Leptonic-Hadronic Tensor Multiplication Proof with Rest and Final Target Momenta

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SUMMARY

The forum discussion focuses on proving the leptonic-hadronic tensor multiplication, specifically with the momenta defined as ##p^\mu=(M,\textbf{0})## for rest target momentum and ##p'^\mu=(E',\textbf{p}')## for final target momentum. The derived equation is ##L_{\mu\nu}T^{\mu\nu}=8\Big(2(k\cdot p)(k'\cdot p)+\frac{q^2}{2M^2}\Big)##, utilizing the hadronic tensor defined as ##T^{\mu\nu}=(p+p')^\mu(p+p')^\nu## and the leptonic tensor as ##L_{\mu\nu}=2\Big(k'_\mu k_\nu+k'_\nu k_\mu + \frac{1}{2}q^2g_{\mu\nu}\Big)##. The author critiques a potential misprint in Greiner's work and confirms their result through relativistic limits.

PREREQUISITES
  • Understanding of leptonic and hadronic tensors in particle physics.
  • Familiarity with four-momentum notation and relativistic kinematics.
  • Knowledge of tensor contraction and its applications in physics.
  • Proficiency in manipulating equations involving Lorentz invariance.
NEXT STEPS
  • Study the derivation of the hadronic tensor in detail.
  • Explore the implications of relativistic limits on tensor equations.
  • Learn about the role of the leptonic tensor in particle interactions.
  • Investigate potential errors in established physics texts, such as Greiner's work.
USEFUL FOR

Physicists, graduate students in particle physics, and researchers focusing on tensor analysis in high-energy physics will benefit from this discussion.

Muh. Fauzi M.
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Homework Statement


Proof the leptonic-hadronic tensor multiplication, with ##p^\mu=(M,\textbf{0})## and ##p'^\mu=(E',\textbf{p}')## is rest target and final target momentum respectively, and ##k^\mu=(\omega,\textbf{k})##, ##k'\mu=(\omega'
,\textbf{k}')## is momenta of incoming and outgoing electron, hence we get

$$L_{\mu\nu}T^{\mu\nu}=8\Big(2(k\cdot p)(k'\cdot p)+\frac{q^2}{2M^2}\Big)$$

Homework Equations



The hadronic tensor,
$$T^{\mu\nu}=(p+p')^\mu(p+p')^\nu$$

The leptonic tensor,
$$L_{\mu\nu}=2\Big(k'_\mu k\nu+k'_\nu k_\mu + \frac{1}{2}q^2g_{\mu\nu}\Big)$$.

The Attempt at a Solution



I try something like this

$$ L_{\mu\nu}T^{\mu\nu} = 2\Big(k'_\mu k_\nu+k'_\nu k_\mu + \frac{1}{2}q^2g_{\mu\nu}\Big) (p+p')^\mu(p+p')^\nu $$

Expanding the second term in the rhs

$$ L_{\mu\nu}T^{\mu\nu} = 2\Big(k'_\mu k_\nu+k'_\nu k_\mu + \frac{1}{2}q^2g_{\mu\nu}\Big) (p^\mu p^\nu + p^\mu p'^\nu + p'^\mu p^\nu + p'^\mu p'^\nu) $$

hence,

$$ L_{\mu\nu}T^{\mu\nu} = 2 \Big[ (k' \cdot p)(k \cdot p) + 2 (k'\cdot p)(k \cdot p') + (k' \cdot p')(k \cdot p') + (k' \cdot p)(k \cdot p) + 2 (k'\cdot p')(k \cdot p) + (k' \cdot p')(k \cdot p') + \frac{1}{2}q^2 (p^\mu p^\nu + p^\mu p'^\nu + p'^\mu p^\nu + p'^\mu p'^\nu) \Big] $$
 
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I think Greiner missprint the result. I've finally got the contraction result:

$$ L_{\mu\nu}=8\Big(2(k\cdot p)(k'\dot p)+\frac{1}{2}q^2 M^2\Big) ,$$

and if I use the relativistic limit, my result is confirmed.
 

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