can we see it #2 ?

[michel123456]

When an object emits light, we can see it.

 

_when the objects moves away at some speed, we observe light coming at us at constant speed = C.

 

_when the object moves away at speed less than C, we observe light at C.

 

_when the object moves away at C, we observe light at C (because C is constant)

 

_when an (hypothetical) object moves away at speed faster than C (?), we must observe (I suppose) light coming at C, because C is constant.

 

So I suppose that even when an object like a galaxy is receding from us at speed faster than C, we are observing the light coming from this object traveling at C. Independently of any "space expansion".

 

Or, in other word: an object receding at FTL, can we see it?

[Nano]

So I suppose that even when an object like a galaxy is receding from us at speed faster than C, we are observing the light coming from this object traveling at C. Independently of any "space expansion".

 

Would the lightbeam ever reach us at all, if space which light is traveling through is expanding faster than the light itself?

[Sisyphus]

An object moving away from us at FTL should still be visible, I think, though I'm not sure what we would actually see. A photon emitted from any particular location it has been would travel towards us at C in our reference frame, and reach us in a finite amount of time.

 

An object receding from us via cosmic expansion at FTL would not be visible, as the distance the photon has to travel would increase faster that it does the traveling, and it would never reach us.

[Mr Skeptic]

An object moving away from us at FTL should still be visible, I think, though I'm not sure what we would actually see. A photon emitted from any particular location it has been would travel towards us at C in our reference frame, and reach us in a finite amount of time.

 

The photon would be red-shifted into negative energy, I think, whatever that means.

 

An object receding from us via cosmic expansion at FTL would not be visible, as the distance the photon has to travel would increase faster that it does the traveling, and it would never reach us.

 

We can still see stuff receding from us at FTL, however it is due to the fact we see photons it emitted while it was still nearer. I think we would always be able to see it due to this (unless it "started out" that far), however the image would be more and more redshifted to the point it would be indistinguishable from the background radiation.

[Sisyphus]

We can still see stuff receding from us at FTL, however it is due to the fact we see photons it emitted while it was still nearer. I think we would always be able to see it due to this (unless it "started out" that far), however the image would be more and more redshifted to the point it would be indistinguishable from the background radiation.

 

I don't think that's right. A photon that is emitted from an object receding FTL will never reach us. It is true that some of the light reaching us now is from objects that are currently receding FTL, but were not when that light was emitted.

[swansont]

An object moving away from us at FTL should still be visible, I think, though I'm not sure what we would actually see. A photon emitted from any particular location it has been would travel towards us at C in our reference frame, and reach us in a finite amount of time.

 

An object in locally flat spacetime can't be moving faster than c, so that question is moot.

[Sisyphus]

An object in locally flat spacetime can't be moving faster than c, so that question is moot.

 

I know, I just thought it might be possible to make physical sense of it anyway, as a hypothetical. But I'm thinking that might not be possible, since as Mr. Skeptic points out, an emitted photon would have "negative energy, whatever that means," which I guess means nothing.

 

And I don't fully understand what a tachyon would actually be except in an abstract sense, and as a colored beam on Star Trek that solves time travel problems.

[Mr Skeptic]

An object in locally flat spacetime can't be moving faster than c, so that question is moot.

 

Tell that to the tachyons.

[michel123456]

The photon would be red-shifted into negative energy, I think, whatever that means.

 

 

 

We can still see stuff receding from us at FTL, however it is due to the fact we see photons it emitted while it was still nearer. I think we would always be able to see it due to this (unless it "started out" that far), however the image would be more and more redshifted to the point it would be indistinguishable from the background radiation.

 

Emphasis mine.

I think it is absolutely correct. I only don't understand the "into negative energy" part.

 

Sysiphus] A photon that is emitted from an object receding FTL will never reach us.

 

That's the question.

If , following Relativity, simple Newtonian summation of speed is wrong, it must be wrong by all means. The concept "a photon that is emitted from an object receding FTL will never reach us" is the intuitive result of simple summation of speed. But we know that Relativity is strongly counter-intuitive.

IMO we cannot assume that such a photon will never reach us. If we stick to the strict fundamental statement that "C is constant", even a photon emitted by an object receding from us at FTL will reach us. And from our FOR, we will see this photon traveling at C, the same as all the other photons emitted by any other object traveling at any other speed.

That's what Sisyphus stated when writing:

 

An object moving away from us at FTL should still be visible, I think, though I'm not sure what we would actually see. A photon emitted from any particular location it has been would travel towards us at C in our reference frame, and reach us in a finite amount of time.

 

On the other hand Swansont wrote:

An object in locally flat spacetime can't be moving faster than c, so that question is moot.

 

IMO it should be stated that "in locally flat spacetime we are always observing objects moving at speed lower than C", which is exactly corresponding to observation. The statement "can't be moving faster than c" is a (rash) conclusion based on this observation.

[Johanluus]

The way i analize the question is to break it down into events and some statments.

 

1. A galaxy can't have a velocity greater than C relative to our spot here on earth.

This would violate Einstein's Special relativity ( one would need infinite energy to achieve this) from our point of view.

 

2. The real question IMO is : WHEN that galaxy emitted that photon LONG AGO . If the distance between us and that galaxy was expanding faster than c AT THAT TIME , we would never see it. It is out of our observable horizon or "future".

 

But if the distance WAS expanding at a rate slower than c , then some time in the FUTURE after that event , the photon will eventually reach us , with a redshift relating to the recession speed.

 

WHEN it does eventually hit US ( SAY NOW) , the galaxy will be much further away and the recession speed could NOW be greater than c .

 

In this case we have detected a photon that WAS emitted some time in the past

from a galaxy that was receeding at a speed less than c at THAT time. But NOW it is receeding faster than c and any photons emitted NOW will not be seen in our future.

 

What i find amazing is this:

If we(hypothetical observer) were hitching a ride on that photon from when it left the galaxy(THEN), to when it hit earth(NOW). We would say that NO time has passed, due to time dilation( I.E the events were spontaneous)!

Furthermore the distance we would have travelled would be zero , due to length contraction.

All because of relativity. Again it depends on the observers point of view!

[Spyman]

An object receding from us via cosmic expansion at FTL would not be visible, as the distance the photon has to travel would increase faster that it does the traveling, and it would never reach us.

This is wrong according to current models of cosmic expansion.

 

I don't think that's right. A photon that is emitted from an object receding FTL will never reach us. It is true that some of the light reaching us now is from objects that are currently receding FTL, but were not when that light was emitted.

So, how fast was the radiating matter receding from us when the CMBR we observe today was emitted?

 

----------

 

Here is a good Cosmos calculator you should test: http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html

 

You need to input these values: Omega=0.27, Lambda=0.73, Hubble=71 and Redshift=1100 for the CMBR.

 

But anything we observe with a redshift higher than 2 was receding from us faster than c when the image we observe now was emitted.

Edited June 10, 2010 by Spyman
Adding Cosmos calculator

[Nano]

Since space-time is getting bigger, shouldn't the EM-field in which the photon is traveling through as a medium also expand proportional to the space-time? Meaning that the photon would have a longer a distance to travel?

[swansont]

Tell that to the tachyons.

 

I can't. I don't know how to interact with a particle that has imaginary energy.

[Spyman]

Since space-time is getting bigger, shouldn't the EM-field in which the photon is traveling through as a medium also expand proportional to the space-time? Meaning that the photon would have a longer a distance to travel?

Yes, photons traversing between two objects in expanding space would need to travel longer than photons in flat space to bridge the distance and the first photon in a lightray has a shorter path than the last photon, causing the lightray to be stretched which we can measure as redshift.

 

Hubble's law of the correlation between redshifts and distances is required by models of cosmology derived from general relativity that have a metric expansion of space. As a result, photons propagating through the expanding space are stretched, creating the cosmological redshift.

 

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

Edited June 10, 2010 by Spyman
Spelling

[michel123456]

But NOW it is receeding faster than c and any photons emitted NOW will not be seen in our future.

 

THAT is the point I am discussing.

I think the above statement is wrong.

[Johanluus]

"I think the above statement is wrong."

explain i'm curious?

[Sisyphus]

Emphasis mine.

I only don't understand the "into negative energy" part.

 

The energy of a photon is proportional to its frequency, and inversely proportional to its wavelength. As the object you're viewing approaches the speed of light, the redshift approaches infinity, and hence the energy of the photons you receive approaches zero.

 

If , following Relativity, simple Newtonian summation of speed is wrong, it must be wrong by all means.

 

And it is.

 

The concept "a photon that is emitted from an object receding FTL will never reach us" is the intuitive result of simple summation of speed.

 

No, it is not the result of a simple Galilean summation of speed. We're not talking about objects moving away from one another, we're talking about the amount of space in between them increasing, and the rate of that increase accelerating.

 

A photon emitted from an object receding at FTL is indeed approaching us at exactly C. However, it will still never reach us.

 

This is an effect that is not covered by special relativity, but it is not contradictory to it, either.

 

This is wrong according to current models of cosmic expansion.

 

No, it is not. Look:

 

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

 

So, how fast was the radiating matter receding from us when the CMBR we observe today was emitted?

 

The early universe is different, because the rate of expansion was decreasing (rapidly). In the intervening time, the Hubble sphere grew faster than the universe and overtook it.

 

When the rate is constant or accelerating (as it is now and will continue to be AFAWK), it remains true that objects receding FTL will never be observed.

 

What I should say is, anything emitted after the first ~ 5 billion years from an object receding FTL will never reach us.

Edited June 10, 2010 by Sisyphus

[michel123456]

"I think the above statement is wrong."

explain i'm curious?

 

In relation with Sisyphus's

No, it is not the result of a simple Galilean summation of speed. We're not talking about objects moving away from one another, we're talking about the amount of space in between them increasing, and the rate of that increase accelerating.

 

Well, from the beginning of this thread, I am speaking about "objects moving away from one another" and not about "the amount of space in between them increasing", following this simple reasoning:

 

Lets take an object moving away (not receding) at speed v < C.

 

This objects emits light that we can see.

The naive concept says that the photons that will reach us must have speed C-v= F. (1)

But Relativity shows it is completely wrong. Photons are reaching us at C. Always.

 

Now, when we take another hypothetical object moving away at v > C, we are used to make the same substraction, putting C-v= F (2) and obtaining negative F, meaning that the photons will go away from us instead of reaching us.

 

IMO it is exactly the same error, because equation (1) is the same erroneous with equation (2). We simply cannot make the naive substraction, it is wrong. In any case, the photons will reach us at C.

[Sisyphus]

Well, from the beginning of this thread, I am speaking about "objects moving away from one another" and not about "the amount of space in between them increasing",

 

Then sorry. I misunderstood. You did say "independent of any space expansion," which I interpreted to mean despite space expansion. I see now you simply meant disregarding it.

 

following this simple reasoning:

 

Lets take an object moving away (not receding) at speed v < C.

 

This objects emits light that we can see.

The naive concept says that the photons that will reach us must have speed C-v= F. (1)

But Relativity shows it is completely wrong. Photons are reaching us at C. Always.

 

Yes, totally correct.

 

Now, when we take another hypothetical object moving away at v > C, we are used to make the same substraction, putting C-v= F (2) and obtaining negative F, meaning that the photons will go away from us instead of reaching us.

 

IMO it is exactly the same error, because equation (1) is the same erroneous with equation (2). We simply cannot make the naive substraction, it is wrong. In any case, the photons will reach us at C.

 

Which I would agree with, except that an object with a relative velocity >C doesn't really make much physical sense, at least not in terms of interaction with us.

[michel123456]

The result is that if such a FTL object exists, we should be able to see it.

[Nano]

Well, from the beginning of this thread, I am speaking about "objects moving away from one another" and not about "the amount of space in between them increasing",

 

Aha, I misunderstood you in the start of the thred. I agree also that the photon would reach us. It would also offcourse have a speed of c when it did.

[Johanluus]

yes then i agree we would be able to see it if such an oblect existed, one that could move faster than c , relative to us, and it emitted photons . Providing the space between us and it was static.i.e no expansion.

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yes then i agree , if such a object existed , and it emitted photons.

we would see it if the space between us was static.

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what would happen if such an object was moving toward you , it would pass you before you "see" it.

[Sisyphus]

And the photons that reach you would have infinite energy if moving towards you, and negative energy if moving away. And the object emitting them would be moving backwards in time, and violate causality.

 

Ignoring most of relativity, though, you could have neat effects! For example, an object moving towards you would appear to be moving away, since the photons it emits last (when closest to you) would reach you first.

[swansont]

Now, when we take another hypothetical object moving away at v > C, we are used to make the same substraction, putting C-v= F (2) and obtaining negative F, meaning that the photons will go away from us instead of reaching us.

 

IMO it is exactly the same error, because equation (1) is the same erroneous with equation (2). We simply cannot make the naive substraction, it is wrong. In any case, the photons will reach us at C.

 

The photon will be redshifted, and simple application of the Doppler shift equation yields an imaginary frequency. There is no spoon photon. Once you hypothesize a physically impossible situation, you can't draw any valid conclusions.

[Johanluus]

"Once you hypothesize a physically impossible situation, you can't draw any valid conclusions"

agreed