Newtonian potential in Helmholtz decomposition

In summary, The conversation is about finding a divergenceless vector field using its curl, through the use of the Helmholtz decomposition. The dimension referred to in the Newtonian potential article is the number of spatial dimensions in which the field exists, and for our world, it is 3. The exponent on the absolute value of x for more than two dimensions is negative, which may seem counterintuitive in the context of the problem being worked on, but it is necessary for the Helmholtz decomposition. The negative exponent bothered the speaker because it could imply that the final field is increasing as distance increases, which contradicts the problem they are solving using Maxwell's equations. The speaker also mentions the limitations of learning advanced materials on the
  • #1
Savant13
85
1
I'm trying to find a divergenceless vector field based on its curl, and discovered that I could use a http://en.wikipedia.org/wiki/Helmholtz_decomposition" , and the article I found on this didn't make much sense to me.

First, can someone confirm that the dimension referred to in the Newtonian potential article is the number of spatial dimensions in which the field exists? If not, what is it?

Next, the exponent on what appears to be the absolute value of x for more than two dimensions can only be negative, which doesn't make sense to me in the context of the problem I am working on. Also, is that an absolute value, or some other notation?
 
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  • #2
The d is the spatial dimension, so for our world d=3.
2-3=-1 which means that G~1/|x| for 3 dimensions.
Why does that bother you.
Having G is trivial, doing the convolution is the whole problem.
You have found out that Wiki is not for learning a subject.
 
  • #3
I am aware of the limitations of learning advanced materials on the internet. However, until I start college next year, it is my only option.

The reason the negative exponent bothered me is because (I thought) it would mean that the final field is increasing as distance increases, which does not make sense in the problem I am doing (I am using Maxwell's equations)

The Helmholtz Decomposition article uses the same notation as the Newtonian Potential article uses for the kernel, as opposed to the convolution. Is this misleading? Do I need to do the convolution for the Helmholtz Decomposition?

If I do have to do a convolution, how will I do this on a three dimensional vector field?
 
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What is the Newtonian potential in Helmholtz decomposition?

The Newtonian potential in Helmholtz decomposition is a mathematical concept used in classical mechanics and electromagnetism. It describes the potential energy of a system of particles or charges in terms of their positions and interactions with each other.

How is the Newtonian potential related to the Helmholtz decomposition?

The Helmholtz decomposition is a mathematical technique used to decompose a vector field into its components. In the case of the Newtonian potential, it can be decomposed into a scalar potential and a vector potential, which represent the conservative and solenoidal parts of the field, respectively.

What is the significance of the Helmholtz decomposition in physics?

The Helmholtz decomposition is an important tool in many areas of physics, including classical mechanics, electromagnetism, and fluid dynamics. It allows for the simplification and analysis of complex vector fields, making it easier to study and understand physical systems.

What are some applications of the Helmholtz decomposition in real-world problems?

The Helmholtz decomposition has numerous applications in various fields, including fluid dynamics (e.g. understanding the flow of air around an airplane), electromagnetism (e.g. analyzing the behavior of magnetic fields), and quantum mechanics (e.g. solving the Schrödinger equation for certain potentials).

Are there any limitations to the use of the Helmholtz decomposition?

While the Helmholtz decomposition is a useful tool in many cases, it has some limitations. For example, it can only be applied to vector fields that are continuous and differentiable everywhere, which may not be the case in certain physical systems. Additionally, the decomposition may not be unique, meaning there could be multiple ways to decompose a given field.

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