Gravity waves analysis opens 'completely new sense'

Noone could know how frequently these waves are generated.

Gravity waves are common like light waves. The problem detecting them is that gravity as a potential field is much weaker, billionths, than electric potential fields. Therefore . . . the effects of gravity waves should be minuscule. It is not surprising that detecting such weak waves is not easy, is it?

"Re: post 52. Based on your equation, gravity is not related to mass. The speed of light is a value that exists outside of the realm of mass."

G=E/M. The M stands for Mass, btw. Since E=MC^2, we know that the speed of Gravity is C squared (E/M=C^2; G=E/M; E/M=E/M; therefore C^2=E/M and E/M=G so G=E/M).

What you've got to understand is that Gravity exists because of motion/speed. Once something moves fast-enough, it begins to emit Gravitons. Even something Massless, such as Light, emits Gravitons so long as it is moving fast-enough. In this way, Light is affected by Gravity. Watch the Light as it bends around the edge of your thumb if you hold it to up to a light-source, for instance. Greater masses bend Light even further, such as planets or Black Holes.

But only up to a certain point. Slow light down to a fast run, for instance, and Gravity no longer bends that light. The recent experiments in which Light has been slowed down to under 30 miles per hour confirm that fact, too.

Okay, I'm with you.

Can you elaborate on how the field changes as the hole absorbs new mass?

I.E. suppose a star falls into the hole, increasing its mass greatly. You are orbiting the hole at a nice safe distance. Does your orbital speed change? (Does the observed gravitational field of the hole change?) If so, how is the change communicated to the outside world?

Presumably by gravity waves. Eventually the field 'settles down' to its new value.

If the source of the field (the new, stronger one) is the event horizon, then the hole cannot appear as a "point source" of gravity (a particle) since the source is distributed. I'm thinking of Lambert's cosine law for radiation.

In other words, I am still confused and need instruction!

--Boris

When an electron has velocity the total electric field increases over that of a stationary electron.

No, it doesn't. That would violate Gauss's Law.

I don't see how the uncertainty principle applies, especially to a field.

Suppose you have an electron in orbit around a nucleus. The HUP states that the uncertainty in energy times the uncertainty in time is greater than some calculable fraction of Planck's constant. If the electron were to continue to radiate energy, its orbit would continually shrink (resulting in a shorter period), and its energy would continually decrease. At some point, the product of these quantities will fall below the stated inequality, which is forbidden. There must therefore be a ground state past which the orbiting electron cannot radiate.

The Earth in its orbit is about 75 orders of magnitude away from this limitation.

If Cavendish could measure g with small lead spheres over 100 years ago, certainly today's physicists could produce gravity waves on the orders of tens of kilohertz and measure them.

Describe how to do it. It's easily worth a Nobel Prize.

Also as the frequency of the wave increases so would its radiated intensity, making the measurement very easy.

That doesn't make sense.

Wouldn't gravity waves cause an effect similar to the Lorentz contraction and hence could never be measured?

The Lorentz contraction can be measured.

If the source of the field (the new, stronger one) is the event horizon, then the hole cannot appear as a "point source" of gravity (a particle) since the source is distributed.

Well, there you go. A spherical distribution of matter exhibits the same field, outside of the sphere, as would a point source of the same mass at the center of the sphere, by symmetry.

In any case, an object falling into a black hole, as viewed from the outside, takes an infinite amount of time to reach the event horizon, owing to the fact that the gravitational time dilation becomes infinite at the event horizon.

"Think of it: Gravity waves come to us from the edge of the universe, from the beginning of time, unchanged,"

I have never been able to understand this thinking. If waves are arriving right here and now from the beginning of time, then why didn't they arrive one minute ago or 2 days from now?

Are they saying that any time you look you will be able to observe the same moment?

I dunno- my head hurts thinking about it. (and I have a BS in Physics AND another one in computer science...)

Oops, I forgot to answer these:

Yes and yes.

Gauss's Law applies to gravitational fields as well as to electromagnetic fields. Imagine the field as a bunch of lines that radiate outwards through space. The stronger the field, the more lines there are. If you draw some surface (say, a spherical shell) enclosing some region of space, and you want to know the integral of the field over that surface, it will be proportional to the amount of charge enclosed within that surface. (The "charge" in the case of gravity is simply the enclosed mass.) The more charge (mass) you throw into the enclosed volume, the more field lines will come out of the surface.

If so, how is the change communicated to the outside world?

All changes in the gravitational field would be communicated in the form of gravitational waves.

Are they saying that any time you look you will be able to observe the same moment?

No, they mean that they can see gravity waves passing by earth now whether the gravity waves started out last week just past Pluto, or 14 billion years ago when some star first came to life. It might be important to note that the Big Bang occurred everywhere, even right here where earth sits today. Old events would be visible now if they were also just far enough away that their light or gravity wave is just arriving.

"Your latest comment is drivel and I suspect nearly everything you think you know about gravity you learned in a comic book. Physicists have yet to understand gravity and do not agree on what they think they know."

First you failed to understand that G=E/M involved Mass (the M stands for Mass, after all), and now you've managed to contradict yourself by first saying that what I know is "drivel" but then pointing out that scientists don't even agree on what they think that they know (about Gravity, anyway).

You poor thing. You must be some failed graduate student stuck in some mundane life.

Buck up! Life will get better for you as you open your eyes. The future is bright, after all!

1. You're right the electric field doesn't increase. Also mass does not increase with velocity either. I forgot about the time derivitives and retardation in Gauss's law.

2. Excellent analogy! What is the charge of the earth and what is the charge of the sun? Do the retarded potentials scale likewise?

3. Easy. Take two lead spheres, each on the end of a single rod. Rotate radially about the center of mass, just like two equal mass planets orbiting about each other (kind of like a Woodward governor). The scale is smaller than planets but the rotation rate is much higher. The gravity signal would be of a frequency much higher than the background noise. To show it is a gravity wave, measure signal intensity as a function of distance. I think I can find some references on this, I'll send them to you if I do.

4. Higher acceleration = higher radiation intensity. Higher acceleration = higher rotation rate. Therefore, higher rotation rate (frequency) = higher radiation intensity.

5. Please cite the experiments showing that the Lorentz contraction can be measured.

It may be very helpful for you to review freshman E&M. Gauss's law applies to electrostatics, not moving charge. Apply Gauss's law to a current carrying wire. The integral of E over the cylindrical volume is zero, is it not? That means there's no net charge enclosed, as is true in an electrically neutral conductor. Yet the measured E field is in the direction of the current travel is it not? E is in the direction of the current density. The wire has a net E field, but net zero charge. Where would this net E field come from if the total electrostatic charge of the wire is zero? Why does it exist only when the charge is moving? Please don't invoke exotic quantum theories to explain this, do it as Einstein would with simple understandable English.

Please cite the experiments showing that the Lorentz contraction can be measured.

Physicist can handle this, I'm sure. But there are demonstrations of other phenomena that involve the Lorentz transformation. Time dilation, using a pair of atomic clocks; and mass increases, using accelerated particles. You're probably aware of those, as they're classics. As for length contractions, nothing rings a bell right now. But I'm betting on Physicist to come up with the answer.

Assume that every point in the universe began with the Big Bang, ignore the effect of the inflation phase and assume the start was right after that. That is, the Big Bang was everywhere, even right here, and there is no point that wasn't involved in the Big Bang. Then it becomes simple. If the Big Bang happened 14 gy ago, and if light or gravity travels at 1 lightyear per year since then, then the light or gravity wave we see now from objects 14 billion lightyears distant would have started on their way 14 gy ago. So objects seen as they were at the time of the Big Bang itself would be found in a spherical shell around us of radius 14 billion lightyears. Other objects are visible closer in, but they would be seen doing things in more recent times, not at the time of the Big Bang.

There are other things happening to the time-space manifold that complicate our model, but we can ignore all that for now in this simple picture. Hope that helps get things started.

Pat, this is an enjoyable discussion. Moving clocks do slow down (clock speed is also a function of gravitational potential), inertial mass does increase etc.
I'd like to see the case where Gauss's law is violated by the additional electric field of moving charge. The divergence of the additional E field of moving charge is always zero, so this shouldn't affect Gauss's law. I guess you could get it if you charged a capacitor that had only one plate.
I'd really like to see where the Lorentz contraction was measured. If it can be measured, one can go back and properly redo Michelson-Morley and measure the anisotropy of c.

I'd really like to see where the Lorentz contraction was measured.

I'm not that familiar with the literature. The only thing that comes to mind is Einstein's thought experiment, involving an observer on a rapidly rotating disk. His measuring rod remains unchanged as he moves out from the center along a radial line (he keeps the rod pointing toward the center), but when he's at the circumfrence (which is spinning rapidly) and he rotates his rod and begins to measure the circumfrence, his rod is shortened by the disk's acceleration, and he discovers that he has measured out a longer circumfrence than the radius would have indicated. But I know of no actual experimental evidence. That would require a very strong disk to withstand such rapid motion -- and you thought your hard drive was high tech.

It may be very helpful for you to review freshman E&M.

I don't know why you feel it appropriate to adopt such a tone with me. I'm starting to suspect that you are less interested in learning the correct answers, and more interested in advancing some sort of kook science agenda.

Gauss's law applies to electrostatics, not moving charge.

Gauss's law is absolutely universal. In fact, it's one of Maxwell's equations.

Apply Gauss's law to a current carrying wire. The integral of E over the cylindrical volume is zero, is it not? That means there's no net charge enclosed, as is true in an electrically neutral conductor. Yet the measured E field is in the direction of the current travel is it not?

Yes, but the integral of the field, as you point out, is zero, thus Gauss's law is satisfied. I'm not sure why you bring this up in any case, as there is no gravitational analogue (any mass current will have a net gravitational charge, as there are no antigravity charges).

E is in the direction of the current density. The wire has a net E field,

No, it doesn't. You correctly stated above that the integral is zero.

but net zero charge. Where would this net E field come from if the total electrostatic charge of the wire is zero? Why does it exist only when the charge is moving?

That's just wrong. Moving current in a wire does not create the electric field; rather, the electric field causes the current to flow.

Disclaimer: Opinions posted on Free Republic are those of the individual posters and do not necessarily represent the opinion of Free Republic or its management. All materials posted herein are protected by copyright law and the exemption for fair use of copyrighted works.