CMB - TV aerial handshake?

[sethoflagos]

I was musing on the life experience of a CMB photon travelling from its source emitter - perhaps some excited hydrogen atom at recombination - to it's absorber which for sake of argument might be a TV aerial.

Am I right in thinking that the spacetime interval between these events is zero? And in particular that the photon 'travels' zero spatial distance in zero time (due to Lorentz contraction)? 

Is there a sense that the photon is sitting for an instant inside some weird tiny spherical(?) surface, in intimate contact with all its possible futures (one of which is said TV aerial), a myriad of even tinier exits to all available absorbers peppered within a matrix of paths forbidden by destructive interference, and/or routes to some eternal void?

Because if source and destination are touching perhaps there's no need for any significant energy investment in filling up its light cone with .... stuff (insert correct jargon). Just pick a destination that matches up and hey presto it happens instantaneously (give or take 13-odd billion years). 

[Mordred]

I truly have a difficult time making sense of your post. So let's deal with what I can make sense of.

A CMB signal such as photons which is the mediator boson for the electromagnetic field. Can be recieved as static on our radio frequencies.

This is due to Cosmological redshift.

If a particle travels at c then the seperation distance is [math]ds^0[/math] however this shows that a reference frame of v=c isn't a valid reference frame.

A different observer ( not v=c can) recognize the constant speed of light value.

If you have some other thought in mind you will need to clarify.

[sethoflagos]

Okay, well the restriction on valid reference frames to v<c is a new one on me, however ...

You agree that the spacetime seperation is zero in the direction of photon travel, which is my starting point. Does this mean that the emitter and absorber are physically adjacent (despite the huge separation in our own spatial and time reference frame), with a consequently strong electromagnetic coupling, and hence that the exchange of a photon between them reduces to a local event. I'm not looking for transfer of information from absorber to emitter as such, but whether the existence of an available absorber (removed in time) can be sensed by the emitter. Or does the emitter simply chuck out a photon irrespective of its ultimate destiny. No handshake in either direction.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

[Mordred]

A V=c is not a valid reference frame as the separation distance is \(ds^2=0\) which would imply the photon exists everywhere in the universe which is obviously nonsense. The only valid reference frame is a v<c observer of the photon path. Transfer of information of any kind is limited to c.

Where you are getting confused is that the v=c isn't a valid reference frame. You must use the separation distance of an observer less than c.

In essence there is no instantaneous communication between any two events in any valid reference frames an observer with v=c isn't a vsid reference frame 

Edited July 2, 2020 by Mordred

[sethoflagos]

Yet from the photon's point of view, all of future spacetime is condensed into a singularity since it can in principle access any of it instantaneously? ie the totality of its future light cone is local.

Edited July 2, 2020 by sethoflagos
clarification

[Mordred]

As stated there isn't a valid reference frame for a v=c observer. The photon reference frame leads to nonsense conclusions hence invalid.

[sethoflagos]

I appreciate that dt perforce =0 but if ds also =0 surely we can consider the measured ratio at v=0.9c, v=0.99c etc and infer that at v=c, 0/0 still =c, can't we?

[Mordred]

Any observer less than c is valid. However a reference frame v=c isn't. Any reference frame less than c will measure c as being constant regardless of the observer velocity.

So you will have a separation distance between two events but never instantaneous.

The v=c does lead to mathematical singularities. Another reason for being invalid as a reference frame ie a particle moving at near c does not form a black hole. Another common misconception.

Edited July 2, 2020 by Mordred

[sethoflagos]

I know when I'm flogging a dead horse. Sleep well, Mordred. Good night.

[Mordred]

Night its a good question that is often confused. Your not the first to hit this stumbling block.

So +1 For the question and attempt to understand.

Any v=c reference frame is invalid because of garbage answers. Ie the object being everywhere in the universe at the same instsnce. An obvious impossibility. A photon for example doesn't exist the entirety of the universe simultaneous.

 

Edited July 2, 2020 by Mordred

[sethoflagos]

Actually, having a point particle filling the observable universe was what I was trying to get around - by taking the point of view of the photon whose spacetime universe is of atomic dimensions - at least looking down the wormhole of its momentum vector. 4D geometries fry neurons

[Mordred]

I understood that from the beginning. You will always have a delay of signal between two events regardless of valid (v<c) observer. A v>c will see time reversal and a v=c observer is invalid. Surprisingly enough GR does handle v>c observers but it suffers the same singularity conditions for v=c.