Post by: CrazyScientist on 19/03/2020 03:04:37
Here's a simple scenario: 4 objects with masses:
m1=4, m2=1, m3=4, m4=1
Objects m1 and m3 move in relation to eachother at v=0,2c (1c=1d/1t)
Distances between m1 and m2, just as between m3 to m4 are equal to 2d. Due to gravitational attraction m1 makes m2 to accelerate at a1=1 (where 1a=0,1d/t^2) and attraction between m3 and m4 is just as strong.
Can you calculate the kinetic energies or acceleration (a2) for object m2 in relation to object m3 or for m4 in relation to m1? I can do it, but I had to find my own way...
(https://i.ibb.co/8KdJHtf/key.jpg)
Frame of m1
(https://i.ibb.co/LRxdXGv/ffg-1.gif)
Frame of m3
(https://i.ibb.co/3pNDZJD/ffg.gif)
I will wait a day or two for you to make any attempt of solving this problem and then I will begin to show you, how to do it my way... :)
Post by: Kryptid on 19/03/2020 04:48:30
If you really want to show me, that theoretical physicists aren't only just a bunch of overconfident snobs
Not a good way to start off your membership here.
Assuming this is new idea, I have moved it to New Theories.
Post by: Bored chemist on 19/03/2020 08:30:39
I would like to present you the only correct model of gravity and energy distribution in relative motion. I've spent around 6 weeks and wasted some 8 pages format a5 on calculations
a bunch of overconfidentAre you seeking to join them?
Post by: Bored chemist on 19/03/2020 08:34:29
Here's a simple scenario: 4 objectsFour objects, a blue one, a green one, a yellow one and two red ones.
I wonder, what then can explain all those generations of professional physicsts, who didn't even think about trying to calculate such things.Because they have proof that it is, in the general case, impossible.
https://en.wikipedia.org/wiki/Three-body_problem
Post by: CrazyScientist on 26/10/2022 16:38:56
This is what happens, when someone like me thinks that he knows how to solve a sophisticated problem to the point, when he doesn't even check, if he isn't just making a fool of himself... So, after I ended up with results that didn't match too well with my own predictions, I decided that it'll better if I rerturn to this subject only after I will be absolutely sure that I have the proper solution this time around...
And just so happens, that around a week ago, I was at last allowed to put the right letters in the right places in my equations and it finally 'clicked'...
But let's get to the point - in order to have any chances to tackle the presented issue, I need to base my model of gravity on my extended edition of mass/energy equivalence formula - who could've guess ..? :P
Below is the link to a thread, in which I tried (with a rather poor outcome) to discuss the general idea behind this formula:
https://www.thenakedscientists.com/forum/index.php?topic=83455.msg659155#msg659155
But I will give you here a short sneak peek of it, so you won't need to go through nultiple threads, to get basic understanding on the mechanics I'll be using here.
Those of you who might rernember me from other threads, which I made here during last couple years, can quess already that my extended formula of mass/energy equivalence is deeply rooted in my (yet another) model of constant c in Galilean relativity. I know, right - one might think, that I did all of this while having some actual reasons. on my mind...
Anyway, here's the link to a thread: where my model of constant c in relative motion is explained in details...
https://www.thenakedscientists.com/forum/index.php?topic=82070.0
Ok, so here is the extended formula of mass/enetgy equivalence
m0 = 
= 
= Pt is for total momentum - m*c
Et is for total energy - m*c^2
And here are 2 images that suppose to visually represent the general ideas behind my extended formula:
(https://i.ibb.co/g758ZzW/pEk.png)
(https://i.postimg.cc/NMNWZL7g/equiv.png)
(https://i.postimg.cc/NMNWZL7g/equiv.png)
Ok, next step is to combine it with relative velocities of objects in motion - but this can't be done without addining couple new letters to my magical fornula of enhanced gravity
The most obvious one is v for relative velocity - but this is just the betinnijg, since together with the v,, we get as well things like p for momentum m*v or Ek for Kinect energy m*v^2 (we can get rid of tthe 1/2 from the classic formula for Ek).
Yet this still isn't enough - Ek makes only part of the total energy that is defined by m*c^2 - but after we subtracf the Ek from Et,, we'll always end up with some amount of energy that is still avaliable for the object in question, before it reaches the speed of lifht c. Where does this energy 'hide'? Is there some term thar describes it? Do we know any formula to calculate it?
Warning! Spoiler alert: the answe for last two questions is: "no". The only term I can think of, would be potential energy Ep - however this term is already used to describe energy that is being added to a system, by applying work - for example by lifting an object in a gravitational field - and this isn't what I'm looking for.
So, it seems that I have no other option, than to make out my own definition of potential energy Ep. Here's how I understand the distribution of energy for objects in relative motion::
Et = Ek + Ep
Total energy is the sun of kinetic and potential energies. As velocity
of a moving body increases, so is it's kinetic energy, but as the relativ velocity keeps getting closer to the constant c, the less potential energy wiill remain avaliable to it, until it finally gets to 0 when a body reaches 100% of c.
In shortcut, for a body moving at c, it's total energy is purely kinettic, while for a body at rest, potential energy makes 100% of it"s total energy - it's actually quite simple...
And Finally, last step is to express such concept of Ep and energy distribution with a mathematically valid formula. How to make it happen? Well, probably as swiftly and erfficiently as we can - if the Ek of a moving body is being defined by iit's relative velocity v, then what should define the Ep. is the velocity that is still needed for that body, to reach 100% c - let's call it for now as \potential velocity\ vp (let's also use the term \kinetic velocity\ vk, to describe the relative velocity of a body in motion - this way it will be much less confusing...
So to wrap this all up - this is, how to calculate the potential velocity: vp = c - vk -
Ahd this is what we get by applying tit to a formula describing the energy distribution for a body in relative motion at kinetic velocity vk::
Et = Ek + Ep = m * vk^2 + m * (c - vk)^2
And wiith all of this being done, we can finally desribe the distribution and relation between Ek and Ep for moving bodies, using my extended frormula of mass/energy equivalence::
m0 = 
+ 
+ And now everything what left for us to do, is to use the formula of energy distribution to describe the gravitational interactions between bodies. in motion - it can't be that hard, right?
You are absolutely correct - it's actually much easier than it sounds. All what is needed, is for us to guess which part of the total energy is responsible for the gravity itself - is it Ek or Ep or maybe both of them that define the magnitude of gravitational attraction between moving bodies?
But maybe this time I'll let you to guess the correct answer. You can't expect that I will always serve you all the answers on a golden plate. It's healthy for rhe brain, to gtive it a small workout from time to time - just try picking out tue answer, that seems to make the most sense tfo ou - and I will be more than happy to see if you are somewhat sensitive to logic. And please: don't be afraid to share gere your answer with the rest of us - I won't laugh even if you'll be wrongl
But if you belong to the group of people, who can only consume the food that is being cooked by others, then you'll learn the correct answer in the thread which is linked below...
https://www.thenakedscientists.com/forum/index.php?topic=83455.msg659155#msg659155
In my next post I will make the actual calulations using the values given by me in the beginning of this thread...
Post by: paul cotter on 26/10/2022 17:30:46
Post by: Origin on 26/10/2022 18:55:13
Ok, so here is the extended formula of mass/enetgy equivalence
That is not a mass/energy equation since your equation is saying mass = mass or energy = energy since you could also write your equation as
Post by: CrazyScientist on 26/10/2022 19:05:51
A moving body may or may not have potential energy in addition to it's kinetic energy( dependent on the observer's frame of reference
True, but energy distribution is actually quite clearly defined in the majority of possible scenarios. Here are the basic rules:
- for every velocity other than constant c and/or 0, total energy will ALWAYS be a mix of potential energy and kinetic energy - there's simply no other option, since...
- ttotal energy of objects that remain at rest (v=0 defines being stationary) has 0% of kinetic energy and 100% of potential energy
- total energy of objects moving at 100% of c includes 0% of potential energy and 100% of kinetic energy
and there are no other options in relative motion - 0 and c set the limits of velocities in each case of relative motion
Quote
However none of this will solve the three( or greater ) body problem.
You're right - but this is not what I've tried to solve here. What I actually did solve here, is the relation between gravity and the relative motion of interacting bodies - and as I wil prove here soon enough, this solves as well the problem with distortions of gravitational fields due to Doppler effect for moving sources pf gravity - you get pretty much the same results by calculating the magnitude of gravitational attraction between moving bodies, and (or) by directly applying the Doppler's effect to the geometry of gravitational fields in motion - iand it's actually a quite important mechanism, that no one ever thpught about before
But don't you worry,- 3 body problem is also on my list of things to solve :)
Post by: CrazyScientist on 26/10/2022 21:05:47
Ok, so here is the extended formula of mass/enetgy equivalence
That is not a mass/energy equation since your equation is saying mass = mass or energy = energy since you could also write your equation as
But Isn't this exactky the thing about equations of equivalence, that makes both sides of such equation equivalent? I'd say that mass end energy being interchangeable is what makes them equivalent. My equation allows me to expres same value in 3 diffferent ways (as mass, energy or momentun) I can easily operate on any of those variables on both sides of equation, while maintaing their numerical values intact - for me it makes my formula a pergect example of equivalence...
You will soon see how useful it is to be able to exprress energy in form of mass - this is what allows me to operate on energy distributions to calculate the properties of mass-driven gravity
It's rather the E=mc^2 which is NOT a real equivalence - because if it is, then p=m*v is an equivalence just as much, and also a=ΔV/t is one and F=m*a is one as well...
Post by: Origin on 27/10/2022 13:23:34
But Isn't this exactky the thing about equations of equivalence, that makes both sides of such equation equivalent?Nope.
equals 
since 
equals 1 so we now have 
and since 
equals m, your equation simplifies to 
. So this is not very enlightening and certainly not an energy/mass equivalency equation.
Post by: CrazyScientist on 27/10/2022 16:44:54
But Isn't this exactky the thing about equations of equivalence, that makes both sides of such equation equivalent?Nope.
It's the mass of an object at rest - so m0=m is actually true. Besides there"s nothing what wouldn't allow us to write this value in form of energy equation:
E=
=m0*c2Or momentum:
p=E/c or p2=E*m0
And I'm sure there are many other ways to juggle with the variables, while keeping them equivalent. Besides, this formula becomes much more uselful while describing a massive body in relative motion...
BTW -;I will have to slightly modify the discussed scenario - it took me 2 years since I made those animations, to conclude that my scenario is far too messy. I dhould make 2 massive sources moving in relation to each other and add 4 smaller and starionary test objects - 2 objects for each moving source , placed at the same distance on both sides of thiose sources
Post by: Origin on 27/10/2022 19:43:56
It's the mass of an object at rest - so m0=m is actually true.Which is obvious to the most casual observer and need not be written in a more complicated form. The equation therefore is not a mass/energy equivalency equation.
Now you should say, "yes that is correct, I will fix the equation so is is actually a mass/equivalency equation".
Here is a hint, here is how to fix the equation, write
(for a mass not moving relative the observer).