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  4. Gravity In Relative Motion - is there any solution?
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Gravity In Relative Motion - is there any solution?

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Offline CrazyScientist (OP)

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Re: Gravity In Relative Motion - is there any solution?
« Reply #40 on: 11/10/2024 17:05:01 »
Quote from: paul cotter on 11/10/2024 12:01:53
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...
« Last Edit: 11/10/2024 17:30:43 by CrazyScientist »
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Offline paul cotter

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Re: Gravity In Relative Motion - is there any solution?
« Reply #41 on: 11/10/2024 18:56:24 »
I was responding to a minor point in one of your contributions and I cannot see how my comment has now been called the best answer to the original question,though I have no objections.
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Re: Gravity In Relative Motion - is there any solution?
« Reply #42 on: 11/10/2024 22:26:08 »
Quote from: alancalverd on 28/09/2024 13:15:47
Quote from: CrazyScientist on 28/09/2024 00:01:42
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.
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...





If you'd use SR, you'd get exactly the same effects of 'time dilation' as those presented on the animations...
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Re: Gravity In Relative Motion - is there any solution?
« Reply #43 on: 11/10/2024 22:54:28 »
Quote from: Halc on 28/09/2024 17:53:33
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)?
« Last Edit: 11/10/2024 23:12:53 by CrazyScientist »
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Re: Gravity In Relative Motion - is there any solution?
« Reply #44 on: 13/10/2024 16:19:41 »
Whooa! It appears that just by accident I managed to discover the "Ultimate Formula For Everything In Relative Motion"... Of couse I'm (still) talking about Doppler's Asymmetry Factor (DAF?), which I decided to call as such. After spending some time on 'mathematical Kama Sutra' (that means forcing intercourse of numbers, symbols and letters in all kinds of different configurations), I think that I start to understand some bits of it - and the more I understand the deeper this rabbi(t) hole seems to reach.

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...



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...






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:


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...


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)...


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 :)


And here is (most likely) mass + kinetic energy distribution (with and without Doppler's Asymmetry Factor)



Also it's probably worth to mention that those 2 functions seem to perfectly align in range -1x to 1x



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:



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?


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... :)
« Last Edit: 13/10/2024 16:29:39 by CrazyScientist »
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Re: Gravity In Relative Motion - is there any solution?
« Reply #45 on: 21/10/2024 20:18:12 »
Ok. So, as promised, time for some small calculations. I actually felt motivated enough to take real, material pen and physical sheet of (natural) paper and spend like 2 or 3 hours of actual calculations. This is what I used as my training ground:

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


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:


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)



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)


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

 

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...


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)]


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


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:


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):
 

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


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



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...
« Last Edit: 21/10/2024 20:22:48 by CrazyScientist »
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Re: Gravity In Relative Motion - is there any solution?
« Reply #46 on: 22/10/2024 10:54:52 »
OK. I think that this time I'm actually prepared to face the 'question of "1-" in "1-√y" in the function which I identified previously as one representing kinetic energy:


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:


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)


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):



In order to make it consistent with the 'looks-like-energy' function we need to square the value of 0.8589326402033412 like this:


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...?
« Last Edit: 22/10/2024 17:42:01 by CrazyScientist »
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