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.postimg.cc/mD784MTX/equi.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 adding couple new letters to my magical formula of enhanced gravity
The most obvious one is v for relative velocity - but this is just the beginning, since together with the v,, we get as well things like p for momentum m*v or Ek for Kinetic energy m*v^2 (we can get rid of the 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 answer 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 will remain available 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 kinetic, 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).
Post by: CrazyScientist on 26/09/2024 19:59:06
To be honest, from the beginning I had a 'gut feeling' telling me that my (new) formula will be somewhat related with the asymmetry of Doppler's effect. This asymmetry is 'universal' and characteristic for every possible velocity between 0 and 100%c. For example, source of light moving at constant velocity 0,5c, will cause Doppler's effect in which space-time in front of that source get's contracted from 1 unit of space to 0,5 units of space (space is 2x shorter), while behind that source, 1 space unit gets 'extended' to 1,5 units - and this is ALWAYS true. For light source moving at velocity 0,8c, space in front is contracted from 1 unit to 0,2 units while behind it's being expanded from 1 unit to 1,8 unit. And this is always true for any other constant velocity - they all lead to Doppler's effect characteristic for given velocity
(https://i.postimg.cc/c1vktp4W/bandicam-2018-11-13-04-33-29-245.jpg)
So it wasn't that hard for me to deduce that this is something with great significance for my model.
(https://i.postimg.cc/pXDZDxww/dh.gif)
I 'simply' started with classic Newtonian formula for gravity and extended it using 'my' version of mass/energy equivalence:
(https://i.postimg.cc/2S0nw52K/equation.png)
And then what left, was to figure a way to calculate the distribution of energy/mass in a moving body. But to do this, couple assumptions are needed:
1. there's a definitive & limited amount of energy in matter (according to Einstein the amount of energy a given amount of matter can gain is unlimited) - equal to (half of) energy released to matter-antimatter annihilation. You simply can't get more energy from mass and that's it.
2. Body moving at 100% c doesn't cause gravitational attraction - any acceleration between it and other body would result in c being exceeded in their total relative velocity.
3. gravitational attraction decreases gradually as source-object accelerates from 0 to 100%c
And now what left, is to find some clever way to distribute total mass/energy into part that is still potential (this one causes gravity) and part which already 'turned kinetic' in a moving body (and doesn't cause gravitational attraction). This is the formula which I figured out:
(https://i.postimg.cc/DyBWVMLW/formula.png)
****
Because I really despise making theoretical calculations with real-life units like (kilo)grams, meters or seconds, I made
myself a simplified system of units based on constant c (moving 1 space unit in 1 time unit). I simply modified it, so that now c = 10 space units (x) in 1 time unit (t) - so velocity V = 5 = 0,5c (V = 10 is 100%c). I also 'made up' a unit of mass (m), where 10m is the mass causing acceleration of 10x/t^2 at distance of 1x, so basically accelerates matter to 100%c - meaning: 'it's a black hole'...
So now let's have a body of mass 2m moving at 0,8c (V=8x/t)
2m*c^2 = 2m*10x/t^2 = 200
p(total) = 2m*c = 2m*10x/t = 20
p(total)^2 = 400
(2m*10x/t)^2 - (2m*8x/t)^2 = 400-256 = 144
Using formula above, we will get 144 / 200 = 0,72m -
And this is the part of mass 2m which will be actually causing gravitational attraction of other bodies at V=0,8c.
-----
So now let's calculate the acceleration between 2 bodies each with mass 2m placed 2x away from each other, if they are (already/initially) moving in relation to each other at V=0,8c
For stationary bodies it would be: (2m*2m) / 2x^2 = 4 / 4 = 1
So, basically bodies would experience acceleration equal to 1x/t^2 (accelerating by 0,1c in each time unit).
Now for body moving at 0,8c: (2m*0,72m) / 2x^2 = 1,44 / 4 = 0,36 (accelerating by 0,036c in each time unit]
***
And now comes the 'banger' - as now I will make a full circle to end up with the Doppler's asymmetry. As I wrote above, for a source moving at V=0,8c this asymmetry is always: 0,2x in front and 1,8x behind
And now, if we try to apply those values in Newton's gravity formula, we'll get:
for distance equal to 2x, Doppler's effect asymmetry will give: 2x * 0,2 = 0,4x and 2x * 1,8 = 3,6x
And then:
(2m * 0,72m) / (0,4x * 3,6x) = 1,44 / 1,44 = 1
So, basically gravitational attraction between stationary bodies can be derived from a scenario with the same bodies in relative motion, by applying the values of asymmetrical Doppler's contraction/expansion of space-time to the potential energy (with 'my' definition of Ep) of moving body/source
-----
I tried couple different values to see if it actually works in all possible cases - and apparently it does... You can try using whatever units/values you want, to test it (I'd be more than happy :P)...
Post by: paul cotter on 26/09/2024 21:52:08
Post by: CrazyScientist on 27/09/2024 02:53:41
Where did you get the idea that Einstein suggested there was unlimited energy in matter? Also you keep referring to massive objects with the speed of light: this is not possible as the factor 1/√1-vsq/csq leads to an increase without bound of the mass involved as the speed asymptotically approaches c.
I don't know - from all kinds of sources? Did you try searching for something on internet? I really recommend it - let me show you an example:
https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_vs._rest_mass
(https://i.postimg.cc/y8ZFZpjr/bandicam-2024-09-27-03-48-21-427.jpg)
Nice, huh :) Some people call it 'basic research'...:)
Post by: CrazyScientist on 27/09/2024 03:34:54
Well, sorry but in my uneducated opinion, matter/antimatter annihilation turns WHOLE mass into energy and there's ALWAYS a finite & measurable amount of it released - and it's theoretically IMPOSSIBLE to get any more energy from matter. Thing is, that for some reason physicists keep thinking that mass has a definitive amount energy at rest (inertial frame) but then you can 'pack it' with potential and kinetic energy to infinity...
And maybe (just maybe) we should try something new - to approach the problem differently? Why won't we assume that there IS limit to energy of given amount of mass and this limit is (in theory) achieved at 100%c - that's how much energy you can theoretically 'pump' into matter. And if the object is stationary in my own frame, this whole energy is 100% potential - while with increasing velocity this potential energy becomes kinetic until theoretical limit at 100%c. Makes sense to me (more than the current model)...
I'm going sleep now, but before I end, let me just pretend that all the time I was fully aware (and not like I've just noticed it by total accident), that all my calculations were nothing but a waste of time... I said that I had a 'gut feeling' about the asymmetry of Doppler's shift being directly related with gravity in relative motion and mass/energy distribution - I just didn't know how close exactly this relation can possibly be...
In my previous calculation for mass/energy distribution in a body moving at 0,8c, I got a result equal to 0,36 to 1 in reduction of gravitational attraction (potential mass (?)) - so, basically if stationary objects would interact with the force of 1 force unit (?), then for the same bodies in relative motion a speed V=0,8c this force would be only 0,36 'force unit'.
Now, assymetry in Doppler's shift at 0,8c is: 0,2 contraction in front and 1,8 expansion behind:
(https://i.postimg.cc/J4dfn0Y5/diagggol.png)
So just for fun , let's try: 0,2 * 1,8 =..... Ohh ...0,36
It's just coincidence (obviously) - so, let's do the same for velocity V=0,4c
(https://i.postimg.cc/xdLhtRf6/uyy.png)
1,4 * 0,6 = 0,84
...I did the math (just couple minutes ago)... It's not coincidence... Good night!
Post by: CrazyScientist on 27/09/2024 12:39:19
Let's take a theoretical comet/asteroid entering edge of Earth's atmosphere at a relatively small (flat) angle - question: Will making the asteroid/comet faster increase or decrease the chance of it hitting Earth?
(https://i.postimg.cc/Zq2pjCFH/comket.jpg)
In Einstein's model increasing speed = increasing energy = increasing mass what should lead to stronger gravitational attraction - so faster comet/asteroid is supposed to have bigger chance to hit Earth in Einstein's model.
My model predicts something directly opposite - the faster the comet moves, the less it is attracted to Earth - and if velocity is high enough, it's path/orbit would only slightly 'bend' towards Earth as it would get 'ejected' from atmosphere by it's own momentum. Of course, making it slower will bend the path stronger and at one point it would collide. The worst thing that could be done, would be to completely stop the comet's motion in relation to Earth at the edge of atmosphere - as it would then literally fall out from the sky...
It's crazy that no one thought about it for over a century... However I'm glad that there's someone intelligent who's (almost always) willing to talk with me about my crazy ideas and help me in my research. Of course, that I'm not talking about any human being...
(https://i.postimg.cc/d3mCqnMN/bandicam-2024-09-27-10-05-55-402.jpg)
(https://i.postimg.cc/RCcKQC31/bandicam-2024-09-27-10-06-07-520.jpg)
Is someoe's willing to stand in defense of Einstein's model?...
Post by: paul cotter on 27/09/2024 18:39:32
Post by: alancalverd on 27/09/2024 19:29:14
And here are 2 images that suppose to visually represent the general ideas behind my extended formula:
and the first one is wrong. Not a good starting point in physics.
Post by: CrazyScientist on 27/09/2024 19:46:06
That just shows the nonsense produced by these chatbots and it also illustrates their suggestibility. The increase in mass associated with speed is utterly negligible at the speed of comets. If an incoming comet/asteroid were to be travelling at a speed where a significant relativistic mass increase did occur it would likely explode/disintegrate on first contact with the atmosphere. There is no need to invoke relativity or any other theory to solve this, simple mechanics answers the question.
Ok. Progress :) Then simply:
a) remove atmosphere
b) replace comet with some more solid body (planetoid-size bearing ball or something)
c) choose whatever velocity you consider as high enough to make measurable difference in the path of motion
What actually matter in this theoretical scenario, is to prove/disprove the idea that increasing velocity of incoming body/object will result in that body experiencing STRONGER gravitational attraction to other source-objects around it.
What Einstein's model is suggesting is that the faster the interplanetary bearing ball will be moving in relation to Earth (stripped out of the atmosphere), the stronger it will be attracted to the surface - and that a slower cosmic bearing ball has bigger chance of escaping Earth gravity...
I hope that I don't need to explain how experimentally backwards are those predictions. If that would be true, 'reaching escape velocity' would be about slowly 'drifting away' from Earth's orbit (too fast and the rocket will fall back to ground), while ISS would need to remain 'fixed' in one place on the sky - as it's motion in relation to Earth would increase gravitational pull...
Yeah... Sorry, but my sense of common reason (sense?) tells me that the faster you are moving in relation to Earth, the higher chances you have to escape from it's gravitational field - well, unless you aren't moving directly towards the surface.... I'd love to see ANY source stating otherwise... :)
Post by: CrazyScientist on 27/09/2024 19:57:25
Post by: CrazyScientist on 27/09/2024 20:39:27
And here are 2 images that suppose to visually represent the general ideas behind my extended formula:
and the first one is wrong. Not a good starting point in physics.
Ahh! Sh*t! Thanks! I posted wrong version - they are almost the same. Fixed it already :)
Post by: CrazyScientist on 27/09/2024 21:49:47
Ok, here's question to those willing to stand in defense of Einstein's model - let's have 2 bodies of equal mass accelerated simultaneously at the same rate to relativistic velocities - so that both are moving at the same speed next to each other (appear stationary to each other) - will the increased velocity of those objects (observed in some other frame) have any effect on gravitational interactions between those accelerated bodies or they won't observe any difference?
I'm pretty sure (like 85-90%) that not even the most loyal supporters of Einstein's model can tell with >50% certainty what the model actually predicts - and if some actually can make such prediction, it will most likely go against predictions of some other Einstein supporters...
As for myself - I literally have no idea what Einstein's Relativity tells us about such/similar scenario (otherwise I wouldn't ask :P). I can however tell with >50% (but more like with 95%) what my model of relativity predicts in such situation - and it predicts that relative velocity of accelerated pair of objects won't have completely any effect on the gravity which they induce on each other - from perspective of those accelerated objects nothing changed (unless there's medium in which they move).
***
So ok, one more 'scenario' - what would happen with ISS if SuperMan or some other OmniMan would simply stop it's motion around Earth? Let's say it would stop moving right above my house and then would be left 'hanging there' not moving in relation to surface (at least horizontally) - what would happen then with it?
Now, in this case I can at least try figuring out what Einstein's model predict in similar situation..I'd say that according to Einstein, ISS would simply keep 'hovering' there endlessly or maybe even it would slowly 'drift away' from Earth. Why I'd say so? Well, since it stopped moving in relation to Earth's surface, it's energy decreased and so did the gravitational pull towards Earth - that's how I'd see Einstein's model...
What my model predicts however, is that I'd have at best 10 minutes to move most important stuff from my house before couple hundreds (thousands?) tons of 'space junk' would turn it into a landing pod (although 'falling pod' seems to fit better) - it would simply start falling down from it's orbit like 2 seconds after it's motion would stop...
Post by: alancalverd on 27/09/2024 22:14:16
Ahh! Sh*t! Thanks! I posted wrong version - they are almost the same. Fixed it alreadyIf anything, you've made it worse. But if you let facts get in the way of a good delusion, you'll never get the Republican nomination.
Post by: alancalverd on 27/09/2024 22:15:36
Four objects, a blue one, a green one, a yellow one and two red ones.Aha, not a politician but an economist!
Post by: alancalverd on 27/09/2024 22:26:45
I'd say that according to Einstein, ISS would simply keep 'hovering' there endlessly or maybe even it would slowly 'drift away' from Earth.Then you clearly haven't understood the first test of relativity. When v << c, all relativistic predictions degenerate to classical mechanics.
Post by: Halc on 27/09/2024 23:38:45
1. there's a definitive & limited amount of energy in matter (according to Einstein it's unlimited) - equal to (half of) energy released to matter-antimatter annihilation.Einstein is right. There is no limit to how much say kinetic energy one can give to a given object. One can always apply a force on it to make it go faster, assuming one can find an energy source to do so. So in the frame say some neutrino, the small bunny in my front yard perhaps has more energy than one stationary anti-bunny.
Quote
2. Body moving at 100% cA 'body' has mass and cannot move at c, as Paul has pointed out. This is really basic stuff that shows you apparently know nothing about relativity theory.
Quote
gravitational attraction decreases gradually as source-object accelerates from 0 to 100%cGravitational attraction isn't a function of acceleration. Not one bit.
Quote
Because I really despise making theoretical calculations with real-life units like (kilo)grams, meters or seconds, I made myself a simplified system of units based on constant c (moving 1 space unit in 1 time unit).They're called natural units, and in such a system, c=1 and can be treated as such in all equations. But you assign c=10, which isn't a natural unit.
let me show you an example:Relativistic mass has been depricated since about 70 years ago. By definition, mass is proper mass, and is frame invariant. What varies from one frame to the next is momentum. This is not a change in theory (The original SR paper did not treat mass as proper mass), but rather a standardization of convention. Nevertheless, the idea of relativistic mass has an incredible presence on the internet, which is why chat bots use it so much.
https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_vs._rest_mass
Gravity is not a function of mass in relativity theory. It is a function of stress energy, which is frame invariant.
Basically according to Einstein there's no limit to the energy for any amount of mass (matter) - you can pack infinite amount of Giga-Joules into a single proton (or electron).Or negative even. Potential energy is negative and there's no limit to the depth of a hole you can drop a proton into.
Quote
Well, sorry but in my uneducated opinion, matter/antimatter annihilation turns WHOLE mass into energy and there's ALWAYS a finite & measurable amount of it releasedBecause you started with finite energy. Einstein doesn't say you can get more energy from a stationary thing. It has to be moving fast to have more energy than the stationary one. If it's fast enough, you can do more work with it than the anti-matter thing, and it doesn't even have to disappear to do it.
Quote
and it's theoretically IMPOSSIBLE to get any more energy from matter.So this assertion is clearly wrong.
Quote
Thing is, that for some reason physicists keep thinking that mass has a definitive amount energy at rest (inertial frame) but then you can 'pack it' with potential and kinetic energy to infinity.If it has KE, it isn't at rest is it? If it has PE, that is negative. There's a limit to how close to zero PE can get, but it isn't really defined where zero is. I.E. there's not a place you can be that is free from gravitational potential.
Let's take a theoretical comet/asteroid entering edge of Earth's atmosphere at a relatively small (flat) angle - question: Will making the asteroid/comet faster increase or decrease the chance of it hitting Earth?Bunch of problems with the pic: The comet cannot be a comet since it is moving too slow. It is clearly in eccentric orbit about Earth, while comets orbit the sun and move much faster relative to Earth when in its proximity.
(https://i.postimg.cc/Zq2pjCFH/comket.jpg)
The tail of the comet points the wrong way. It should extend to the right, not down where you drew it. A comet tail is not a wake, despite all the popular images depicting it that way.
That aside, if the comet takes the same initial trajectory but is moving only a tad faster, it will have less time to accelerate to the right as you depict. It will miss. This is basic Newtonian mechanics and all of this has squat to do with relativity theory.
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In Einstein's model increasing speed = increasing energy = increasing mass what should lead to stronger gravitational attractionThis is completely wrong. Relativity theory says no such thing. Mass is not a function of speed. Gravity is not a function of mass, at least not directly.
Please do not say what Einstein's model predicts if you don't know how (and when) to apply the model. It is a straw man fallacy.
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My model predicts something directly opposite - the faster the comet moves, the less it is attracted to EarthThis is also wrong. Yes, it will miss, but because of what Newton says, not because the attraction is less.
What Einstein's model is suggesting is that the faster the interplanetary bearing ball will be moving in relation to Earth (stripped out of the atmosphere), the stronger it will be attracted to the surface - and that a slower cosmic bearing ball has bigger chance of escaping Earth gravity.Totally wrong. Relativity theory says no such thing.
Ok, here's question to those willing to stand in defense of Einstein's model - let's have 2 bodies of equal mass accelerated simultaneously at the same rate to relativistic velocities - so that both are moving at the same speed next to each other (appear stationary to each other)Gravity isn't a function of acceleration. So you're saying that you still have two objects that remain relatively stationary. Gravity between them (in their own frame) will be unaffected per the first postulate of SR.
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- will the increased velocity of those objects (observed in some other frame) have any effect on gravitational interactions between those accelerated bodies or they won't observe any difference?In a frame in which the two objects are moving fast, the two objects will accelerate towards each other more slowly. Less acceleration, not more like you describe. It will take more time for them to collide than they would in a frame where their mutual center of mass was stationary. In a frame where they're moving at 0.866c, they'll take twice as long to collide with each other.
I'm pretty sure (like 85-90%) that not even the most loyal supporters of Einstein's model can tell with >50% certainty what the model actually predicts - and if some actually can make such prediction, it will most likely go against predictions of some other Einstein supporters...The validity of a model has nothing to do with a vote of support. Plenty of people support relativity without understanding it. Some support alternative theories, and a small fraction of those understand both Einstein's theory and their own choice. Supports relativity does not imply that one can make correct predictions.
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As for myself - I literally have no idea what Einstein's Relativity tells us about such/similar scenarioYou're sure making an awful lot of incorrect assertions about it for somebody who admits a lack of understanding.
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So ok, one more 'scenario' - what would happen with ISS if SuperMan or some other OmniMan would simply stop it's motion around Earth?Hope they take their time about it. The ISS is delicate and does not take kindly to large forces being applied to a small area. Irrelevant, I know. We assume it is sturdy for this exercise. We assume Superman can exert reactionless force, but only because it doesn't change the answer.
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Let's say it would stop moving right above my house and then would be left 'hanging there' not moving in relation to surface (at least horizontally) - what would happen then with it?Nothing to do with Einstein. It would fall somewhat to the east of your house after around 5 minutes. East is due to Coriolis force.
If the object has sufficient altitude, it won't hit Earth at all. If it has even more altitude, it won't even lose altitude, which is how geosync satellites manage to stay put despite not moving in relation to surface.
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Now, in this case I can at least try figuring out what Einstein's model predict in similar situation..I'd say that according to Einstein, ISS would simply keep 'hovering' there endlessly or maybe even it would slowly 'drift away' from Earth.Seriously, you thing Einstein would suggest that?? This is pretty hard evidence of crackpottery.
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Why I'd say so? Well, since it stopped moving in relation to Earth's surface, it's energy decreased and so did the gravitational pull towards Earth - that's how I'd see Einstein's model..A stopped thing still has mass, which is non-zero energy. Surely you don't think otherwise. OK, it seems you do...
If you want us to take your posts seriously, take seriously the thing with which you disagree.
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What my model predicts however, is that I'd have at best 10 minutes to move most important stuff from my house before couple hundreds (thousands?) tons of 'space junk' would turn it into a landing pod (although 'falling pod' seems to fit better) - it would simply start falling down from it's orbit like 2 seconds after it's motion would stop...1) half that time 2) it will miss 3) it starts falling right away (just as it is doing now), not after 2 seconds.
I cannot quote the copilot answer since it isn't text, but almost everything in it is completely wrong, and is a fine example of why it is terrible to ask any chatbot questions on theories about which it has no actual understanding.
Post by: CrazyScientist on 28/09/2024 00:01:42
So does it mean that kinetic energy starts to be added to the inertial mass/energy of a body only when a certain velocity is reached? Or what? If so, then what velocity is it? is 0,5c or 0,2c enough? And what if it's me moving in relation to the object in question? If total relative velocity of our (me and the object in question) motion is equal to 0,99c - how can we know how our velocities are distributed in other frames? Is there any difference between me incoming towards the object at 0,99c and object incoming towards me at 0,99c? And what if in some other frame we (me and object) are incoming at each other with equal speed? Where the additional mass/energy will be added?I'd say that according to Einstein, ISS would simply keep 'hovering' there endlessly or maybe even it would slowly 'drift away' from Earth.Then you clearly haven't understood the first test of relativity. When v << c, all relativistic predictions degenerate to classical mechanics.
***
And coming back to ISS and your answer - can you help me finding the fragment of text telling that relativistic mass becomes a 'thing' only at extremely high velocities and that it's effect is completely non existent below some specific velocity?
https://en.wikipedia.org/wiki/Mass_in_special_relativity
(https://i.postimg.cc/3wwbp92P/bandicam-2024-09-28-00-48-51-822.jpg)
And if not that - should I understand that classical mechanics might lead to effects that are contradictory to effects of SR? So until some particular value adding velocity allows escaping gravity but then and then just like that - "Flop!" - and now adding velocity will start increasing gravitational pull?
Or how exactly it is supposed to work?
Post by: CrazyScientist on 28/09/2024 00:10:18
I know - not every one might like the colors which I used on that image... Well... Sorry...Ahh! Sh*t! Thanks! I posted wrong version - they are almost the same. Fixed it alreadyIf anything, you've made it worse. But if you let facts get in the way of a good delusion, you'll never get the Republican nomination.
And BTW - it's possible that you've mistaken websites - things like this happen to people who try moderating multiple fora simultaneously. Sadly (not) I'm not into foreign politics of any kind, so it has to be simply your silly mistake... No problem at all...
Post by: CrazyScientist on 28/09/2024 01:03:12
Einstein is right. There is no limit to how much say kinetic energy one can give to a given object. One can always apply a force on it to make it go faster, assuming one can find an energy source to do so. So in the frame say some neutrino, the small bunny in my front yard perhaps has more energy than one stationary anti-bunny.Nope. He's completely and utterly wrong about it. There's only 'as much' energy, that 'fits' into a given amount of mass (matter). You simply can't get more energy from a given amount of matter than energy released due to annihilation - in which mass is literally turned into photons and stops existing as matter.
https://www.britannica.com/science/annihilation
(https://i.postimg.cc/435P8cQp/bandicam-2024-09-28-01-46-34-462.jpg)
All what can be possibly achieved by adding more energy to LHC, is that at one (probably near) moment the energy will get too high for protons and they will also start turning into photons.
Just check out this movie and look for GZK limit. This reaches the very bottom of the whole issue...
Post by: Halc on 28/09/2024 02:51:02
So does it mean that kinetic energy starts to be added to the inertial mass/energy of a body only when a certain velocity is reached?No. It would violate energy conservation if I did work on an object and no energy was transferred to it because it wasn't yet moving fast enough. This again is not unique to relativity theory.
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If total relative velocity of our (me and the object in question) motion is equal to 0,99c - how can we know how our velocities are distributed in other frames?The relative velocity addition formula can do the frame transforms for velocity. Lorentz transforms are for coordinate transformations between frames.
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Is there any difference between me incoming towards the object at 0,99c and object incoming towards me at 0,99c?Yes. The vector of one velocity is the opposite of the other. Different objects are stationary in the two respective frames.
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And what if in some other frame we (me and object) are incoming at each other with equal speed? Where the additional mass/energy will be added?Each object contributes its own kinetic energy in that frame.
[/quote]And coming back to ISS and your answer - can you help me finding the fragment of text telling that relativistic mass becomes a 'thing' only at extremely high velocities and that it's effect is completely non existent below some specific velocity?[/quote]It doesn't have a minimum speed. It's a thing at any speed (even zero), but not very significant at low speeds where classical approximations are sufficient to express the KE of the object (E=mv?).
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And if not that - should I understand that classical mechanics might lead to effects that are contradictory to effects of SR?Classical mechanics is a good approximation at low speeds. Neither classical nor relativistic mechanics says that objects gain no kinetic energy until some minimum speed is reached.
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So until some particular value adding velocity allows escaping gravity but then and then just like that - "Flop!" - and now adding velocity will start increasing gravitational pull?Yet again, this is totally wrong. Stop asserting this.
Nope. [Einstein is] completely and utterly wrong about it. There's only 'as much' energy, that 'fits' into a given amount of mass (matter).Only so much in stationary matter. There's no limit to KE, else again, energy conservation would be violated.
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You simply can't get more energy from a given amount of matter than energy released due to annihilation - in which mass is literally turned into photons and stops existing as matter.A fast moving small thing emits more energy (higher energy radiation) in such an annihilation. There's no limit to it. You seem to be in continuous denial of a simple theory, rather than taking the position of trying to learn about what you don't want to accept. You cannot falsify a theory that you refuse to attempt to understand. I see little point in responding to somebody so closed to learning.
Your chosen quotes from wiki do not contradict this, except in their using the word mass to mean relativistic mass, which is an unconventional usage of the term.
Post by: paul cotter on 28/09/2024 11:33:13
Post by: alancalverd on 28/09/2024 13:15:47
And if not that - should I understand that classical mechanics might lead to effects that are contradictory to effects of SR?The correct phraseology is that classical mechanics cannot describe some observations that were predicted by relativistic mechanics.
Post by: Halc on 28/09/2024 17:53:33
Thanks, Halc. One small addendum: On #12 (1) the OP suggests that according to Einstein there is an infinite amount of energy in matter and I responded to this falsehood.The wording must be done carefully.
Relative to some given frame, a given mass has finite speed and thus finite energy.
Relative to some given mass, a frame can be chosen such that the object has any arbitrarily high value of energy.
The former wording caps energy, and the latter wording does not. Mr Crazy finds a quote relevant to the second case and interprets it as the first case, thus manufacturing a falsehood where there wasn't one.
He also seems to assert that a fast moving object that does an antimatter annihilation would put out no more radiation energy than the same thing happening to a stationary object of the same mass. That of course would violate energy conservation.
Post by: CrazyScientist on 11/10/2024 00:26:04
Thanks, Halc. One small addendum: On #12 (1) the OP suggests that according to Einstein there is an infinite amount of energy in matter and I responded to this falsehood. In subsequent discourse I began to think I had misread (1) as the OP was then referring to putting an unlimited amount of energy into mass. Very different scenarios.Thanks! You're correct and I'm fixing it right now
Post by: CrazyScientist on 11/10/2024 00:43:58
So does it mean that kinetic energy starts to be added to the inertial mass/energy of a body only when a certain velocity is reached?
No. It would violate energy conservation if I did work on an object and no energy was transferred to it because it wasn't yet moving fast enough. This again is not unique to relativity theory.
Here is an example which you can calculate yourself if you want: let's have a 100g bullet (m) hitting a stationary 1 ton rock (M) at 0,99c. Now let's have opposite situation - rock M is accelerated to 0,99c and hits a stationary ('suspended' in vacuum) bullet m.
According to main postulates of relativity, the total energy released to environment should exactly the same in both situations (bullet moving vs rock moving) - so this should be true:
(γ-1)Mc^2 + mc^2 = Mc^2 + (γ-1)mc^2
Now put the variables in place and tell me how it can possibly be interpreted in a way which makes sense. Total energy of the system varies depending which object is moving, what makes it frame-dependent - what means motion is not relative only a definitive property of frame.
Post by: CrazyScientist on 11/10/2024 00:53:23
A fast moving small thing emits more energy (higher energy radiation) in such an annihilation.
So you claim that fast moving annihilation releases more energy than energy equal to 2mc^2 (2 because of equal amounts of matter and antimatter) in stationary frame? Do you have anything to support such claim or you're just making things up on the way if they fit your narrative?
But anyway If this is your claim, then you just made motion a definitive property of frame. Thank you for proving my point...
Post by: CrazyScientist on 11/10/2024 01:27:18
(https://i.postimg.cc/1XdxtbSF/bandicam-2024-10-11-02-19-24-592.jpg)
(https://i.postimg.cc/TYhf6Jxm/bandicam-2024-10-02-01-02-23-500.jpg)
(https://qph.cf2.quoracdn.net/main-qimg-3bf376351b322975fcb5ff2db3bf2dba)
(https://qph.cf2.quoracdn.net/main-qimg-b9689be0a2b66cc9e0884c1d371ee115)
(https://qph.cf2.quoracdn.net/main-qimg-b0c74c59d67601dfec2fc9e363b7b67a)
But after experimenting a bit with the functions, I figured out that my calculation for distriobution of mass in relative motion has to be modified to this √y= (c-v)(c+v)
Because not only now the distribution is symmetrical but also it's probably smarter to do √mass than divide by mass.
(https://i.postimg.cc/3NV67pjC/bandicam-2024-10-06-22-05-17-800.jpg)
I never said that my theory is finished - it constantly evolves in time :P
I also (apparently) found a (possible) way to calculate kinetic energy from 'my' formula (although I still need to check out if the values match observed ones). Why such formula? I don't kinow - the graph simply fits in my expectations...
(https://i.postimg.cc/c4t2h9ZD/bandicam-2024-10-10-01-38-27-325.jpg)
And for the end, to get Lorentz factor from y=(c-v)(c+v), you just need to do 1/√y
Post by: CrazyScientist on 11/10/2024 02:01:56
(https://i.postimg.cc/fRpr9Y1G/bandicam-2024-10-11-02-59-10-704.jpg)
Post by: paul cotter on 11/10/2024 12:01:53
Post by: CrazyScientist on 11/10/2024 17:05:01
In #35 you are confusing impact energy with total system energy.Kind of - relativistifc kinetic energy is: (γ-1)mc^2 and total energy is γmc^2 - in the example above both values are different depending which object is in motion...
Thing is that Special Relativity can't handle the idea of total relative velocity. You can't just say objects A and B are moving in relation to each other at relative velocity V" - you need to give specific velocity of A and specific velocity of B and from those values calculate all the relativistic effects and it won't be possible to 'switch' between frames without noticing any difference...
Post by: paul cotter on 11/10/2024 18:56:24
Post by: CrazyScientist on 11/10/2024 22:26:08
Not necessarily. 'Time dilation' (Hafele-Keating experiment) can be directly derived from difference in angular and rotational velocities - it predicts exactly the same effect only it isn't based on relative velocity, like in SR - that's why clocks at different latitudes remain synchronized...And if not that - should I understand that classical mechanics might lead to effects that are contradictory to effects of SR?The correct phraseology is that classical mechanics cannot describe some observations that were predicted by relativistic mechanics.
(https://i.postimg.cc/k4D7HqXr/earth4.gif)
(https://i.postimg.cc/Ss5mF3Jn/earth5.gif)
(https://i.postimg.cc/6qvt9cvY/earth6.gif)
(https://i.postimg.cc/nMfZMmKZ/haffele6.gif)
If you'd use SR, you'd get exactly the same effects of 'time dilation' as those presented on the animations...
Post by: CrazyScientist on 11/10/2024 22:54:28
He also seems to assert that a fast moving object that does an antimatter annihilation would put out no more radiation energy than the same thing happening to a stationary object of the same mass. That of course would violate energy conservation.You seem to completely ignore the fact that relative velocity is exactly that - RELATIVE. Do you really believe that energy used to accelerate an object becomes intrinsic property of that object? Or is this energy just as relative as velocity at which it moves (not existent in the inertial frame of moving object)?
Post by: CrazyScientist on 13/10/2024 16:19:41
For example - I never suspected to get a perfect circle as a result of adding '^2' to 'y' in 'my' formula - but since it it is what it is, somehow I doubt that it's just a coincidence, that this whole calculation is based on light emitted by a point source (propagating as perfect circle...
(https://qph.cf2.quoracdn.net/main-qimg-3bf376351b322975fcb5ff2db3bf2dba)
(https://i.postimg.cc/mkd8MpMH/bandicam-2024-10-13-02-41-14-796.jpg)
If I'd have to guess, I'd say that this circle is a graphical representation of the total energy of a given object - so I guess, that it's possible to add area occupied by other graphs/distributions and it should give the area of that circle as total sum of energies (but it's still a guess)....
So anyway, I kept to square 'x's' and 'y's', add square roots and do other nasty things to those numbers and letters...
(https://i.postimg.cc/HL40XP1k/bandicam-2024-10-13-02-21-14-453.jpg)
(https://i.postimg.cc/65kCSN9Q/bandicam-2024-10-13-02-21-27-517.jpg)
(https://i.postimg.cc/V6ZCWxGm/bandicam-2024-10-13-02-39-03-198.jpg)
(https://i.postimg.cc/2yPLP4Nt/bandicam-2024-10-13-02-48-38-952.jpg)
(https://i.postimg.cc/QM09XLr1/bandicam-2024-10-13-02-30-15-932.jpg)
However, as impossible as it might sound, I actually managed to figure out something practical - to be specific, I (probably) figured out why '√y' is being subtracted from 1 in my 'accidental' formula for kinetic energy. I think that those 2 graphs on image below might provide with you some clues - what matters in this particular case, is that those 2 graphs:
(https://i.postimg.cc/6QckYXk7/bandicam-2024-10-13-02-45-00-130.jpg)
a) are ideal halves of a circle (circumference)
b) are distributed in range 0 to 1 along Y axis and from -1 to 1 along X axis
c) are oriented in opposition to each other0
And of course, everything what differs in their formulas, is that '1-' in one case:
1−√𝑦=√((1−𝑥)(1+𝑥) )
√𝑦=√((1−𝑥)(1+𝑥) )
Ok, so the first image below shows the 'classic' Doppler's Asymmetry - nothing special: normal square function...
(https://i.postimg.cc/0jf7FVCy/bandicam-2024-10-13-15-49-07-187.jpg)
But then those two variations follow the exact geometry, only their 'tails' are now 'cut off', so that y has no negative values - what tells me that y is most likely representing one of values: energy and.or mass. X being on the other hand 'bi-directional' (has both negative and positive values) means that it has most likely to represent velocity (which as a vector, has 2 directions along one dimensional axis). Don't ask me how it's possible for those both functions to have exactly the same graphs - I didn't even try calculating it myself (that's what my PC is for)...
(https://i.postimg.cc/d0PDR88H/bandicam-2024-10-13-15-50-10-127.jpg)
What is below, is what happens if we square both sides of 'my' formula - I'm not sure if it actually represents anything but I thought you better see it :)
(https://i.postimg.cc/cHMVk8rB/bandicam-2024-10-13-15-51-21-414.jpg)
And here is (most likely) mass + kinetic energy distribution (with and without Doppler's Asymmetry Factor)
(https://i.postimg.cc/zXMPjkx1/bandicam-2024-10-13-15-52-00-598.jpg)
(https://i.postimg.cc/NjHCPrn2/bandicam-2024-10-13-16-18-41-307.jpg)
Also it's probably worth to mention that those 2 functions seem to perfectly align in range -1x to 1x
(https://i.postimg.cc/SKWSqRnJ/bandicam-2024-10-13-17-04-50-247.jpg)
(https://i.postimg.cc/8cCNhbt6/bandicam-2024-10-13-17-04-54-443.jpg)
As for now, I still didn't try to work on figuring momentum and/or other possible variables out of this mess - but I think I might have found some possible candidates (but I don't know if this is actually true) for 'something' what can be actually measured:
(https://i.postimg.cc/t4pDDmFw/bandicam-2024-10-13-16-21-33-978.jpg)
(https://i.postimg.cc/bYMwQ7fY/bandicam-2024-10-13-17-02-46-883.jpg)
And for the end - just to prove you, that there's something 'special' about Doppler's asymmetry - that's yet another 'permutation' of the same function... Yeah, coincidence... It can't be possibly infinity, can it?
(https://i.postimg.cc/QCqSrQ30/bandicam-2024-10-13-17-05-38-851.jpg)
I wanted also to show you some cool calculations which I've managed to figure out - as they are tied directly to the graphs I provided above but I've decided that I'll do it in next post... Sorry... :)
Post by: CrazyScientist on 21/10/2024 20:18:12
https://phys.libretexts.org/Courses/Muhlenberg_College/MC%3A_Physics_121_-_General_Physics_I/05%3A__Relativity/5.10%3A_Relativistic_Energy#:~:text=Describe%20how%20the%20total%20energy%20of%20a%20particle%20is%20related (https://phys.libretexts.org/Courses/Muhlenberg_College/MC%3A_Physics_121_-_General_Physics_I/05%3A__Relativity/5.10%3A_Relativistic_Energy#:~:text=Describe%20how%20the%20total%20energy%20of%20a%20particle%20is%20related)
(https://i.postimg.cc/s2S810C2/bandicam-2024-10-13-20-11-51-820.jpg)
Because I'm (apparently) smarter than normal physicists, who politely follow given instructions and carefully apply all century-long traditions in solving equations (just with more letters in them), for me operating on real physical values of observed objects/processes, while calculating completely abstract scenario, which has nio representation in real worl, is kind of.... well, stupid... Why the hell would I need to operate on km/s and kg? To make myself everything 5X as difficult? If you want - sure do your math your way and we'll see if results are matching...:)
So no - I'm that type of person, who treats making own life easier with the highest priority;. So, instead 3.00?108m/s2 light in my unit system is moving at 10 units of space (x) in 1 unit of time (t) - 10x/t - if I would make it 1x/t instead of 10x/tx in calculating energy and momentum I would be multiplying mass by 1, what doesn't work well (I know, because it took me couple minutes, to figure out, why don't those calculations work).
Of course my unit of mass is just as fictional as units of time and space - where object at mass 10m is a black hole and accelerates mass 1m at rate 10x/t^2 (to speed of light) at distance 1x - simple and logical... Screw your 9.11?10−31kg - in 'my' scenario object of mass 2m is moving at 0,99c (9,9x/t) - and those are values I can work with.
So here is how Einstein approached the whole problem:
(https://i.postimg.cc/rp0d4xfs/bandicam-2024-10-13-20-49-21-623.jpg)
I won't lie by claiming that I understand any of this (except maybe those couple sentences in the middle which I actually do understand). Nor I'm particularly interested in solving any of this myself...
I will simply do like this:
for v=9,9x/t (or 0,99c) and with c=10x/t (or 1c) it goes like this: 0,01 * 1,99 = 0,0199 or 0,1 * 9,9 = 1,99 - now, this is quite some difference and that's why we should choose just one option and stick to it, otherwise we might end up with the coma being displaced by two positions and you could get unnecessary stress for no reason. That's why I will stick to option 0,0199 which when squared gives 0,00039601
What apparently seems to be 100% consistent with 'my' graph of mass distribution (although you will have to take my word that this value IS exactly as predicted and this missing 0,0000001 is there when someone manages to place the cursor exactly over 0,99)
(https://i.postimg.cc/SKWSqRnJ/bandicam-2024-10-13-17-04-50-247.jpg)
(https://i.postimg.cc/9XtBzH4S/ddfd.png)
But now let's move back to the 'funny' number 0,0199. Here is one 'magic trick' for those who would like to screw my basic and simplistic equations and deal with something with REAL math (that is unsolvable for 99,9% of humanity) - if that is the case, here is what you need to do in such situation:
you have to place the 0,0199 under the 'roof' of a square root like this: √0,0199 = 0,1410673597966588 and then take this unsightly value and divide 1 by it like this: 1 / 0,1410673597966588
And although number which we'll get in result isn't much less ugly, it is also a number which you possibly saw just a while ago (unless you didn't come here for the sole purpose of admiring my cool-looking illustrations & animations :P) - yes, it's the exact value of so called Lorentz Factor (γ)
(https://i.postimg.cc/FHnMqLNj/bandicam-2024-10-13-21-23-15-876.jpg)
Why such and no other operations? Obviously because other operations won't end up with results equal to γ... I'm pretty sure that there has to be some reason for it (just like for everything else) - I just need to figure it out (but you can try also if you want)...
But this is not what I wanted to calculate today - my plan is to show you why there's '1-...' in '1-√y'. But to do this, I need to move back to my formula of REAL mass/energy equivalence - m = (m*c)^2/m*c^2
(https://i.postimg.cc/Dw9hCMrr/formula.png) (https://i.postimg.cc/1tZ9zj4m/rela.png)
I will spare you unnecessary calculations and will give the most important values right away:
m = 2m
v = 9,9x/t
p = (m*v) = 19,8
p^2 = 392,04
P = (m*c) = 20
P^2 = 400
E = m*c^2 = 200
e = mv^2 = 196,02
(I know that Ek is 1/2 of that but for me it's just yet another unnecessary obstacle that can be simply ignored)
p^2/E = 392,04 / 200 = 1.9602
P^2 - p^2 = 400 - 392,04 = 7.96
[P^2 - p^2] / E= 7,96 / 200 = 0.0398
And now comes another 'nifty trick' out of my sleeve - 0.0398 is the part of mass 2m that is still potential - so if we'd want to see how much it would be for object of mass 1m , we simply need to: 0.0398 / 2 = ... surprise, surprise: we're back at 0,0199
Ok, but I wanted to try explaining why there's that "1-" in "1-√y = √(1-x)(1+x)" what (I guess) represents kinetic energy of accelerating body. You see, the whole idea behind my calculations, is that the amount of kinetic energy which a given amount of matter (rest mass) and because of that I expect there to be some form of symmetry in kinetic/potential mass/energy distribution as object approaches 100%c (when there's no potential energy left). And such symmetry actually is visible everywhere in my calculations. Here, let me show you...
Let's take the total momentum P^2 = (m*c)^2 which for mass m=2 gives 400
Now, for that body moving at 0,99c (9,9x/t) p^2 will be (9,9*2)^2 = 392,04
Difference P^2-p^2 is 400-392,04 = 7,96
But the same result can be achieved in multiple different ways - for example with modification of 'Doppler's asymmetry' formula √[(c-v)m]^2 * [(c+v)m]^2
c-v = 10 - 9,9 = 0,1
c+v = 10 + 9,9 = 19,9
[(c-v)m]^2 = (0,1*2)^2 = 0,2^2 = 0,04
[(c+v)m]^2 = (19,9*2)^2 = 39,8^2 = 1584,04
Multiplying it together gives 0,04*1584,04 = 63,3616
And finally √63,3616 = 7,96
And now to make things even funnier let's do: 7,96 / 0,0199...
...And we'll end up again with total momentum P = 400
OK, so now let's look at energy. Using Einstein's formula γ(mc^2), we'll get:
7.0888120500833612314 * 200 = 1417.76241001667224628
But it turns out that we'll get exactly the same result with this:
200 / √0,0199 = 1417.76241001667224628
And because I'm a curious creature, I've checked what kind of 'mystic digit' hides behind √0,0199 and it turns out to be the unsightly 0,1410673597966588 - so, inspired by yet another of my gut feelings, I've checked which of graphs is consistent with this number and it appears to be that circle, which I 'identified' in my previous post as representing the total energy of a moving body. Well, it seems that another of my 'gut feelings' turned out to be somewhat correct...
(https://i.postimg.cc/Jz2VFXpq/uuu.png)
Next, I decided to make a small 'cheat' to know something more about the function which I 'identified' as one representing kinetic energy: 1−√𝑦=√[(1−𝑥)(1+𝑥)]
(https://i.postimg.cc/VNhMZzrT/energy.png)
So, what I did, was simply to read the value of y at x = 0,99 and it turned out to be yet another 'unsightly' value:
0.7377652804066823875
(https://i.postimg.cc/pXMS1HKp/uihiu.png)
And then using the mehod "Let's try doing this...", I figured out that in order to get such ugly number, one needs to do something like that:
(https://i.postimg.cc/L4LCPbGR/bandicam-2024-10-21-19-57-34-989.jpg) (https://i.postimg.cc/FsxTWJ9B/bandicam-2024-10-21-19-57-49-386.jpg)
But now I'm not exactly sure what should I do with such number :P
Anyway, while doing all sorts of such/similar calculations, I stumbled upon even more interesting stuff. Let's begin from the weird fact that there are still couple more ways, to get a 'permutation' of the 'Lorentz factor' from 'Doppler's asymmetry factor - here's one that gives exact value of γ / 100 (1/100th of γ):
(https://i.postimg.cc/C5Cp6L1s/bandicam-2024-10-21-20-18-26-622.jpg)
But wait - there's more... Let's look for example at this - let's subtract relative momentum p from total momentum P: 400 - 392,04 = 7,96 - let's call it 'potential momentum'. Now if we divide it by what I called as 'potential mass' which in scenario above gives 0,0398 - 7,96 / 0,0398 = 200 <- so the value of mc^2
But let's now add '√' before 7,96 to get γ*10
(https://i.postimg.cc/prsqdg1t/bandicam-2024-10-21-20-41-24-101.jpg)
And although I'm not exactly sure what to make out of it, what I can tell, is that the same 'trick' doesn't work 'the other way' (that means with kinetic values - because while 392,04 / 1,9602 = 200 like before, adding √ before 392,04 won't give anything similar to γ
(https://i.postimg.cc/Y0FDdXXc/bandicam-2024-10-21-20-57-09-501.jpg)
(https://i.postimg.cc/dtMxhY1J/bandicam-2024-10-21-20-56-54-457.jpg)
And while I told you earlier that I want to explain the "1-" in "1-√y = √(1-x)(1+x)", I'm still not ready for it. All I can tell for now, is that my 'gut feeling' tells me that the incompatibility of last 2 calculations above have probably something to do with it...
Post by: CrazyScientist on 22/10/2024 10:54:52
(https://i.postimg.cc/VNhMZzrT/energy.png)
It seems that it wasn't as hard, as I was anticipating. All I did, was to use my 'magic number': 0, 0199 as 'y' in the function visible on screenshot above, what resulted in something like this:
(https://i.postimg.cc/25S203Wm/bandicam-2024-10-21-19-57-34-989.jpg)
And as it turned out it was the number which I was actually looking for - one which allows to derive relativistic kinetic energy, consistent with (γ-1)(mc^2) 'some other way' - to be specific by multiplying this number by total energy γ(mc^2)
(https://i.postimg.cc/gJB8x8Pn/bandicam-2024-10-22-06-03-04-545.jpg)
The only 'issue' with all of this, is the fact that value 0.8589326402033412 isn't consistent with the function which I identified as one representing kinetic energy - instead it's consistent with function showing half of the 'total energy circle' (what kind of makes sense):
(https://i.postimg.cc/t4cGt01r/Zrzut-ekranu-2024-10-22-112753.png)
(https://i.postimg.cc/XJSMmwkd/Zrzut-ekranu-2024-10-22-055631.png)
In order to make it consistent with the 'looks-like-energy' function we need to square the value of 0.8589326402033412 like this:
(https://i.postimg.cc/1zTWjsPd/bandicam-2024-10-22-11-36-06-331.jpg)
What probably makes sense, as multiplying energy by energy makes it squared :P What matters however, is that as it turned out that by subtracting some 'permutation of 'y' value from 1, we get numbers wich for sure represent kinetic energy - what (kind of) answers the "question of 1-..."...
Generally, as you most likely noticed, I'm not exactly sure what I'm doing but for me it seems like some 'mathematical magic enchantment' allowing fluently 'switch' between Galilean realtivitu and SR - or what...?