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I’ve made a series of the biggest discoveries in physics.


OlegGorokhov

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Hello.
Recently I’ve made a series of the biggest discoveries in physics.


First of all I proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy.

Please, read about it on Medium:

link removed (3 min read)
link removed  (5 min read)

 

Also, please, read “The biggest discoveries in physics” on Medium:

link removed (30 min read)

 

These discoveries completely change our understanding of the universe and give us access to a new colossal clean Energy.
 

I will answer all the questions that arise.

 
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I found a logical mistake at the very beginning of physics.
I will explain this with a simple example.

Imagine that you in a distant cosmos (far from all the planets) threw a stone and this stone detached from your hand and flew away at a volocity of 9.8m/s.

0*8q0I-tgU-cwuZILJ.

In this case, the stone will fly by inertia with this volocity (9.8m/s) infinitely — i.e. every second the stone will move through a displacement of 9.8m!

And now imagine that since the 2nd second, this stone happened to be being pushed by Carlson flying in the same direction (but this Carlson is not with a propeller, but with a jet engine, since in space there is no atmosphere and the propeller does not have anything to repulse from)

0*flrhkTOHsZNyztkX.

and now this Carlson every second, pushing the stone, increases the volocity of the stone by 9.8m/c.

So, to the beginning of the 3rd second, the volocity of the stone will be already 19.6m/s (and the total movement passed by the stone during the 2nd second will be: 14.7m = 9.8m + 4.9m).

Importantly, as a result of Carlson’s pushing, during this 2nd second, the stone will only move the 4.9m displacement (of 14.7m), and the 9.8m displacement will not be the result of Carlson’s pushing, but only will be the result of moving by inertia, since exactly with this velocity the stone was separated from you. And on this displacement of the stone (9.8m), during the 2nd second, Carlson’s Energy will not be spent!

The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes.

0*NGqOJ6KWRekKao8o.

These two processes are completely identical, because in both cases:
1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second);
2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second);
3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement).

Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second. Please, read “The erroneousness of the Joule” on Medium: https://goo.gl/YffRQL

Also, please, read “The biggest discoveries in physics” on Medium: https://goo.gl/1q9HaU

These discoveries completely change our understanding of the universe and give us access to a new colossal clean Energy.

As I said before 

Recently I’ve made a series of the biggest (but simple) discoveries in physics.

One of these discoveries is that, I, through the simplest mathematics, have proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy.

 
1*BgkZ-WVWOGod45fq76BQxw.png

The proof of the erroneousness of the Joule.

First of all, it is important to understand the meaning of the word “inertia”.
“Inertia” means “inaction” (or inactivity).

According to Newton’s 1st law, if the object has already received any velocity, then this object can move with this velocity by inertia, i.e. even in the absence of any additional pushing (i.e., with total inactivity of all Forces).

Let’s consider a simple example: a 20kg object is falling freely down.

0*aMO15_D3kwZ1Sxlj.

The 20kg object is falling down because this object is pushed down by the gravitational Force. The gravitational Force (F) can be imagined as a pushing by “the invisible hand”. The green arrow shows the direction of this pushing.

1*myJPjhcZNRgtxVRnD_UdVA.png

All that the gravitational Force (F) can do is to push this 20kg object down. The Work of the gravitational Force is precisely the pushing. As a result of this pushing the 20kg object evenly increases the velocity (uniformly accelerated).

Thus, by creating a formula for the calculation of the Work of any pushing Force (including the gravitational Force), we (people) are trying to calculate exactly the size of the done work (the size of the done pushing).

But the “current” formula for Work and Energy

1*r3gUnkkyL0SIQ5ExydlaaQ.png

is erroneous, since it relates the Work (W) of the gravitational Force (F) to the displacement of the pushed object (D).

And as a result, using this erroneous formula, it turns out that if the object falls freely, the Work (W) of the gravitational Force (F) increases every second, because the object moves the increasing distance (D) each subsequent second.

And this is the error of this formula.

The fact is that through a free fall, the movement of the 20kg object, during every second, starting from the second second, is the result of two movements: uniformly accelerated motion and uniform motion due to inertia.

 
1*YYwzC3gcy9rnC_geQ1bevA.png

Consider the Work (Pushing) of the gravitational Force in detail.

So, by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves a total distance D (= 4.9 m) for only one reason:
- the object moves the displacemen D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and
- the object does not move the displacemen (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v).

D = D1 + D2 = 4.9m + 0m = 4.9m

By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the total displacement D (= 14.7 m) already for two reasons:
- the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and
- the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force.

D = D1 + D2 = 4.9m + 9.8m = 14.7m

By the beginning of the 3rd second of the fall, the 20kg object had already accumulated the velocity v (= 19.6 m/s). Therefore, during the 3rd second of the fall, this object moves the total displacement D (= 24.5 m) also for two reasons:
- the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and
- the object moves the displacement D2 (= 19.6 m) as a result of movement by inertia with a velocity v (= 19.6 m/s), i.e. without the Work (Pushing) of the gravitational Force.

D = D1 + D2 = 4.9m + 19.6m = 24.5m

And so on.

Thus, for this 20kg object in order to move every second the displacement (D2), there is no need for Work (Pushing) of the gravitational Force, because the object moves by inertia, i.e. without the Work (Pushing) of the gravitational Force!

It is very important to understand that in this “old” formula for Work and Energy,

0*qGJ-sQ_EDR08CM5K.

in the calculation of Work and Energy, the total displacement (D) was erroneously taken into account, in which 2 completely different displacements were summarized:
- non-inertial displacement (D1), which the object moves as a result of the Work (Pushing) of the gravitational Force; and
- inertial displacement (D2), which the object moves as a result of motion by inertia.

We can not “give credit” to the gravitational Force for the inertial displacement (D2).

Thus, the using of inertial displacement (D2) in the calculation of Work and Energy (that is, the calculation of the Joule) is absolutely erroneous!

So, I mathematically proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy.

This discovery has tremendous significance.

Since the concept of Energy is basic in physics and since other physical concepts are based on it, even this one discovery changes all modern physics (both quantum and classical) and makes all the existing physics textbooks irrelevant and obsolete!

Please, read “The biggest discoveries in physics” on Medium: https://goo.gl/1q9HaU

These discoveries completely change our understanding of the universe and give us access to new colossal clean Energy.

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3 hours ago, OlegGorokhov said:

I will answer all the questions that arise.

What is your (new) equation for work-energy? (I may have missed it in your rambling description)

What is the experimental evidence that your equation is correct?

Edited by Strange
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33 minutes ago, Strange said:

What is your (new) equation for work-energy? (I may have missed it in your rambling description)

What is the experimental evidence that your equation is correct?

Please, read “The biggest discoveries in physics” on Medium: https://goo.gl/1q9HaU

It's a fragment of the text:

The New Concept of Work and the New Concept of Energy.

I consider that in the most general sence:
Work is any activity that requires effort;
Energy is an ability to do Work.

In other words,
Work is any activity that requires Energy (Work is any activity Energy is spent on).

Then in physical terms:
Work is a pushing an object for some time;
Energy is an ability to push (an object) for some time.

Some Works, at first glance, may not look like pushings. But this is only at first glance. Any Work is always just a pushing for some time or a combination of several pushings. I will give one example.

A heating-up is also the result of pushings, because a heating of a body is due to the increase in the velocities of particles inside the body. And the increase in the velocities of particles occurs only due to additional pushings of these particles. Therefore, a heating-up is the result of a lot of pushings. And this means that a heating-up is Work and that Energy is being spent in this process.

There are the Target Works and the Non-target (but inevitable) Works. The Energies that are spent on the Non-target (but inevitable) Works, we always consider our losses.

For example, the Target Work of the electric motor is to push the motor shaft. At the same time, the heating of the electric motor is the Non-Target (but inevitable) Work. The Energy that is spent on this Non-Targeted (but inevitable) Work is regard as our losses.

It is important to understand that when any Work (any Pushing)is being done, Energy is always being spent on this Work (on this Pushing). Always.

When some closed system (for example, some device) pushes an object, simultaneously the ability of this system (this device) to push this object decreases.

If Energy is spent on something, then this something is Work.

If something can push other thing then this something is Energy.
If something can turn into one that can push other thing then this somethingis Energy.

The New Universal Formula for Work and Energy.

Work (W) is the pushing an object with a Force (F) for a time (t);
Energy (E) is the ability to push an object with a Force (F) for a time (t).

 
0*OGkhKN-W783oLben.

Although this new formula of Work and Energy looks like the formula for Momentum (p),

 
1*f5wdas8z55N7ffwAVxK6xA.png

in fact, the concepts of these formulas are totally different. All this will be explained in detail several sections lower.

Inserting the product (m*a) instead of the Force (F) in W(E)=F*t, we get the expanded Formula for Work and Energy.

 
0*L8w9MvGCkThfiGPB.

Now, this is the New Universal Formula for Work and Energy.

 
0*Dv-jgweMPRHbknAK.

The New Unit of measure for Work and Energy.

I propose to name the new unit of measurement for Work and Energy the Energy. The dimensional formula of the Energy is:
Energy = Newton*second = kilogram*meter/second

Please, read the whole text “The biggest discoveries in physics” on Medium: https://goo.gl/1q9HaU

Edited by OlegGorokhov
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50 minutes ago, OlegGorokhov said:

Please, read “The biggest discoveries in physics” on Medium

No.

 

50 minutes ago, OlegGorokhov said:

Now, this is the New Universal Formula for Work and Energy.

And where is the experimental evidence that this is correct?

The current equation works, i.e. matches the experimental results, so it is hard to see how yours can too. Please show the evidence for this.

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30 minutes ago, Strange said:

No.

And where is the experimental evidence that this is correct?

The current equation works, i.e. matches the experimental results, so it is hard to see how yours can too. Please show the evidence for this.
 

Why not?    The text is too big to be put here.

The experimental evidences have been presented -- see the tables and the numbers carefully.
Engage the logiсs.

The conservation of the New Energy has already been proved by a lot of experiments with the Momentum.

One of the most powerful laws in physics is the law of momentum conservation. ... For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

 

Edited by OlegGorokhov
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28 minutes ago, OlegGorokhov said:

The experimental evidences have been presented -- see the tables and the numbers carefully.

Can you explain how you measured these?

Ideally you would show a table with the predictions of the old equation, those of your equation and the measured values so we can see which equation best matches the measurements. Can you do that?

28 minutes ago, OlegGorokhov said:

Why not?    The text is too big to be put here.

You read the rules when you joined, no?

Edited by Strange
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20 minutes ago, Strange said:

Can you explain how you measured these?

Ideally you would show a table with the predictions of the old equation, those of your equation and the measured values so we can see which equation best matches the measurements. Can you do that?

You read the rules when you joined, no?

I am Oleg Gorokhov. 
My first account (OlegGorokhov) was banned. Why?

What evidences and measurments do you need?

Don't you know that the free-falling acceleration is 9.8(m/s^2)?

Are here smart people?

Edited by Oleg_Gorokhov
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13 minutes ago, Oleg_Gorokhov said:

What evidences and measurments do you need?

You know how science works?

You create a model (eg your new equation) then you test the predictions of that model against experiment to see if they match and if they are better than other models. Have you done that?

You said you have experimental evidence. Can we see it, and a description of how you made the measurements.

Thank you.

Edited by Strange
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8 minutes ago, Strange said:

You know how science works?

You create a model (eg your new equation) then you test the predictions of that model against experiment to see if they match and if they are better than other models. Have you done that?

You said you have experimental evidence. Can we see it, and a description of how you made the measurements.

Thank you.

I again say to you , --my first account was banned!

I created the model
1*YYwzC3gcy9rnC_geQ1bevA.png

What specific figures do you not agree with?

Don't you know that the free-falling acceleration is 9.8(m/s^2)?
Engage the logics!

Read the whole text carefully!

Are here smart people?

Edited by Oleg_Gorokhov
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4 minutes ago, Oleg_Gorokhov said:

I again say to you , --my first account was banned!

Nothing to do with me. Try reporting one of the posts under your old account to the moderators to find out (click the Report Post link at the top of the post).

5 minutes ago, Oleg_Gorokhov said:

What specific figures do you not agree with?

Which of these are measured?

For example, your D column (total displacement per this second) is different from that predicted by standard physics.

So, you need to measure the displacement after each second and compare it with your equation. Have you done this?

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5 hours ago, OlegGorokhov said:

the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force.

But that is just movement - inertia is a resistance to the change - it doesn't apply any force in itself so adds nothing to the acceleration in it self....  is that right? The free fall under gravity has an acceleration of 9.8 m/s^2 as you pointed out  -  this is measured, so any perturbation from other influences (if they were there) have been taken into account when it was measured to be 9.8 m/ss. That is the acceleration due to 'gravity' as we have measured it.

 

Maybe I am missing something. Engage with Strange though - he will test your theory for you if you answer his questions. Although, I suspect, that there is nothing wrong with the current theory.  

 

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16 minutes ago, Strange said:

Nothing to do with me. Try reporting one of the posts under your old account to the moderators to find out (click the Report Post link at the top of the post).

Which of these are measured?

For example, your D column (total displacement per this second) is different from that predicted by standard physics.

So, you need to measure the displacement after each second and compare it with your equation. Have you done this?

I got you.
You don't understand the fact that there the standart formula is used for the displacement  -- D=a*t^2/2

Please add all the digits in the column (D) and you will get your figure (in each line the movement is indicated for exactly this one second, and not for the total during during this time) So the total displacement during all 5 secons (i.e. the displacement for 1st second (4.9m)+the displacement for  2nd second (14.7m) + so on )=122.5m! (according to the standard current formula) = 122.5m ! (according to the standard current formula)

Edited by Oleg_Gorokhov
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5 minutes ago, Oleg_Gorokhov said:

So the total displacement during all 5 secons (i.e. the displacement for 1st second (4.9m)+the displacement for  2nd second (14.7m) + so on )=122.5m! (according to the standard current formula) = 122.5m ! (according to the standard current formula)

So you seem to have a more complicated way of reaching the same result. That doesn't seem very useful.

Unless your model can produce a different prediction that standard physics, then it is just standard physics.

So, what predictions does your model make that are different from standard physics? How do measurements compare between your model and standard physics?

Edited by Strange
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51 minutes ago, Oleg_Gorokhov said:

I again say to you , --my first account was banned!

!

Moderator Note

You were not banned. New members have a 5-post limit their first day, as an anti-spamming measure.

 
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18 minutes ago, Strange said:

So you seem to have a more complicated way of reaching the same result. That doesn't seem very useful.

Unless your model can produce a different prediction that standard physics, then it is just standard physics.

So, what predictions does your model make that are different from standard physics? How do measurements compare between your model and standard physics?

Here is Oleg Gorohov. My 1st and my 2st accounts are banned so I can not answer through them!

Dear Strange, Please wake up!
I don't disagree with the standard formula of the displacement.  (D=a*t^2/2)
I agree with it!

I say that the current formula for Work and Energy is erroneous !!!

Now erroneously Work is counted that was not done. Also now erroneously Energy is considered, which was not spent !!!!

I proved (through the simplest math and elementary logics) that now erroneously Work is counted that was not done. Also now erroneously Energy is considered, which was not spent !

 

TO THE ADMINS!
IF MY 1ST ACCOUNT WAS NOT BANNED
WHY CAN NOT I ANSWER???

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2 minutes ago, Oleg__Gorokhov said:

Here is Oleg Gorohov. My 1st and my 2st accounts are banned

Only the second one. See the post above.

4 minutes ago, Oleg__Gorokhov said:

I don't disagree with the standard formula of the displacement.  (D=a*t^2/2)
I agree with it!

Then I'm not sure whey you mentioned it.

2 minutes ago, Oleg__Gorokhov said:

I say that the current formula for Work and Energy is erroneous !!!

Then please show how the predictions of your new equation for energy match with experiment.

For example, please show the prediction of the old ("wrong") equation, the prediction of your equation and the experimental results. Then we can see which equation best matches the evidence. (A brief description of how you make the measurements would be useful too.)

Thank you.

Edited by Strange
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3 minutes ago, Strange said:

Yes, thank you.

Can you show how well the predictions of your new equation match experiment?

Once again.

What evidences and measurments do you need?
Don't you know that the free-falling acceleration is 9.8(m/s^2)?

1*YYwzC3gcy9rnC_geQ1bevA.png

What specific figures do you not agree with?  Pay the attention to the RED column (D2). Understand the point.

The experimental evidences have been presented -- see the tables and the numbers carefully. Also  RREAD THE TEXT CAREFULLY!
Engage the logiсs.

The conservation of the New Energy (counted by my new formula for Energy) has already been proved by a lot of experiments with the Momentum.

One of the most powerful laws in physics is the law of momentum conservation. ... For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Read the link 
 https://goo.gl/1q9HaU
As I has said before, -- Unfortunately the text is too big to be put here.

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18 minutes ago, Oleg__Gorokhov said:

Here is Oleg Gorohov. My 1st and my 2st accounts are banned so I can not answer through them!

TO THE ADMINS!
IF MY 1ST ACCOUNT WAS NOT BANNED
WHY CAN NOT I ANSWER???

!

Moderator Note

This was answered. New users have a 5-post limit on their first day. If you make another sockpuppet account, you WILL be banned.

Your other accounts have been (or will be) deactivated, since you are only allowed one account, and I am temporarily locking this to keep you from breaking more rules.

 
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