Timeline for Product of delta distributions
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2022 at 11:02 | comment | added | Andrew Be | Okay, I see, now your answer is clear, but I need to understand the specific case of a product between two deltas: one of a surface and the other of a curve, in $R^3$. My idea was to start from a discrete intersection, then we have the scalar product in the point of the infinitesimal vector tangent of the curve and normal vector of the surfaces. But following the idea the integral should be equal to $1 \cdot \cos \theta$ with $\theta$ the angle between the two vectors. For this reason, I don't understand the $1/(\sin \theta)$ | |
| Oct 7, 2022 at 19:02 | comment | added | Arturo don Juan | @AndrewBe I believe it's because the document you provided defines delta-functions differently according to the dimensionality of the submanifold, so that the total integral over that submanifold gives you $1$, rather than the volume which for $p<N$-dimensional submanifolds have zero $N$-volume. Therefore the formulas I wrote in my answer, as well as the formulas you wrote in your question, apply to $3$-dimensional submanifolds living in a $3$-dimensional space. The formula showing $1/\sin (\theta)$ is for the product of delta functions lower-dimensional submanifolds (i.e. lines or surfaces). | |
| Oct 7, 2022 at 6:16 | comment | added | Andrew Be | Thank you for your interpretation, but as u can see in the referente that I have put, in the context of a single intersection the result is equal to $1/\sin\theta$ The angle between the tangent vectores, so it is not the volume | |
| Oct 6, 2022 at 18:58 | history | made wiki | Post Made Community Wiki by Qmechanic♦ | ||
| Oct 6, 2022 at 18:57 | history | answered | Arturo don Juan | CC BY-SA 4.0 |