I Solve Mystery Equation - Get Insight into Relevance & Contents of Brackets

[emrock]

TL;DR Summary

Looking for context around an equation

I am hoping that people here might be able to provide insight into
what context/s this equation might be relevant, particularly the contents of the brackets.

I am aware it is a strange request, related to puzzle solving,
but perhaps someone can help guide me in an interesting direction.

 

[davenn]

Summary:: Looking for context around an equation ...

So give us all some context ... where did you see this ?

 

[Delta2]

Usually in equations that arise from problems in physics we have derivatives up to second, but here a third derivative w.r.t to $r$ will appear because the operator of gradient contains the first derivative w.r.t $r$.
So this equation is most likely from a mathematical problem.

However it could be that $\phi_E$ is the scalar potential of the electric field (commonly known as voltage) and the electric field is then $\vec{E}=\nabla\phi_E$ (this last equation holds only in the electrostatic case ) , so this equation takes the second derivative $\frac{d^2}{dr^2}$(w.r.t $r$) of the electric field $\vec{E}$ inside some dielectric material. The greek letter epsilon $\epsilon$ that appears next to it could be the electric permittivity of the material.

 

[Vanadium 50]

This is not an equation. An equation has something on both sides of the equals sign.

 

[Delta2]

This is not an equation. An equation has something on both sides of the equals sign.

Strictly speaking you are right, it is not an equation, it is the LHS of an equation.

 

[kuruman]

This is not an equation. An equation has something on both sides of the equals sign.

That is exactly the mystery. If the right hand side of the equation were known, the mystery would cease to exist.

 

[jedishrfu]

What if one assumes the right hand side is zero?

 

[omega_minus]

Could it be the spatial component of a wave equation in terms of an electric field? Ignoring the epsilon (permittivity I guess) if you solve Maxwell’s equations you can get a wave equation which (for the one dimensional case) is the second derivative with respect to space of the E field minus the second derivative with respect to time of the E field. The time component is multiplied the reciprocal the square of the wave speed when the spatial component has a coefficient of unity. This is equal to zero in the case of empty space with no charges present. I’ve never seen it solved for the D field as opposed to the E field but the epsilon in this case looks like it would make it so. This of course would make it not in empty space since it’s not epsilon naught. So if I had to put some thing on the right hand side of that equation it would be one over the square of the speed of light times the second time derivative of a displacement current. This may need a constant added to it if it is not an empty space due to possible charges present.

 

[vanhees71]

The problem is that the right-hand side of the equation is not clearly defined. Any equation making sense for vector fields must be formulated in terms of covariant objects, i.e., for spatial derivatives it must somehow be built with $\vec{\nabla}$.

 

[A.T.]

I am hoping that people here might be able to provide insight into
what context/s this equation might be relevant,

Designing a nerdy t-shirt?

 

[omega_minus]

The problem is that the right-hand side of the equation is not clearly defined. Any equation making sense for vector fields must be formulated in terms of covariant objects, i.e., for spatial derivatives it must somehow be built with $\vec{\nabla}$.

Thanks for the information @vanhees71. In EE we don’t discuss Maxwell’s equations in tensor notation (unfortunately) so I’ve not been formally educated on the meaning of “covariant”. Our notation of the spatial derivative part is “del squared” and is taken as the 2nd partial derivative of each vector component along an axis in Cartesian coordinates. (Occasionally we’d see a d’Alembertian for the whole wave equation). Could you help me understand what the difference is here? It sounds like something I’d like to know more about.

 

[jedishrfu]

d'Alembertian is the four dimensional analogue to the Laplacian ie it has a time derivative component added.

https://en.wikipedia.org/wiki/D'Alembert_operator

Here's something on covarience and contravarience notice where the basis vectors come from:

https://en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors

the tangent lines for vectors vs the normals to the tangent planes for covectors.

 

[Vanadium 50]

It may be worth pointing out that this was a drive-by posting. The OP hasn't been here since he asked his question.

 

[omega_minus]

Thanks @jedishrfu for the info. The d’Alembertian part I’m familiar with, I only mentioned it because that’s as “fancy” as we got in my grad school. But the tensor links are helpful to review for me since in orthogonal coordinates covariant and contravariant are the same. But I’m always left confused by how one knows what to assign a given vector quantity. For example why did @vanhees71 say “covariant” objects and not contravariant ones? Are E fields always covariant? Thanks.