Author Topic: New Sungenis film: The Fool on the Hill (Read 32312 times)

Struthio · « **Reply #75 on:** July 30, 2019, 07:23:41 AM »

Quote from: Stanley N on July 29, 2019, 11:08:17 AM

And the way we've been discussing sagnac devices is by way of a time difference in the lab frame.

Sorry Stanley, but no, not in the lab frame. We are discussing the apparatus at mathpages.com which is in motion with respect to the lab, and is being described by the equation

The "stationary circular loop" mentioned at the beginning of the article measures a null fringe shift indicating no time difference.

And yes, Stanley, a time difference. But the time difference is neither measured by clocks in different frames, nor by clocks in the same frame, nor by clocks at all, nor is it measured at all. What is measured by the detector of the interferometer is a fringe shift, not a time difference.

With an interferometer one does not need to synchronize clocks, since none are used. The physical quantity measured is not a time difference, but a fringe shift. The measured fringe shift is a unitless fraction of the wavelength of the light beam used, and not a time difference. The time difference is calculated from the fringe shift.

Obviously, not only your idea of clocks in Special Relativity is flawed (as shown in a recent comment), you also seem to not even know how the interferometers work, which are used in the experiments we discuss.

Struthio · « **Reply #76 on:** July 30, 2019, 07:57:41 AM »

Quote from: King Wenceslas on July 17, 2019, 01:15:03 PM

He [Sungenis] holds a "doctorate" from Calamus International University in the Republic of Vanuatu, a noted diploma mill.

It looks like PhD titles from Calamus International University are better than their fame.

At least, Sungenis does not fall for flawed relativistic physics.

Quote from: King Wenceslas on July 17, 2019, 01:15:03 PM

Anyone with half a brain would run away from this guy. He is literally a carbuncle on the rear end of Tradition.

He does not seem to realize that the Robber Council is full of heresies. But, on the other hand, he shows how relativistic physics are flawed. One may assume that this devaluates quite a lot of PhD titles from generally respected institutions, but I think the problem is less a lack of ability than an abundance of iniquity.

Stanley N · « **Reply #77 on:** July 30, 2019, 01:07:05 PM »

Quote from: Struthio on July 30, 2019, 07:23:41 AM

Quote from: Stanley N on July 29, 2019, 11:08:17 AM
And the way we've been discussing sagnac devices is by way of a time difference in the lab frame.
Sorry Stanley, but no, not in the lab frame. We are discussing the apparatus at mathpages.com which is in motion with respect to the lab, and is being described by the equation

So yes, a time difference. In the lab frame. For a rotating sagnac ring. Just as I said.

Quote

The "stationary circular loop" mentioned at the beginning of the article measures a null fringe shift indicating no time difference.

Obviously, if it's stationary in the lab frame.

Quote

Obviously, not only your idea of clocks in Special Relativity is flawed (as shown in a recent comment), you also seem to not even know how the interferometers work, which are used in the experiments we discuss.

Oh dear, Struthio says my understanding is flawed. Whatever shall I do?

There's no point further discussing SR because you apparently refuse to use a standard conceptual tool in SR analysis. This is one of the ways I've tried to help you understand Sagnac devices in SR, but if you don't want the help, I'm done trying.

So fine, you win. I'll stipulate Sagnac devices successfully falsify Struthio's theory of relativity. Unfortunately, Struthio's theory of relativity has little to do with Special Relativity, and you can't falsify a theory if you don't apply it correctly.

On to classical mechanics:

Quote from: Struthio on July 25, 2019, 08:51:04 PM

Mass of the sun yes, but not squared.

If Newton's formula has the mass of the sun, then Popov's too: Popov "postulates the existence of vector and scalar potentials caused by the simultaneous motion of the masses in the Universe, including the distant stars." These potentials are designed to produce forces which make the objects move in the same way Newton saw them. Nothing to worry about the mass of the sun.

I've also given you several days to think about the Popov formula with Msun squared. So far the response has been: <crickets>.

The last time we had this discussion (a year ago?) I said you can make a geocentric system that was kinematically equivalent. But that doesn't mean they are kinetically equivalent. Objects in a geocentric system accelerate differently. Do you allow for any device capable of measuring acceleration?

Struthio · « **Reply #78 on:** July 30, 2019, 01:46:15 PM »

Quote from: Stanley N on July 30, 2019, 01:07:05 PM

Quote from: Struthio
Sorry Stanley, but no, not in the lab frame. We are discussing the apparatus at mathpages.com which is in motion with respect to the lab, and is being described by the equation

So yes, a time difference. In the lab frame. For a rotating sagnac ring.

No, not in the lab frame.

The above formula describes the process with respect to that reference frame where the light source and the detector are fixed and not moved.

How do we know that? Well, the travelled distance in the formula is the circuмference of a whole circle: 2 pi R.

In the lab frame, the travelled path begins North and ends North-East, thus resulting in a travelled distance which is greater than 2 pi R for one beam and less than 2 pi R for the other beam.

In the rotating frame the travelled distance is 2 pi R, since the path from the light source to the detector is a whole circular turn.

That's how we know that you don't know what you're talking about.

Quote

There's no point further discussing SR [...]

Indeed, Stanley. Not even classical physics.

Stanley N · « **Reply #79 on:** July 30, 2019, 02:18:05 PM »

Quote from: Struthio on July 30, 2019, 01:46:15 PM

So yes, a time difference. In the lab frame. For a rotating sagnac ring.

No, not in the lab frame.

The above formula describes the process with respect to that reference frame where the light source and the detector are fixed and not moved.

How do we know that? Well, the travelled distance in the formula is the circuмference of a whole circle: 2 pi R.
...
In the lab frame, the travelled path begins North and ends North-East, thus resulting in a travelled distance which is greater than 2 pi R for one beam and less than 2 pi R for the other beam.

The last paragraph - exactly. In the lab frame the travelled distance is greater in one direction and less in the other direction. That's in the lab frame, again. And as I have already shown, the formula can be derived from the lab frame, with the travelled distance greater in one direction and less in the other direction.

Why are you still fighting this?

Struthio · « **Reply #80 on:** July 30, 2019, 03:13:43 PM »

Quote from: Stanley N on July 30, 2019, 02:18:05 PM

Quote from: Struthio
So yes, a time difference. In the lab frame. For a rotating sagnac ring.

No, not in the lab frame.

The above formula describes the process with respect to that reference frame where the light source and the detector are fixed and not moved.

How do we know that? Well, the travelled distance in the formula is the circuмference of a whole circle: 2 pi R.
...
In the lab frame, the travelled path begins North and ends North-East, thus resulting in a travelled distance which is greater than 2 pi R for one beam and less than 2 pi R for the other beam.

The last paragraph - exactly. In the lab frame the travelled distance is greater in one direction and less in the other direction. That's in the lab frame, again. And as I have already shown, the formula can be derived from the lab frame, with the travelled distance greater in one direction and less in the other direction.

Why are you still fighting this?

The formula has only one travelled distance, which is 2 pi R:

Hence, the formula describes the situation in the rotating frame and not in the lab frame.

And you have not shown or presented any formula which would be valid in the lab frame.

And we do not need such a formula. The formula above says that the two light beams travel at c + v and c - v respectively from the light source to the detector along a path of length 2 pi R (in the rotating frame).

I say, that the formula is not valid in Special Relativity, since you need relativistic addition instead of c + v and c - v.

You and the mathpages guy say that the formula is valid in Special Relativity.

If you were right, then light would travel faster/slower than c. Which is known to be not the case following Special Relativity.

If I am right (relativistic addition), then light travels at c. Which is known to be the case following Special Relativity.

Stanley N · « **Reply #81 on:** July 30, 2019, 03:29:39 PM »

Quote from: Struthio on July 30, 2019, 03:13:43 PM

The formula has only one travelled distance, which is 2 pi R:

Go back to the derivation. Neither of the distances travelled are 2 pi R. And light travels at c. In the lab frame.

Struthio · « **Reply #82 on:** July 30, 2019, 07:32:26 PM »

Quote from: Stanley N on July 30, 2019, 03:29:39 PM

Neither of the distances travelled are 2 pi R. And light travels at c. In the lab frame.

Even assumed that your statement was correct. So what? Special Relativity is falsified already, if it is falsified in the rotating frame, where the light source and the detector are at rest. That is the reason why you can save the effort to transfrom the mathpages equation to the lab frame.

Special Relativity fails in the rotating frame, where the mathpages formula says that light travels at speeds other than c, which is illegal for Special Relativity. The mathpages formula says that one beam travels at c + v and the other at c - v:

Both beams travel a distance of 2 pi R, and both beams travel at a speed other than c, thus falsifying Special Relativity, thus proving that light can move faster or slower than c.

The experiment refutes the mantra of the Einsteinian relativists.

Stanley N · « **Reply #83 on:** July 30, 2019, 11:20:32 PM »

Quote from: Struthio on July 30, 2019, 07:32:26 PM

Even assumed that your statement was correct.

Wait, what? You still don't get this?

So here you go, relativistic.

The light is emitted inside the ring. The ring and emitter move at radial speed v.
Let's also account for the medium, so light (or the "signal") travels at some speed s <= c.

The signal speed in the lab frame for signal taking less time is u-.
So u- = (s-v)/(1-sv/c^2). If we had s=c, then u- = c.
We still have u-t- = L - vt- , so t- = L/(u- + v)
Thus t- = L/ [ s (1-v^2/c^2) / (1-sv/c^2) ] = L (c^2 - sv) / [s (c^2 - v^2)]
Similarly, u+ = (s+v)/(1+sv/c^2), and t+ = L/(u+ - v)
So t+ = L/(u+ - v) = L (c^2 + sv) / [s (c^2 - v^2)]
And therefore t+ - t- = L (2sv) / [s(c^2-v^2)] = 2 L v / (c^2- v^2)
So with L=2 pi R, you get the same formula.

Isn't it interesting that the signal speed s drops out entirely?

Perhaps I should make youtube videos.

Struthio · « **Reply #84 on:** July 30, 2019, 11:57:43 PM »

Quote from: Stanley N on July 30, 2019, 11:20:32 PM

Wait, what? You still don't get this?

So here you go, relativistic.

The light is emitted inside the ring. The ring and emitter move at radial speed v.
Let's also account for the medium, so light (or the "signal") travels at some speed s <= c.

The signal speed in the lab frame for signal taking less time is u-.
So u- = (s-v)/(1-sv/c^2). If we had s=c, then u- = c.
We still have u-t- = L - vt- , so t- = L/(u- + v)
Thus t- = L/ [ s (1-v^2/c^2) / (1-sv/c^2) ] = L (c^2 - sv) / [s (c^2 - v^2)]
Similarly, u+ = (s+v)/(1+sv/c^2), and t+ = L/(u+ - v)
So t+ = L/(u+ - v) = L (c^2 + sv) / [s (c^2 - v^2)]
And therefore t+ - t- = L (2sv) / [s(c^2-v^2)] = 2 L v / (c^2- v^2)
So with L=2 pi R, you get the same formula.

Isn't it interesting that the signal speed s drops out entirely?

Perhaps I should make youtube videos.

So you are now saying that the mathpages formula is valid in both frames. In the lab frame and in the rotating frame? One and the same formula for both frames?

Stanley N · « **Reply #85 on:** July 31, 2019, 12:14:24 AM »

Quote from: Struthio on July 30, 2019, 11:57:43 PM

So you are now saying that the mathpages formula is valid in both frames. In the lab frame and in the rotating frame? One and the same formula for both frames?

I said lab frame in my post. I's an analysis in the lab frame using relativistic velocity addition. You could also add length contraction to L for a second order effect.

Struthio · « **Reply #86 on:** July 31, 2019, 07:27:07 AM »

Quote from: Stanley N on July 30, 2019, 11:20:32 PM

We still have u-t- = L - vt- , so t- = L/(u- + v)

Resolving u- t- = L - vt- for t- requires to divide by u-. The result is

Code: [Select]

u- t- = L - vt- <=> t- = (L - vt-) / u-

Whatever you calculate for the lab frame, in the rotating frame both beams travel the same distance. Given Special Relativity and thus a constant speed of light c, a zero fringe shift should be measured. Special Relativity is falsified by the experiment whatever you calculate for the lab frame.

Stanley N · « **Reply #87 on:** July 31, 2019, 09:28:17 AM »

Quote from: Struthio on July 31, 2019, 07:27:07 AM

Resolving u- t- = L - vt- for t- requires to divide by u-. The result is

Code: [Select]
u- t- = L - vt- <=> t- = (L - vt-) / u-

Whatever you calculate for the lab frame, in the rotating frame both beams travel the same distance. Given Special Relativity and thus a constant speed of light c, a zero fringe shift should be measured. Special Relativity is falsified by the experiment whatever you calculate for the lab frame.

Repeated claims do not make it so.

SR analysis of a Sagnac device works in the loop frame, too. But you have rejected using the conceptual device of clocks to understand that. And you can't even follow simple algebra. If you're unwilling to listen, there's nothing further I can do here.

So go look this up at a reliable site. They are out there.

Struthio · « **Reply #88 on:** July 31, 2019, 09:52:27 AM »

Quote from: Stanley N on July 31, 2019, 09:28:17 AM

And you can't even follow simple algebra.

I am sure that most readers are able to see that it's your algebra that is flawed:

Quote from: Stanley N on July 30, 2019, 11:20:32 PM

We still have u-t- = L - vt- , so t- = L/(u- + v)

Quote from: Struthio on July 31, 2019, 07:27:07 AM

Resolving u- t- = L - vt- for t- requires to divide by u-. The result is

Code: [Select]
u- t- = L - vt- <=> t- = (L - vt-) / u-

It's not even necessary to know what the quantities u-, t-, L, and v are.

Stanley N · « **Reply #89 on:** July 31, 2019, 02:26:45 PM »

Quote from: Struthio on July 31, 2019, 09:52:27 AM

I am sure that most readers are able to see that it's your algebra that is flawed:

It's not even necessary to know what the quantities u-, t-, L, and v are.

I'm going to drop the - indicators for this.
The equation is UT = L - VT
The solution, using just algebra, is T =L/(U+V)

Author Topic: New Sungenis film: The Fool on the Hill (Read 32312 times)

Struthio

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

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Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill

Struthio

Re: New Sungenis film: The Fool on the Hill

Stanley N

Re: New Sungenis film: The Fool on the Hill