# Derivative of a function of a lorentz scalar

1. Sep 24, 2011

### 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:

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

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?

2. Sep 24, 2011

### 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.

3. Sep 24, 2011

### 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:

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

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 $\frac{1}{8} \frac{\partial}{\partial p_\nu}\frac{\partial}{\partial p^\nu}\Sigma(p^2)|_{p^2=\mu^2}$, 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).

4. Sep 24, 2011

### 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.