# Debunked: Telescope Distances of Billions of Light Years are impossible

#### Justatruthseeker

##### New Member
[Admin: The following post argues that the inverse square law makes it impossible to see long distances (billions of light years) with a telescope, however is off by several orders of magnitude, primarily because because it neglects the effects of long exposure. See full discussion in the thread that follows this post]

------------------------------------------------------------------------------
How far do you believe the Hubble allows us to see?

It's mirror has a surface area of 49,062.5 cm^2 the average human eye has a surface area of 0.38465 cm^2.

In comparison with the lens of the eye it has an objective diameter of 357.14 times larger than the human eye. It's surface area is 127,551 times that of the human eye.

This means Hubble collects 127,551 times more light than the human eye, so can make objects appear 127,551 times brighter than with the human eye.

Now, being the inverse square law of light says that the apparent intensity of the light of a point source is inversely proportional to the square of its distance to the observer. This means that if the distance of a star is doubled its apparent light is reduced four times. If its distance is increased three times its apparent luminosity is reduced nine times. See following link for the inverse square law.

http://hyperphysics.phy-astr.gsu.edu/hbase/vision/isql.html

Assuming that a star is so far away that it is barely visible to the naked eye, we know that the Hubble telescope can make the star appear 127,551 times brighter. Does this mean that the Hubble telescope enables an observer to see the star if it were 127,551 times farther away? The answer is no. The Inverse Square Law says that the light that we receive from a star is inversely proportional to the square of its distance. According to this law, at that distance, the light of the star becomes 127,551^2 or 16,269,262,700 times dimmer, far too dim for us to see with the telescope.

This raises the question: What is the maximum distance an object can be seen through the Hubble telescope? The answer is 357.14 times the distance that the naked eye can see. The reason is that an object 357.14 times farther away, its light becomes 127,551 times dimmer. Since the Hubble telescope can make a star appear 127,551 times brighter, then looking through the telescope the star would be barely visible.

Of course this does not take into account long exposures to film or digital media which would increase the distance several times, but not the claimed billions.

So would someone taking the inverse square law of light into effect show me how we can see galaxies a claimed 13.7 billion light years away? Remember - magnification spreads out the light received and so does not make a star appear brighter, but actually dimmer. Because this would mean the human eye can see 38,360,306 million light years????? See following link for the furthest object that the human eye can see.

http://www.physics.ucla.edu/~huffman/m31.html

The Andromeda Galaxy is the most distant object you can see with your naked eyes, two million light years away.
Content from External Source
The very math itself for the inverse square law of light debunks the claim of telescopic distances. But then that is why even professional astronomers will never discuss the inverse square law of light in connection with telescope distances. Every one I have contacted has refused to answer my question without ignoring the inverse square law.

Last edited by a moderator:

#### Chew

##### Senior Member.
Of course this does not take into account long exposures to film or digital media which would increase the distance several times.

"several times"?

Nope. Look into that and you'll find the answer to your question.

#### Chew

##### Senior Member.
Remember - magnification spreads out the light received and so does not make a star appear brighter, but actually dimmer

Oh, and this is mostly wrong. Higher magnification enables dimmer stars to be seen.

#### Justatruthseeker

##### New Member
"several times"?

Nope. Look into that and you'll find the answer to your question.

Yep.

And of course you ask we ignore the inverse square law in going beyond those "several times", right?

https://en.wikipedia.org/wiki/Inverse-square_law

The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

For example, the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation.
Content from External Source
But if you have something besides a claim that it can occur, please let me know.

Thanks

#### Justatruthseeker

##### New Member
Oh, and this is mostly wrong. Higher magnification enables dimmer stars to be seen.

Incorrect. Magnification defeats the purpose of making an image brighter. A common misconception among the populace is that magnification helps.

https://starizona.com/acb/basics/equip_magnification.aspx

Another reason for keeping the magnification low has to do with image brightness. An unfortunate law of physics dictates that when the magnification is doubled, the image gets four times dimmer.
Content from External Source

#### Balance

##### Senior Member.
Have you taken into account the image sensors? I'm sure they're not directly comparable to the human eye, which seems to be the basis of your query?

#### Justatruthseeker

##### New Member
Have you taken into account the image sensors? I'm sure they're not directly comparable to the human eye, which seems to be the basis of your query?

A telescope works on the same principle as the human eye - the ability to gather light.

https://starizona.com/acb/basics/observing_theory.aspx

How much more light a telescope gathers compared to the unaided eye is determined by the ratio between the light-gathering area of the telescope and the light-gathering area of the eye.
Content from External Source
https://en.wikipedia.org/wiki/Optical_telescope

A telescope's light gathering power and ability to resolve small detail is directly related to the diameter (or aperture) of its objective (the primary lens or mirror that collects and focuses the light). The larger the objective, the more light the telescope collects and the finer detail it resolves.
Content from External Source
It is only the aperture size of the primary mirror that determines it's light gathering power. The sensors have nothing to do with how much light it can gather. The sensors only determine the upper level of resolution. Example a 4 megapixel camera verses an 8 megapixel camera. The 8 megapixel camera has finer resolution but collects no more light than the 4 megapixel camera, assuming both have the same lens size opening.

#### Chew

##### Senior Member.
And of course you ask we ignore the inverse square law in going beyond those "several times", right?

lol. The inverse square law is inapplicable to the exposure time. Nice try though.

For an example of long exposure times, check out the Hubble Deep Field image. It is made of 342 separate stacked images taken for a minimum of 30 hours of exposure.

#### Chew

##### Senior Member.
Incorrect. Magnification defeats the purpose of making an image brighter. A common misconception among the populace is that magnification helps.

I must have magic eyes then. Whenever I zoom in on a star the background gets darker and the star stands out more.

#### Chew

##### Senior Member.
The sensors have nothing to do with how much light it can gather.

Sure. But it has everything to do with how much light is stored.

#### Justatruthseeker

##### New Member
lol. The inverse square law is inapplicable to the exposure time. Nice try though.

For an example of long exposure times, check out the Hubble Deep Field image. It is made of 342 separate stacked images taken for a minimum of 30 hours of exposure.

Which enables it to see several times further. But each and every one of those exposure times is limited by the inverse square law. It gathers no more light in one exposure than it does in another. They are stacked on top of one another to increase DETAIL.

A one hour exposure on Sunday gathers no more light than a one hour exposure on Tuesday.

#### Justatruthseeker

##### New Member
I must have magic eyes then. Whenever I zoom in on a star the background gets darker and the star stands out more.

Your eye does not magnify - you are simply focusing on one star and therefore gathering more light from that star and less from the surrounding stars. Don't confuse focusing on a star with magnification.

#### Justatruthseeker

##### New Member
Sure. But it has everything to do with how much light is stored.

Which can increase the distance several times. Yet again a one hour exposure on Sunday gathers no more light than a one hour exposure on Tuesday. It just allows you to increase the resolution (DETAIL) and only increases the distance by a factor of several times - not billions. The inverse square law ALWAYS applies, that's why we call it a law.

#### Justatruthseeker

##### New Member
If you are going to claim the inverse square law does not apply - you must first open a thread and debunk the inverse square law.

#### Balance

##### Senior Member.
A telescope works on the same principle as the human eye - the ability to gather light.

I understand that, and I apologise if I'm off target on what you're saying but am willing to learn
I'm just asking if electronic/optical imagery sensors would distort your comparison to the human eye?

#### Chew

##### Senior Member.
Your eye does not magnify - you are simply focusing on one star and therefore gathering more light from that star and less from the surrounding stars. Don't confuse focusing on a star with magnification.

You thought I was referring to increasing the magnification of my eye??? Are you for real?

#### Chew

##### Senior Member.
A one hour exposure on Sunday gathers no more light than a one hour exposure on Tuesday.

Do you think your eye can take a one hour exposure? Your OP compared the human eye to the Hubble.

And the inverse square law has nothing to do with exposure time. The inverse square law only determines how many photons will enter the telescope; it tells us nothing how long those photons are collected. It is an entirely different parameter than the inverse square law.

#### Chew

##### Senior Member.
It just allows you to increase the resolution (DETAIL)

No, it doesn't. The diameter of your optics determines resolution.

and only increases the distance by a factor of several times - not billions.

The distance doesn't need to be increased a billion-fold. As you said in the OP, M31 is 2 million light-years away.

[...]

Last edited by a moderator:

#### Mick West

Staff member
The most distant galaxies are only visible with gravitational lensing, and would be invisible to the Hubble alone.
http://www.space.com/18502-farthest-galaxy-discovery-hubble-photos.html

The new record holder is the galaxy MACS0647-JD, which is about 13.3 billion light-years away. The universe itself is only 13.7 billion years old, so this galaxy's light has been traveling toward us for almost the whole history of space and time.

Astronomers spotted the object using NASA's Hubble and Spitzer space telescopes, with the aid of a naturally occurring cosmic zoom lens as well. This lens is a huge cluster of galaxies whose collective gravity warps space-time, producing what's called a gravitational lens. As the distant galaxy's light traveled through this lens on its way to Earth, it was magnified.

"This cluster does what no manmade telescope can do," Marc Postman of the Space Telescope Science Institute in Baltimore, Md., said in a statement unveiling the discovery today (Nov. 15). "Without the magnification, it would require a Herculean effort to observe this galaxy." Postman leads the Cluster Lensing And Supernova Survey with Hubble (CLASH), which performed the study.
Content from External Source
Although the infrared James Webb Space Telescope apparently should be able to see things at that distance.
http://imagine.gsfc.nasa.gov/features/cosmic/farthest_info.html
(Note: JWST will be able to see these first galaxies without the aid of gravitational lensing; gravitational lensing might allow us to see them better, but would not necessarily let us see further back in time.)
Content from External Source

#### Justatruthseeker

##### New Member
I understand that, and I apologise if I'm off target on what you're saying but am willing to learn
I'm just asking if electronic/optical imagery sensors would distort your comparison to the human eye?

The optical sensors simply affect the detail that can be attained. And all fall far short of the human eye.

http://www.cambridgeincolour.com/tutorials/cameras-vs-human-eye.htm

Most current digital cameras have 5-20 megapixels, which is often cited as falling far short of our own visual system. This is based on the fact that at 20/20 vision, the human eye is able to resolve the equivalent of a 52 megapixel camera (assuming a 60° angle of view).
Content from External Source

The Wide Field Camera 3 (WFC3), which is to be installed during the servicing mission in 2009, will also have 2 CCD chips each of 2048 x 4096 pixels for a total of 16 mega-pixels.
Content from External Source
Your camera sees nothing your eyes do not see. If the CCD affected what you see then cameras would not take the same image our eyes see.

So in reality the Hubble has less resolution of detail than is capable by the human eye. But again - resolution of detail has nothing to do with it's ability to collect and gather light which is determined solely by its aperture and the inverse square law of light. As stated it would only increase the distance several times, because each exposure is limited in distance by the inverse square law. Multiple exposures do not enable it to see the billions of light years that is claimed. A point source would have to be metaphysical in brightness to attain the distances claimed by modern cosmology since the intensity of the light is proportional to the square of the distance.

That is - the sun at 3 times the distance would be nine times dimmer.

Since further galaxies are claimed to be young - they should therefore have less stars and be less bright - not increase in brightness which would be required under current belief in distances of telescopes. This is why metaphysical processes are declared for those further galaxies, because in order to be seen at the distances claimed - they would have to have a light source beyond anything even imagined.

Distances are calculated by cosmological redshift - which is the biggest blunder in all of modern cosmology.

http://www.newtonphysics.on.ca/hubble/

It is known that many astronomical observations cannot be explained by means of the ordinary Doppler shift interpretation. The mere examination of a recent catalog of objects having very large redshifts shows that among 109 quasi-stellar objects for which both absorption and emission lines could be measured, the value of the absorption redshift of a given object is always different from the one measured in emission for the same object. It is clear that such results cannot be explained as being due solely to a Doppler redshift.
A new mechanism must be looked for, in order to explain those inconsistent redshifts and many other observations related to the “redshift controversy”.
Content from External Source

#### Mick West

Staff member
lol. The inverse square law is inapplicable to the exposure time. Nice try though.

For an example of long exposure times, check out the Hubble Deep Field image. It is made of 342 separate stacked images taken for a minimum of 30 hours of exposure.

The effective exposure time of the human eye is under 0.1 seconds (try opening and closing your eyes within 0.1 seconds, you can still see stars). The longest Hubble UDF exposure was effectively ~350,000 seconds. So the exposure multiplication is (conservatively) 3.5 million.

So if we accept the OPs figure of 127,551, that means the Hubble gets 450 billion times as much light as the human eye during the combined exposure.

#### Justatruthseeker

##### New Member
No, it doesn't. The diameter of your optics determines resolution.

The distance doesn't need to be increased a billion-fold. As you said in the OP, M31 is 2 million light-years away.

[...]

And at 6 million light years away would be nine times dimmer.

[...]

Last edited by a moderator:

#### Justatruthseeker

##### New Member
The effective exposure time of the human eye is under 0.1 seconds (try opening and closing your eyes within 0.1 seconds, you can still see stars). The longest Hubble UDF exposure was effectively ~350,000 seconds. So the exposure multiplication is (conservatively) 3.5 million.

So if we accept the OPs figure of 127,551, that means the Hubble gets 450 billion times as much light as the human eye during the combined exposure.

And the inverse square law does not allow it to see any further than 357.14 times the distance of the human eye, because at 357.14 times the distance the light is 127,551^2 or 16,269,262,700 times dimmer, far too dim for us to see with the telescope.

[off topic material removed]

Last edited by a moderator:

#### Mick West

Staff member
And the inverse square law does not allow it to see any further than 357.14 times the distance of the human eye, because at 357.14 times the distance the light is 127,551^2 or 16,269,262,700 times dimmer, far too dim for us to see with the telescope.

With inverse square, the fraction of light (F) of an object at distance Dmax relative to the max distance we can see by eye (Deye). We know long exposure Hubble can see something with 1/(450 billion) times the light.

F = (Dmax/Deye)^2 (inverse square law)

Hence
Dmax = Deye*Sqrt(F)
Dmax = Deye * Sqrt(450,000,000,000)
Dmax = Deye * 670,820

Which is considerable more than 357. So please address that before proceeding.

#### Justatruthseeker

##### New Member
With inverse square, the fraction of light (F) of an object at distance Dmax relative to the max distance we can see by eye (Deye). We know long exposure Hubble can see something with 1/(450 billion) times the light.

F = (Dmax/Deye)^2 (inverse square law)

Hence
Dmax = Deye*Sqrt(F)
Dmax = Deye * Sqrt(450,000,000,000)
Dmax = Deye * 670,820

Which is considerable more than 357. So please address that before proceeding.

In comparison with the lens of the eye it has an objective diameter of 357.14 times larger than the human eye. It's surface area is 127,551 times that of the human eye.

http://www.physics.ucla.edu/~huffman/m31.html

The Andromeda Galaxy is the most distant object you can see with your naked eyes, two million light years away.
Content from External Source

[...]

#### Chew

##### Senior Member.
which is determined solely by its aperture and the inverse square law of light.

Your belief that those are the only two factors is apparently unshakable. Good luck with that.

#### Mick West

Staff member

Dmax = Deye * 670,820

Dmax = 2 million light years * 670,820 = ~1.2 trillion light years.

#### Justatruthseeker

##### New Member
Your belief that those are the only two factors is apparently unshakable. Good luck with that.

Oh there is also dust extinction, but it seems that is to be ignored since it's off topic somehow in a discussion about telescopic distances.

#### Justatruthseeker

##### New Member
Dmax = Deye * 670,820

Dmax = 2 million light years * 670,820 = ~1.2 trillion light years.

Whats F, since that's the fraction of light we can see at Dmax, but you can't know that without assumption.

Since Dmax = Deye*Sqrt(F) you must first deduce F. But since F relies on DMax, your assuming quite a few parameters without proving any of them.

#### Chew

##### Senior Member.
Oh there is also dust extinction, but it seems that is to be ignored since it's off topic somehow in a discussion about telescopic distances.

From the same website you quoted earlier:

https://starizona.com/acb/ccd/projectsfaint.aspx
With a CCD camera on the same 8" telescope it is possible to reach the same 16th magnitude limit with only a few seconds exposure. In 60 seconds, magnitude 18 can be reached -- approximately the visual limit on a 30" scope! Longer exposures will reach fainter magnitudes, but there are limitations -- and ways around them!
Content from External Source

#### Mick West

Staff member
F was calculated earlier as 450 billion by multiplying the incident light area scale (127,551) by the exposure time scale (3.5 million)

The effective exposure time of the human eye is under 0.1 seconds (try opening and closing your eyes within 0.1 seconds, you can still see stars). The longest Hubble UDF exposure was effectively ~350,000 seconds. So the exposure multiplication is (conservatively) 3.5 million.

So if we accept the OPs figure of 127,551, that means the Hubble gets 450 billion times as much light as the human eye during the combined exposure.

#### Mick West

Staff member
Whats F, since that's the fraction of light we can see at Dmax, but you can't know that without assumption.

Since Dmax = Deye*Sqrt(F) you must first deduce F. But since F relies on DMax, your assuming quite a few parameters without proving any of them.

We know F for the specific situation we are talking about (the Hubble), as we know the relative size of the mirror and exposure time, so we can use F and Deye to calculate Dmax.

#### Mick West

Staff member
Dmax = Deye * 670,820

Dmax = 2 million light years * 670,820 = ~1.2 trillion light years.

And if a Dmax of 1.2 trillion light years seems rather large, as the universe is only 13 billion, remember that the distant objects are generally (relatively) very small. The Andromeda galaxy is 220,000 light years across, but the farther object, MACS0647-JD, is only 600 light years across.

So Dmax in that equation above is the maximum distant that andromeda would be visible. You'd ned to scale that by the relative brightness of the object.

#### Justatruthseeker

##### New Member
F was calculated earlier as 450 billion by multiplying the incident light area scale (127,551) by the exposure time scale (3.5 million)

The 127,551 already includes the calculation of light gathered by the mirror. Exposure time does not increase that amount of light given because you must first also apply the inverse square law to any distance. At 357.14 times the distance of the human eye the light is 127,551^2 or 16,269,262,700 times dimmer. You can't multiply exposure time by the light collected by the amount of light gathered. It doesn't work that way. You are violating the inverse square law. The same point source 357.14 times the distance of the human eye becomes 127,551^2 or 16,269,262,700 times dimmer. All you can do is multiply that 2 million by 357.14 (the difference in the objective diameter of the telescope compared to the human eye).

This gives you 714,280,000 light years max. No this does not account for long exposures which increase it several times, but you cant support billions. You know this as well as I do. You can't ignore objective diameter which is 357.14 times that of the eye allowing it to collect 127,551 times more light at 2 million light years. Exposure time does not increase the objective diameter nor the surface area. Nor does it increase the distance except to make fainter objects brighter.

Then you clipped the part of dust extinction because you knew it shattered your beliefs and you had no work around to that problem. Your only solution was to clip dust extinction from the debate and claim it is off-topic in a topic about astronomical viewing range - of which dust extinction is paramount.

All I am getting from this is you will declare whatever you can't answer as off-topic to avoid having to deal with it.

#### Mick West

Staff member
Exposure time does not increase that amount of light given
yes it does, more time = more photons.

Exposure time does not increase the objective diameter nor the surface area. Nor does it increase the distance except to make fainter objects brighter.

The inverse square law says an object that is X times as far away as another object will require X^2 as much light to be equally visible.

So, for 13 billion vs 2 million, you need (13,000)^2, or 169,000,000 times as much light.

You can do that with a 127,000x larger area (mirror vs. eye) and a 1330 times longer exposure. All quite possible with the Hubble.

[Edit: Correction for value of 13 billion / 2 million]
So, for 13 billion vs 2 million, you need (6,500)^2, or 42,250,000 times as much light.

You can do that with a 127,000x larger area (mirror vs. eye) and a 332 times longer exposure. All quite possible with the Hubble.

Last edited:

#### Mick West

Staff member
Then you clipped the part of dust extinction because you knew it shattered your beliefs and you had no work around to that problem. Your only solution was to clip dust extinction from the debate and claim it is off-topic in a topic about astronomical viewing range - of which dust extinction is paramount.

Dust is irrelevant in the calculation of Dmax, because it's a linear factor that is already included in the Andromeda observation.

#### Mick West

Staff member
I think the root misunderstanding here is that you don't understand how longer exposure can make a distant object visible.

Would you agree though that a ten second exposure of the sky at night would show a lot more stars than a 1/10th second exposure. All other things being equal?

#### Justatruthseeker

##### New Member
yes it does, more time = more photons.

The inverse square law says an object that is X times as far away as another object will require X^2 as much light to be equally visible.

So, for 13 billion vs 2 million, you need (13,000)^2, or 169,000,000 times as much light.

You can do that with a 127,000x larger area (mirror vs. eye) and a 1330 times longer exposure. All quite possible with the Hubble.

More photons that still obey the inverse square law.

But Hubble does not have a mirror 127,000x larger, but only 357.14 times as large. It's objective diameter is only 357.14 times that of the eye, not 127,000 times larger.

No because at 13 billion the light would be (13,000,000,000)^2 not 13,000^2 or 1.69x10^20th dimmer

#### Justatruthseeker

##### New Member
I think the root misunderstanding here is that you don't understand how longer exposure can make a distant object visible.

Would you agree though that a ten second exposure of the sky at night would show a lot more stars than a 1/10th second exposure. All other things being equal?

It would increase the amount of stars visible, but not the distance that those stars could be seen by any drastic measurement. It would make the fainter ones more visible but not by any measurement of just a few parsecs, if that.

And we have yet to add light extinction into the equation by dust, which is 30 times more abundant then previously believed.

#### Mick West

Staff member
No because at 13 billion the light would be (13,000,000,000)^2 not 13,000^2 or 1.69x10^20th dimmer

But we are calculating the relative brightness of things at 13 billion and 2 million, not 13 billion and 1.

I made a mistake above, the relative brightness would be
(13,000,000,000/2,000,000)^2
or 6500^2, hence 42 million times dimmer

You understand this? 13 billion light years is only 6500x times as far away as 2 million light years. Right?

So if something is 6500x as far way, it would be 42,250,000 times dimmer, right?

So to see it, you just need to get 42,250,000 as much light onto the sensor, right?

Replies
21
Views
2K
Replies
5
Views
519
Replies
69
Views
6K
Replies
0
Views
972