Timeline for When can I set $d=4$ in dimensional regularization?
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| Oct 21, 2019 at 0:12 | comment | added | Pxx | I see. If you want the divergence, you may take some shortcuts, but if you want the finite part you must set $\omega=2$ as late as possible. Thanks again, this discussion has been amazingly helpful, and I hope that it will be useful for others! | |
| Oct 20, 2019 at 22:29 | comment | added | Seth Whitsitt | Oh yes, if you want the finite part of the integral you absolutely need to obtain the function $f(x_1,x_2,x_3,x_4)$, and you shouldn't set $\omega = 2$ right away. These finite parts contribute the important logarithmic dependence on the external coordinates. I was assuming you were only interested in extracting the divergence (which is all you need to get beta functions for example), but if the whole analytic dependence of this diagram on the coordinates is your interest, you need to keep the terms I called $+ \, \mathrm{finite}$ in my answer. | |
| Oct 20, 2019 at 21:06 | comment | added | Pxx | Sorry to bother you again with that topic, but I am still not totally convinced that it is okay to set $\omega=2$, because of the following reason: imagine that I expand the integrand of $(1)$ at $\omega=2$, then I get $I\propto\epsilon^{-1}\int \lbrace x_{13}^{-2}x_{24}^{-2}+\epsilon f(x_1,x_2,x_3,x_4) + \mathcal{O}(\epsilon^2) \rbrace$. The first term is the same as if I had set $\omega=2$ earlier, but what about the 2nd term? The $\epsilon$'s cancel and I am left with a rather strange additional finite part. On the other hand, it seems strange that setting $\omega=2$ would be wrong... | |
| Oct 11, 2019 at 6:49 | comment | added | Seth Whitsitt | In any case such as this, you will need to be careful to properly evaluate the limit $\lim_{\omega \rightarrow 2} \Gamma(\omega - 2) g(\omega)$. If $g(\omega = 2) = 0$ results in an indeterminate form, you need to know its analytic properties in order to properly perform the limit and determine whether you have a pole or not. For example, if $g(\omega) = \log(3 - \omega)$, then $\lim_{\omega \rightarrow 2} \Gamma(\omega - 2) g(\omega) = 1$ and there is no pole. | |
| Oct 10, 2019 at 19:36 | comment | added | Pxx | On the other hand, if that is true, you could say $\Gamma(\omega-2) \cdot (\text{something finite} + 0)$ is indeterminate, since the second term is indeterminate, and obviously this is non-sensical. Do you know of an obvious resolution to this apparent paradox? | |
| Oct 10, 2019 at 19:36 | comment | added | Pxx | Actually, I just thought about another case, in which I think that the $\omega=2$ substitution is particularly unwise: what if, after I extracted a divergence in the form of a factor $\Gamma(\omega-2)$, I have an integral that gives $0$ at $\omega=2$? As far as I know, I cannot say that the whole thing is zero, since $0\cdot\infty$ is an indeterminate form. If I encounter that case, does that mean that this is also an instance in which setting $\omega=2$ should not be done? | |
| Oct 10, 2019 at 3:30 | history | edited | Seth Whitsitt | CC BY-SA 4.0 |
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| Oct 9, 2019 at 22:32 | comment | added | Pxx | Yes I see, that’s a good point! Thanks again! | |
| Oct 9, 2019 at 20:26 | comment | added | Seth Whitsitt | Yes, if it was really a double-pole in $(\omega-2)$, then you would find a divergence if you set $\omega = 2$ too early, so you're correct that the substitution can be used to check the total divergence. My point was just that one also needs the residue of the order-1 pole in $(\omega-2)$. So for example, if you only set $\omega=2$ in one part of the integrand and found a double pole due to other parts, you would miss part of the residue of $(\omega-2)$ which comes from expanding the part of the integrand where you set $\omega=2$. | |
| Oct 9, 2019 at 19:42 | comment | added | Pxx | I mean, if I were to set $\omega=2$ wrongfully, wouldn’t anyway a divergence show up while integrating? I would think that, in general, one can always set $\omega=2$ and see if the integral converges that way. And if not, do it again with dimensional regularization. Does that make sense? | |
| Oct 9, 2019 at 19:10 | vote | accept | Pxx | ||
| Oct 9, 2019 at 19:10 | comment | added | Pxx | Dear Seth, thank you so much for a very nice answer. I have added details about the notation in the body of the question, but I think I am fine setting $\omega=2$ in the remaining integral. | |
| Oct 8, 2019 at 6:14 | history | edited | Seth Whitsitt | CC BY-SA 4.0 |
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| Oct 8, 2019 at 3:55 | review | First posts | |||
| Oct 8, 2019 at 7:01 | |||||
| Oct 8, 2019 at 3:53 | history | answered | Seth Whitsitt | CC BY-SA 4.0 |