By Dwayne McDuffie
Back to the longer term 2 was once the second one factor of the again to the longer term comedian sequence via Harvey Comics.
The Dinosaurs upward push again!
Perhaps the strangest event of all of them! As document sends the group again to the Cretaceous Period... the 3rd geologic part of the Mesozoic period and the tip of the Dinosaur Age!
In an attempt to avoid wasting earth from the striking destruction of a large meteor, document transforms historical past and retains the dinosaurs alive. "Ouch-a-ma-goucha!" as document may well say, this example now threatens to break all mankind...TODAY!
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Extra info for Back to the Future v2
200) 2M Now we are in trouble! Not only do we have to face the problem of singularities in G(x, x|A), but we also have to give a meaning to the derivative at this singular coincidence point. The ﬁrst guess is to introduce an appropriate point splitting regarding the two arguments of the Green’s function. But then, we have to keep in mind that G(x, x |A) is a gauge variant quantity while G(x, x|A) is not. In order to ensure gauge invariance, we have to replace the ordinary derivative with covariant derivatives.
90) The scalars v 2,⊥ simply become v 2,⊥ = ω 2 sin2 θ. 92) sin eBs sin eBs eBs cos νeBs νeBs sin νeBs cot eBs + N⊥ = − sin eBs sin eBs 2eBs(cos νeBs − cos eBs) . + sin3 eBs Here, we have arrived at the representation of the polarization tensor found by Tsai and Erber , who employed this expression to obtain refractive indices for a magnetic background over a wide range of frequency and ﬁeld strength. We shall conﬁne ourselves to the case of low-frequency photons, ω m, which allows us to approximate the function φ0 in the exponent by Π0 → m2 .
155) (−) R(s)µ ν = ˆ R(s) ≡ e2C s 1 0 0 0 0 cos 2eBs sin 2eBs 0 = 0 − sin 2eBs cos 2eBs 0 . 157) where ˆ I(s) I(s)µ ν = 2s 0 0 0 0 (sin 2eBs)/(eB) (1 − cos 2eBs)/(eB) 0 = 0 (cos 2eBs − 1)/(eB) (sin 2eBs)/(eB) 0 . 39) of Sect. 1. In the case of a purely magnetic ﬁeld, the expression reduces to x(0) (s)|x(0) (0) ≡ x(0) | U (0) (s) |x(0) eBs i ei(e/2)σF s . e. the constant-ﬁeld part of the problem. Our next intermediate step is characterized by ﬁnding the relations between the zeroth-order and exact quantities.