Derivative of a function of a lorentz scalar

[RedX]

This is probably a dumb question, but I have a book that claims that if you have a function of the momentum squared, f(p²), that:

$$\frac{d}{dp^2}f=\frac{1}{2d}\frac{\partial }{\partial p_\mu} <br /> \frac{\partial }{\partial p^\mu}f$$

where the d in the denominator is the number of spacetime dimensions, so for 4-space the numerical factor would be 1/8.

But this seems to only be true if your function is the identity $f(p^2)=p^2$, and doesn't hold for all functions f(p^2).

So is the book wrong?

 

[Ben Niehoff]

That is a bit of a strange formula, since it sets a first derivative equal to a second derivative. I suppose it's technically a differential equation, in which case it would not be surprising to find it has only a few linearly-independent solutions.

Using the chain rule, one can write

$$\frac{d}{d(p^2)} = \Big( \frac{\partial (p^2)}{\partial p^\mu} \Big)^{-1} \frac{\partial}{\partial p^\mu} = \Big( \frac{\partial (p^\nu p_\nu)}{\partial p^\mu} \Big)^{-1} \frac{\partial}{\partial p^\mu} = \frac{1}{2 p_\mu} \frac{\partial}{\partial p^\mu}$$

which does not agree with the formula you gave.

 

[RedX]

Normally I would just dismiss the formula, but I found it in two different sources (both particle physics sources though). One book talked about the vacuum bubble expansion of the integral:

$$\int \frac{1}{[k^2-m^2][(k-p)^2-m^2]}=\int \frac{1}{[k^2-m^2]^2}<br /> -\int \frac{p^2}{[k^2-m^2]^3}<br /> +\frac{4}{d}\int \frac{k^2p^2}{[k^2-m^2]^4}+O[(p^2)^2]$$

where the integrals are over k, and p is an external momentum. I can only get the RHS assuming that the formula for the derivative works, where I just Taylor expand the LHS about p^2=0 (I just set p=0 after taking derivatives of the LHS, since p=0 satisfies p^2=0).

The other book was a well regarded book long ago, Gauge Theory of Elementary Particle Physics by Cheng and Li, where they write the self-energy of a scalar in phi^4 theory as:

$$\Sigma(p^2)=\Sigma(\mu^2)+\Sigma'(\mu^2)(p^2-\mu^2)+\tilde{\Sigma}(p^2)$$

Cheng and Li claim that the first term on the RHS is quadratically divergent, and that the second term is logarithmically divergent rather than linear divergent, because

$\Sigma'(\mu^2)$ can be written in the form $\frac{1}{8}<br /> \frac{\partial}{\partial p_\nu}\frac{\partial}{\partial p^\nu}\Sigma(p^2)|_{p^2=\mu^2}$, and each differentiation with respect to external momentum $p_\mu$ reduces the degree of divergence of the integral by one (so one derivative takes it from quadratic divergence to linear divergence, and the other from linear to logarithmic).

 

[Ben Niehoff]

The

$$\left. \phantom{\frac12} \right|_{p^2 = \mu^2}$$

part might be important. That's all I can think of at the moment.