The Sun and its Gravity

[Johnny5]

I want to know if the photons which the sun emits, are responsible for the path of earth's orbit around it.

 

The sun is for certain emitting photons, and they must leave the sun at a speed 299792458 meters per second relative to the sun.

 

They are emitted in many directions, roughly in a spherical shape.

 

So the region of space surrounding our massive sun, is filled with the photons which the sun is sending out.

 

So could these photons be solely responsible for the orbital path's of the planets?

 

Thank you

 

PS: If the answer is yes, then the speed of gravity is c.

[Meir Achuz]

No. The emitted photons exert a very small repusive force on the Earth.

This "radiation pressure" is not usually considered in calculating the orbit of the Earth.

[Johnny5]

No. The emitted photons exert a very small repusive force on the Earth.

This "radiation pressure" is not usually considered in calculating the orbit of the Earth.

 

The radiation pressure yes I've seen it. Perhaps the photons carry with them two effects?

[Martin]

In 1919 Eddington checked to see by what angle light passing close to the sun was bent by the sun's gravity

 

he verified an interesting formula. are you comfortable using radians?, the formulas for angles are always simpler using radians instead of degrees.

 

BTW no, AFAIK the photons on their way out from sun have a negligible effect on the planet orbits (they have some mass but it is miniscule compared with the sun mass, they have some radiation pressure but it is also very slight)

 

But this formula is interesting, there is a key distance R₀

= GM/c²

 

And if a ray of light is approaching the sun like it is going to pass by at a distance R from center

then the angle the ray will be bent by gravity is equal to 4R₀/R

 

Now Eddington observed some light from the Pleiades stars which was approaching as if to pass by the sun at a distance of 800,000 km.

the radius of the sun is 700,000 km, so you see the light was going to just barely squeak by.

 

I ask you J5, BY WHAT ANGLE WAS THE LIGHT that Eddington observed in 1919 BENT BY THE SUN'S GRAVITY?

[Johnny5]

In 1919 Eddington checked to see by what angle light passing close to the sun was bent by the sun's gravity

 

he verified an interesting formula. are you comfortable using radians?' date=' the formulas for angles are always simpler using radians instead of degrees.

 

BTW no, AFAIK the photons on their way out from sun have a negligible effect on the planet orbits (they have some mass but it is miniscule compared with the sun mass, they have some radiation pressure but it is also very slight)

 

But this formula is interesting, there is a key distance R[sub']0[/sub]

= GM/c²

 

And if a ray of light is approaching the sun like it is going to pass by at a distance R from center

then the angle the ray will be bent by gravity is equal to 4R₀/R

 

Now Eddington observed some light from the Pleiades stars which was approaching as if to pass by the sun at a distance of 800,000 km.

the radius of the sun is 700,000 km, so you see the light was going to just barely squeak by.

 

I ask you J5, BY WHAT ANGLE WAS THE LIGHT that Eddington observed in 1919 BENT BY THE SUN'S GRAVITY?

 

yes i can use radians.

 

Even steradians

 

You say that the photons from the sun have a negligible effect on an orbital path, and of course my question is whether perhaps it is those photons which are responsible for that path somehow.

 

Immediately, you started talking about photon mass, and how it's so miniscule that a photon cannot be responsible for gravitational force. I know radiation pressure is slight. When I turn on the lights in my house, it is not the case that the expanding sphere of light pushes my head backwards.

 

Yes radiation pressure is slight, but nonzero.

 

As I read that, the principle of equivalence came to mind.

 

There is the weak equivalence principle, and the strong equivalence principle, I don't remember them usually.

 

But of course I am trying to learn something which I don't yet know, so i have been reading up on GR lately.

 

Here is where I remember reading about them last: Wikipedia article on equivalence principle

 

(by the way, the article has changed since the last time I read it)

 

It's the strong equivalence principle I'm thinking of. Wikipedia has it in words.

 

I believe it means the same thing as

 

SEP: Gravitational mass = Inertial mass

 

So Martin here is my question...

 

If the principle of equivalence is false, might it be the case that photons emitted by the sun are responsible for curving the path of the earth around it?

 

Now I will try and answer your question.

 

 

[math] R_0 = \frac {GM}{c^2} [/math]

 

Let me check the units first

 

G has units mmm/Kgss

M has units Kg

c has units mm/ss

 

GM has units mmmKg/Kgss = mmm/ss

 

GM/c^2 has units (mmm/ss)(ss/mm) = m

 

Ok yes units of distance, meters. So R₀ is a length.

 

You say that the angle it will be bent is 4R₀/R.

 

Hmm.

 

Well assuming that this angle formula thing is right, the answer is:

 

[math] G = 6.672 \times 10^{-11} \frac{m^3}{Kg s^2} [/math]

[math] M_{sun} = 1.988 \times 10^{30} Kg [/math]

[math] c = 299792458 \frac{m}{s} [/math]

 

Therefore:

 

[math] \frac{GM}{c^2} = \frac{(6.672 \times 10^{-11})(1.988 \times 10^{30})}{(299792458)^2} = 1475.8 [/math]

 

So the units are meters, R₀ for our sun, is 1475 meters.

 

You give the radius of the sun as 700,000 Km.

You give the perpendicular distance at which photons from the Plieades passed by our sun as 800,000 Km.

 

So in that angle formula thing, which you stated without proof, the value of R would be umm,

 

R is the distance from the center of our sun so...

 

R = 800,000 Km, = 800,000 X 1000 meters

 

[math] 4R_0 = 4(1475) = 5900 [/math]

 

[math] \frac{4R_0}{R} = \frac{5900 }{800,000 X 1000 } = \frac{5.9 }{800,000 } = 7.375 \times 10^{-6}[/math]

 

That is the answer I get to your question Martin.

[[Tycho?]]

How would the photons alone be responsible for the orbital paths?

 

Photons have never been obvserved to have any attractive force, that would be easily tested here on earth with a laser for example.

 

Plus we know the moon orbits around the earth, but the earth certainly emits photons in comparitavely small amounts, and only on its day side.

[Johnny5]

']

Photons have never been obvserved to have any attractive force' date=' that would be easily tested here on earth with a laser for example.

 

[/quote']

 

Not too long ago, I read about lasers beams that did attract/repel.

 

I think the beams had to be oriented perpendicularly to one another, though.

 

I was a bit startled when I read it, which is why I still remember it.

 

As for your other questions...

 

I don't know how they alone do it mathematically yet, I am only asking if it's even possible at this point. I've not yet even tried to come up with a theory of it. Theories can be such a waste of time when they are wrong.

 

Best to falsify the main idea first, if that can be done you save yourself work. Of course, if you have the right idea, then it cannot be falsified.

 

Actually, in what i was conceiving, there would be some kind of interaction between what the sun emits (which are photons), and the material in the earth, and after this interaction occurs, the earth's trajectory is curved according with reality.

 

Moon and earth interaction may be a counterexample I will give it some thought.

[J.C.MacSwell]

']How would the photons alone be responsible for the orbital paths?

 

Photons have never been obvserved to have any attractive force' date=' that would be easily tested here on earth with a laser for example.

 

Plus we know the moon orbits around the earth, but the earth certainly emits photons in comparitavely small amounts, [b']and only on its day side[/b].

 

Has temperature/emits photons.

[Martin]

... I've not yet even tried to come up with a theory of it. Theories can be such a waste of time when they are wrong.

 

Best to falsify the main idea first' date=' if that can be done you save yourself work. Of course, if you have the right idea, then it cannot be falsified.

 

Actually, in what i was conceiving, there would be some kind of interaction between what the sun emits (which are photons), and the material in the earth, and after this interaction occurs, the earth's trajectory is curved according with reality.

 

Moon and earth interaction may be a counterexample I will give it some thought.[/quote']

 

Congratulations J5, 7 millionths of a radian is the right answer for the bendinbg of light just barely passing by the sun (like at distance 800,000 km)

 

the distance you calculated R₀ = around 1500 meters is interesting for other reasons.

 

the radius of a black hole with sun's mass would be 2R₀

which comes to about 3 km.

 

A LIGHTRAY PASSING BY THE SUN AND A PLANET in the arc of its orbit around the sun are doing essentially the same thing.

 

the sun does not have to communicate (at the speed of light) with them while they are doing it, because it has already curved the space they run in.

 

gravity SIGNALS travel at speed c. Like if today the sun was abruptly annihilated the earth would not instantly leave its orbit and continue off into space in a straight line, space where we are would still be curved and, for 8 minutes, we would follow the accustomed track.

 

THEN the news would reach us and the earth would stop following a curve around where the sun used to be.

 

that is what is meant by gravity propagating at speed c. Changes in the gravitational field travel at speed c. News, information, travels at speed c.

 

When a lightray passes the sun in a curve it does not have to "talk back and forth" and "exchange gravitons" etc with the particles of matter in the sun. that is a massively complicated and I think ugly picture of gravity.

the picture I like better is that it does not have to talk back and forth because the gravitational field (the geometry of space) is already there to show it the curve it should travel

[swansont]

Not too long ago' date=' I read about lasers beams that did attract/repel.

 

I think the beams had to be oriented perpendicularly to one another, though.

 

I was a bit startled when I read it, which is why I still remember it.

 

[/quote']

 

Yes, there is the dipole force that can be generated if the light has an intensity gradient. The intense light has an electric field, and this induces a dipole in the atom, and thus a force. Depending on whether the laser is tuned to the red or blue of the resonance, you will get an attractive or repulsive force. The laser need not be close to resonance for this to happen, so the atom doesn't have to absorb photons and have to deal with the recoil which would tend to knock it out of the beam. You can trap already-cold atoms this way - known as a FORT (Far Off Resonance Trap), and force atoms to be in either an intense region or a dark region, depending on the detuning.

[Johnny5]

You guys are great.

 

I have to go now.

 

This might just work... something...