# Amount of energy inside a sphere of radius 1au around the Sun

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• confusedhome
In summary, the conversation discusses a thought experiment about placing a sphere around the sun with a radius equal to the orbit of Earth. It also explores the concept of the entire universe being a medium through which light travels and suggests that the universe should have a refractive index. The conversation also touches on the idea that the total mass of photons present in the universe is not taken into account in our current understanding of cosmology.
confusedhome
TL;DR Summary
A thought experiment
Suppose a sphere were to be placed around our Sun with a radius equal to the radius of the orbit of Earth, it would have a volume of 6.266x1022 m3. At any instant in time there would be 3.846x1027 Watts of energy within this volume in the form of photons from the Sun.

Since watts= joules/sec. this gives us 3.846x1027 Joules of energy at any instant in time, which according to the Theory of Relativity (E=MC2) would give us an equivalent mass of 1.2828x1021 Kg. That is approximately equal to the mass of a small planet (Pluto =1.25 x 1022 Kg.).

You will notice I emphasize ‘instant in time’ and do not take into consideration the photons from other stars and galaxies which are traveling through this volume at the same instant.

Now expand that same concept to the entire Universe, even if only the Universe we can observe. You might notice that the Universe itself is a medium through which light travels and should be recognized to have a ‘Refractive Index’

Science ‘assumes’ a refractive index of the universe to be equal to 1. But we have no way to be certain with our current level of technology or knowledge. All we really know for certain is that our universe is one huge glob of photons and mass each acting on the other through gravity yet we do not seem to be taking into account the total mass of the total photons present everywhere, constantly.

Just a thought,
Bob

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You should really use latex, or make your notation more readable otherwise. (I assume that e.g. the power output you write is ##3.846\times 10^{27}## W). Based on wikipedia you are off by an order of magnitude (should be ##10^{26}## W). Anyway regarding this

confusedhome said:
Summary: A thought experiment

Since watts= joules/sec. this gives us 3.846x1027 Joules of energy at any instant in time, which according to the Theory of Relativity (E=MC2) would give us an equivalent mass of 1.2828x1021 Kg. That is approximately equal to the mass of a small planet (Pluto =1.25 x 1022 Kg.).

the mass of the system would probably be basically the mass of the sun, at ##\approx 10^{30}## kg (and of course not change). Over time, this would be converted into energy of radiation inside your sphere (though at very slow rate, you would probably not notice a significant difference in the mass of the sun over its lifetime).

I don’t get what you are asking in the second part of your post.

In mks c2≅1017

confusedhome
confusedhome said:
At any instant in time there would be 3.846x1027 Watts of energy

Watts aren't a unit of energy, they are a unit of power--energy per unit time.

confusedhome said:
within this volume

No, there are not that many watts within the entire volume inside the sphere. There are that many watts passing through the sphere itself.

confusedhome said:
3.846x1027 Joules of energy at any instant in time,

No, ##3.846 \times 10^{27}## Joules passing through the sphere over the course of one second. Not at an instant.

confusedhome said:
which according to the Theory of Relativity (E=MC2) would give us an equivalent mass of 1.2828x1021 Kg

Check your math. In SI units, ##c^2## is about ##9 \times 10^{16}##. That gives a much smaller number when divided into ##3.846 \times 10^{27}##.

confusedhome said:
All we really know for certain is that our universe is one huge glob of photons and mass each acting on the other through gravity yet we do not seem to be taking into account the total mass of the total photons present everywhere, constantly.

That's because that total mass is miniscule compared to the mass of ordinary matter, dark matter, and dark energy. If you do the math correctly it will make that obvious.

confusedhome
confusedhome said:
All we really know for certain is that our universe is one huge glob of photons and mass each acting on the other through gravity yet we do not seem to be taking into account the total mass of the total photons present everywhere, constantly.

As a statement about our actual theory of cosmology, this is false. Cosmology does take the energy contained in photons into account--that's why our model of the universe says it was radiation dominated (which means the energy in radiation, like photons, was the largest component of the total energy in the universe) prior to a hundred thousand years or so after the Big Bang.

confusedhome
confusedhome said:
Science ‘assumes’ a refractive index of the universe to be equal to 1. But we have no way to be certain with our current level of technology or knowledge. All we really know for certain is that our universe is one huge glob of photons and mass each acting on the other through gravity yet we do not seem to be taking into account the total mass of the total photons present everywhere, constantly.
Sure we do. It's approximately 0.006% of the total energy density. It's typically ignored because it's so small as to be negligible in almost every context.

Hello Kimbyd,
Yes, it may seem small from localized viewpoints but when taken overall cumulatively it amounts to much more than that for it exists even in the vast empty regions.
The total energy density is not what I am referring to, it is only the mass equivalent of the photons which needs to be considered seriously as small differences can account for large differences, much like the ''Butterfly Effect"
Thanks for giving it some thought.
Bob

Dr.AbeNikIanEdL said:
You should really use latex, or make your notation more readable otherwise. (I assume that e.g. the power output you write is ##3.846\times 10^{27}## W). Based on wikipedia you are off by an order of magnitude (should be ##10^{26}## W). Anyway regarding this
the mass of the system would probably be basically the mass of the sun, at ##\approx 10^{30}## kg (and of course not change). Over time, this would be converted into energy of radiation inside your sphere (though at very slow rate, you would probably not notice a significant difference in the mass of the sun over its lifetime).

I don’t get what you are asking in the second part of your post.
Hello Dr.,
I apologize for the format, it was correct until I pasted into the post so yes your assumption was correct on that but the value stated was for the entire volume of the sphere, not only the surface area which includes the 8.19 second time it takes light to traverse that distance.
Bob
PeterDonis said:
Watts aren't a unit of energy, they are a unit of power--energy per unit time.
No, there are not that many watts within the entire volume inside the sphere. There are that many watts passing through the sphere itself.
No, ##3.846 \times 10^{27}## Joules passing through the sphere over the course of one second. Not at an instant.
Check your math. In SI units, ##c^2## is about ##9 \times 10^{16}##. That gives a much smaller number when divided into ##3.846 \times 10^{27}##.
That's because that total mass is miniscule compared to the mass of ordinary matter, dark matter, and dark energy. If you do the math correctly it will make that obvious.

The math is correct for it includes the entire 8.19 seconds of time from point A to point B as a volume, not a surface.
Also viewed this way Watts is a unit of energy unit of Joules/meters/ (8.19 sec.)
Thanks for considering this,
Bob
PeterDonis said:
As a statement about our actual theory of cosmology, this is false. Cosmology does take the energy contained in photons into account--that's why our model of the universe says it was radiation dominated (which means the energy in radiation, like photons, was the largest component of the total energy in the universe) prior to a hundred thousand years or so after the Big Bang.
A Theory is only a Theory until it is proven, which it isn't.
Bob

confusedhome said:
The math is correct

What math? You didn't derive anything. You just threw out numbers. Please show your work.

confusedhome said:
Also viewed this way Watts is a unit of energy unit of Joules/meters/ (8.19 sec.)

This is nonsense. Watts are units of power, not energy. Joules/meters/8.19 sec is a unit of Watts per meter (times a constant because of the 8.19 in the denominator), which is not a unit of energy.

confusedhome said:
A Theory is only a Theory until it is proven, which it isn't.

Our best current model of the universe in cosmology has plenty of evidence in its favor. You don't get to ignore it just because you have a highly idiosyncratic interpretation of the word "theory".

confusedhome said:
3.846x1027 Watts

I believe this is too high by an order of magnitude; it should be ##3.846 \times 10^{26}## Watts.

confusedhome said:
the entire 8.19 seconds of time from point A to point B

If you mean 8.19 seconds as the time it takes for light to travel the distance from Sun to Earth, it should be minutes, not seconds.

confusedhome said:
the value stated was for the entire volume of the sphere, not only the surface area which includes the 8.19 second time it takes light to traverse that distance.

That does make even less sense, can you show what you are actually trying to calculate and where your numbers are from?

Folk,
It is a thought experiment to obtain a different view point, the math is simple, as is the thought so you can think about the point I'm trying to make or don't.

confusedhome said:
the math is simple

Sure, if you get the numbers right. I pointed out two wrong ones in previous posts. And if you do the math right, you will see, as has already been pointed out, that the average energy density of photons in the universe now is negligible. Which addresses the issue you raise in your OP.

hutchphd

## 1. How is the amount of energy inside a sphere of radius 1au around the Sun calculated?

The amount of energy inside a sphere of radius 1au around the Sun, also known as the solar constant, is calculated by dividing the total solar irradiance (TSI) by the surface area of the sphere. TSI is the amount of solar energy that reaches the Earth's atmosphere per unit area per unit time.

## 2. What is the value of the solar constant?

The current estimated value of the solar constant is approximately 1361 watts per square meter. However, this value can vary slightly depending on factors such as the Sun's activity and Earth's distance from the Sun.

## 3. How does the amount of energy inside a sphere of radius 1au around the Sun affect Earth's climate?

The solar constant plays a crucial role in Earth's climate as it determines the amount of solar energy that reaches the Earth's surface. Changes in the solar constant can lead to changes in Earth's temperature and climate patterns.

## 4. How does the solar constant vary throughout the year?

The solar constant is not constant throughout the year. It varies slightly due to the Earth's elliptical orbit around the Sun, with the highest values occurring when Earth is closest to the Sun (perihelion) and the lowest values occurring when Earth is farthest from the Sun (aphelion).

## 5. How is the amount of energy inside a sphere of radius 1au around the Sun measured?

The amount of energy inside a sphere of radius 1au around the Sun is measured using instruments such as satellites and ground-based telescopes. These instruments measure the total solar irradiance and calculate the solar constant using the formula mentioned in the first question.

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