"Science vs. Religion: What Scientists Really Think"

I made a presentation video of the new gravity model, added some music that I had created.

Classical Theory of Nonlinear Universal Relativity:
https://www.youtube.com/watch...

humble brother wrote:

<quoted text>
And how do you think m relates to m1 or m2 above?

Say we are talking about earth/ sun.

There is one force acting: F.

The expression mv^2 will be the same for the earth as the sun, but in the earth's case, m will be relatively small and v will be large, which in the sun's case, m will be large and v smaller.

Fine as far as relative masses goes. But to determine actual mass you will still need the G on the other side of the equation.

humble brother wrote:

<quoted text>
Do you understand what that actually means when you interpret the language of mathematics?
Now, tell me what is m in relation to m1 and m2?

And this is your problem - you cannot interpret the language of mathematics. You cannot relate it to the actual physical quantities and units of mass, distance, and time, that the equations are referring to, which is why you believe your own nonsense. Instead you try to redefine the fundamental units to try and make reality fit your equations. Wrong end of the stick.

http://3.bp.blogspot.com/-Z3L5tqWQiME/T2aod_o...

Chimney1 wrote:

Say we are talking about earth/ sun.
There is one force acting: F.
The expression mv^2 will be the same for the earth as the sun, but in the earth's case, m will be relatively small and v will be large, which in the sun's case, m will be large and v smaller.
Fine as far as relative masses goes. But to determine actual mass you will still need the G on the other side of the equation.

You're going in circles. The original question was about how you can mathematically express the counter effect that rotation has to gravity.

Can you mathematically explain through your gravity models how the rotation of Earth counters some of the gravitational acceleration at the equator?

Have anything?

Chimney1 wrote:

And this is your problem - you cannot interpret the language of mathematics. You cannot relate it to the actual physical quantities and units of mass, distance, and time, that the equations are referring to, which is why you believe your own nonsense. Instead you try to redefine the fundamental units to try and make reality fit your equations. Wrong end of the stick.

Once more you come here with an opinion and not an actual answer to the question presented. How long will it take for you to understand that your opinions are worthless?

How far would science have got from the beginning if whining people like you had managed to prevent the creation of new units and concepts in science?

With only people like you science would be dead.

humble brother wrote:

Here's what they should observe according to the logic of relativity:
Lets say that the emitting of the photons of the blink begins when both observers are 5.0 light seconds away from the object. This gets us to:
- both observers will observe the first photon of the signal at the same time (in five seconds)

You are unclear whether you are using the beacon frame or the observer frame here.
If the beacon frame: this is wrong. In fact, the two will NOT see the photon at the same time. Think of it this way: in 5 seconds, the light is 5 light seconds from the beacon, but one observer is 2.5 light seconds from the beacon and the other is 7.5 light seconds from the beacon. One has already seen the light and the other has yet to see it. In fact, the first sees it after 5/1.5=3.33 seconds and the other doesn't see it until 5/.5=10 seconds.
If, instead, you are talking about the observer's frame, you are correct.

- when the last photons are emitted one observer is 4.5 light seconds away and the other 5.5

In either frame, this is correct, but in the observer's frame, the light has moved closer to the approaching observer. So the time it takes for that observer to see the end is less than the time for the receding observer to see it.
In the beacon's frame, we still have to determine when the end actually is seen by the observers as above.

- both observers will observe the 1.0 block's frame seconds as 0.87 seconds due to time dilation

Other way around: it should be 1/.87=1.15 as the time dilation factor.
I deleted the rest of your calculation because it is based on these incorrect statements.

It is actually funny how the theory of relativity dictates this. As soon as a photon is emitted it jumps into the observers frame and the relative speed between the emitter and the observer no longer matters.

Yikes! This is simply wrong. Furthermore, it shows some basic misunderstanding of what frames are. The photons are not 'in' either frame. But their motion can be described in either frame.

The observer will observe that photon travel to him in the same duration even if the observer accelerated to 0.9*c to try an escape the photon. That's quite some magic.

Wrong yet again.

humble brother wrote:

I made a presentation video of the new gravity model, added some music that I had created.
Classical Theory of Nonlinear Universal Relativity:
https://www.youtube.com/watch...

What a load of garbage.

humble brother wrote:

<quoted text>
You're going in circles. The original question was about how you can mathematically express the counter effect that rotation has to gravity.
Can you mathematically explain through your gravity models how the rotation of Earth counters some of the gravitational acceleration at the equator?
Have anything?

Already dealt with.

Easily, one calculates the centripetal force of an object at the equator using F= mv^2 / r = 440^2 / 6,380,000 = 0.03 N

Then one subtracts that amount from the gravitational force expressed as G m1m2/ r^2 = 9.8 N

So 9.8 - 0.03 = 9.73 N per kg of an object on the equator

Now the question WE can answer easily using G but YOU still cannot:

How do you determine the mass of the sun? As in, how many of those lumps of platinum iridium sitting in that vault in Paris would equal the mass of the sun?

See? You are no closer to answering it than you were a month ago, but we already have the answer because Cavendish measured G directly and we can use that known constant to pin down the necessary variables, one at a time, in a series of consistent equations that reveal the mass of the sun. That can only be done with one unknown variable at a time, i.e. 1 unknown per equation. Do you understand that?

humble brother wrote:

Once more you come here with an opinion and not an actual answer to the question presented. How long will it take for you to understand that your opinions are worthless?

You came here with equations you claim are revolutionary and eliminate the need for G, but cannot answer the simple question - how do you determine the mass of the sun using them? You have been unable to provide an answer to this simple question for a month, instead wasting your efforts on avoiding the question. How long will it take for you to understand that your opinions on the value of your equations are worthless when you can demonstrate nothing useful with them?

polymath257 wrote:

You are unclear whether you are using the beacon frame or the observer frame here.
If the beacon frame: this is wrong. In fact, the two will NOT see the photon at the same time. Think of it this way: in 5 seconds, the light is 5 light seconds from the beacon, but one observer is 2.5 light seconds from the beacon and the other is 7.5 light seconds from the beacon. One has already seen the light and the other has yet to see it. In fact, the first sees it after 5/1.5=3.33 seconds and the other doesn't see it until 5/.5=10 seconds.
If, instead, you are talking about the observer's frame, you are correct.

You're giving light a variable relative speed again. The travel distance of the first photons to both ships is known to be 5.0 light seconds.

Both ships KNOW that they are 5.0 light seconds away from the object/beacon when the emitting beings. How could they observe the first photons after any other duration than exactly 5.0 seconds?

polymath257 wrote:

Yikes! This is simply wrong. Furthermore, it shows some basic misunderstanding of what frames are. The photons are not 'in' either frame. But their motion can be described in either frame.

A single photon can only be consumed by one observer.

polymath257 wrote:

Wrong yet again.

So the speed of light is not constant for all observers regardless of their speed relative to anything?

Chimney1 wrote:

So 9.8 - 0.03 = 9.73 N per kg of an object on the equator

WOW :D That's a mighty big error you have there. You might want to recalculate. Or is that the best you can do?

Chimney1 wrote:

Now the question WE can answer easily using G but YOU still cannot:
How do you determine the mass of the sun? As in, how many of those lumps of platinum iridium sitting in that vault in Paris would equal the mass of the sun?
See? You are no closer to answering it than you were a month ago, but we already have the answer because Cavendish measured G directly and we can use that known constant to pin down the necessary variables, one at a time, in a series of consistent equations that reveal the mass of the sun. That can only be done with one unknown variable at a time, i.e. 1 unknown per equation. Do you understand that?

You are quite confused. Apparently the size of the arbitrarily chosen relative lump for a common measure has some emotional value for you.

Instead of the kilogram lump for this I prefer to use the whole Earth as the lump. Do you understand that?

The Earth is the arbitrarily chosen lump and within the new model the Sun is 1.68*10^5 times the size of that lump when talking about mass. We can call that earth-grams.

The Sun is 1.68*10^5 eg in mass, are you happier with that?

Chimney1 wrote:

<quoted text>
You came here with equations you claim are revolutionary and eliminate the need for G, but cannot answer the simple question - how do you determine the mass of the sun using them? You have been unable to provide an answer to this simple question for a month, instead wasting your efforts on avoiding the question. How long will it take for you to understand that your opinions on the value of your equations are worthless when you can demonstrate nothing useful with them?

He seems to be trying to equate mass as a force of motion only.
The problem with it is mass doesn't have to be in motion to exert itself. He is calculating virtual mass but negating real time mass.
I can't see how that would ever work, except in his mind?

humble brother wrote:

<quoted text>
WOW :D That's a mighty big error you have there. You might want to recalculate. Or is that the best you can do?
<quoted text>
You are quite confused. Apparently the size of the arbitrarily chosen relative lump for a common measure has some emotional value for you.
Instead of the kilogram lump for this I prefer to use the whole Earth as the lump. Do you understand that?
The Earth is the arbitrarily chosen lump and within the new model the Sun is 1.68*10^5 times the size of that lump when talking about mass. We can call that earth-grams.
The Sun is 1.68*10^5 eg in mass, are you happier with that?

http://www.atlantic.edu/program/academic/stra...

Location


Value, m/s2

Average value at the equator


9.78036

Average value at the poles


9.83208

Average value over the Terrestrial Ellipsoid


9.7978

Aura Mytha wrote:

http://www.atlantic.edu/program/academic/stra...
Location
Value, m/s2
Average value at the equator
9.78036
Average value at the poles
9.83208
Average value over the Terrestrial Ellipsoid
9.7978

Oh, you're confused again. Chimney was calculating forces and you present numbers on acceleration.
It's all the same, eh?

Chimney's calculation, when corrected, reveals a considerable error.

humble brother wrote:

<quoted text>
WOW :D That's a mighty big error you have there. You might want to recalculate. Or is that the best you can do?
<quoted text>
You are quite confused. Apparently the size of the arbitrarily chosen relative lump for a common measure has some emotional value for you.
Instead of the kilogram lump for this I prefer to use the whole Earth as the lump. Do you understand that?
The Earth is the arbitrarily chosen lump and within the new model the Sun is 1.68*10^5 times the size of that lump when talking about mass. We can call that earth-grams.
The Sun is 1.68*10^5 eg in mass, are you happier with that?

We already have a definitions to define lager bodies of mass.

1 solar mass = 1.9891 × 10^30 kilograms
The Earth would be a fraction of that or 1 M = 5.9722 × 10^24 kg All based on that lump in Paris , or more correctly for me the one NIST holds.

Aura Mytha wrote:

We already have a definitions to define lager bodies of mass.
1 solar mass = 1.9891 × 10^30 kilograms
The Earth would be a fraction of that or 1 M = 5.9722 × 10^24 kg All based on that lump in Paris , or more correctly for me the one NIST holds.

So?

We're talking about a different model where all masses are different.

humble brother wrote:

<quoted text>
So?
We're talking about a different model where all masses are different.

Then explain how you can calculate the mass of Earth or the Sun without a physical measure to multiply.

humble brother wrote:

<quoted text>
Oh, you're confused again. Chimney was calculating forces and you present numbers on acceleration.
It's all the same, eh?
Chimney's calculation, when corrected, reveals a considerable error.

that was to am, and this is off topic. just to say hi and check in on our mutual acquaintance! has he apologized yet? are you really in Finland, by the way? and what nationality are you? As I watch Olympics I pay more attention to nationality of folks on topix, but some could be Amricans abroad.

humble brother wrote:

<quoted text>
You're giving light a variable relative speed again.

No, I am not. I am requiring the same speed of light.

The travel distance of the first photons to both ships is known to be 5.0 light seconds.

In which frame? The beacon's frame or the observer's frame? it makes a big difference.

Both ships KNOW that they are 5.0 light seconds away from the object/beacon when the emitting beings. How could they observe the first photons after any other duration than exactly 5.0 seconds?

This would be true if we are talking about the observer's frame, but it is false if we are talking about the beacon's frame. In the beacon's frame, it is easy to see it is wrong. The light has moved 5 light seconds and the ships have both moved 2.5 light seconds in that 5 seconds. So the light has already passed one and has yet to pass the other.

In the observer's frame, the story is different. When the *end* of the signal is sent, the beacon has moved .5 light seconds closer to the approaching observer and so has only 4.5 light seconds to go. It has moved to 5.5 light seconds from the other observer, so now takes 5.5 seconds to reach the receding observer.

In any case, it takes a different amount of time for the signal to pass the two observers. And this is all before any effects from time dilation are considered. Again, you have shown how the Doppler effect is *predicted* by relativity.

A single photon can only be consumed by one observer.

Fine, so talk about the wavefront rather than a specific photon.*sigh*

So the speed of light is not constant for all observers regardless of their speed relative to anything?

Yes, the speed of light is constant in all reference frames and has the same value in all of them. Nothing you have said contradicts this.

humble brother wrote:

<quoted text>
So?
We're talking about a different model where all masses are different.

It's not a 'different model'. It's a pile of cráp that doesn't work anywhere and can't predict anything.