Twin paradox (Pete and MacM)

[Paul T]

MacM,

V = Vo(1-Vs^2/c^2)^.5/(1-Vs/c)

V is Observed Frequency
Vo is Source frequency
Vs is Source Velocity
c is velocity of light in a vacuum

Now you have the standard frequency of the initial beam when viewed at rest (Vo). You have the observed frequency (V).

Your above doppler shift equation was originated from SR. Since, you wanted to think not based on SR (in fact, you thought that SR is wrong), how do you ascertain that this equation is correct?

 

[MacM]

Do I have to spell everything out for you?? .75+.75=1.50
Without Relativity, one would conclude that the two crafts are separating
from one another at more than lightspeed and so no signals could be
interchanged . 'Got it now, sir??

Well, well, well. Based on the smug and arrogant attitude I see developing
here I think it is time I stop being polite and step back and let some of that air out of some over inflated sails. Any grade schooler should recognize your failed logic.

1 - If you choose to disregard the Velocity Addition Formula then you must
also disregard the v = c limit. That is if you assume an actual1.5 c
relative velocity you must also not consider v = c a limit and assume all is
well. If you keep the speed of light invariant then the speed of relative
communication would still be at v=c to each observer. You see there is no
basis to view the invariance of light speed as a consequence of Relativity
it is the foundation of Relativity.

Put in plain english ignoring all relavistic functions which stem from the
observation of the invariance of light does not alter the observed
invariance. Hence communication would continue to exist. Applying
Relativity to such a case you only have a relative velocity of 0.96 c, and
the v = c would persist.

2 - If I assume no Relativity and I observe an object coming toward earth at
0.5c and it is 10 ly away, what you see is what you get.

3 - If I assume Relativity then the object is actually 11.547 ly away and
has been reduced to 10 ly by Lorentz Contraction.

4 - Further since such an object moving toward the earth is under the
enfluence of gravity it is going to be accelerating. At that distance the
acceleration is substantially linear and I'll not bother computing an actual
accurate conversion using the inverse square, since Newton's formula's are
not considered to hold beyond 0.1 ly anyhow.

5 - Without Relativity if the object were to have an acceleration of 0.1 c
per ly then in one years time it will have reached 9.45 ly distance. i.e.
0.5c to 0.6c per year = (.5c+.6c)/2 = 0.55c average velocity.

6 - But if you assume Relativity then at the end of one year the distance
will be 11.547 ly/gamma = 9.2. (This is only generally true because it is a
nonlinear average and I don't feel like wasting the time to be precise
here.

The point being that you claim that Relativity is protecting us is
bas-ackwards. You are full of it. Relativity accelerates the closure rate
and increases impact velocity. Before becoming an advocate you should learn
Relativity.

Again, do I have to spell everything out for you?? Without Relativity
and
its c=cosmic speed limit, then the meteor/asteroid could impact seconds
after it is perceived to be a trillion miles away!! 'Got it
now, sir??

Got it now ,. sir?? If not See above.

Without Relativity, what would light propagate relative to? Under
relativity
it always propagates at speed c relative to each and every observer. Without
Relativity, it propagates relative to an absolute space/grid, another name
for which is "aether".

Well I think it should be clear that this statement is outright horseshit. the impact timing will be more linear without Relativity. Relativity will accelerate (reduce) the time to impact and due to contract dimension appear to have a higher velocity.

Ether? Well lets see what Einstein had to say about that issue.

********************************************************
Ether and the Theory of Relativity
Albert Einsteinan address delivered on May 5th, 1920, in the University of
Leiden

************** Extracts from Einstein's Speech******************

More careful reflection teaches us, however, that the special
theory of relativity does not compel us to deny ether. We may assume the
existance of an ether................

Recapitulating, we may say that according to the general theory of
relativity space is endowed with physical qualities; in this
sense, therefore, there exists an ether.

According to the general theory of relativity space without
ether is unthinkable; for in such space there not only would be
no propagation of light, but also no possibility of existence for standards
of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense.

***************************************************

So not only has he said Relativity requires an ether to function he has said space-time does not exist without it.

Seems it is back to school for you but at what grade level? Hmmmm.

 

[MacM]

MacM,

Your above doppler shift equation was originated from SR. Since, you wanted to think not based on SR (in fact, you thought that SR is wrong), how do you ascertain that this equation is correct?

Not to disappoint you but I have no intention of playing circle jerk with you. I explained the posting above which was at the request of James R.

The simple point being if he want to vouch for SR then the doppler shift is a perfectly valid method of determining relative velocity. Stick it in your ear Paul T. Unless you have an actual contribution to make you will be ignored.

 

[Paul T]

Not to disappoint you but I have no intention of playing circle jerk with you. I explained the posting above which was at the request of James R.

The simple point being if he want to vouch for SR then the doppler shift is a perfectly valid method of determining relative velocity. Stick it in your ear Paul T. Unless you have an actual contribution to make you will be ignored.

I learned from your previous comment on my assessment that "one cannot prove a theory correct based on the theory itself". We concern about SR here and your proposed test to disprove SR. Based on your so-called rule, we cannot prove SR based on SR. Can we prove another theory based on SR? Yes, we can, provided that SR must be proven correct at least in the respective domain. Now, you just simply grab SR's doppler effect equation and used it as the base for your analysis without first verify that such equation is valid. Is that equation valid in this respect? The answer should be NO, since your proposed test in the end supposed to tell us something about SR's validity. Assuming that in the end the test does not disprove SR, we still have question whether the basic equation (doppler effect) used in the analysis is valid.

 

[Paul T]

I have many comments for your answer to distram, however I believe I should let him address them.

There is one general comment though. You need to be more straight forward on which part of SR incorrect and which part correct, according to you and then from there the disagreement -- hopefully -- can be resolved. On second thought, this may not work as you have been discussing about SR incorrectness for months or probably years but nobody (at least I don't) seem to see what exactly in SR you disagree with.

 

[Pete]

Doopler formulas:
Mac's formula is correct.
The convention in that formula is that negative velocity indicates a receding source, positive velocity an approaching source.
JR's formula is also correct, but uses positive velocity for a receding source.

 

[Paul T]

Doopler formulas:
Mac's formula is correct.
The convention in that formula is that negative velocity indicates a receding source, positive velocity an approaching source.
JR's formula is also correct, but uses positive velocity for a receding source.

I am pretty sure the formula is correct. However, since MacM believes that SR is fundamentally flaw, I do not know how he could apriori adopt this formula as correct. He demanded others not to use SR's formula to assess SR. I just do not understand why he didn't demand himself to do the same. He should use formula which has no connection to SR, unless he checked carefully that SR's formula that he adopt -- accidentally -- correct although the theory behind it (SR) fundamentally flaw as he believe.

 

[MacM]

I have many comments for your answer to distram, however I believe I should let him address them.

There is one general comment though. You need to be more straight forward on which part of SR incorrect and which part correct, according to you and then from there the disagreement -- hopefully -- can be resolved. On second thought, this may not work as you have been discussing about SR incorrectness for months or probably years but nobody (at least I don't) seem to see what exactly in SR you disagree with.

Fair post so I will try to respond. You have an incorrect view. There is no crusade against SR. Much of what I do believe has aspects of SR in it. The point is to get away from the pure mathemactical view and to see a broader picture.

There are some aspects of SR that I contend are merely perception and do not affect reality, where current physics holds they are reality. I guess that is probably the greatest disagreement.

 

[Pete]

Mac, if I may? I'm not sure of James' electronics background.
Note - The figures used in this post are fairly arbitrary.

Modulation:
A digital signal may be transmitted on a light beam as follows - a zero is indicated by slightly varying the beam frequency up and down 100 times at a rate of 1/10 the carrier frequency.
A one is indicated by slightly varying the beam frequency up and down 50 times at a rate of 1/20 the carrier frequency.


For example:
If a light beam is transmitted at 10¹⁴Hz, a signal "101" may be transmitted like so:

Repeat 50 times:
(
Transmit at 1.01x10¹⁴Hz for 2x10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 2x10^-13 seconds
)
Repeat 100 times:
(
Transmit at 1.01x10¹⁴Hz for 10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 10^-13 seconds
)
Repeat 50 times:
(
Transmit at 1.01x10¹⁴Hz for 2x10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 2x10^-13 seconds
)
Continue transmission at 10<sup>14<sup>Hz.​

In practice, the frequency modulation might be sinusoidal rather than stepped (I'm not sure), but it makes no difference.

 

[Pete]

Just to clarify, Mac - the ships start together, then accelerate away from each other at some agreed rate for some agreed subjective time?

 

[MacM]

I am pretty sure the formula is correct. However, since MacM believes that SR is fundamentally flaw, I do not know how he could apriori adopt this formula as correct. He demanded others not to use SR's formula to assess SR. I just do not understand why he didn't demand himself to do the same. He should use formula which has no connection to SR, unless he checked carefully that SR's formula that he adopt -- accidentally -- correct although the theory behind it (SR) fundamentally flaw as he believe.

Actually Paul T it is the best kind of evidence. By using accepted relavistic formulas and then to get different conclusion. Anyone can get different conclusion if you don't use the relavistic mathematics.

 

[MacM]

Doopler formulas:
Mac's formula is correct.
The convention in that formula is that negative velocity indicates a receding source, positive velocity an approaching source.
JR's formula is also correct, but uses positive velocity for a receding source.

Not sure you are right here Pete. You seem to be addressing the "+" / "-" sign issue but if you will note it is the V^2 term in the one that I jposted that seems to be the problem. Just not sure what that formula is there. James' formula was right next to it but the two do not equate. It howver came from the Hyperphysics link.

 

[MacM]

Mac, if I may? I'm not sure of James' electronics background.
Note - The figures used in this post are fairly arbitrary.

Modulation:
A digital signal may be transmitted on a light beam as follows - a zero is indicated by slightly varying the beam frequency up and down 100 times at a rate of 1/10 the carrier frequency.
A one is indicated by slightly varying the beam frequency up and down 50 times at a rate of 1/20 the carrier frequency.


For example:
If a light beam is transmitted at 10¹⁴Hz, a signal "101" may be transmitted like so:

Repeat 50 times:
(
Transmit at 1.01x10¹⁴Hz for 2x10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 2x10^-13 seconds
)
Repeat 100 times:
(
Transmit at 1.01x10¹⁴Hz for 10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 10^-13 seconds
)
Repeat 50 times:
(
Transmit at 1.01x10¹⁴Hz for 2x10^-13 seconds
Transmit at 0.99x10¹⁴Hz for 2x10^-13 seconds
)
Continue transmission at 10<sup>14<sup>Hz.​

In practice, the frequency modulation might be sinusoidal rather than stepped (I'm not sure), but it makes no difference.

I had specified AM and you are using FM. I suspect FM would be harder to take advantage of since it appears as doppler (a shifting frequency). By using AM whatever the doppler shift in the carrier beam it is also directly proportional to the sideband frequency and the digital relationship (ratio) between carrier and modulation becomes easy to detect. i.e one modulation envelope contains 100 carrier wave peaks, etc.

 

[MacM]

Just to clarify, Mac - the ships start together, then accelerate away from each other at some agreed rate for some agreed subjective time?

All those stipulations aren't required. Since the transition from acceleration to constant velocity (coasting) is evidenced by the doppler shift stabilizing.

Once that happens you can convert doppler into relative velocity so it doesn't matter at what rate and for how long you were under acceleration, just the end result.

 

[Paul T]

Not sure you are right here Pete. You seem to be addressing the "+" / "-" sign issue but if you will note it is the V^2 term in the one that I jposted that seems to be the problem. Just not sure what that formula is there. James' formula was right next to it but the two do not equate. It howver came from the Hyperphysics link.

MacM's formula is for condition where the relative motion is closing one to another while James R's formula for the opposite direction of movement. Just change (1-v/c) to (1+v/c) in MacM's formula and James R's one will be obtained.

 

[Pete]

Try it.
Your formula with Vs = -0.9c results in V = 0.2294 Vo

James's formula with Vs = 0.9c results in V = 0.2294 Vo

James's formula is not quite the same as the one listed at Hyperphysics - the signs for source velocity are reversed.

James has provided the usual formula for relativistic Red-shift (ie positive velocity is motion away).
The one at Hyperphysics is the formula often used for relativistic Blue-shift (positive velocity is motion towards).

The two are easy to confuse, because they are identical except for the sign of Vs.
Blue-shift: V = Vo [ (1 + Vs/c)/(1 - Vs/c) ] ^ 0.5
Red-shift: V = Vo [ (1 - Vs/c)/(1 + Vs/c) ] ^ 0.5


Note that this shows the importance of understanding the meaning of equations in a model - a lack of understanding often results in using equations in the wrong way, which leads to nonsensical results that are not what the model predicts at all.

 

[James R]

MacM:

Thanks to Pete, I have sorted out the Doppler shift formula problem.

The formula you gave was

V = Vo sqrt(1-(v²/c²)/(1-v/c)

where v is the source velocity and it is assumed the source is approaching the receiver. For a source receding from the receiver we must replace v with (-v), which gives:

V = Vo sqrt(1-(v²/c²)/(1+v/c)

This is identical to my formula, as I will now show.

V = Vo sqrt(1-(v²/c²)/(1+v/c)

Multiply the top and bottom of the fraction by c:

V = Vo c sqrt(1-(v²/c²)/(c+v)

Put the c on top into the square root:

V = Vo sqrt(c² - v²)/(c+v)

Factorise the stuff in the square root:

V = Vo sqrt((c-v)(c+v))/(c+v)

Put the demoninator into the square root:

V = Vo sqrt[(c-v)(c+v)/(c+v)²]

Cancel the (c+v) term:

V = Vo sqrt[(c-v) / (c+v)]

Divide top and bottom of the fraction in the square root by c:

V = Vo sqrt[(1 - v/c) / (1 + v/c)]

This is the formula I quoted originally.

I assume we are now in agreement that this formula can be used by B to calculate his velocity based on the Doppler shift of the signal received from A. Correct?

Please tell me if you have any issues with this. In my next post I will address the other remaining issues I have with your proposal.

 

[Pete]

All those stipulations aren't required. Since the transition from acceleration to constant velocity (coasting) is evidenced by the doppler shift stabilizing

Since simultaneity is the issue, I think the stipulations are required otherwise there's no clear relationship between the times at which each ship stops accelerating.

We can still do it if you like, but it will be harder to visualize.

 

[MacM]

MacM:

Thanks to Pete, I have sorted out the Doppler shift formula problem.

The formula you gave was

V = Vo sqrt(1-(v²/c²)/(1-v/c)

where v is the source velocity and it is assumed the source is approaching the receiver. For a source receding from the receiver we must replace v with (-v), which gives:

V = Vo sqrt(1-(v²/c²)/(1+v/c)

This is identical to my formula, as I will now show.

V = Vo sqrt(1-(v²/c²)/(1+v/c)

Multiply the top and bottom of the fraction by c:

V = Vo c sqrt(1-(v²/c²)/(c+v)

Put the c on top into the square root:

V = Vo sqrt(c² - v²)/(c+v)

Factorise the stuff in the square root:

V = Vo sqrt((c-v)(c+v))/(c+v)

Put the demoninator into the square root:

V = Vo sqrt[(c-v)(c+v)/(c+v)²]

Cancel the (c+v) term:

V = Vo sqrt[(c-v) / (c+v)]

Divide top and bottom of the fraction in the square root by c:

V = Vo sqrt[(1 - v/c) / (1 + v/c)]

This is the formula I quoted originally.

I assume we are now in agreement that this formula can be used by B to calculate his velocity based on the Doppler shift of the signal received from A. Correct?

Please tell me if you have any issues with this. In my next post I will address the other remaining issues I have with your proposal.

I agree. It ws my stipulation that doppler was used to extract relative velocities.

I was concerned about the V^2 term and I believe you got a different answer using it and yet you just showed the two should be the same?

 

[James R]

MacM:

I wrote:

Sure, if clock A jams 10 waves into the modulated component of its carrier wave in 1 second, as measured by A's clock, then we can establish that, according to A, A's clock has 1 tick for every 10 waves. But how does B use that information when he receives it?

When B receives 10 waves in a carrier wave from A, B does not have a copy of A's clock to know how long it took to put those 10 waves into the carrier. B only has his own clock, which may or may not run at a different rate (that is one thing the test aims to determine, right?) As far as I can tell, all B can do is measure how long it takes to receive those 10 waves according to his own clock, and that tells him nothing about the actual rate of A's clock.

Your reply was:

Simply false. A and B are initially calibrated to the same frequency carrier beams. Local operation by both clocks have no proper time shift or doppler shift in their beam transmitter. Both clocks hence have the standard on board to convert the data to a real clock rate.

I believe we have now agreed that if A sends out a 1 MHz signal to B, then B receives the signal at 0.229 MHz due to the Doppler shift. Is that correct?

Now, if such a signal is used as a "carrier beam", then A and B are NOT calibrated, because when A sends out 1000 wavelengths to B in one second as measured on A's clock, then when B receives the signal he only receives 229 waves per second, as measured on his own clock.

You are right to say that A sees no Doppler shift in his own emitted beam. But B DOES see a different frequency due to the Doppler shift.

So, I still need to know how clock B is going to measure the rate of clock A.

Let me make a suggestion. A and B agree in advance that A will send out a 1 MHz signal, as measured on A's clock. When B actually receives this signal, he measures the frequency and finds it to be 0.229 MHz. Knowing this represents a 1 MHz signal for A, B sets his monitor of A's clock so that for every second that B's clock ticks off, B's monitor of A's clock only ticks off 0.229 seconds. In other words, B models A's clock as running slower than B's clock, and this keeps everything properly synchronised.

Do you agree? If not, why not?

Lets try this. Do you agree that I can pick up a Motorola catalog and purchase a frequency to digital converter?. Do you agree I can also purchase a digital to frequency converter?. Do you agree that I can purchase today (if they are in stock) a frequency to digital counter and display? Yes or No?

Yes. That's irrelevant, as far as I can see. This is not a technological issue, but an issue of how your test is to be carried out. What is relevant is what I have written above.

If this hasn't done it then express your conceerns further. We have in fact made jprogress in that you now recognize that Relativity is not ging to shift my reading by clock "B" of clock "A's" local proper recorded time.

I'm not sure we're in agreement on this. It depends on how you respond to what I've written above.

Both clocks are going to remain synchronized and agree on 36,000 seconds at the end of the test. Time dilation did not occur in absence of a delayed information system.

I won't discuss the actual result of the test until you have responded to what I have written above, because I think we most likely still have mutual misunderstandings about the procedure.