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Old 5th December 2019, 07:24 PM   #81
Reality Check
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Exclamation Cannot understand how his "geostationary altitude" question is nonsense

Originally Posted by philippeb8 View Post
What defines the geostationary altitude of this lonely star?
6 December 2019 philippeb8: Cannot understand how his "geostationary altitude" question is nonsense and irrelevant.
Geostationary is a term for Earth orbits. The proper term for a general body is a stationary orbit. To define the altitude of a stationary orbit we would follow the Derivation of geostationary altitude and get the same result !
We have his "The star can spin on its own" but it will be impossible in his incoherent scenario to measure the T period needed for the altitude of a stationary orbit because he has no observer. Add an observer and the task is possible.The irrelevance is that he replied to a post about his "no fire will be ejected" and "perfectly spherical" delusions about stars

Last edited by Reality Check; 5th December 2019 at 07:40 PM.
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Old 5th December 2019, 08:05 PM   #82
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Originally Posted by philippeb8 View Post
Let me ask you differently (what Reality Check cannot answer):
- What defines the geostationary altitude of this lonely star?
You are not asking differently, you are asking a different question.
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Old 5th December 2019, 08:27 PM   #83
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Originally Posted by Reality Check View Post
Correct. Ignorant nonsense about an incoherent scenario is a deadlock to an conversation.

A simple question in case you want to actually start a conversation
What does your "Kinematical Time Dilation" predict for the Frisch-Smith experiment?
Measurement of the Relativistic Time Dilation Using μ-Mesons Frisch, David H. ; Smith, James H. (1963).
So for a time dilation factor of 8.83:
- if we use SR, the particle needs to go at: v≈0.993566 c;
- if we use FT, the particle needs to go at: v≈0.941674 c.

Where:
c = 299792458 m/s

On page 351, they use SR to convert the energy into speed in the first place!?! I must conclude this paper is nonsense.

Now please answer my question. Thank you.
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Old 5th December 2019, 08:33 PM   #84
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Originally Posted by Matthew Ellard View Post
Santilli has been sloppy this time. I can explain the anti Einstein stance.

" In 1985 he (Santilli) published a book, Il Grande Grido: Ethical Probe on Einstein's Followers in the U.S.A, an Insider's View, in which he said that in many institutions there is an effective conspiracy to suppress or not investigate novel theories which may conflict with established scientific theories, such as Einstein's theory of relativity. He has complained that papers he has submitted to peer-reviewed American Physical Society journals were rejected because they were controlled by a group of Jewish physicists led by Weinberg
https://en.wikipedia.org/wiki/Ruggero_Santilli
Hahaha.

Since you're here why don't you answer the simple question nobody else is able to answer? Thank you.
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Old 5th December 2019, 08:33 PM   #85
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Question What does your "Kinematical Time Dilation" predict for the Frisch-Smith experiment

Originally Posted by philippeb8 View Post
....
Did not answer my question: What does your "Kinematical Time Dilation" predict for the Frisch-Smith experiment?
Take the real speeds of the muons in that experiment. Plug then into your equation. Give the result. Compare it to the SR result.

6 December 2019 philippeb8: A "I must conclude this paper is nonsense" delusion about the Frisch-Smith paper because they use SR.
The Frisch-Smith paper is an application of textbook physics to the real world. They are allowed to use textbook physics such as SR !

Last edited by Reality Check; 5th December 2019 at 08:37 PM.
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Old 5th December 2019, 08:35 PM   #86
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Originally Posted by Reality Check View Post
Did not answer my question: What does your "Kinematical Time Dilation" predict for the Frisch-Smith experiment?
Take the real speeds of the muons in that experiment. Plug then into your equation. Give the result. Compare it to the SR result.
I specifically said the speed of the particle is calculated using SR so obviously the predicted time dilation will match SR's predictions only.
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Old 5th December 2019, 08:46 PM   #87
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Time's up. So I conclude you all failed to defend 300 years of physics!
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Old 5th December 2019, 08:58 PM   #88
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This is like the person who came last in the Boston Marathon declaring victory.
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Old 5th December 2019, 09:04 PM   #89
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Originally Posted by arthwollipot View Post
This is like the person who came last in the Boston Marathon declaring victory.
Or like a person who stayed at home in a basement and having little idea what a marathon was or where Boston is declaring victory in the Boston Marathon.
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Old 5th December 2019, 09:07 PM   #90
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Originally Posted by philippeb8 View Post
Time's up. So I conclude you all failed to defend 300 years of physics!
So an hour and a half after asking me a question you conclude I cannot answer it.

Good reasoning ability there.

You didn't even acknowledge my answer to your first question.
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Old 5th December 2019, 09:10 PM   #91
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Originally Posted by Robin View Post
So an hour and a half after asking me a question you conclude I cannot answer it.

Good reasoning ability there.

You didn't even acknowledge my answer to your first question.
Reality Check already threw in the towel so I wish you good luck; I'll wait.
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Old 5th December 2019, 09:23 PM   #92
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Originally Posted by philippeb8 View Post
Reality Check already threw in the towel so I wish you good luck; I'll wait.
We have already dealt with your faulty assumption that without a point of reference, there would be no difference between between a stationary reference frame and a spinning reference frame.

That being out of the way, the answer to your question seems to be to supply you with a calculation that you could easily look up on Google yourself.

If you are looking for an answer that goes beyond what you might easily google, please clarify.
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Old 5th December 2019, 09:29 PM   #93
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Originally Posted by Robin View Post
We have already dealt with your faulty assumption that without a point of reference, there would be no difference between between a stationary reference frame and a spinning reference frame.
I never said that. I said the reference frame will be spinning with the star explaining why the fire (or plasma) cannot be ejected.

Quote:
That being out of the way, the answer to your question seems to be to supply you with a calculation that you could easily look up on Google yourself.

If you are looking for an answer that goes beyond what you might easily google, please clarify.
If you think it's possible then the answer regarding the geostationary altitude will be complicated indeed, if not impossible to find.
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Old 5th December 2019, 09:44 PM   #94
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Originally Posted by philippeb8 View Post
I never said that. I said the reference frame will be spinning with the star explaining why the fire (or plasma) cannot be ejected.
And I pointed out if there was a spinning reference frame then there would be a strangely shaped gravitational field pulling things outward.

Think of the two particles. Suppose they are motionless in one frame. How do we know whether they are spinning or not? If they are spinning then there is an outward force exerted on the bond.

If they are stationary and there are no external gravitational effects then there is no outward force exerted on the bond.

So if your star is in a spinning reference frame then this is equivalent to a stationary frame where there is a gravitational pull outwards radially from a line through the middle of the staras well as the gravitational pull of the star, so your question simply resolves to the question of which point is the pull of each gravitational field equal on a stationary object.

In a non-spinning frame of reference there is no gravitational field pulling outward then your question can be answered by the equation that you can find on google.

I am not sure what else you are looking for.
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Old 5th December 2019, 09:46 PM   #95
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Even in a hypothetical universe with only one star, the geostationary orbit altitude is 22,258 miles.

If you disagree, you might need to look up what "geostationary" actually means.
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Old 5th December 2019, 09:57 PM   #96
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Originally Posted by Myriad View Post
Even in a hypothetical universe with only one star, the geostationary orbit altitude is 22,258 miles.

If you disagree, you might need to look up what "geostationary" actually means.
I am interpreting here. I assume he is talking about the analogous point for this star.
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Old 6th December 2019, 12:37 AM   #97
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Originally Posted by arthwollipot View Post
No fire will be ejected because there is no fire on or around a star.
But also the fire cannot escape the iron shell of the sun, ...
or am I mixing up ludicrous ideas right now :-)
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Old 6th December 2019, 01:34 AM   #98
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Old 6th December 2019, 03:42 AM   #99
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Originally Posted by Myriad View Post
I think using a thought experiment that contradicts the results of a widely accepted real experiment should be worth more than a piddling 5 points.
It seems more of an injustice that there are no points available for declaring victory in an argument that nobody else is choosing to engage in.

Originally Posted by Myriad View Post
Anyhow, again, centrifugal force in rotating reference frames contradicts the claim that no mass (plasma or "fire") would be ejected or that the shape would be unaffected.
I think the point the OP is trying to make is either that there is no such thing as a rotating reference frame, or that rotating and inertial reference frames are equivalent. Either way, he seems to be claiming that, in the same sense that there is no such thing as absolute linear motion, there is also no such thing as absolute rotation. As a postulate, it seems a useless one, in that it inevitably leads to results that contradict observation; but in crackpot physics that's not really a negative.


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Old 6th December 2019, 04:24 AM   #100
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Originally Posted by philippeb8 View Post
I never said that. I said the reference frame will be spinning with the star explaining why the fire (or plasma) cannot be ejected.
If you accept that in my spinning pair of particles, there will still be a force on the bond in a spinning frame that keeps the particles still, then it follows that a spinning star in a spinning reference frame will still eject mass.
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Old 6th December 2019, 05:18 AM   #101
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Originally Posted by philippeb8 View Post
The star can spin on its own but no fire will be ejected as fast as it can spin and the shape of the star will remain perfectly spherical at all times.
Why in the blue hell are you talking about fire?
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Old 6th December 2019, 05:20 AM   #102
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Originally Posted by philippeb8 View Post
Time's up. So I conclude you all failed to defend 300 years of physics!
Time's up in 13 minutes? Do you have any idea how forums work?

Besides, many people have answered your questions and claims, so it's a lie to say no one has. Why do you lie?

Quote:
fire (or plasma)
Those are two COMPLETELY different things. If this is the level of your knowledge, then it inspires even less confidence in your ability to formulate even a basic solution to any problem, let alone one of the most elusive one in physics.
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Old 6th December 2019, 07:47 AM   #103
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philippeb8's thought experiment universe is not a toy universe, nor a hypothetical universe; rather it's an entirely magical universe (so it seems to me).

Just as the 1 star in it is magical too; it spits fire (like a dragon).

Why do I say this?

Because of the answers philippeb8 has given (or, in many cases, not given) to direct, pertinent questions about them.
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Old 6th December 2019, 10:08 AM   #104
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Originally Posted by Belz... View Post
Originally Posted by philippeb8 View Post
The star can spin on its own but no fire will be ejected as fast as it can spin and the shape of the star will remain perfectly spherical at all times.
Why in the blue hell are you talking about fire?
Time's up. So I conclude you have failed to overturn 300 years of physics!
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Old 6th December 2019, 11:11 AM   #105
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Originally Posted by Belz... View Post
Time's up. So I conclude you have failed to overturn 300 years of physics!


Lol. I was on a lunch break so that doesnít count.

Fire or plasma, you get the idea.
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Old 6th December 2019, 01:04 PM   #106
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Originally Posted by Robin View Post
If you accept that in my spinning pair of particles, there will still be a force on the bond in a spinning frame that keeps the particles still, then it follows that a spinning star in a spinning reference frame will still eject mass.
So I don't understand where your "bond" is coming from?
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Old 6th December 2019, 01:11 PM   #107
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Originally Posted by Dave Rogers View Post
I think the point the OP is trying to make is either that there is no such thing as a rotating reference frame, or that rotating and inertial reference frames are equivalent. Either way, he seems to be claiming that, in the same sense that there is no such thing as absolute linear motion, there is also no such thing as absolute rotation. As a postulate, it seems a useless one, in that it inevitably leads to results that contradict observation; but in crackpot physics that's not really a negative.


Dave
- All bodies got their own frame of reference
- The importance of the frame of reference is given by the amplitude of the mass of the body

So:
- if you have only 1 star then there will be only 1 frame of reference;
- if you have 1 galaxy in the real visible universe then the galaxy's frame of reference will influence the rotation curve of the galaxy.

That is why:
- it's impossible to have a geostationary altitude with only 1 star;
- it's impossible the plasma will be ejected by a spinning lonely star.
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Old 6th December 2019, 01:14 PM   #108
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Originally Posted by philippeb8 View Post
Lol. I was on a lunch break so that doesnít count.
For a full day?

But the 13 minutes you gave another post did count?

Quote:
Fire or plasma, you get the idea.
As I told you before the two are so fundamentally different that, no, I don't get the idea. It seems to me like your grasp of astronomy is pretty damned weak.
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Old 6th December 2019, 01:17 PM   #109
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Originally Posted by Myriad View Post
Even in a hypothetical universe with only one star, the geostationary orbit altitude is 22,258 miles.

If you disagree, you might need to look up what "geostationary" actually means.
I know about:
https://en.wikipedia.org/wiki/Geostationary_orbit

But in the case of a lonely star the conditions will be much different.
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Old 6th December 2019, 01:21 PM   #110
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Originally Posted by Belz... View Post
For a full day?

But the 13 minutes you gave another post did count?
I know everybody was online so that's why it counted.

Quote:
As I told you before the two are so fundamentally different that, no, I don't get the idea. It seems to me like your grasp of astronomy is pretty damned weak.
Chemical reactions that emit energy are the same thing.
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Old 6th December 2019, 01:21 PM   #111
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Originally Posted by philippeb8 View Post
I know about:
https://en.wikipedia.org/wiki/Geostationary_orbit

But in the case of a lonely star the conditions will be much different.
Why?
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Old 6th December 2019, 01:40 PM   #112
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Originally Posted by philippeb8 View Post
So I don't understand where your "bond" is coming from?
So in this hypothetical universe you are talking about there are no forces?

How can you have a star in this hypothetical universe if you can't have a bond between particles?
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Old 6th December 2019, 01:45 PM   #113
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Originally Posted by phunk View Post
Why?
You need to take the "spacetime curvature" out of your mind 1 sec. And also the kappa fudge factor to make Einstein's calculations work correctly:
https://en.wikipedia.org/wiki/Einstein%27s_constant

Now we're in the normal classical physics world. Why would there be any predefined altitude if there is no interaction with other bodies?
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Old 6th December 2019, 01:52 PM   #114
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Originally Posted by Robin View Post
So in this hypothetical universe you are talking about there are no forces?

How can you have a star in this hypothetical universe if you can't have a bond between particles?
Ah ok so a bond is a force...

There is indeed a force between particles. The group of particles create a stronger force and so on. If you see the star as a giant particle then it's exactly the same thing.

But the difference is at the atomic level, the subatomic particles aren't linked with gravity but with nuclear forces.
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Old 6th December 2019, 02:31 PM   #115
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Thank you for playing Space Quest 1!

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Old 6th December 2019, 02:35 PM   #116
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Thanks!

Originally Posted by philippeb8 View Post
- All bodies got their own frame of reference
- The importance of the frame of reference is given by the amplitude of the mass of the body

So:
- if you have only 1 star then there will be only 1 frame of reference;
- if you have 1 galaxy in the real visible universe then the galaxy's frame of reference will influence the rotation curve of the galaxy.

That is why:
- it's impossible to have a geostationary altitude with only 1 star;
- it's impossible the plasma will be ejected by a spinning lonely star.
Now I know why you didn't answer my question about your "1 star". Because it cannot be anything like any star that astronomers have studied. Instead it's a magic star, composed of "1 stuff", not things like protons, electrons, or iron ions.

Glad you clarified that. And showed clearly that the topic of this thread is philippeb8 magic, not physics.
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Old 6th December 2019, 02:39 PM   #117
JeanTate
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Originally Posted by philippeb8 View Post
<snip>

Chemical reactions that emit energy are the same thing.
But your magic "1 star" cannot undergo chemical reactions. So this is irrelevant.

Your magic star can only emit "fire". Just like dragons.
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Old 6th December 2019, 02:42 PM   #118
philippeb8
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Originally Posted by JeanTate View Post
Thanks!


Now I know why you didn't answer my question about your "1 star". Because it cannot be anything like any star that astronomers have studied. Instead it's a magic star, composed of "1 stuff", not things like protons, electrons, or iron ions.
Exactly. The star I am referring to is not "peer reviewed"...
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Old 6th December 2019, 03:17 PM   #119
phunk
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Originally Posted by philippeb8 View Post
You need to take the "spacetime curvature" out of your mind 1 sec. And also the kappa fudge factor to make Einstein's calculations work correctly:
https://en.wikipedia.org/wiki/Einstein%27s_constant

Now we're in the normal classical physics world. Why would there be any predefined altitude if there is no interaction with other bodies?
I didn't say anything about spacetime curvature.

You don't need anything other than the star itself to have rotation. For one, a star is not a solid body, and it's parts rotate at different speeds, so there is rotation regardless of any external reference.

Secondly, rotation is absolute. When an object rotates, it's parts have motion relative to its axis of rotation, and each other. And every part is in constant acceleration. No external frame of reference is needed to measure rotation.

A star, completely alone in its universe, would not be spherical if it were rotating, it would be an oblate spheroid.

A stationary orbit would absolutely be definable. If you want to argue it couldn't exist because there is nothing else in the universe to be IN that orbit, then you're just being silly.
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Old 6th December 2019, 03:32 PM   #120
MRC_Hans
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Originally Posted by philippeb8 View Post
In other words the star cannot eject fire because the reference frame spins with the star.
No. Spin can exist within a reference frame. It requires an inertial energy and a centripetal force (in the case of a star, that would be gravity).

Spin is one of the few kinds of motion that CAN be detected entirely within a reference frame.

Let me make it easy for you:

Imagine a centrifuge spinning in deep space (but in this universe).

Its spin is easily observable and stuff sticks to the perimeter because of the centrifugal force. Even if you are on board the centrifuge, you will easily observe the spin (centrifugal force, Coriolis force).

Now build a box around the centrifuge. This is a virtual universe; nothing of the outside can be observed from within, nothing of the inside can be observed from without.

Do you think the conditions inside the centrifuge will be changed by that?

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