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So...Why can't a singularity exist?


Obnoxious

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I've been told it's because it somehow violates the uncertainty principle, but I'm pretty sure that's not a valid reason (as according to uncertainty, the more time we spend on determining the energy, the more accurate it'll be. And we've had a long time, but we still don't know exactly how much infinity is)

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The uncertainty principle does not (only) state that dE*dt >= hbar. In fact, the dE*dt thing is one that does actually not follow directly from the uncertainty principle, at least in classical QM. The question to what extend it still is correct and how it might actually do derive in relativistic QM aside, I think that the "it [a singularity] somehow violates the uncertainty principle" rather relates to dp*dx >= hbar. So given you have something in a fixed position (dx=0) it can have any momentum. You´d need a mechanism that lets all possible momenta lead to the same velocity for the particle to remain at a fixed location.

 

I don´t really understand what you´re talking about because I neither know what you call a singularity (a point in spacetime with infinite energy density?) nor what the "it´s impossible" argument relates to or how it really works. I hope that my above statement still helps you.

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The uncertainty principle does not (only) state that dE*dt >= hbar. In fact, the dE*dt thing is one that does actually not follow directly from the uncertainty principle, at least in classical QM.

 

?

 

Energy and time are conjugate variables and are thus Fourier transforms of each other, which is the source of the uncertainty principle.

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That's not fair, Swansont. The energy-time uncertainty does not come out of QM the same way as the x,p relation. Yes, energy (frequency) and time are canonical conjugate variables (fourier transforms of each other), and this provides an argument for the existence of an energy-time uncertainty relation. You can't do the same thing that you do with position-momentum, because time is not an observable in NRQM. The way I arrive at the HUP relation for x,p is by plugging in for <[x,p]> and <{x,p}> in the Schwarz Inequality and juggling with the math (of hermitian and anti-hermitian operators), but there exist no canonical commutation relations for E,t so I can't do the same with them.

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Talk about moving off the topic a bit.

 

Maybe Obnoxious you should state what you mean by singularity...

Mathematical singularity :-where maths breaks down e.g. 1/0

Gravitational singularity:- Infinite spatial curvature (same maths as above)

(Black holes and big bang being examples).

Technological singularity:- A future time we cannot comprehend.

 

I presume you mean Gravitational singularity.

 

The problem as I stated is that equations end up similar to 1/0. the maths behind general realtivity break down as mass is compressed to a single point. It is obvious that the current theory of gravity is inadequate to describe what happens to objects, space or time at these extremely small, dense, high temperature situations.

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Sorry for the "extended digression"...this will be the last of it (and it's only for the sake of completeness).

 

I recall arriving at the energy-time uncertainty by thinking about correlation amplitudes (courtesy Sakurai).

 

For a continuous spectrum you can write :

 

[math]C(t) = \langle \alpha,0|\alpha,t \rangle = \int dE |g(E)|^2 \rho(E) exp \left( \frac{-iEt}{\hbar} \right) [/math]

 

Now for a real, physical problem let [imath]|g(E)|^2 \rho(E)|[/imath] be peaked at some E = U. Then writing

 

[math]C(t) = exp \left( \frac{-iUt}{\hbar} \right) \int dE |g(E)|^2 \rho(E) exp \left( \frac{-i(E-U)t}{\hbar} \right) [/math]

 

For large t, the integrand oscillates rapidly unless |E-U| is small compared to [imath]\hbar /t [/imath]. So to not see significant deviations from C(t) = 1, you need [imath]t |E-U| \equiv t~\Delta E < \hbar [/imath].

 

And this is not true only in the case of a continuous spectrum.

 

Okay, now back to the naked singularity....

 

It has been shown repeatedly, I believe, (and Hawking has had to eat his words on this) that observable singularities can exist (from the collapse of a scalar field, for instance). So I don't see the validity of the question.

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Naked singularity is one without an event horizon right?

I'm not sure on the details on naked singularities but it was a computer simulation that showed they could exist right. Is there any proper theory behind them and of course they have not been observed.

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What's wrong with infinity density?

 

Infinity doesn't exist in the real world (NOTHING can be infinte in the real world).

 

Aside from which, a point, which has dimentions of 0mx0mx0m cannot have a volume. Mass requires volume to be contained in, and since a singularity has no volume in which to contain it's mass, it cannot exist.

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Infinity doesn't exist in the real world (NOTHING can be infinte in the real world).

 

Aside from which' date=' a point, which has dimentions of 0mx0mx0m cannot have a volume. Mass requires volume to be contained in, and since a singularity has no volume in which to contain it's mass, it cannot exist.[/quote']

 

This is an oversimplification, and hence quite inaccurate. A rule "nothing can be infinite" certainly does not exist. Its just that the maths that lead to gravitational singularities seems to be just that; math. Zero volume and infinite density are extremely weird properties, and ending up with infinities generally means that there is a problem. So its not that singularities are necessarly impossible, its that a lot of people think with a more complete theory we wont get the singularity result. Just speculation as far as I know though.

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