Genady

3 minutes ago, dNY said:

Formula for my case, I mean. I don't get how your answer applies to my question, so a formula would show that. 

Pick any: Calculate Pi with Python - GeeksforGeeks

Just now, dNY said:

Do you have a formula? 

There are many formulas and algorithms for calculating \(\pi\). See, e.g., here: Approximations of π - Wikipedia

57 minutes ago, dNY said:

You can't know result1 on the calculation of result2, and vice versa.

Each one happens in their own separate environment at runtime, so to say that result2 is equal to result1 in result2's environment, you'd need to know result1 beforehand, thus breaking this invariant.

Let "result" to be a b's digit of the number \(\pi\).

I don't know it beforehand, but they will be equal.

1 minute ago, dNY said:

Perhaps that something that's missing is the clarification that a1 and a2 are the only actual variables (while b and c are the constants)? 

You can also introduce more variables/constants in your end, if that helps to achieve the invariant of result1 === result2. 

So, what is wrong with defining, say, result1==result2==5?

9 minutes ago, MigL said:

Yes they do.
But such interactions tend to make the accretion disk planar; see Saturn's rings for an example.
Further, such interactions would be random, and would produce random radiation.

The inspiralling accretion disk produces a very specific radiation pattern, polar jets, of the type associated with orbiting charges.
All 'active' Black Holes display such jets; some around early massive galactic center BHs produce jets that outshine the parent galaxy, and are known as Quasars.

Since the infalling plasma of the accretion disk is in free fall, why is the radiation apparent/accessible to an outside observer ?

I remember that charged particles moving in rotating magnetic fields are involved in the description of this, but it was too long ago (and not in my main line of study) to recall details. Hope to see some expert answers.

Philosophy is certainly neither pseudoscience nor science. It is also not football, chess, music, engineering, cooking, etc. I think that "love of wisdom" is a good definition.

40 minutes ago, swansont said:

But that’s not the reason for the radiation. The radiation is along the direction of motion, perpendicular to the instantaneous acceleration.

Well, the instantaneous change of acceleration, in the case of circular motion, happens to be also "along the direction of motion, perpendicular to the instantaneous acceleration". However, I don't know how they (i.e., the time derivative of acceleration and the radiation) are related and don't claim anything in this regard. My point in this post was that AFAIK, a constant acceleration of charged particle can be insufficient to cause radiation.

3 minutes ago, swansont said:

Moving in a circle is not a changing acceleration. It’s always in the radial direction

The direction of the acceleration changes. The vector rotates.

Something is missing in this definition. As it is described, the simple solution would be just to have a "result" some constant or, more generally, independent on the variable "a".

22 minutes ago, swansont said:

But there is radiation, so how is this an example of an acceleration not resulting in radiation?

It is not. It is an example of not every acceleration resulting in radiation, i.e., in these examples a changing acceleration results in radiation.

1 hour ago, KJW said:

A charged object sitting on a table doesn't radiate even though it is accelerating.

AFAIK, not every acceleration causes radiation from a charged particle. For example, in synchrotron radiation the acceleration is perpendicular to the particle velocity. In some other cases, a magnitude of acceleration is variable. Is it correct?

19 minutes ago, MigL said:

These charged particles are in free fall

Are they? Don't they collide with each other and affect each other non-gravitationally?

If the light orbiting the BH is not a test particle but rather has the energy as described in the OP, then I think it is not on a photon sphere anymore. The "photon sphere" of the original black hole would be inside the new black hole.

Is there a stable circular orbit for a massless test particle around a Schwarzschild black hole? Photon sphere is unstable, AFAIK.

16 minutes ago, Capiert said:

I guess you mean,

You are free to guess.

45 years ago President Carter has helped me to escape from the USSR. 

Moreover, it is not necessary for c to be a speed of anything. It is a coefficient in the spacetime metric.

10 minutes ago, John Cuthber said:

I'm really glad someone knew the answer.
Thanks

I am sorry, I didn't pay attention and did not notice that the question was directed to you. Deleted.

3 minutes ago, joigus said:

I think this is amazing even after one learns about GR.

Another way to look at it is that mass of an affected body was introduced into the equation to make it fit into the force-based model. Then, it disappears when the model is not force-based anymore.

deleted

In QFT, photon does not have dimensions.

A gravitational acceleration being independent of a body's mass also follows from the Kepler's third law.

1 hour ago, joigus said:

There it is. There's your standard of force if you want to make the definition operational. The fixed spring is your standard of force. It's actually inescapable that one needs the other, as F=ma involves both and neither F, nor m, is a primitive concept with a direct observational interpretation, like time or space have.

Sure, we need a standard of force, but I don't see how it makes F and m interdependent. I see that F and m are defined separately while F=ma sets their interdependence.

2 hours ago, joigus said:

you need a standard of force to define mass

I am not sure it is necessary. We only need to define that in any given conditions (i.e. under given albeit undefined forces) mass is inversely proportional to acceleration. We take a spring, again, stretch it by 10 cm, attach a "unit" of mass, release and measure the acceleration. Then we attach another body to the same spring stretched by the same amount and measure its acceleration. We get the mass of the second body. Etc.

Now we have independent definitions of mass and of force and can put them together without any circularity. We could've discovered that \(F \propto m^2a^2\) or \(F \propto e^{ma}\), ... 

PS. To establish that "mass is inversely proportional to acceleration under given conditions" we can do, e.g., the following. First, the same spring, body, stretch, release, measure acceleration procedure. Then, halve the body and define that its half has half the mass. Repeat the procedure and find that the acceleration doubles. ...

3 hours ago, joigus said:

F=ma is the definition of force. It's a definition, rather than an equation really. 

Are you sure? I seem to remember (I think, from Shankar's lectures, but would need to look back to make sure), that force is defined first statically, like the following. Take a spring. Some force needs to be applied to stretch it say by 10 cm. Take two identical springs like that attached parallelly. A force needed to stretch both of them together by 10 cm is defined as double the first force. Etc. After force is defined in such a way, its use in dynamics is a law / equation.