usenet weirdness

If I reply to posts from the pointy-ear guy, they go to sci.astro, even though his posts are in sci.electronics.design.

This is agent/eternal september.

The issue was

3. By contrast to light, gravitational waves are quadrupole waves that are
> only emitted when the spacetime curvature changes in a non-spherically
> symmetric way. They are also emitted by objects and systems which do not
> emit light.

Suppose that somewhere out in free space a cannonball somehow appeared. Wouldn't that create a symmetric, spherical gravitational wave?

Microphones some distance away, in any direction, would hear a click some time later.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

So Agent does not give you an opportunity to edit which groups you are replying to when replying to a cross posted message? You could try a currently maintained newsreader based on Agent which works properly:

Glen

I would assume that by the time you had the tech to make a cannonball "somehow appear" in free space, you'd already know the answer to this question ;-)

Agent lets me edit the reply groups, but the post was so s.e.d. but the reply group was the astro thing.

Now I know to look for this weirdness.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Making a sphere of mass suddenly appear is a separate issue. There are ways it might be done.

But the gravitational effect will spread spherically and symmetrically at the speed of light, from the instant of mass creation/destruction. Is that a gravitational wave?

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

There is nothing odd about it at all. He has set followups to sci.astro which is now a hell hole where all the lunatic fringe nutters hang out.

sci.astro.amateur and .moderated was created to separate the deranged nutters with their NEW THEROY (sic) OF THE UNIVERSE all in CAPS from people who wanted to talk about astronomy. It worked for a while too.

Your Usenet client is doing *exactly* what it is supposed to do.

E = mc^2

It is not possible to violate conservation of mass energy in the way that you propose with enough mass to have an appreciable gravitational effect. Quantum mechanics uncertainty principle puts very strict bounds on how much mass/energy you can spirit into existence and for how long.

The simplest system that can create gravitational waves is a close binary star system. Ones with two pulsars in are very handy tests of relativity. I was at the seminar where that discovery was announced.

If you could do it and on a large enough scale then it would perturb the orbits of objects but only after enough time has elapsed for its influence to reach them (and it would switch on suddenly in a step).

However, conservation of mass-energy prevents it from being possible.

This was always the problem with Newtonian gravitation - to avoid having the Earth spiral into the sun in classical mechanics the speed of Newtonian gravity has to be infinite. IOW It has to be an exactly central force between the two centres of mass at every instant in time.

GR gets around this by redefining what is a straight line so that the influence of the sun is handled by the curvature of spacetime itself.

Exactly. Mass/energy conversion is one way to create or destroy some mass in my cannonball.

Once that is done, a g-field change extends radially at the speed of light, and could be detected as it passes sensors.

If that is not a spherically symmetrical gravitational wave, what should we call it?

Any nonzero effect, no matter how small, would still be real.

My instant cannonball would obviously lose energy from pushing out its spherical g-thing-that-must-not-be-named.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

john larkin snipped-for-privacy@glen--canyon.com wrote or quoted:

"E = mc^2" does not mean, "Mass can be converted to energy and/or vice versa". It means, "When the momentum of a system is zero, its energy is its mass (up to a conversation factor of c^2).".

In general, the formula for a non-zero momentum p is (with c=1):

E^2 = m^2 + p^2,

and since E and p are conserved, m must also be conserved. So this tells us, "Mass is conserved.", exactly the opposite of, "Mass can be converted . . .".

Well, I mentioned "Boltzmann brains" here (in sci.electronics.design) recently, but their energy would not suddenly come from "nothing" or violations of conservation but be redistributed from the surrounding equilibrium state.

|It is often said that the uncertainty principle means energy |is not strictly conserved in quantum mechanics - that you're |allowed to "borrow" energy delta E, as long as you "pay it |back" in a time delta t approximately hbar / (2 delta E); |the greater the violation, the briefer the period over which |it can occur. Now, there are many legitimate readings of the |energy-time uncertainty principle, but this is not one of |them. Nowhere does quantum mechanics license violation of |energy conservation, and certainly no such authorization |entered into the derivation of Equation 3.76. |But the uncertainty principle is extraordinarily robust: It |can be misused without leading to seriously incorrect |results, and as a consequence physicists are in the habit of |applying it rather carelessly. | Section "3.5.3 The Energy-Time Uncertainty Principle" in "Introduction to Quantum Mechanics" (2018) - David J. Griffiths

Newsgroups: sci.electronics.design,sci.physics Followup-To: sci.physics Subject changed

Doesn’t work though—the energy has mass too.

Cheers

Phil Hobbs

Phil Hobbs snipped-for-privacy@electrooptical.net wrote or quoted:

Energy does not always have mass. Counterexample: A photon - it has energy, but no mass.

Newsgroups: sci.electronics.design,sci.physics Followup-To: sci.physics Subject changed

A photon has no *rest mass* - which not quite the same thing.

Doesn't a nuclear fuel rod (or a firecracker) lose mass when it delivers energy?

Two gamma rays (zero mass total) can collide to produce a particle pair (which has mass.)

After the collision, the particles create gravitational attraction to everything else in the universe. That spreads out from the new mass at the speed of light.

Why not?

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

john larkin snipped-for-privacy@glen--canyon.com wrote or quoted:

Let's call the momenta of the two photons p0 and p1.

We may assume p1 = -p0 as the two photons are moving towards each other from opposite directions. Let's call the momentum of the system of these two photons "p", then we have:

p = p0 + p1 = p0 +( -p0 )= p0 - p0 = 0

. Let's call the energy of this pair "E" and its mass "m". From

E^2 = m^2 + p^2

(in units with c=1) and

p = 0

, we get,

E^2 = m^2

for the pair. I.e., all its energy is mass. And this is the mass the particle pair has after the collision.

Newsgroups: sci.electronics.design,sci.physics Followup-To: sci.physics Subject changed.

The closest that you could come in principle to doing what he wants would be in reverse and would pose some serious technical difficulties.

LEAR used to be able to hold about 10^9 antiprotons which sounds like a lot until you compare it with Avogadro's number 6x10^23 (the number of hydrogen atoms in 0.5g of hydrogen gas as H2).

Matter and antimatter both have a positive gravitational mass. Matter is much more common though and the antimatter would need to be specially made. Exactly why this matter anti-matter asymmetry exists in nature remains unclear.

Start with two equal solid cannon balls of matter and antimatter. Slam them together to generate annihilation radiation and as the gamma wavefront expands at the speed of light it leaves behind a region with no mass or energy in it. The gravitational attraction ceases in an expanding shell that expands at the speed of light.

There is a step change in the gravity as the light energy goes past equivalent to the mass that was destroyed in the annihilation reaction.

Maxwell's Demon would have to work double overtime and some to run this experiment in reverse! ;-)

No *rest* mass. But a photon is never at rest.

Dang, I was going to say that.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

The gamma rays had no gravitational effect on the rest of the mass in the universe, up until the instant that they collided to form a particle pair. Then a gravitational object magically appeared.

Does't that create a spherical, symmetric, expanding bubble of gravity?

I'm just an engineer, but I think this is real. People say it's not technically possible, or that the efffect is too small to worry about, or some other excuse for not saying that it could happen.

There is an electrical equivalent. A metal sphere could suddenly become charged, and it would create a symmetric e-field pulse that expands at the speed of light, like a wave, but it isn't electromagnetic.

I guess that you can't make an antenna that radiates em waves symmetrically in all directions.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Energy, like mass, is a physical quantity; a property of a physical system e.g. of an object. It is not an object itself, so it *cannot* "have mass",

More precisely, for various good reasons its mass is assumed to be zero, and therefore is zero in the Standard Model of particle physics. (Its

The concepts of "rest mass" and "relativistic mass" have been obsolete since

In modern physics, by "mass" we mean that which was called "rest mass" before. (That) mass is a scalar quantity, therefore frame-invariant: it does NOT depend on the choice of rest frame of reference, and therefore does NOT change as the relative speed of an object changes:

m = 1/c √(p_b p^b),

where p^b is a component of the four-momentum, and Einstein summation is to be used.

See also:

Thomas Lahn: Special Relativity (YouTube playlist) Fermilab: Is relativistic mass real?

F'up2 sci.physics.relativity

The gammas together had the same mass as the particle pair.

I know of no mechanism for that to occur either. Charge is conserved, so you’d have to transport it to or from the sphere.

Either way, it’s an electric current that’s measurable from the time it starts out to the time it gets there.

Depends on your definition. An antenna with one pair of wires couples to one mode of the EM field in free space, and there are no monopole modes. I haven’t done the math, but I expect that three crossed dipoles driven 120 degrees apart would be isotropic.

The phase would be different in different directions. (Note that this assumes very small dipoles—the usual half-wave wire things would talk to each other, which might screw things up.)

Cheers

Phil Hobbs