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View Full Version : If we live in a 3-sphere, would we be able to see stars on the other side?

Frog march
2009-Aug-17, 08:06 AM

http://farm4.static.flickr.com/3423/3828847825_87a32c6eba_o.jpg

ok, it's a bit homemade, but it shows a 2-sphere with a star near the top giving off light, that radiates out as yellow circles.

There is an "equator" which represents the halfway point around the 2-sphere.

There is an observer on the opposite side of the 2sphere, from the star.

My question is: will he/she be able to see the star, as the light entering their eye will be coming from a wide area of the sky...if you see what I mean.

This should still apply to a 3-sphere, so I was wondering if we are limited, in the stars/galaxies we see, to those that would be on "our side" of the 3-sphere(if we do live in a 3sphere).

hhEb09'1
2009-Aug-17, 10:27 AM
You have stumbled upon a very interesting concept.

Going to your 2-D diagram, should the light from the star ever reach the opposite side, it would be as if the entire output of the star hit at once--allowing for whatever dissipation had occurred along the way.

Earth moves enough each year that it'd probably miss us--after all, we manage to avoid most of Sol's output. :)

The same thing happens with earthquakes upon the earth--there is a strong focussing effect at the antipodes. It's not as if there'd be another earthquake at the antipodes, but it is measureable--just hard to measure because it is strongly local, and by coincidence, most seismic monitoring stations are not directly opposite strong seismic zones.

Fred Sagen
2009-Aug-17, 10:35 AM
This strikes me as a very hypothetical question analogous to a Flatlander attempting to understand the hypothetical, to him, third dimension.

Correct me if I'm wrong but your diagram of a 2-sphere suggests that it's 'universe' is restricted to the two-dimensional surface of a three-dimensional sphere and your question relates to a 3-sphere that occupies the three-dimensional 'surface' of a four-dimensional (hyper)sphere.

If I am correct in this interpretation then, plainly, an occupant of the 2-sphere would see the whole 'sky' intensely illuminated by the diametrically opposed star.

Any other stars not diametrically opposed to the observer would illuminate the whole 'sky' but more intensely in the direction of the shortest path to that star.

Of course, if these 2-sphere stars were anything like the stars we are familiar with and the observer anything like us s/he would be instantly vapourised by the focussing of the stars radiation on the observation point.

Intervening gas and dust may prevent this from occurring but the problem of an evenly illuminated 'sky' would persist.

If the 2-sphere were an expanding balloon the radiation from the diametrically opposed star would be red-shifted (assuming a fixed and constant speed of light) and may (depending on the rate of inflation of the 2-sphere) present itself as uniform background microwave radiation.

I trust that you can see how this might translate to a higher dimensional space but as I said, this is a very hypothetical scenario and just because it serves as an analogue of some of our real-world observations we should not accept it as an accurate representation of reality as it will very probably fall to pieces on closer examination like an overworked metaphor.

Frog march
2009-Aug-17, 01:04 PM
just to make it clear; the observer can be anywhere in the opposite half of the 2-sphere, not just directly opposite the star.

hhEb09'1
2009-Aug-17, 01:17 PM
Or, even the nearside half--there is still a path that reaches all the way around the opposite way.

Born mentions this possibility in his popularization, Einstein's Theory of Relativity, that a certain configuration of stars in one direction might be detected in the opposite direction.

a1call
2009-Aug-17, 01:40 PM
That's a very good question. I think I have heard the concept before on this board but it took your :liar:homemade:liar: rendering to make me grasp it.

The way I see it nothing changes further from the equator. The assumption that space is curved does not imply that there is more than one straight (curving along the skin) line of sight anywhere except precisely at the anti-pole.

In your rendering only one great-circle/meridian (equivalent to straight line of sight) will pass through both points. Any other circle/arc would not be a straight path.

Frog march
2009-Aug-17, 02:00 PM
In your rendering only one great-circle/meridian (equivalent to straight line of sight) will pass through both points. Any other circle/arc would not be a straight path.

but to form an image on the retina(or whatever) would take more than one great circle.
There would be a great circle for every light-path, that entered through the eye, or telescope, lens.

My question was whether, you would even get an image on the opposite half of the 2-sphere(or 3-sphere).
I know individual paths exist, but the information it would take to make an image of the star(even as a small dot) would be spread out across part(half?) of the sky.

If you see what I mean.

On the opposite half of the 2-sphere, the light circle(sphere for a 3-sphere) would be contracting---the light coming from different parts of the sky, not appearing to come from a small location.

eta: I dunno; maybe it would form an image...

hhEb09'1
2009-Aug-17, 02:16 PM
eta: I dunno; maybe it would form an image...It would.

The great circle analogy is similar to our real world straight line. There is only one straight line between any two points but we can still see objects. On your sphere, there are two straight lines, one from the near direction, the other from the far, opposite direction.

a1call
2009-Aug-17, 02:17 PM
but to form an image on the retina(or whatever) would take more than one great circle.
There would be a great circle for every light-path, that entered through the eye, or telescope, lens.

There is no difference in a curved or flat space. Light from a point/object emanates in all direction along a straight-line/great-circle. Whatever reaches the lens of the eye along these lines is focused on the retina/film.

My question was whether, you would even get an image on the opposite half of the 2-sphere(or 3-sphere).

Yes, as long as light has the time to travel that far and the red shift does not make it invisible.

I know individual paths exist, but the information it would take to make an image of the star(even as a small dot) would be spread out across part(half?) of the sky.

Individual paths don't exist. Only a single path exists anywhere except precisely at the anti-pole where unlimited paths exist.

On the opposite half of the 2-sphere, the light circle(sphere for a 3-sphere) would be contracting---the light coming from different parts of the sky, not appearing to come from a small location.

You are assuming that light emanating from a point on the sphere in all directions will reach any point on the other side along many paths. This is a false assumption. It will reach any other point other than the anti-pole only along a single great-circle.

hhEb09'1
2009-Aug-17, 02:20 PM
Individual paths don't exist. Only a single path exists anywhere except precisely at the anti-pole where unlimited paths exist.
There is the other path the "long way around" the great circle.

a1call
2009-Aug-17, 02:22 PM
I stand corrected.:)

Frog march
2009-Aug-17, 03:19 PM
It would.

The great circle analogy is similar to our real world straight line. There is only one straight line between any two points but we can still see objects. On your sphere, there are two straight lines, one from the near direction, the other from the far, opposite direction.

I'm still not sure...

Could a lens focus converging light paths?

hhEb09'1
2009-Aug-17, 03:24 PM
Could a lens focus converging light paths?At the distances we are talking about, the paths are essentially parallel, near hemisphere or far hemisphere.

a1call
2009-Aug-17, 03:33 PM
Frog march,

How many straight paths are there along the surface of the earth (not through it) between [New York and Sydney-Australia] or [New York and Santiago-Chile] ?

Frog march
2009-Aug-17, 03:33 PM
At the distances we are talking about, the paths are essentially parallel, near hemisphere or far hemisphere.

yes, I know, but maybe the fact that they are not quite parallel means that stars on this side of the 3-sphere could be focused, whereas the convergent nature of light from the other side might not be able to focus.

I don't know that much about lenses.

Frog march
2009-Aug-17, 03:35 PM
Frog march,

How many straight paths are there along the surface of the earth (not through it) between [New York and Sydney-Australia] or [New York and Santiago-Chile] ?

an infinite number, as those cities aren't single points.

which is true of a star too.

DrRocket
2009-Aug-17, 03:43 PM

http://farm4.static.flickr.com/3423/3828847825_87a32c6eba_o.jpg

ok, it's a bit homemade, but it shows a 2-sphere with a star near the top giving off light, that radiates out as yellow circles.

There is an "equator" which represents the halfway point around the 2-sphere.

There is an observer on the opposite side of the 2sphere, from the star.

My question is: will he/she be able to see the star, as the light entering their eye will be coming from a wide area of the sky...if you see what I mean.

This should still apply to a 3-sphere, so I was wondering if we are limited, in the stars/galaxies we see, to those that would be on "our side" of the 3-sphere(if we do live in a 3sphere).

That is an interesting question.

I think the answer is that your analogy does work on the 3-sphere. That is because if one looks at spherical coordinates for the 3-sphere, the parameterization by latitudinal angle would still apply with that angle running from 0 to pi, with a convergence at pi.

http://en.wikipedia.org/wiki/N-sphere

It is also important to the argument that Huygen's principle holds in dimension with 3 spatial and one time dimension.

What is also interesting is that the argument for the 2-sphere does not work. That is because the expanding circular wave fronts in your picture do not contain all of the energy of the propagating wave, and that is in turn because Huygen's principle does not hole with an even number of spatial dimensions (in this case 2). In two dimensions there would be energy left behind the advancing wave fronts.

http://www.mathpages.com/home/kmath242/kmath242.htm

One reason that we do not actually see the convergence of light from antipocal points on the sphere, assuming that space is a 3-sphere,may be that the analysis depends on a sphere of fixed radius, and it seems that the universe is expanding.

a1call
2009-Aug-17, 03:45 PM
an infinite number, as those cities aren't single points.

which is true of a star too.

True, So let me rephrase the question:

How many straight paths along the surface of the Earth, are there between any given point in New York to any given point in Santiago-Chile?

Frog march
2009-Aug-17, 04:06 PM
True, So let me rephrase the question:

How many straight paths along the surface of the Earth, are there between any given point in New York to any given point in Santiago-Chile?

well, that is one; ie the shortest route along the surface.

a1call
2009-Aug-17, 04:36 PM
well, that is one; ie the shortest route along the surface.

Now, for the sake of argument let's say that light can travel along the surface of earth and in a straight line on this surface. Let's call the any-given-point-in-New-York A and Any-given-point-in-the-southern-hemisphere-other than the anti-pole B.
If Light emanated rays in all directions from point A, How many of these rays would pass through B?

So why do you talk about converging light rays at the southern hemisphere?
The convergence/focus of rays in a lens is not relevant. Any light passing through the area of a lens is focused. This happens in a curved space as well as a flat space. The point is that light from A does not approach B (or any given point on the surface of a lens at B) from all directions but from a single direction (not considering the opposite path on the great circle :) ) along the great circle that passes through both points.

hhEb09'1
2009-Aug-17, 04:43 PM
So why do you talk about converging light rays at the southern hemisphere?When light leaves a point on an object, the rays diverge. When they get to your eye, they are still (slightly) divergent. However, in the opposite hemisphere, they will be (slightly) convergent. That's the only concern. My comment was that the "slightly" is really "slightly" and doesn't make much of a difference.

Frog march
2009-Aug-17, 05:07 PM
When light leaves a point on an object, the rays diverge. When they get to your eye, they are still (slightly) divergent. However, in the opposite hemisphere, they will be (slightly) convergent. That's the only concern. My comment was that the "slightly" is really "slightly" and doesn't make much of a difference.

but it would make a big difference at or near the exact opposite side of the star, at which point the light would appear to be coming from all directions.

So surely that would still apply as one moved away from the opposite point(to the star), just in a decreasing manner.

hhEb09'1
2009-Aug-17, 05:16 PM
So surely that would still apply as one moved away from the opposite point(to the star), just in a decreasing manner.It's just inverted. So, on the other side of the earth (or even on the other side of this building), the difference would be decreased to almost nothing. I mean, the difference between convergent, and parallel. The energy density is another matter.

a1call
2009-Aug-17, 06:10 PM
When light leaves a point on an object, the rays diverge. When they get to your eye, they are still (slightly) divergent. However, in the opposite hemisphere, they will be (slightly) convergent. That's the only concern. My comment was that the "slightly" is really "slightly" and doesn't make much of a difference.

Ok, Now I see what you mean. But the divergence of more than one ray turning to convergence after travelling a 90 degrees arc still does not change the fact that from any point on the emmiter there is only one ray path hitting any given point on the lens. The facts remain:

*- A convex lens will focus diverging, converging, as well as parallel rays. The image will form. It will form closer than parallel or diverging rays but it will form. Since the deviation from parallel rays of distant objects is negligible this is not an issue for any imaginable lens size.

*- Light does not approach any point-on-the-other-side from all-sides anywhere other than the anti-pole of the emitting point.

*- Any other point than the anti-pole of the emitter will have one and only one great circle passing through both points (emitter and receiver) regardless of how close or far it is from the said anti-pole

a1call
2009-Aug-17, 09:32 PM
After some thinking, There might be a very important question raised in this thread:

*- Assuming a curved spatial universe of radius A

*- Ignoring the dispersion due to scattering phenomenon of light

*- Considering a star light at a distance of greater than (2A*3.14)/4 (i.e arc of 90 degrees)

*- We would expect a very close to parallel but converging star light which would probably be beyond detection

*- But might be detectable if we factored in the expected dispersion due to scattering and developed a precise-enough/large-enough instrument/area for measuring the dispersion angle of the said star light

*- Then Assuming the star distance is known this might be a means of "observing" the curvature of Space or lack thereof

DrRocket
2009-Aug-17, 10:59 PM
True, So let me rephrase the question:

How many straight paths along the surface of the Earth, are there between any given point in New York to any given point in Santiago-Chile?

Two.

Both along the same great circle.

Straight lines on the sphere are great circles. There is a unique great circle connecting two distinct points except when they are opposite points on a diameter of the sphere.

Antipodal points are joined by infinitely many straight lines, just as the north and south poles are joined by any longitude.

a1call
2009-Aug-18, 04:02 PM
Some more personal thoughts seeking expert input:

*- Assuming a spatially curved universe

*- Assuming that the universe is populated by stars homogeneously throughout including on the other-side

*- Assuming that the universe is old enough for the star light to reach this-side from the other-side

*- Assuming no expansion/red-shift

*- Assuming uniform intensity to all stars

*- Assuming a perfectly transparent space with no light attenuation due to interspatial absorption

*- disregarding scattering of light

**- Then one would expect that the dimmest stars from our vantage point would be the ones on the "equator" i.e. half way to the other-side and the stars would be getting brighter towards us and also away from the "equator" towards our anti-pole on the other-side. This, considering the convergence of light passing through the "equator" as pointed out by hhEb09'1

***- :)

WayneFrancis
2009-Aug-19, 03:05 AM

http://farm4.static.flickr.com/3423/3828847825_87a32c6eba_o.jpg

ok, it's a bit homemade, but it shows a 2-sphere with a star near the top giving off light, that radiates out as yellow circles.

There is an "equator" which represents the halfway point around the 2-sphere.

There is an observer on the opposite side of the 2sphere, from the star.

My question is: will he/she be able to see the star, as the light entering their eye will be coming from a wide area of the sky...if you see what I mean.

This should still apply to a 3-sphere, so I was wondering if we are limited, in the stars/galaxies we see, to those that would be on "our side" of the 3-sphere(if we do live in a 3sphere).

I think the issue here is 2 fold.
1) what is the odds that we would be exactly on the opposite side of the universe/3Sphere.
2) what is the size of the 3Sphere if we do live on the surface of one.

The 2nd is the more relevant. As far as I've read the universe seems flat thus our observable universe, if on the surface of a 3shpere, would be analogise to a fly sitting on the the north pole. Another problem is that if we are living on the surface of a 3Sphere it seems to be inflating at a rate where light would never get to the equator.

phunk
2009-Aug-19, 03:41 PM
Some more personal thoughts seeking expert input:

*- Assuming a spatially curved universe

*- Assuming that the universe is populated by stars homogeneously throughout including on the other-side

*- Assuming that the universe is old enough for the star light to reach this-side from the other-side

*- Assuming no expansion/red-shift

*- Assuming uniform intensity to all stars

*- Assuming a perfectly transparent space with no light attenuation due to interspatial absorption

*- disregarding scattering of light

**- Then one would expect that the dimmest stars from our vantage point would be the ones on the "equator" i.e. half way to the other-side and the stars would be getting brighter towards us and also away from the "equator" towards our anti-pole on the other-side. This, considering the convergence of light passing through the "equator" as pointed out by hhEb09'1

***- :)

With all of those assumptions, plus a smooth enough universe that gravitational lensing didn't blur out the effect, at the antipode of every star would be a mirror image of the star (delayed by the light travel time). If you flew through one of these, you'd be vaporized as surely as if you flew through the actual star.

WayneFrancis
2009-Aug-20, 03:19 AM
Now, for the sake of argument let's say that light can travel along the surface of earth and in a straight line on this surface. Let's call the any-given-point-in-New-York A and Any-given-point-in-the-southern-hemisphere-other than the anti-pole B.
If Light emanated rays in all directions from point A, How many of these rays would pass through B?

So why do you talk about converging light rays at the southern hemisphere?
The convergence/focus of rays in a lens is not relevant. Any light passing through the area of a lens is focused. This happens in a curved space as well as a flat space. The point is that light from A does not approach B (or any given point on the surface of a lens at B) from all directions but from a single direction (not considering the opposite path on the great circle :) ) along the great circle that passes through both points.

Exactly why I stated the question about what is the odd that you are exactly opposite a star at the other end of the universe.

WayneFrancis
2009-Aug-20, 03:37 AM
After some thinking, There might be a very important question raised in this thread:

*- Assuming a curved spatial universe of radius A

*- Ignoring the dispersion due to scattering phenomenon of light

*- Considering a star light at a distance of greater than (2A*3.14)/4 (i.e arc of 90 degrees)

*- We would expect a very close to parallel but converging star light which would probably be beyond detection

*- But might be detectable if we factored in the expected dispersion due to scattering and developed a precise-enough/large-enough instrument/area for measuring the dispersion angle of the said star light

*- Then Assuming the star distance is known this might be a means of "observing" the curvature of Space or lack thereof

Ok, let me explain this another way.

Say we where in an universe, does that actually read weird to anyone else?, was a 3 sphere of 1by in circumference.

A star exists a point A
Point B is the polar opposite of A

What happens to light coming from A?

Well for all other spots beside B the light from A look fairly normal
For someone at B there might be problems in 1by when the first light gets to it.

To other observers point B would appear to be a odd ghostly mirror of the start at point A. as the photons go over that point and star heading back to point A.

The issue really is if we live on the surface of a 3sphere how big is it?
Well current measurements show that the universe appears to be flat which means that if we are really on the surface of a 3sphere that 3sphere is orders of magnitude larger then the observable universe.

This means we wouldn't expect to make any of these measurements for orders of magnitude longer then the current age of the universe.

To make things worse it appears that the universe is expanding at a rate that decreases the amount of stuff we can observe within the universe. So even if there was light that hit that opposite pole and was travelling back it would never get here.

Finally you have to assume that the universe is a perfectly spherical 3sphere. The universe might be some wrinkly 4 dimensional raisin type
shape and that due to the expansion we just see flat space where we are but on the larger scale it still has many folds and creases. This would also be very bad for your experiment because even it means light would probably never hit point B at the same time.

Frog march
2009-Aug-20, 04:05 AM
I wasn't really trying to deal with the exact opposite side from the star; what I was interested in was just observations in the opposite half of the 3sphere.

Even just being around the antipode(to the star) you would see light coming from all parts of space, and I was suggesting that as one moved away from the antipode, that that would still, to some extent, apply, even moving a long way from the antipode, and that stars, in the other half of the 3sphere, would appear so blurred that they wouldn't be seen....

Another question came to mind: if the Universe is expanding in a way that light can't keep up, would those photons even be released in the first place? As a photon needs a start point AND an end point, in order to exist, I believe.

Maybe that last question needs a new thread...

DrWho
2009-Aug-20, 04:16 AM
Another question came to mind: if the Universe is expanding in a way that light can't keep up, would those photons even be released in the first place?
If a tree falls... :)

You could say yes, the photon is emitted even though no one can ever see it. Or you could say the question is moot because the photon cannot be said to exist if no one can ever see it.

WayneFrancis
2009-Aug-20, 04:26 AM
If a tree falls... :)

You could say yes, the photon is emitted even though no one can ever see it. Or you could say the question is moot because the photon cannot be said to exist if no one can ever see it.

I'm not so philosophical. The photons are emitted. Think of it this way. Do you believe all of space is full of the CMBR or only the bits where there is something for the CMBR to run into?

With the universe expanding...the photons just red shift. Indications are that you would never get photons to start to converge because the rate of expansion at that scale would be many orders of magnitude greater then the speed of light.

Nice thought experiment but when I do it I quickly come to the realisation that it isn't, and never will be, testable given our understanding of the universe.

DrWho
2009-Aug-20, 04:43 AM
I'm not so philosophical. The photons are emitted.
I wasn't trying to be philosophical, just trying to present two possible answers to the question. I would tend to lean to the realist side of the fence, however, saying anything with absolute certainty about something we can never observe does fall in the realm of metaphysics.

Frog march
2009-Aug-20, 03:39 PM
but if a star was unable to emit most of its photons, due to any potential photons having no receiver, then that star will get hotter and hotter, which should be measurable, shouldn't it?

although most potential photons, for our star, would have a receiver.

phunk
2009-Aug-20, 04:05 PM
Even just being around the antipode(to the star) you would see light coming from all parts of space

No you wouldn't, unless you were closer to the antipode than the star's radius, you'd only see it from 2 opposite directions. Remember that the light rays in that 'ring' in the first image in the thread are traveling perpendicular to the ring. Anyone not at the focal point will only see the rays from 2 points on the ring, the ones that hit before the ring reaches the antipode, and the ones from the other side after the ring passes the antipode and inverts.

Looking directly away from the star's antipode point, you'd see the star far away. Looking towards the star's antipode, you'd see a false image of the star up close. The close up false image would be delayed relative to the 'real' view of the star by 2x the light travel time from you to the antipode.

So if you were in a tiny perfectly spherical universe, say 1000 light years in diameter, and there was a star 999 light years away from you, you'd have 2 views of the star. One 999ly away view as the star was 999 years ago, and in the opposite direction you'd see it 1ly away but it would be an image of the star from 1001 years ago. It would be a confusing universe for astronomers. :)

hhEb09'1
2009-Aug-20, 05:06 PM
So if you were in a tiny perfectly spherical universe, say 1000 light years in diameter, and there was a star 999 light years away from you, you'd have 2 views of the star. One 999ly away view as the star was 999 years ago, and in the opposite direction you'd see it 1ly away but it would be an image of the star from 1001 years ago. It would be a confusing universe for astronomers. :)It's a confusing universe for me too. :)

It's 1000 light years in diameter? so its circumference is 3142 light years?

And, if a star is 1 light year away, wouldn't its light reach us in 1 year?

phunk
2009-Aug-20, 05:20 PM
It's a confusing universe for me too. :)

It's 1000 light years in diameter? so its circumference is 3142 light years?

And, if a star is 1 light year away, wouldn't its light reach us in 1 year?

You're right, I should have said circumference!

In the example, the star is 999 ly away. The false image at it's antipode would be 1 ly away. It would take light 999 years to reach us the short way, and 1001 to go the other way around, through the antipode, then to us.

DrWho
2009-Aug-21, 02:46 AM
but if a star was unable to emit most of its photons, due to any potential photons having no receiver, then that star will get hotter and hotter, which should be measurable, shouldn't it?

Why would a photon care if it has a 'receiver'?

Frog march
2009-Aug-21, 03:07 AM
Why would a photon care if it has a 'receiver'?

Well, I thought light was a wave-probability thing, and would only be seen as a photo, when it got somewhere.

To light, there is zero time between emission and reception, so really wouldn't it have to be received at some stage?

WayneFrancis
2009-Aug-21, 05:55 AM
but if a star was unable to emit most of its photons, due to any potential photons having no receiver, then that star will get hotter and hotter, which should be measurable, shouldn't it?

although most potential photons, for our star, would have a receiver.

Actually I'd say most photons for our star would not have an observer. If you think of the universe in a fractal sense there seems to be more empty space then actual matter. For example look at the Hubble deep field. Even way back then, when the galaxies where closer together, there is more space then galaxies. This means the photons from much of our sun will shoot off in a direction where they will never hit anything because at a certain point they'll end up infinitely red shifted before hitting anything.

WayneFrancis
2009-Aug-21, 05:59 AM
It's a confusing universe for me too. :)

It's 1000 light years in diameter? so its circumference is 3142 light years?

And, if a star is 1 light year away, wouldn't its light reach us in 1 year?

I got what he meant.

He meant 2000ly in circumference. The false image of the star would indeed reach you 1001 years after actually being emitted from the real star but it would look like it came from only 1ly away.

So the "real star" is 999ly away and there would be a false star in the opposite direction 1ly away.

Frog march
2009-Aug-21, 09:50 AM
. This means the photons from much of our sun will shoot off in a direction where they will never hit anything because at a certain point they'll end up infinitely red shifted before hitting anything.

So where would the energy from the photon have gone?

a1call
2009-Aug-21, 01:33 PM
Well, I thought light was a wave-probability thing, and would only be seen as a photo, when it got somewhere.

To light, there is zero time between emission and reception, so really wouldn't it have to be received at some stage?

The "wave-probability thing" means that the exact/pinpoint location of the photon/particle is not known but it exists and is located somewhere where the light as a wave has traveled. The wave characteristic of the light disappears when the light is absorbed/detected as a photon/particle.

*- bottom line: light exists even if it is not detected. It's just that we can only approximate it's precise position as a point-particle.

The resident gurus on the subject might want to clarify further.

WayneFrancis
2009-Aug-24, 03:00 AM
So where would the energy from the photon have gone?

That is like saying if you speed up a ship to .5c where does all the energy come from to make all the photons in front of you blue shift. It hasn't gone anywhere.