how can mass be converted into energy?

Discussion in 'Physics & Math' started by StMartin, May 4, 2008.

1. MontecRegistered Senior Member

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248
If photons exert a gravitational attraction on other objects then how can a photon leave a gravity well like the Sun? Now if photons have a different form of (or method to generate) gravity that allows them to generate a gravity field that does not affect them (sounds like dark matter) then there would be no problem.

For me, the warping of space and the slowing of time still makes the most sense for the behavior of light (massless and non-gravitational photon) near a gravity well. (This view maybe altered if the evidence for an alternate view is strong enough).

3. ZardoziIsvara.... . 1S Evil_LauRegistered Senior Member

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germs taken into space come back even deadlier after the force of fluid shear takes effect on cytoplasmic formation which has been altered because of movement from a microgravitional focus point (earth) into a clinostatic vessel which produced g forces up to and greater tha 1G.

5. rich68Registered Senior Member

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to all that ponder the thought of the post....the answer is within.. gamma rays!!can you elaborate?

7. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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the sun is a minor star - light leaves much more massive ones. Every spherical mass has a relatively easy to calculate "escape velocity" (for anything leaving from the surface or some greater distance from the mass center). Light has much more than the required escape velocity from almost all finite radius mass distributions.

As any fixed mass becomes more "point like" (the object becomes more dense) the inverse square law makes escape from the smaller radius surface more difficult - higher escape velocity. Even a very small mass like the Earth, if it were to shrink to very small size would have a tiny radius and surface from which light can not escape. The sun is not very dense (lower density than the Earth, if memory serves me) not hard fro light to get away from the sun's surface.

See the following table well organized at:
http://en.wikipedia.org/wiki/Escape_velocity
Hint for here: the values after "at" are for escape from the sun (solarsystem) if starting from spot on that planet's orbit, where planet is NOT. I.e. if Earth distance from the sun but far from Earth you need 42.1 km/s to leave solar system.

on the Sun, the Sun's gravity: 617.5 km/s (Compare this to speed of light 300,000km/s I.e. Light has about 500 times the required velocity.)
on Mercury, Mercury's gravity: 4.4 km/s at Mercury, the Sun's gravity: 67.7 km/s
on Venus, Venus's gravity: 10.4 km/s at Venus, the Sun's gravity: 49.5 km/s
on Earth, the Earth's gravity: 11.2 km/s at the Earth/Moon, the Sun's gravity: 42.1 km/s
on the Moon, the Moon's gravity: 2.4 km/s at the Moon, the Earth's gravity: 1.4 km/s
on Mars, Mars's gravity: 5.0 km/s at Mars, the Sun's gravity: 34.1 km/s
on Jupiter, Jupiter's gravity: 59.5 km/s at Jupiter, the Sun's gravity: 18.5 km/s
on Saturn, Saturn's gravity: 35.5 km/s at Saturn, the Sun's gravity: 13.6 km/s
on Uranus, Uranus's gravity: 21.3 km/s at Uranus, the Sun's gravity: 9.6 km/s
on Neptune, Neptune's gravity: 23.5 km/s at Neptune, the Sun's gravity: 7.7 km/s
in the solar system, the Milky Way's gravity: ~1,000 km/s (The solar system is "far out" - much harder to leave our galaxy is were closer to the center of galaxy.)

Last edited by a moderator: May 9, 2008
8. MontecRegistered Senior Member

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248
Hello Billy T

The point is that zero mass means zero inertia which means an instant stop for a photon that reacts to a gravity field like an equivalent energy level fermion.

If photons do contribute to the stress-energy tensor (by some method) then the contribution must be weaker than the contribution of an equivalent energy level fermion.

9. MetaKronRegistered Senior Member

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5,502
The gravitational attraction of a photon to anything is an almost negligible amount. It has no trouble escaping from any normal gravity well.

The typical photon is actually a bit of kinetic energy shed by an electron as it drops from a higher to a lower energy level. It starts out with no "substance." However, that packet of energy has an equivalent mass that can be computed.

10. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Its been a long time, but I think photons are bosons, not fermions and that electrons are fermions. Fermions obey the Pauli exclusion principle, which basically means in any system that has quantized energy levels like an atom only one to each set of quantum numbers. That is what builds up the periodic table - in some deep sense - explains all of chemistry.

Photons can be quantized in a system also.
For example bouncing back and forth between to perfect mirrors perfectly parallel (very closely spaced in this example you can have one whose wave length is twice the mirror separation, one where the wave is half as large (so equal to the mirror separation) etc. to ever shorter wavelengths. I.e. the permitted "Photon Quantum Numbers" PQN, are: 1/2, 1, 3/2, 2, 5/2 .... however, if you like, you can have more than one photon with the same PQN, for example PQN = 735 you (or the system really) can have 3,482 photons with PQN = 745.5 Photons are bosons and bosons are very social, like company, not exclusive.

I do agree "zero mass means zero inertia" but the whole concept of inertia does not make much sense (unless you want to say that photons have infinite inertia - in vacuum they cannot be speeded up or slowed down. They do not really slow down in glass etc even though the index of refraction may be 1.5. What they do in a classical POV is "wiggle" the bound electrons and these accelerated electrons radiate at the same frequency but their radiation has a phase lag wrt to the driving photons. The total field when you add two sin waves of same frequency but different phase is a new sig wave of the same frequency but the +1 peak of the new wave is shifted from where the +1 peak of the original wave would have been if the electrons could not "wiggle." That is the heart of it, but it is more complex read about "phase velocity" and "group velocity." There you will find, if memory serves that the product of these two must be c^2 and the math I verbally described will if done correct make the phase velocity greater than c so the group goes less than c as the photons go thru glass etc. This has all gotten much more complex as the phase of the radiated waves, if you drive then near a natural resonance changes rapidly and in the case of extremely "slow light" you can even kill the original photons and store their energy and phase in the coherent oscillation of the bound electrons and have the photons "reborn" again as if they had hardly advanced thru the material. (Or something like that) A few years ago, someone made system in which the photons appeared to go thru it slower than a fast bicycle! But I cling to the idea that in truth, photons (the originals) always go at the speed of light, c, everywhere, and these interactions with matter are sort of "magicians tricks" to make you think otherwise.

:bawl:

11. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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From second sentence here, it seems safe to assume that you are speaking of a "free photon" not one confined between perfect mirrors in a box. If that is the case, then I think "almost negligible" is wrong. It has ZERO gravitational attraction to anything, IMHO, but admittedly others do hold a different POV. Sort what this thread has become to be about is this difference in POVs. In my POV the apparent motion of the distant star that had light passing near the edge of sun during solar eclipse (made Einstein sleep well that next night)was due to the sun's "warping of space" nothing to do with the sun attracting the mass-less photon by "gravitational force."

Free photons do not make or respond directly to gravity. When one tries to leave a black hole it is not "pulled back" into the BH by the force of gravity. It simply loses momentum as it "Climbs up the step space hill" the BH's mass has made. When it has lost all the momentum it had, it falls back into the BH recovering its momentum as it slides down the steep space gradient. That verbal nonsense is about the best I can do - the tensor math is the correct description but I can no longer do that. (I never was much good at it.)

12. VkothiiBannedBanned

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3,674
If you have two gamma photons inside a box that has perfect mirrors for walls, then again, won't they eventually interact, or collide with each other at some point?

Is this not significant?

The other observation with this "box" is that there's no way to see inside it, or it implies the outer surface transmits nothing whatsoever from the inner cavity. It doesn't change momentum if the particle/antiparticle has mass or they annihilate each other, because they can pair up again and gain mass.

13. PeteIt's not rocket surgeryRegistered Senior Member

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10,167
Photons don't collide, but it shouldn't be significant anyway. It would make a detailed analysis more involved, but the end result should be the same.

You could go way down deep with this thought experiment... there are thermodynamic implications for perfectly reflecting walls (the photons give the box interior a temperature, which implies that the box must radiate)... and there are quantum implications of how precisely the momentum of the box can be defined.