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# Mass of Light

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I believe you but I don't understand your calculation. The sun radiates in all directions. What the Earth receives is a tiny part of it. Where do the 500s come from?

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mm. the 500s is the orbital radius. So you are estimating the volume of the cylinder from Earth to the Sun.

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Isn't there something missing?

Sphere. Newtonian gravity depends on the mass inside the radius of the sphere — it acts as if all the mass were at the origin. That's the relevant mass equivalent (the energy) you'd have in general relativity.

"The sun converts about 4 kg/s into other forms of energy" - I am not disputing that number but it does blow my mind. I have looked after oil-tankers that burn almost as kilograms of fuel oil per second just to move across the ocean! It just shows the complete difference in scale between chemical energy burning and the energy release in nuclear reactions. All life on earth (well barring hydrothermal vents etc) - that lovely sunshine...waxes lyrical for a while about nature... all from 4kg per second. Amazing

I misread the source — truncation error. It's 4 billion kg. Doesn't affect the overall argument for ignoring it.

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"The sun converts about 4 kg/s into other forms of energy" - I am not disputing that number but it does blow my mind. I have looked after oil-tankers that burn almost as kilograms of fuel oil per second just to move across the ocean! It just shows the complete difference in scale between chemical energy burning and the energy release in nuclear reactions. All life on earth (well barring hydrothermal vents etc) - that lovely sunshine...waxes lyrical for a while about nature... all from 4kg per second. Amazing

From ramblings over the net there may be a 10^9 missing.

??

Found this:

kg/sec = Solar luminosity/c² = 3.939E26 W/3E8² = 4.377E9 kg/sec
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From ramblings over the net there may be a 10^9 missing.

??

Found this:

Already noted and fixed.

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cross posting

But there are a lot of other photons coming from all over the universe passing by this sphere, going in all directions that do not cross our planet.

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cross posting

But there are a lot of other photons coming from all over the universe passing by this sphere, going in all directions that do not cross our planet.

Many less than from the sun. It is dark at night.

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Many less than from the sun. It is dark at night.

That's what i asked in another thread.

If there is only one photon for each star in the sky, that makes billions of billions at any place.

I mean, if I look at a star and change position taking care to keep my eye on the star, there is no gap, there is no place where there is no photon coming from this star.

Photons from any faint star are literally everywhere.

The vacuum must be full like an egg.

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That's what i asked in another thread.

If there is only one photon for each star in the sky, that makes billions of billions at any place.

I mean, if I look at a star and change position taking care to keep my eye on the star, there is no gap, there is no place where there is no photon coming from this star.

Photons from any faint star are literally everywhere.

The vacuum must be full like an egg.

It's not quite that simple. Photons are not particles in a classical sense they also have wave properties.

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The sun converts about 4 billion* kg/s into other forms of energy, which is eventually radiated away at c.

2e+30 / 4e+9 = 5e+20 seconds needed to convert everything.

5e+20 / 3600 / 24 / 365 = almost 16e+12.

16 trillion years to convert everything.

But nuclear reactions are supposed to end after 4e+9 years.

Space should be full of such objects that don't radiate anything anymore, or so little they're not detectable by devices.

In the volume inside the earth, that's 500 s of travel, or the equivalent of 2 x 10^12 kg. The mass of the sun is 2 x 10^30 kg.

That's true, sphere with radius sun+150 million km is containing just "how massive sun was 500 seconds ago".

In mine post I was thinking about 14 billion * 365 * 24 * 3600 * 300,000 km spheres.. radius = 1.32e+23 km

It's not quite that simple. Photons are not particles in a classical sense they also have wave properties.

So the same regular particles, if believing Broglie

http://en.wikipedia.org/wiki/Matter_wave

mm. the 500s is the orbital radius. So you are estimating the volume of the cylinder from Earth to the Sun.

Not cylinder but rather cone.

Cylinder would be correct only when radius of Earth and Sun would be equal.

Edited by Przemyslaw.Gruchala
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That nobody cared making an evaluation of the gravitational effect of light.

That's not what happened. That's not what a physicist generally means by "ignore" in this context.

The sun converts about 4 billion* kg/s into other forms of energy, which is eventually radiated away at c. In the volume inside the earth, that's 500 s of travel, or the equivalent of 2 x 10^12 kg. The mass of the sun is 2 x 10^30 kg. Ignoring the effect of the light gives an error of a part in 10^18 in terms of effect on the earth. At that point you ignore it, unless you are doing an experiment where a part in 10^18 matters.

Not including in a calculation is not the same as refusing to consider.

The immediate effect is significantly smaller than one part in 10^18. The solar system is nearly Newtonian in behavior. One way to account for relativistic effects is a parametric post-Newtonian formalism, which models relativistic effects as perturbations of Newtonian gravity.

The light that is still in transit from the Sun to the Earth is essentially indistinguishable from mass at the center of the Sun. It is completely indistinguishable in a Newtonian context thanks to Newton's shell theorem. There's a slight relativistic effect, but now we're looking at a tiny perturbation of any already tiny perturbation.

The same applies to the light that was emitted more than eight minutes ago. In a Newtonian context the contribution of this older light is identically zero, once again thanks to Newton's shell theorem. There once again is a slight relativistic effect, but once again this is but a tiny perturbation on top of an already tiny perturbation.

There is a cumulative effect due to the Sun losing mass. The Sun also loses mass in the form of the solar wind. Both of these effects are very small, less than 10-13 solar masses per year. This mass loss should change the orbits of planets slightly, but the effect is negligibly small (i.e., much less than measurement error).

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$\text{Since light has energy, there is a small amount of mass equal to that energy.}$

$Given: f=1.0\times10^{15} \text{ Hz}, h=6.626\times10^{-34} J\cdot s, c=299792458 \text{ m/s}$

$Find: \text{mass of photon } m_p$

$E=hf, E=mc^2$

$\text{Equate the 2 values of E:}$

$hf=mc^2$

$m_p=\frac{hf}{c^2}$

$=\frac{\left(6.626\times10^{-34}\right)\left(1.0\times10^{15}\right)}{\left(299792458\right)^2}$

$=7.37242\times10^{-36} kg$

$\text{Compare with mass of electron }m_e=9.10939\times10^{-31} \text{ kg}$

$m_e/m_p=123,560$

The modern definition of mass is given as

$m \equiv \frac{\sqrt{E^2 - p^2 c^2}}{c^2}$

For a photon $E_\gamma = pc$ therefore $m_\gamma=0$ and this is the well-known result that photons $\gamma$ are massless particles. Applying the definition for an electron we obtain the well-known value for $m_e$ reported below

Note that the value $9.10939\times10^{-31} \mathrm{kg}$ that you give for the electron is inaccurate in its last figure. It is $9.10938\times10^{-31} \mathrm{kg}$ because the figures beyond the 8 are 291(40) and this is less than 500... The last recommended value for electron mass is found in the next table of constants

http://juanrga.com/en/knowledge/scientific-constants.html

Edited by juanrga

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