I don't think space is expanding.

[hecd2]

That is true.

My point was the photon count in transit is not a linear relationship to distance.

No, but it’s a constant in time for a given separation and that’s all we need to know to understand that it cannot compensate for the accumulated difference in clock ticks which does increase as a function of time without limit. It is an unavoidable consequence of your scheme that the accumulated time on clocks that have ever been at any separation from your own clock will be less than your own clock, even when the clocks are brought together. Which leads a logical inconsistency which in turn proves that your scheme cannot work.

As we have shown that this scheme is internally inconsistent, you are bound to abandon it. Have you got anything else?

 

[Mike Helland]

It is an unavoidable consequence of your scheme that the accumulated time on clocks that have ever been at any separation from your own clock will be less than your own clock, even when the clocks are brought together.

That's not what would happen.

Any clock ticks, or light pulses, ever sent would still be received before the clocks are brought together.

If the clock signal leaves the distant galaxy at 0.5 c, and arrives at c, there will be more photons still in the first half of their journey (distance wise) than photons in the second half.

When the clocks come together, two things happen, the light signals from my past get closer to my present, and the light signals from your past start arriving at a higher rate.

Like burning a rope from both ends, the each observer will see the other's clock rapidly accelerate as it you gobble up the signals in transit.

 

[hecd2]

That's not what would happen.

Any clock ticks, or light pulses, ever sent would still be received before the clocks are brought together.

If the clock signal leaves the distant galaxy at 0.5 c, and arrives at c, there will be more photons still in the first half of their journey (distance wise) than photons in the second half.

When the clocks come together, two things happen, the light signals from my past get closer to my present, and the light signals from your past start arriving at a higher rate.

Like burning a rope from both ends, the each observer will see the other's clock rapidly accelerate as it you gobble up the signals in transit.

Write down an expression for the time accumulated on the other clock from separation to coming together again and you will see that I am right. As you know, whenever I have challenged you to do something like this, I have been right and you have been wrong, and this will be the case again.

It is also true that any ticks ever sent during the process should be received. That is inconsistent with the number received if you calculate it according your own scheme. That inconsistency arises through the fundamental internal inconsistency in your idea. The fact that the two results will not match is another nail in its coffin.

So, off you go. Write down an expression for the accumulated pulses received versus sent. I’ll help you if you get stuck - when have you heard that before?

 

[Mike Helland]

Write down an expression for the time accumulated on the other clock from separation to coming together again and you will see that I am right. As you know, whenever I have challenged you to do something like this, I have been right and you have been wrong, and this will be the case again.

It is also true that any ticks ever sent during the process should be received. That is inconsistent with the number received if you calculate it according your own scheme. That inconsistency arises through the fundamental internal inconsistency in your idea. The fact that the two results will not match is another nail in its coffin.

So, off you go. Write down an expression for the accumulated pulses received versus sent. I’ll help you if you get stuck - when have you heard that before?

They only way this would end up wrong is if you traveled faster than photons you emitted.

Do it where the observer never moves faster than c on their clock.

 

[hecd2]

They only way this would end up wrong is if you traveled faster than photons you emitted.

Do it where the observer never moves faster than c on their clock.

No you are wrong. Write down the the expressions.

 

[Mike Helland]

No you are wrong. Write down the the expressions.

I'm trying to make sure the scenario is clear.

Bob and Alice have a plan. They're going to sync up their clocks, then Bob is going to take off for a distance of z=1. When there, Bob is going to note the tie and wait 100 ticks, 100 seconds. Then Bob will begin the return journey.

Alice will have observed Bob to reach z=1, wait 200 seconds, and then begin return to come back.

In that 200 seconds, Bob's clock will only have ticked 100 times.

 

[Reality Check]

An ignorant "Bob and Alice have a plan" scenario from Mike Helland

I'm trying to make sure the scenario is clear.
...

19 March 2021: An ignorant "Bob and Alice have a plan" scenario from Mike Helland.
Being at z = 1 does not say anything about time dilation because it is caused by speed, not distance. However, getting to z = 1 in an expanding universe means that Bob (like a galaxy at the distance) is physically receding away from the Earth with a speed giving a Lorentz factor of 2. Bob can stay there for 200 seconds, Alice will measure he stayed there for 100 seconds.
This does not happen in his static universe.

If Bob travelled back to Alice, Bob's clock will read less than Alice's clock. This is just adding a delay to the resolved and measured twin paradox.
19 March 2021: Ignorance about SR from Mike Helland (the twin paradox)

 

[Reformed Offlian]

I'm trying to make sure the scenario is clear.

Bob and Alice have a plan. They're going to sync up their clocks, then Bob is going to take off for a distance of z=1. When there, Bob is going to note the tie and wait 100 ticks, 100 seconds. Then Bob will begin the return journey.

Alice will have observed Bob to reach z=1, wait 200 seconds, and then begin return to come back.

In that 200 seconds, Bob's clock will only have ticked 100 times.

What will Bob have observed about Alice's clock in the same scenario?

 

[Crossbow]

I'm trying to make sure the scenario is clear.

Bob and Alice have a plan. They're going to sync up their clocks, then Bob is going to take off for a distance of z=1. When there, Bob is going to note the tie and wait 100 ticks, 100 seconds. Then Bob will begin the return journey.

Alice will have observed Bob to reach z=1, wait 200 seconds, and then begin return to come back.

In that 200 seconds, Bob's clock will only have ticked 100 times.

Oh gee whiz!

You are quite unable to keep your own nonsense correct.

Earlier, you were saying that the clocks would resynchronize when they were brought together after such a trip, and now you are saying that the clocks will be out of synchronization after such a trip.

 

[Mike Helland]

What will Bob have observed about Alice's clock in the same scenario?

Bob will have observed Alice's clock running twice as slow as his.


So here we go.

time=0
distance=0
photons emitted (A/B = 0
photons received (A/B) = 0

Now Bob is going to travel at the speed of light to z=1.

How far is that, and how long would it take?

From my "lab" I get:

v=HD, t=9170 d=13229.593
v=c-HD, t=9170 d=6614.623
c / (1 + H × D)^2, t=15238 d=10356.065

In an expanding universe, and in v=c-HD, you would get there in the same time, 9.17 billion years. You would be either 13.2 Bly away or 6.6.

But I'm going with the inverse square law after seeing how it fits the evidence for dark energy.

So what happens now?

Bob needs to reach of distance of 10.3 Bly.

So let's say Bob heads off at speed c.

How is Bob going to know he's gone far enough?

He can't just watch his clock for 10.3 billion years, because his clock should be moving at 0, if he's moving at c.

So that's a detail I'd like to know.

But let's say he knows, so he goes out there, now we're at:

t=15.2 billion years, d=10.3 billion light years

Alice will have transmitted 15.2 billion * seconds in a year clock pulses.

Bob will have received zero, since he's traveling with the first clock signal. Since his watch hasn't moved yet, he won't have have sent any clock pulses.

When he stops at his destination, his clock moves. He sends out 60 pulses over 60 seconds, then heads back to Alice at the speed of light.

If Bob's clock is moving slower, Alice will see those clock pulses over 120 seconds.

Alice will have sent out 120 pulses while Bob was observed to be stopped.

Bob would have observed 30 pulses while stopped. The other 90 would still be in transit when he started moving. He would encounter them all on his return trip.

 

[jadebox]

I haven't been able to follow most of the more advanced physics in this thread, but as I hinted at in an earlier post, I have a very hard time (pun) with the idea of photons slowing down.

When a photon is emitted, it moves at the speed of light and, from its perspective, it arrives immediately at its destination. How can it slow down as time passes when no time ever passes?

It's enough to make a Star Trek computer explode.

 

[Mike Helland]

I haven't been able to follow most of the more advanced physics in this thread, but as I hinted at in an earlier post, I have a very hard time (pun) with the idea of photons slowing down.

When a photon is emitted, it moves at the speed of light and, from its perspective, it arrives immediately at its destination. How can it slow down as time passes when no time ever passes?

It's enough to make a Star Trek computer explode.

I think the idea I'm on now (it changes day to day, sorry everyone), is that the photon has a clock, that ticks verrrry slow.

If it's first tick is after, say a couple million of our years, the redshifts wouldn't appear until millions of light years away.

*edit* Yep, that works too:

        photon: {x: 0, dx: c, f: 6e5, w: 500.68, f_start: 6e5, clock: 0},
        next: function () {
            this.photon.dt = 1 / Math.pow(1 + this.H / 1000 * this.photon.x, 2)
            this.photon.clock += 0
            this.photon.dx = c * this.photon.dt
            this.photon.x += this.photon.dx
            this.photon.f = (this.photon.dx * 299792458 / this.photon.w) 

        }

 

[hecd2]

Oh gee whiz!

You are quite unable to keep your own nonsense correct.

Earlier, you were saying that the clocks would resynchronize when they were brought together after such a trip, and now you are saying that the clocks will be out of synchronization after such a trip.

That's because there is an inconsistency at the heart of his proposal. If you think abouuut it one way the clocks re-synchronise. If you think about it a different way they don't. So they resynchronise and they don't resynchronise.

 

[steenkh]

That's because there is an inconsistency at the heart of his proposal. If you think abouuut it one way the clocks re-synchronise. If you think about it a different way they don't. So they resynchronise and they don't resynchronise.

Is this inspired by quantum mechanics so that the clocks are indefinite until you actually look at them, and they will then either be resynchronized or not?

 

[Mike Helland]

That's because there is an inconsistency at the heart of his proposal. If you think abouuut it one way the clocks re-synchronise. If you think about it a different way they don't. So they resynchronise and they don't resynchronise.

Right.

So that idea's out.

Here's what I did now.

Instead of decelerating the photon's, I give them a clock that ticks at 1/(1+HD)².

Then I move the photon at c, according to that clock.

It redshifts analogously to the decelerating photon.

I think this solves the reflection problem.

Before when slow light was hitting the mirror, the longer it took to reach the mirror counted against it, because it was moving slow.

In this theory, it's moving at c (just by a slowed clock), and the time it takes to reach the observer point after mirror counts against it in the least time principle, because the clock is running full speed counting more time.

Works?

 

[hecd2]

Is this inspired by quantum mechanics so that the clocks are indefinite until you actually look at them, and they will then either be resynchronized or not?

Nothing as sophisticated as that. Whenever you look at them they are both synchronised and desynchronised.

 

[hecd2]

Right.

So that idea's out.

Shucks! And I was just about to show you how to calculate the relative clock rates according your scheme while they are in motion, how to allow for red/blue shift, and how to integrate the rates over the movement to get an accumulated clock difference. But now you'll never know.

What makes you think you can come up with a different scheme every other day which can replace all of physics?

 

[theprestige]

He's still trying to come up with a scheme that replaces his previous scheme.

 

[Mike Helland]

Shucks! And I was just about to show you how to calculate the relative clock rates according your scheme while they are in motion, how to allow for red/blue shift, and how to integrate the rates over the movement to get an accumulated clock difference. But now you'll never know.

I made a little computer program that sort of does all that.

What makes you think you can come up with a different scheme every other day which can replace all of physics?

Probably black mold.

 

[hecd2]

I made a little computer program that sort of does all that.

I doubt it. I think your program (or at least what you reported above) says the clock will resynchronise. But according to your description of the scenario (distant clocks are observed to run slower as a function of distance) the clocks would not. But anyway you have abandoned it for something that is completely incoherent. I have no idea what you are talking about in your latest iteration.