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Constancy of SOL from all objects


michel123456
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Maybe this should be posted in cosmology.

We know about the constancy of Speed Of Light.

Whatever the relative velocity of 2 objects, light coming from one object and reaching the other will always be observed traveling at C.

 

In cosmology, when objects are receding at velocity greater than C because of the expansion of space, we are observing light coming still arriving at velocity of C, but redshifted.

 

And we say that the velocity grater than C is not a regular velocity (it is caused by expansion of space) because velocity greater than C is not physically possible.

One of the reason is that (correct me if I am wrong) IF an object ever could go faster than SOL, light would never reach us. As if light would have traveled backwards.

And I wonder why? I wonder because we accepted the statement that C is a constant. It does not depend from the velocity of the source. IOW even for hypothetical faster-than-light objects, this statement should hold. It means that even faster than light objects would be visible without having to invoke expansion of space. That is what constant means.

Edited by michel123456
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Recessive velocity in Hubbles law is an apparent velocity. Its a consequence of the formula used and the seperation distance to the observer.

 

Locally though a galaxy gains no momentum due to expansion. Though the distances between two galaxies does increase which gives the illusion of inertia if you look at how Newtons laws of inertia work and consider a uniform fluid with a uniform pressure surrounding a galaxy. Then realize a body in space requires a difference in force upon a facing, we can't apply Newtons laws of inertia to galaxies.

 

For that matter we can't even state expansion is due to pressure. This requires a difference in pressure from one region to another.

 

A good paper covering the seperation distance details is

 

http://tangentspace.info/docs/horizon.pdf :Inflation and the Cosmological Horizon by Brian Powell

 

Dr Powell is a professor in Philosophies of Cosmology. His specialty is inflation, though he used to visit forums and wrote the above article with forum members in mind.

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Recessive velocity in Hubbles law is an apparent velocity. Its a consequence of the formula used and the seperation distance to the observer.

 

Locally though a galaxy gains no momentum due to expansion. Though the distances between two galaxies does increase which gives the illusion of inertia if you look at how Newtons laws of inertia work and consider a uniform fluid with a uniform pressure surrounding a galaxy. Then realize a body in space requires a difference in force upon a facing, we can't apply Newtons laws of inertia to galaxies.

 

For that matter we can't even state expansion is due to pressure. This requires a difference in pressure from one region to another.

 

A good paper covering the seperation distance details is

 

http://tangentspace.info/docs/horizon.pdf :Inflation and the Cosmological Horizon by Brian Powell

 

Dr Powell is a professor in Philosophies of Cosmology. His specialty is inflation, though he used to visit forums and wrote the above article with forum members in mind.

Thank you for your comment

Here is what I don't understand, from your link page pdf 4

post-19758-0-34794600-1468952751_thumb.jpg

In a world where there is no expansion (Vrec=0), why wouldn't be able to observe an object going away at 2times SOL?

I mean, light coming from an object getting closer at 0,1 C is observed coming at C

Light coming from an object getting away at 0,00000001 Cis observed coming at C

Light coming from any object, coming closer or getting away, is always observed at C.

That is what a constant means: it is constant.

 

Then, why do we have to invoke Vrec in order to explain the observation of objects getting away at 5 times SOL. In my understanding, if C is constant, then it is constant. Point.

Edited by michel123456
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Ok first off you need to look at how v_rec is defined.

 

[latex]v_rec=H_od [/latex]

 

so if you have no expansion H becomes zero and so does the recessive velocity.

 

Now in a static universe with no recessive velocity. The observable universe would be defined as c*age of the universe. This is the Hubble horizon where the recessive velocity equals the speed of light. (keep in mind this depends on seperation distance from the observer to horizon).

 

Now recessive velocity does not state the speed of light is greater than c. The speed of light stays constant. The peculiar velocity of the object being measured decreases the closer the observer gets to the object. See formula above.

 

Light beyond Hubble radius where recessive velocity 》c doesn't go faster than c to reach us. It still travels at c.

 

So how does it reach us? This is where you have to recognize that peculiar velocity depends on the distance of separation from the emitter and observer.

 

If you change the observer to one that is beside the light ray. There is no recessive velocity.

 

Locally to the leading edge of the light path the rate of expansion is a measly 70 km/sec/Mpc. Light has no problem transversing this change in distance.

 

Now due to expansion both the path ahead and behind the leading lightray edge increases. This leads to the Cosmological redshift.

 

Its unfortunate Hubble used the term recessive velocity. The terminology leads to confusion. Would have been better to call it Hubble recession.

Edited by Mordred
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Ok first off you need to look at how v_rec is defined.

 

[latex]v_rec=H_od [/latex]

 

so if you have no expansion H becomes zero and so does the recessive velocity.

 

Now in a static universe with no recessive velocity. The observable universe would be defined as c*age of the universe. This is the Hubble horizon where the recessive velocity equals the speed of light. (keep in mind this depends on seperation distance from the observer to horizon).

 

Now recessive velocity does not state the speed of light is greater than c. The speed of light stays constant. The peculiar velocity of the object being measured decreases the closer the observer gets to the object. See formula above.

 

Light beyond Hubble radius where recessive velocity 》c doesn't go faster than c to reach us. It still travels at c.

 

So how does it reach us? This is where you have to recognize that peculiar velocity depends on the distance of separation from the emitter and observer.

 

If you change the observer to one that is beside the light ray. There is no recessive velocity.

 

Locally to the leading edge of the light path the rate of expansion is a measly 70 km/sec/Mpc. Light has no problem transversing this change in distance.

 

Now due to expansion both the path ahead and behind the leading lightray edge increases. This leads to the Cosmological redshift.

 

Its unfortunate Hubble used the term recessive velocity. The terminology leads to confusion. Would have been better to call it Hubble recession.

Why?

Why would the observable universe defined as c*age of the universe?

Is that because "nothing can travel faster than c"?

Or because of the belief that things that travel faster than c are outside the observable universe?

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if you have zero expansion and nothing travels faster than c. The observable universe would be the Hubble horizon. Which is c×age of universe. However we have expansion so the observable universe is larger.

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if you have zero expansion and nothing travels faster than c. The observable universe would be the Hubble horizon. Which is c×age of universe. However we have expansion so the observable universe is larger.

Would there be objects beyond the observable universe in this case?

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Presumably, but since it's unobservable we could never know for sure.

So my question again.

Why would there be unobservable objects?

Why would we think that light coming from some objects would suddenly go backwards? What is the principle that says that light would suddenly have a negative velocity? Especially when we we know for sure that SOL is a constant.

Edited by michel123456
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Not enough time has passed.

 

 

And, because of the increasing distance between those objects and us, there never will be enough time.

Especially when we we know for sure that SOL is a constant.

 

As Markus pointed out, it is invariant not necessarily constant.

 

It is a constant locally, but in an expanding universe if you divide the distance when the light was emitted by the travel time, you will not always get c. So, for example, the light from the most distant observable objects was emitted when they were about 4.5 billion light years away but has taken nearly 14 billion years to reach us.

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And, because of the increasing distance between those objects and us, there never will be enough time.

 

 

As Markus pointed out, it is invariant not necessarily constant.

 

It is a constant locally, but in an expanding universe if you divide the distance when the light was emitted by the travel time, you will not always get c. So, for example, the light from the most distant observable objects was emitted when they were about 4.5 billion light years away but has taken nearly 14 billion years to reach us.

Well, but it still travelled 14 billion light years. It's not that dividing the distance by the time doesn't give you c. It's that dividing the starting distance by the time may not give you c, because the distance between two points when the light is first emitted may not be the total distance that exists between those points by the time the light arrives, and therefore not the total distance the light actually travels while en route.

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So my question again.

Why would there be unobservable objects?

Why would we think that light coming from some objects would suddenly go backwards? What is the principle that says that light would suddenly have a negative velocity? Especially when we we know for sure that SOL is a constant.

Anyone here that does understand the question?

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OK

You have an object A moving away from you at velocity V. This object A emits light. What is the velocity of light (SOL)for the followings:

 

V=0, SOL=c

V=1m/s, SOL=c

V=10m/s, SOL=c

V=100000m/s, SOL=c

V=299999km/s, SOL=c

 

Is the above correct?

 

And then why would we think that suddenly the value of V matters when V gets higher than that?

 

In a world without expansion.

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In a world without expansion.

 

 

So that is a completely different question. In this case, the light from an object 4 billion light years away would take 4 billion light years to get here. No matter how fast that object was moving. And, of course, that object could not be travelling at c.

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If there was a large explosion 720 mi away, 30 min ago, you wouldn't have heard it yet, as it would take another 30 min for the sound to reach you

If you got in a plane and started moving at 720 mi/hr in the opposite direction you will never hear it, as it will never reach you.

Keep in mind that in this case, these are actual velocities subject to relativity. Universal expansion is NOT a velocity and is not subject to relativity ( in the same way ). We can have superluminal expansion.

Some thing or effect outside the observable universe is causally disconnected because its 'information' can never reach us or affect us. The speed of light is the speed of information, and causality.

 

This is one of the reasons for postulating the original inflationary period.

The homogeneity and isotropy of the universe could not be assured unless there was ability to exchange information between distant parts of the universe ( i.e. a causal connection ) at some early stage.

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So that is a completely different question. In this case, the light from an object 4 billion light years away would take 4 billion light years to get here. No matter how fast that object was moving. And, of course, that object could not be travelling at c.

OK.

That is what our Theory says.

Now we are looking in our telescopes and discover that objects go away from us at more-than-c velocities.

IOW we have a confirmation of "No matter how fast that object was moving".

My question is: why don't we stick to this.

Why do we have to introduce other kind of displacement (expansion)?

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Is the question: why is the big bang model currently accepted as the best description of the universe?

 

In which case the answer is that is was predicted by theory and it is the best match for all the evidence. Various other models were tried to explain the observed red-shifts, none worked as well. And no other model can also explain the CMB.

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In which case the answer is that is was predicted by theory ...

Under some mind assumptions it is known that a static Universe is very very unstable. It would require perfect balance and any small changes would result in a collapse. This even led Einstein to propose his cosmological constant to provide this balance - but it does not really solve give a static Universe and it is still unstable. So, we have to think about an expanding or contracting Universe... and then we have the evidcnce for an expansion.

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Is the question: why is the big bang model currently accepted as the best description of the universe?

 

In which case the answer is that is was predicted by theory and it is the best match for all the evidence. Various other models were tried to explain the observed red-shifts, none worked as well. And no other model can also explain the CMB.

No no.

I simply mean:

You can keep c as an invariant, you can keep "No matter how fast that object was moving". You can keep c as the velocity of causality. You can keep all of Relativity and keep being consistent with observation.

Don't you agree?

The only thing that you must drop off is "nothing can go faster than c" and replace it with "c is the max velocity of causality".

IOW the universe does not care about the relative velocity of anything, there is no upper bound.

Is that so dramatic?

Edited by michel123456
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