Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If you have a field which value is just Gaussian noise plus a constant do you call it isotropic?

  • there is no preferred direction
  • however it is not "the same" in all directions if "the same" means "constant"

The question is about a definition of a word, thus obviously conflicting answers can arise. However what definition do you use or have seen used: "no preferred direction" or "constant wrt translation"?

share|improve this question
    
I have a hard time distinguishing between your two alternatives as it is not clear what you mean by constant. Isotropy is pretty well defined from it's Greek origin, i.e. equal in all directions. In a way that implies constant, so the property cannot be a function of the direction. Though it can still vary in time or distance. –  Alexander Jun 27 '12 at 16:30
    
How can something vary in distance and be isotropic? Central symmetry is not isotropy. And my question is whether you call something with no preferred direction isotropic even if it contains random fluctuations? –  Krastanov Jun 27 '12 at 19:51
    
An example is an isotropic radiator (en.wikipedia.org/wiki/Isotropic_antenna), the signal strength varies with distance, but not direction. Isotropy does not imply translational invariance. –  Alexander Jun 27 '12 at 21:12
    
Please be clearer about the type of field: what is constant? Do you have a constant vector field plus gaussian fluctuations? I use "isotropic" as a synonym for translational invariant, and use "rotationally invariant" for "rotationally invariant". I don't know if this is universal usage. –  Ron Maimon Jul 28 '12 at 5:13
    
There seems to be some confusion in the answers. Are you referring to a scalar field? Or a vector field? –  Colin K Sep 26 '12 at 2:15
show 1 more comment

4 Answers

The noise may have no preferred direction, but for large enough volumes the average $$\langle\mathbf{F}\rangle_V=\frac{1}{V}\int_V \mathbf{F}(x) d^3x$$ will approach the constant term. Thus the field is not isotropic and there is a preferred direction: that of averages over sufficiently large volumes. (Here, of course, specifying the direction to arbitrary precision and confidence requires sufficiently large volumes, and one has to assume this is not a problem or the question becomes ill-defined.) Although the field is not constantly in this preferred direction, this direction does exist - the field is just more likely to be in this direction.

share|improve this answer
    
Why is this downvoted (+1)? The OP seems to be asking about a case where there is a vector which is mostly one direction, with an additional random noise that prevents you from measuring the global average. Without more detail from OP regarding what is the vector in question (is he talking about E or the Poynting vector), it is hard to be more precise. –  Ron Maimon Jul 28 '12 at 21:31
    
What if the average is zero? –  Krastanov Oct 26 '12 at 22:31
    
If the average is zero then the field is indeed isotropic. –  Emilio Pisanty Oct 26 '12 at 22:47
add comment

The way I would approach this is to consider blackbody radiation, where a spherical blackbody radiator is isotropic, e.g. it emits radiation equally in all directions. If we were to model the noise introduced by such a blackbody, it would be additive Gaussian white noise. In this example, it is called white noise because it has a flat power spectrum density, and it is Gaussian because the noise has Gaussian amplitude distribution. It is additive because it can be added linearly to the desired signal.

In this case isotropy is a characteristic of the emitter. If I would interpret isotropy in terms of noise, I would understand it as being the same in all directions with respect to the receiver, e.g. regardless of direction, the received signal would have identical noise functions that were added to the signal. If the noise had some constant associated with it, as long is the constant was additive and the same in all directions with respect to the receiver, I would consider that isotropic as well.

share|improve this answer
add comment

You say:

•there is no preferred direction

but in your example this is true only at a single point i.e. at the centre of the gaussian. At every other point in your space the grad of your field is non-zero so the field at that point can't be isotropic (because the gradient will vary depending on the direction you look). An isotropic field must have zero gradient everywhere. A consequence of this is that an isotropic field is necessarily homogeneous.

share|improve this answer
    
I mean Gaussian noise for the field, not an amplitude that has the same functional dependence of $\exp(x^2)$ in space. I will make the text clearer. –  Krastanov Jun 27 '12 at 19:48
add comment

It depends how you define your Gaussian noise. A Gaussian random field can be modelled as having a distribution which is a multi-variate Gaussian of the Fourier modes. The field is isotropic if the distribution only depends on the magnitude of the Fourier mode and the Fourier modes are uncorrelated with each other.

In cosmology the primordial perturbations are thought to be such a field and current observations apear to be consistent with this, although some argue that there may be violation on the largest scales. In the comsology case the constant part is the background density and it is spatially uniform so it does not effect the isotropy. In theory one could have a constant part that varies spatially. But if your observations only consist of one realization of the field, then it is not really possible to distinguish between a spatially varying constant part and the random fluctuations.

An example of an article discussing these issues is available at: http://arxiv.org/abs/astro-ph/0509301 .

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.