Laser + Lens = Black hole?

Laser + Lens = Black hole?

Arthur Kalliokoski

Second in Command

February 2005

"Weight" refers to the gravitational attraction of mass. If you could show that they weren't proportional, you'd be in line for a Nobel Prize.

They all watch too much MSNBC... they get ideas.

type568

Member #8,381

March 2007

Arthur Kalliokoski said:

"Weight" refers to the gravitational attraction of mass. If you could show that they weren't proportional, you'd be in line for a Nobel Prize.

Right.. I'm either insane or Evert is saying it isn't so. And actually, I've that feeling he's right. At least he has that diploma.

Billybob

Member #3,136

January 2003

Evert said:

Anyway, I forgot the details but apparently you can't build a self-gravitating object out of photons. Having a high photon density means photon-photon (weak) interactions become important and that means you'll probably have large energy losses (from neutrinos) that the effect goes against you.

That's a shame :'(

Another crazy question! How about a Bose-Einstein Condensate? Most of the particles are in the same state, and so if you had enough of them would they collapse into a black hole?

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And the difference is...?

To nitpick, I believe black holes only refer to singularities with event horizons. So a singularity is not a black hole, unless it also has an event horizon. i.e. there are naked singularities, and they are not called black holes.

type568: A long time ago Isaac Newton came up with what we call Newtonian Physics, which includes the formula for gravity you are familiar with. In particular, that two masses attract one another with a force proportional to their masses and inversely proportional to the distance between them squared.

Then the question came up, if light has no mass, is it affected by gravity? Does it attract other objects?

Einstein said yes, and that has been one of the big deals in physics recently. Newton was not wrong, Einstein's contribution merely says that it isn't just mass that creates gravitational attraction. Energy does too! So light is affected by gravity, and it also attracts other light and even attracts other masses!

So, gravity isn't caused by only mass. It can also be caused by energy. So, if you take the Earth, and move it reallllly fast (close to the speed of light) it will create a much stronger gravitational field then if it was just sitting still.

And finally, and this one is probably a little weirder, is energy really does, for all intents and purposes, act just like mass. So, as you accelerate Earth towards the speed of light, since it is gaining energy, it is getting heavier. Literally, the planet will feel heavier. Which means it will be more difficult to acceleration (Inertia). This is why it is impossible to accelerate past the speed of light. As an object's velocity approaches the speed of light, it becomes heavier and heavier such that it eventually, at the extreme, is infinitely heavy.

Now, let's talk about your experiment, and let's try to make it simple. Instead of burning something, let's just take up two containers of matter and anti-matter into space. They both have mass, and both create a gravitational field. Now, release the matter and anti-matter. They will annihilate one another, and release pure energy. No mass is left, whatsoever (ignore the containers). What is the gravitational field like? Exactly the same! Because Energy is a form of Mass, and Mass is a form of Energy.

Evert

Member #794

November 2000

type568 said:

Some url as backup & more detail please?

However.. I was said it is impossible.

No you weren't. Energy is conserved. If you heat something up, then you're adding energy to the system - in other words, it's not a closed system.
And it's perfectly right. It's the same thing as I said: what we call mass is a form of energy and energy is the source for the gravitational field.

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Fine, let it not be gas but wood or whatever.

You're missing the point. It doesn't matter what you're burning. Burning wood is a chemical reaction just as much as burning a gas is. It's exactly the same question and therefore gets exactly the same answer.

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If I burn something IN that isolated system, will not it's mass reduce?

Of course not. It's a closed system, where would that mass go? Try it out.
Questions to help you think about this: what weighs more, a lego construct or the loose bricks that make up the construct? An ice cube or the water resulting from it after it melts? A kilo lead or a kilo of feathers?

type568 said:

Right.. I'm either insane or Evert is saying it isn't so.

Neither, I should hope.
Colloquially, "weight" refers to "mass", or people say "weight" when they mean "mass". Technically, it's the apparent acceleration of a unit mass under a gravitational field, so it's a force. The gravitational acceleration (for instance on Earth) is the same for all objects, so an object's weight is proportional to its mass.
That's when talking about Newtonian dynamics. I don't think the concept of "weight" in the same sense is ever applied in a relativistic context and I'm not even sure whether it makes sense. Either way, it's confusing and in my opinion the term should be avoided.

Billybob said:

How about a Bose-Einstein Condensate? Most of the particles are in the same state, and so if you had enough of them would they collapse into a black hole?

Probably, but that doesn't have anything to do with Bose-Einstein statistics, since you can do the same thing with Fermi-Dirac statistics. Neutron stars are kept stable by FD statistics, but if you make them too heavy they collapse into black holes.

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Yes, ok, that's right. It depends on whether you think naked singularities can exist or not ("cosmic sensorship"). Personally I think a singularity is a mathematical outcome that tells you that the theory breaks down for the phenomenon you're trying to describe.

Arthur Kalliokoski

Second in Command

February 2005

Evert said:

if you make them too heavy they collapse into black holes.

If the escape velocity exceeds the speed of light, you wouldn't be able to "see" if it's a singularity or not. Is there some reason to suppose the exclusion principle can't stand up?

[EDIT]

To say that an object made of matter becomes more massive with increased temperature is necessary since the increased thermal motion means the atoms are traveling faster, and increased speed increases mass all on its own. I'd suppose that's not the whole story though.

They all watch too much MSNBC... they get ideas.

type568

Member #8,381

March 2007

Billybob said:

Then the question came up, if light has no mass, is it affected by gravity? Does it attract other objects?

Why just not to look at it as mass is producing a gravity field, which applies acceleration towards it to all around it(and even really far from it), both energy & mass?

Evert said:

No you weren't. Energy is conserved. If you heat something up, then you're adding energy to the system - in other words, it's not a closed system.

Got it.. And I suppose if I let some chemical reaction occur inside it'll be just change of forms of energy, which won't any help..

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Got it..

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Of course not. It's a closed system, where would that mass go? Try it out.

I've never heard of a "mass-saving-law"..
But anyways...

Billybob said:

Einstein said yes, and that has been one of the big deals in physics recently. Newton was not wrong, Einstein's contribution merely says that it isn't just mass that creates gravitational attraction. Energy does too! So light is affected by gravity, and it also attracts other light and even attracts other masses!

Oooh..
I think I am beginning to understand.. Yeah.
Here I go:
Mass is just a form of energy(hmm, makes sense with the big boom)?
& in detail:
A)
If we pick some mass, which is producing a grav field & isolate it.
Then, regardless of it's inner reactions it's total energy & mass are going to stay the same.
B)
If we pick another mass, and.. Send some photons in, which are to be caught by solar batteries that are going to catch it & have some use to it- we will increase the total mass & total energy there(assuming the system doesn't leak out).

P.S:
But that actually proves photons have mass, and we had some thread stating they don't, didn't we?
However, they DO have impulse don't they? And why don't we count up their mass by dividing their impulse by C(if that's the constant for speed of light)?

Ooh, and that Energy of universe equals to the product of total mass of universe & square of speed of light.. Somehow related to the said above. By the way, why not divided by two(as formula for kinetic energy), is it not derived from there?

Perhaps if I get some clue on what actually the time is, I can get some basic understanding of the uppers the theory of relativity.

Evert

Member #794

November 2000

type568 said:

I've never heard of a "mass-saving-law"..

Conservation of mass, see here.

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But that actually proves photons have mass, and we had some thread stating they don't, didn't we?

No, photons don't have mass. Photons have energy, but they don't have rest energy (ie, mass).

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However, they DO have impulse don't they?

Do you mean momentum?
Yes, they do have momentum, which is related to their wavelength.

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And why don't we count up their mass by dividing their impulse by C(if that's the constant for speed of light)?

That gives you something with the dimension of mass, but it's really just the energy of the photons.
Several things. First of all, the relation $<math>\vec p = m \vec v</math>$ for momentum only holds for non-relativistic dynamics, in other words, for things moving much slower than the speed of light. That's the same thing as saying that the kinetic energy is much smaller than the rest energy. But photons (almost by definition) move with the speed of light, and they have no rest energy.
Second of all, if you did try to calculate the "mass" of a photon this way, then you'd find that its mass depends on its wavelength, in other words, you'd find that photons don't have a well-defined mass.

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Energy of universe equals to the product of total mass of universe & square of speed of light..

It ought to be, yes.

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By the way, why not divided by two(as formula for kinetic energy), is it not derived from there?

Within the framework of special relativity, the expression for total energy is $<math>E = \sqrt{m^2 c^4 + p^2 c^2}</math>$ . Obviously the rest energy is $<math>E = m c^2</math>$ ; a Taylor expansion for small velocities gives $<math>E = m c^2 + \frac{p^2}{2m}</math>$ , which is just the classical expression for the kinetic energy, with an offset (the rest energy).

type568

Member #8,381

March 2007

Evert said:

Conservation of mass, see here

Cool thanks..

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No, photons don't have mass. Photons have energy, but they don't have rest energy (ie, mass).

Okay, so are they producing gravity?

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Second of all, if you did try to calculate the "mass" of a photon this way, then you'd find that its mass depends on its wavelength, in other words, you'd find that photons don't have a well-defined mass.

I've no problem with that, why can't the photos have different mass.. ?

However, is this statement true:
"Between any two wavelengths A & B(A shorter than B), there's another wavelength C which is shorter than B but is longer than A."?

One more thing(I'm afraid will be many more afterwards though ) :
What is the P in your latest few lines?

Evert

Member #794

November 2000

type568 said:

Okay, so are they producing gravity?

Probably, but I'm not sure there's a good way to calculate it (I don't know). The problem is that there is no inertial frame of reference where photons are at rest, which I think one normally uses to calculate gravity between particles.

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I've no problem with that, why can't the photos have different mass.. ?

Eh... because in order for mass to be a sensible property (on the particle level) all particles of the same type need to have the same mass. But as I said, that expression for momentum isn't valid for particles moving at the speed of light anyway.

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However, is this statement true:
"Between any two wavelengths A & B(A shorter than B), there's another wavelength C which is shorter than B but is longer than A."?

You mean, given $<math>x_1, x_2 \in \mathbb{R}</math>$ with $<math>x_1 < x_2</math>$ , is there a $<math>y \in \mathbb{R}</math>$ such that $<math>x_1 < y < x_2</math>$ ? Of course that's true. It's also true if you replace $<math>\mathbb{R}</math>$ with $<math>\mathbb{Q}</math>$ .

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What is the P in your latest few lines?

What it usually is, momentum.