in a series of terms and solving for each term. Other methods, such as renormalization group techniques, can also be used to solve the equation.

4. What are some applications of the Callan-Symanzik equation?

The Callan-Symanzik equation has many applications in theoretical physics, including in the study of phase transitions, critical phenomena, and the behavior of quantum field theories. It is also used in particle physics to study the behavior of fundamental particles and their interactions.

5. Are there any limitations to the Callan-Symanzik equation?

While the Callan-Symanzik equation is a powerful tool in theoretical physics, it does have some limitations. It is most accurate for systems with weak interactions and small energy scales, and it becomes less accurate at higher energy scales. Additionally, the equation can be difficult to solve for complex systems with many interacting particles.

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