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.

In the context of this question should mass distribution $\rho(r,t)$ and momentum distribution $p(r,t)$ be well behaved ? By 'well behaved' it is meant that derivatives of all orders exist everywhere.

I don't see any reason why they should be well behaved. For the equations (given in the answer (by Mark Eichenlaub) to the above mentioned question) to be consistent the conditions required are

  1. $\vec{p}$ should be continuously differentiable in order that $\nabla\cdot\vec{p}$ exists.

  2. $\rho(r,t)$ should be partially differentiable wrt $t$ i.e., $\frac{\partial\rho(\vec{r},t)}{\partial t}$ exists.

    Please give strong reasons as to why it is required that $\rho(r,t)$ and $\vec{p}(r,t)$ to be always well behaved ?

share|improve this question
    
Are you stating that the distributions should be well behaved, or asking whether they should? Because I don't think they necessarily are, at least not if you allow point particles. –  David Z Nov 29 '10 at 17:59
    
@David Zaslavsky : I am asking whether they should be ? what are the consequences if they are not well behaved ? –  Rajesh D Nov 29 '10 at 18:01
    
@David Zaslavsky : Apart from satisfying the two conditions If they are not well behaved then why would it lead to existence of point masses ? –  Rajesh D Nov 29 '10 at 18:08
    
Consider editing your question to end the first sentence with a question mark, then. Also, I mentioned point particles as an example of a mass distribution that would not be well-behaved (or smooth, to use the technical term). –  David Z Nov 29 '10 at 18:25
add comment

2 Answers 2

up vote 2 down vote accepted

Will $\rho$ always behave well? No, not at all.

This is physics, not mathematics -- if something is wrong with solutions, the explanation is simple: too bad for equations, probably approximations behind them lack some process or the mathematical description starts to be invalid.

In your case, it is obvious that a cloud with no angular momentum will be collapsing into one point, yet this is not a problem -- it is just "physical" that there will be some repulsion that would start to be significant in high density regions invalidating this simplified model.

share|improve this answer
    
@mbq: the intention here is not to explain any physical reality. Its just a question. –  Rajesh D Nov 29 '10 at 20:34
    
@mbq: It is not necessary that a cloud collapsing into a point implies that $\rho(r,t)$ is not well behaved. –  Rajesh D Nov 29 '10 at 20:44
    
@mbq: existence of repulsion in high density regions is something outside the current theory. –  Rajesh D Nov 29 '10 at 20:47
1  
No. It is predicted to reach zero in a finite amount of time in both Newtonian and Relativity theory, unless there is a pressure that can stop the collapse. If you have access to it, look up "Oppenheimer-Snyder collapse" in Misner, Thorne and Wheeler's 'Gravitation'. –  Jerry Schirmer Nov 30 '10 at 5:02
1  
@Rajesh Your problem is that there is no physical system that just fully follows those equations (they don't form any "newtonian gravitation theory" -- NG is only $Gm_1m_2/r^2$). Physicist talk about reality, not models on their own -- that's why you are still getting answers jumping beyond your model. –  mbq Nov 30 '10 at 9:20
show 11 more comments

They are not necessary well behaved. But if they're not you will use distributions instead of functions. For example, if $\rho$ is not continuous, $\partial\rho/\partial t$ will behave like a Dirac distribution. The maths are just more complex, rarely bring new insight about the physics, that's why physicists often assume the functions to be regular enough.

share|improve this answer
    
Apart from satisfying the two conditions mentioned should they behave well ? this is the question i intended to ask. –  Rajesh D Nov 29 '10 at 18:03
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.