Is Local the Same as Isotropic in Physics?

[Niles]

Hi

Say I have two expressions of the form

$$F(r, t) = \int{dr'\,dt'\,\,x(r,r',t,t')g(r',t')}$$

and

$$F'(r, t) = \int{dt'\,\,x'(r,t,t')g'(r, t')}$$

It is clear that F' is local in space, whereas F is non-local in space. Is it correct of me to say that F' describes an isotropic object? I.e., does isotropic = translational invariance?


Niles.

 

[mathman]

You should define the various symbols to make it a physics question. Right now they are mathematical expressions.

 

[Niles]

Good point, thanks. Say "x" denotes the susceptibility and "g" the electric field.

 

[mathman]

How about r, r', t, and t'.

 

[pmacias]

Hello Niles,

Thank you for your question. In physics, isotropy refers to a property of a system that remains unchanged under rotations or translations. In your example, F' is indeed local in space and therefore it is also isotropic, as it remains unchanged under translations. However, it is important to note that being isotropic does not necessarily imply translational invariance. Translational invariance refers specifically to the property of a system that remains unchanged under translations in space, whereas isotropy includes both rotations and translations.

In summary, F' can be described as both local and isotropic, but it is not necessarily translational invariant. I hope this helps clarify the relationship between local and isotropic in your example. Keep questioning and exploring the properties of your system to deepen your understanding of physics. Keep up the good work!