On Friday, February 5, 2021 at 6:13:32 AM UTC-7 [email protected] wrote:
> On Thu, Feb 4, 2021 at 9:13 PM Alan Grayson <[email protected]> wrote: > > >>In relativity mass and energy are the same thing, remember E=MC^2, so >>> the kinetic energy needed to do work comes from the mass/energy released by >>> vacuum potential energy falling outward. In a similar way a hydroelectric >>> dam produces electrical energy that can do work from the potential energy >>> released by water falling inward. >>> >> >> *> But rest energy is positive whereas potential energy is negative. How >> do you expect negative potential energy to transform into positive rest >> energy? AG* >> > > You've asked that exact same question before and I've answered it before, > it does it the same way a hydroelectric dam transfers negative potential > energy into positive energy that can do work; in the case of normal matter > like water that's done by falling inward, in the case of vacuum energy > that's done by falling outward. > > *>>> Is this the GR expression for PE, which you earlier stated is >>>> different from Newtonian physics? * >>>> >>> >>> >> No. The formula for gravitational potential energy is the same in >>> both Newtons and Einstein's theory. >>> >> >> *> I could swear you posted the opposite recently. When I have the >> motivation, I'll try to find it. AG * >> > > The formula for gravitational potential energy is the same in both > theories, although Newton didn't know about E=MC^2 or vacuum energy so the > calculations sometimes differed, sometimes only slightly sometimes by a > lot. For example Newton would've said that 2 hot iron cannonballs placed > one foot apart and 2 cold cannonballs at the same distance would have > exactly the same gravitational potential energy, but Einstein would say > they would not because the hot iron cannonballs had more energy and thus > have more mass than the cold iron cannonballs. > > >> If vacuum energy really does exist then It's an intrinsic property of >>> space itself and so it doesn't move, it always stays the same, so I >>> guess you could call that rest mass if you want but I don't know why >>> you'd want to. Light moves as fast as things can go and has zero rest mass, >>> but even a photon of light has a gravitational field, in fact if you >>> concentrated light enough into a small enough volume it would turn into a >>> Black Hole. Such a ball of light is called a "Kugelblitz". >>> >> >> *> Didn't you assume your sphere has some initial mass in the form of >> "sand"? * >> > > I gave two examples, the first was a sphere made of normal matter like > sand, as the radius R of the sphere got larger the mass stayed the same, > so according to the formula for gravitational potential energy PE= > (-G*M^2)/R becomes less negative and more positive, and that means it's > uphill and so would need work to accomplish. In my second example I > considered an expanding sphere of vacuum energy, in that case M does not > stay the same but increases to the cube of R, So by using the same formula > that means it would be downhill and can produce work. > >> >> *> Or is it now light?* >> > > As I said Newton didn't know about E=MC^2 so he would've said light > wouldn't produce a gravitational field no matter how intense it became, > Einstein > would say something different. The gravitational field produced by an > expanding ball of light would behave differently than either a ball of sand > particles or a ball of vacuum energy because as R got larger the number of > photons in this sphere would remain the same but each individual photon > would get stretched, it would get red shifted to a longer wavelength and > longer wavelength photons have less energy, so as R increases the mass M > would not stay the same as a sphere of sand of would or get larger as a > ball of vacuum would but the mass would actually get smaller. > > >> > *Doesn't really matter, except you have to account for positive rest >> and kinetic energies equating to negative potential energy, * >> > > The engineers who design hydroelectric dams seem to have no difficulty > accounting for that, and neither do I. > *In your model, you offer no clue about the motion of the particles involved, and therefore no calculation of kinetic energy; hence, no argument, plausible or otherwise, that the net total energy is "precisely zero". AG* > > > -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/bb373f33-a2a8-46a7-b4f8-0be730b1b41an%40googlegroups.com.
