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Rustify QED #416
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Rustify QED #416
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1762846
start implementation of aem1
tgiani 61220f8
complete aem1.py
tgiani a40908b
unit tests
tgiani f0edfae
add struct for charge combinations
tgiani a8d72c0
some work on gamma_ns_qed
tgiani e8a3a8b
start as1aem1.rs
tgiani 136dbe8
gamma_qph as1aem1
tgiani ce11227
some work on as1aem1
tgiani 659fe31
addingv recursive harmonics to cache
tgiani 4007f0d
S1p2 and G3p2. Check that this is correct
tgiani 7c160f1
gamma_nsp
tgiani 37a65e8
rewrite ChargeCombination struct
tgiani fc5de81
missing entries in as1aem1
tgiani 2a911b1
adding tests
tgiani bad124a
some work on spacelike.rs and recasting from list to vec
tgiani d50e9c7
valence qed
tgiani b0f6722
some unit tests
tgiani e4c621b
use Vec<> only when dimension is not known, else use normal list
tgiani 81ece22
fix notation row/columns
tgiani 0def938
add unravel functions for qed case. To be checked
tgiani 9b58455
modifying rust_quad_ker_qcd, PyQuadKerQCDT and QuadQCDargs to include…
tgiani 588496a
add qed valence option in rust_quad_ker_qcd
tgiani 3452243
remove useless flag is_qed
tgiani 9335b97
extend QuadQCDargs to include arguments for c_quad_ker_qed
tgiani 3390aca
setting up input vectors for cb_quad_ker_qed
tgiani 6a3af19
cb_quad_ker_qed
tgiani f0a727d
fix modes for singlet QED and use == in if statements
tgiani 398c89a
small fix
tgiani ea44cee
cargo fmt
tgiani 5721326
forgot pre-commit
tgiani 83af836
updating patch file
tgiani 1af5cbf
Update crates/ekore/src/constants.rs
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I'm not sure we want to keep those - can we inline them?
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what do you mean? Could you please provide an example?
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instead of
we do
where$f = g_3(N+2)-g_3(N)$ is computed analytically beforehand
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I am afraid I don t understand. How do you compute f analytically beforehand? Do you mean to move
g_functions::g3(self.n + 2., self.get(K::S1p2))
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yes
Mathematica 🙃 let's try to use a short cut: @giacomomagni can you compute quickly this$f$ ? (also feel free to give your opinion if we should keep that cache entry or not)
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so I think
$$f = \frac{6 (n+1) (2 n+1) H_n+n \left(-\pi ^2 (n+1)^2-6 n\right)}{6 n^2 (n+1)^3}$$ $N=1$ , $N=2$
which I computed half with pen and paper and half Mathematica and checked for
to be more precise: for$g_3$ we could use the above equation and for the others we inline the expressions
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ok thank you! I ll first try to get things working as they are and then I will add this