- #1
CAF123
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Consider a d dimensional integral of the form, $$\int \frac{d^d \ell}{(2\pi)^d} \frac{\ell^{\sigma} \ell^{\mu}}{D}\,\,\,\text{and}\,\,\, \int \frac{d^d \ell}{(2\pi)^d} \frac{\ell^{\sigma}}{D}$$ where ##D## is a product of several propagators. One can reduce this to a sum of scalar integrals by writing e.g the 1 tensor ##\ell^{\mu}## in terms of two linearly independent vectors p and n and the 2 tensor ##\ell^{\sigma} \ell^{\mu}## as a linear combination of all possible lorentz structure with two indices in the basis {p,n} (so we have e.g ##g^{\sigma \mu}, p^{\mu}p^{\sigma}, p^{\sigma}n^{\mu}, ##etc...
My question is simply, why bother with this 2 tensor decomposition if I can perhaps simply rewrite ##\ell^{\sigma} \ell^{\mu}## as a multiplication of two 1 tensors? This avoids having to deal with a 2 tensor decomposition and just using the 1 tensor decomposition repetitively?
Explicitly, can I not write ##\ell^{\sigma} \ell^{\mu} = (p^{\sigma} (\ell \cdot n) + n^{\sigma} (\ell \cdot p)) ( \sigma \rightarrow \mu)## thereby avoiding the use of the 2 tensor decomposition and only dealing with the 1 tensor decomposition ##p (\ell \cdot n) + n(\ell \cdot p) ? ##
Thanks!
My question is simply, why bother with this 2 tensor decomposition if I can perhaps simply rewrite ##\ell^{\sigma} \ell^{\mu}## as a multiplication of two 1 tensors? This avoids having to deal with a 2 tensor decomposition and just using the 1 tensor decomposition repetitively?
Explicitly, can I not write ##\ell^{\sigma} \ell^{\mu} = (p^{\sigma} (\ell \cdot n) + n^{\sigma} (\ell \cdot p)) ( \sigma \rightarrow \mu)## thereby avoiding the use of the 2 tensor decomposition and only dealing with the 1 tensor decomposition ##p (\ell \cdot n) + n(\ell \cdot p) ? ##
Thanks!