Post by: A-wal on 21/05/2025 13:22:06
A. Velocity
For the speed of light v = x/t = c where x is space, t is time and v is velocity.
For c to have the same value in both frames the value of x/t varies between them.
For light's speed relative to a body in motion relative to an observer c = (x/t)+or−v.
For light's two way speed c = ((x/t)+v)+((x/t)−v) → t = (x/(c+v))+(x/(c−v))
t = x((1/(c+v))+(1/(c−v))) = x(c+v+c−v)/((c+v)(c−v)) = x((2c)/(c^2−v^2)
t = ((2xc)/(c^2−v^2)) → for two way speed (2x)/t = (2x)/(((2xc)/(c^2−v^2))
c' = (c^2−v^2)/c and for x to alter to keep light's speed constant x' = x(c'/c).
x' = x(((c^2−v^2)/c)/c) → x' = x((c^2−v^2)/c^2) for length contraction alone.
c' = (c^2−v^2)/c and for t to alter to keep light's speed constant t' = t(c'/c).
t' = t(((c^2−v^2)/c)/c) → t' = t((c^2−v^2)/c^2) for time contraction alone.
Combining the contraction of space and contraction of time use the square roots.
x' = x(square root of((c^2−v^2)/c^2)), t' = t(square root of((c^2−v^2)/c^2)).
So dilation in the observer's frame compared to that of a body in relative motion:
x = x'(square root of(c^2/(c^2−v^2))), t = t'(square root of(c^2/(c^2−v^2))).
(x'/t')^2+(t'/t)^2 = c^2 is a body's overall speed through space and time = c.
An asymmetry not of space and time but how /t reflects how we perceive time.
x'/t' cancels to x/t so (x/t)^2+(t(square root of((c^2−v^2)/c^2))/t)^2 = c^2
This means velocities don't add linearly, velocities are reduced by this factor.
v = v'+v'' where v' is the initial velocity and v'' is the velocity to be added on.
The correcting factor applies once to v so v' and v'' must first be uncontracted.
v' = x'/t', y = t'(square root of((c^2−v'^2)/c^2)).
v'' = x''/t'', z = t''(square root of((c^2−v''^2)/c^2)).
Sum the uncontracted relative accelerations then contract the total to find v.
v = (x'+x'')/((y+z)(square root of(c^2/(c^2−((x'+x'')/(t'+t''))^2)))).
Dilating into raw velocity, summing the velocities and then recontracting gives:
v = (c^2(v'+v''))/(c^2+(v'v'')) or v = v'+v''/(1+((v'v'')/c^2)).
B. Acceleration
The acceleration of a body must be reduced in an inertial frame to keep v<c.
If the proper acceleration is constant the addition formula and time lag are given.
To get proper acceleration in an inertial frame uncontract the velocity difference.
a = v''−v' where v' is the initial velocity and v'' is the velocity after acceleration.
y =t'(square root of(c^2/(c^2−v'^2))), z =t''(square root of(c^2/(c^2−v''^2))).
a = (x''/z)−(x'/y) = p/q → t = q(square root of((c^2−(p/q)^2)/c^2)).
t gives the time duration on the accelerating clock compared to an inertial clock.
Time contraction is t/d where d is duration of acceleration in the inertial frame.
p/q(square root of((c^2−(p/q)^2)/c^2)) is the inertial frame velocity increase.
For acceleration addition take the relative velocity differences to combine them.
a = a'+a'' where a' is the 1st acceleration and a'' is the 2nd acceleration.
The correcting factor applies once to a so a' and a'' must first be uncontracted.
a' = v''−v' where v' is the initial velocity, v'' is the velocity after 1st acceleration.
a'' = v'''−v'' where v''' is the final velocity after the 2nd acceleration is applied.
a' = v''−v' = x'/t', y = t'(square root of((c^2−v'^2)/c^2)) to uncontract t'.
a'' = v'''−v'' = x''/t'', z = t''(square root of((c^2−v''^2)/c^2)) to uncontract t''.
Sum the uncontracted relative accelerations before contracting the total to find a
a = (x'+x'')/((y+z)(square root of(c^2/(c^2−((x'+x'')/(t'+t''))^2))))
Dilating into raw velocity, summing the velocities and then recontracting gives:
a = (c^2(a'+a''))/(c^2+(a'a'')) or a = a'+a''/(1+((a'a'')/c^2))
Inertial coordinate acceleration is given by A = a/t^2 where A = V''−V' (where V' is the velocity before the acceleration as measured in the inertial frame and V'' is the velocity after the acceleration as measured in the inertial frame), a = v''−v' (where v' is the velocity before the acceleration as measured in the accelerating frame and v'' is the velocity after the acceleration as measured in the accelerating frame) and t = c/square root of(c^2−2a^2).
Raw acceleration is given by R = at^2 where R = V''−V' (where V' is the uncontracted velocity before the acceleration and V'' is the uncontracted velocity after the acceleration), a = v''−v' (where v' is the velocity before the acceleration as measured in the accelerating frame and v'' is the velocity after the acceleration as measured in the accelerating frame) and t = c/square root of(c^2−2a^2).
R/A = t^4 because R/a = t^2 and a/A = t^2.
A = a/t^2 and R = at^2 so A = R/t^4 and R = At^4.
I wrote a paper https://zenodo.org/records/15409966 that's been rejected by a journal, exactly what I was expecting for multiple reasons. It makes several bold and grandiose claims in the abstract that were always going to be hard to swallow, I'm not affiliated which might even prevent the abstract from being read and the writing style probably needs to conform with scientific norms regardless of the content.
Everything in this post is equivalent to standard relativity, in fact I'd argue everything in the paper is, it just goes further. I'd like some advice on how to proceed with this. Should I break it up into more bite sized chunks maybe submitting this as an alternative derivation of the velocity addition formula (I'm not even sure there is an existing standard formula for acceleration addition) or maybe just rearrange it so that each bold claim is first explicitly derived, at the moment the derivation is at the very end.
I've rewritten the abstract to this: "This paper reexamines relativity from somewhat altered first principles to build an alternative perspective that conforms with experimental results and is entirely equivalent to Special and General relativity in its physical application, although not in its conclusions. The intent is to show that a more streamlined formulation that recognises certain key equivalences is able to produce a far simpler model that while matching known physics, reframes some aspects of the interdependent relationships to provide a clearer, more precise and far broader picture and in doing so challenges some of the conclusions that have previously been made."
It probably still needs toning down. I'm generally quite blunt but I think I need to be very delicate with how I present this. I'd welcome any suggestions on how to proceed or just discuss the model in the context of already established relativity formulations.
The full formulas (just for giggles) look like this:
v = (x'+x'')/(((t'(square root of((c^2−(x'/t')^2)/c^2)))+(t''(square root of((c^2−(x''/t'')^2)/c^2))))(square root of(c^2/(c^2−((x'+x'')/((t'(square root of((c^2−(x'/t')^2)/c^2)))+(t''(square root of((c^2−(x''/t'')^2)/c^2)))))^2))))
a = (x'+x''+x''')/(((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))(square root of(c^2/(c^2−((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))^2))))
Edit: Hmm, I can't see the LaTeX formulas?
Post by: paul cotter on 21/05/2025 14:20:28
Post by: A-wal on 21/05/2025 14:36:22
(1) there is no time contraction in SR.
Not true, it's just presented as time dilation (the dilated time on the watch of the observer as compared to that of a moving body). This is a very silly way of looking at it. Length contraction (the decreased spatial length of a body with respect to the observer) does the exact opposite. It's senseless to use contraction for space but dilation for time. It's time contraction, as in the contracted time on the watch of a moving body compared to that of the observer.
(2) an inertial frame of reference is one without acceleration.
Huh? I'm defining an inertial frame as one without acceleration.
(3) it is possible to accelerate uniformly and stay v<c.
Well yes, that's why the acceleration addition formula.
I think a move to "new theories" is on the cards, where's Halc?
This is in no way a new theory! This is all entirely consistent with and equivalent to standard relativity formulations.
Post by: paul cotter on 21/05/2025 16:46:17
B. Acceleration
The acceleration of a body must be reduced in an inertial frame to keep v<c.
Post by: A-wal on 21/05/2025 17:00:31
Then I show that the acceleration addition formula takes a form identical to the velocity addition formula. Acceleration is usually different units than velocity but I'm just treating it as the change in relative velocity, so it has the same units and cancels back to the exact same formula.
Post by: Eternal Student on 22/05/2025 04:46:15
I either don't understand the first few lines or don't agree with them.
For the speed of light v = x/t = c where x is space, t is time and v is velocity.
For c to have the same value in both frames the value of x/t varies between them.
You write c= x/t, so if x/t varies between frames then c is not the same. Maybe you meant "x and t vary" when you wrote "x/t varies" - but presumably they vary in such a way that the ratio x/t does not vary.
? Maybe, I've just misunderstood ?
For light's speed relative to a body in motion relative to an observer c = (x/t)+or−v.
I'm not entirely sure how many frames of reference there are, possibly 2 inertial frames? There's an observer, presumably they're at rest in some inertial frame. There's a body in motion relative to that observer, so that body presumably has another inertial rest frame. Then, there's light travelling which should not have any rest frame.
You've written that c, which is presumably the speed of light, = (x/t) +/- v. According to prior lines you have set v = x/t , so you're saying that c = 0 (if we take the - v) or c = 2v (if we take the +v). But then v was set equal to c in the prior lines as well, so c= 2v is just c = 2c, so that also gives c=0.
So you're saying c = 0 always, however I suspect I have just misunderstood what you were trying to write. Perhaps the first few lines where v, x, t, c were defined, just doesn't apply to the next few lines where v, x, t, c get used again. I don't know.
- - - - - - - - - - -
I think a lot of the problem is that I and presumably most of us just can't understand what you were trying to write. You mentioned later that your derivation is shorter than the standard approach. Maybe it is but perhaps that's only because you've skipped a lot of lines of discussion that really do need to be there if anyone else is going to make sense of what you've written.
As it stands, especially without any LaTeX support for the formulas, your work is very difficult ("painfully difficult") to read and I'm afraid I have barely done much more than glance over it.
- - - - - - - - - - -
You also made this comment about why your paper may have been rejected:
Quote
.....and the writing style probably needs to conform with scientific norms regardless of the content....One example that @paul cotter has already mentioned is that you can't make-up new terminology and expect anyone else to know what it is. "Time dilation" is a recognised piece of scientific terminology, "time contraction" is not. You might have good reason for using alternative terminology but you will then HAVE to put in a line or two explaining what your terminology means. Once again, this means that your shorter derivation may not end up actually being that much shorter after you've spent the necessary lines to educate the readers about the terminology you wish to use.
Since it doesn't seem essential or important to change the terminology, it's probably best that you just don't. Use the existing terminology and everyone will understand. Terminology isn't there just to be fancy or to be "elitist", every branch of knowledge has its own specialist terminology and there is some good reason for having it and using it: When you say, for example, "gamma factor", I know precisely what that is and why it's important in special relativity. It communicates information to the reader very precisely and very succinctly. The only other way we could communicate that information is by writing several lines of text and just hoping the reader understands it.
- - - - - - - - - - - -
Quote
.....Hmm, I can't see the LaTeX formulas? ...
LaTeX support has stopped working on this forum. It is a sad loss. The only work-around I have is to create your formulas on something else and then take a screenshot. You can insert picture files of various file formats straight into the forum post - although, to be honest, that task is a little bit fiddly. For a first experiment with it, just add some pictures as attachments and leave the readers to click on those attachments.
With a bit more practice and the notes written here: https://www.thenakedscientists.com/forum/index.php?topic=45718.msg397740#msg397740 you can insert these pictures directly into the right places in a forum post.
- - - - - - - - - - -
Quote
I'd welcome any suggestions on how to proceed (with publishing a paper)The advice I would always give is to take your ideas to an establishment like a University. You'll have to do the usual stuff, like convince them that the ideas have some potential. Depending on which country you're in, this may also cost you some money. For example, you may need to register as a research student.
To the best of my knowledge, The Naked Scientists (TNS) Forum is not registered with the Arxiv print server. So TNS could not even get your paper up on the Arxiv website let alone recommend it for publication in some recognised journal.
I AM NOT A MODERATOR OR TNS STAFF, that is merely how it is to the best of my knowledge.
There's only a small number of regular users but some of them would probably be willing to discuss your work. That's pretty much all the forum can do. However, my post is already quite long so I'm going to end here.
Best Wishes.
Post by: A-wal on 22/05/2025 17:34:21
Hi.Hello
I either don't understand the first few lines or don't agree with them.For the speed of light v = x/t = c where x is space, t is time and v is velocity.You write c= x/t, so if x/t varies between frames then c is not the same. Maybe you meant "x and t vary" when you wrote "x/t varies" - but presumably they vary in such a way that the ratio x/t does not vary.
For c to have the same value in both frames the value of x/t varies between them.
? Maybe, I've just misunderstood ?
This is just the starting point that the two way speed of light is always c, Galilean relativity is assumed. The speed of light is v = x/t = c where x is space, t is time and v is velocity, for c to have the same x/t value in all frames the value of x/t must vary between them, in other words if the x/t of light is always c the the value of x and/t has to change between frames.
No this is just light's two way speed relative to a body in motion as measured by the observer. If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c. x/t is the speed of light c in the observer's frame.For light's speed relative to a body in motion relative to an observer c = (x/t)+or−v.I'm not entirely sure how many frames of reference there are, possibly 2 inertial frames? There's an observer, presumably they're at rest in some inertial frame. There's a body in motion relative to that observer, so that body presumably has another inertial rest frame. Then, there's light travelling which should not have any rest frame.
You've written that c, which is presumably the speed of light, = (x/t) +/- v. According to prior lines you have set v = x/t , so you're saying that c = 0 (if we take the - v) or c = 2v (if we take the +v). But then v was set equal to c in the prior lines as well, so c= 2v is just c = 2c, so that also gives c=0.
So you're saying c = 0 always, however I suspect I have just misunderstood what you were trying to write. Perhaps the first few lines where v, x, t, c were defined, just doesn't apply to the next few lines where v, x, t, c get used again. I don't know.
I think a lot of the problem is that I and presumably most of us just can't understand what you were trying to write. You mentioned later that your derivation is shorter than the standard approach. Maybe it is but perhaps that's only because you've skipped a lot of lines of discussion that really do need to be there if anyone else is going to make sense of what you've written.Yea without the LaTeX formulas it's like trying to read base code rather than script. You can click on the link to read the LaTeX formatted PDF, but I take your point. I was thinking that the whole derivation section should be removed anyway and placed separately in the main body of the paper to derive each step.
As it stands, especially without any LaTeX support for the formulas, your work is very difficult ("painfully difficult") to read and I'm afraid I have barely done much more than glance over it.
You also made this comment about why your paper may have been rejected:In the rest of the paper I do refer to it as time dilation but I can't in the derivation. I can't use contraction for space and dilation for time because contraction is distance in space or time being shortened for a body in motion compared to the observer and dilation is distance in space or time being lengthened for the observer compared to a body in motion. Time dilation doesn't really make sense, the rate of the passage of time on the observer's watch is obviously constant, it's the contraction of the rate of the passage of time on a moving clock as measured by the observer.Quote.....and the writing style probably needs to conform with scientific norms regardless of the content....One example that @paul cotter has already mentioned is that you can't make-up new terminology and expect anyone else to know what it is. "Time dilation" is a recognised piece of scientific terminology, "time contraction" is not. You might have good reason for using alternative terminology but you will then HAVE to put in a line or two explaining what your terminology means. Once again, this means that your shorter derivation may not end up actually being that much shorter after you've spent the necessary lines to educate the readers about the terminology you wish to use.
Since it doesn't seem essential or important to change the terminology, it's probably best that you just don't. Use the existing terminology and everyone will understand. Terminology isn't there just to be fancy or to be "elitist", every branch of knowledge has its own specialist terminology and there is some good reason for having it and using it: When you say, for example, "gamma factor", I know precisely what that is and why it's important in special relativity. It communicates information to the reader very precisely and very succinctly. The only other way we could communicate that information is by writing several lines of text and just hoping the reader understands it.
Even the term length contraction doesn't really work, length could mean space or time. It should be space and time contraction because that describes how lengths of moving bodies are measured by observers.
Thank's. That's a bit annoying. There's a lot of formulas, the link to the paper should do.Quote.....Hmm, I can't see the LaTeX formulas? ...LaTeX support has stopped working on this forum. It is a sad loss. The only work-around I have is to create your formulas on something else and then take a screenshot. You can insert picture files of various file formats straight into the forum post - although, to be honest, that task is a little bit fiddly. For a first experiment with it, just add some pictures as attachments and leave the readers to click on those attachments.
With a bit more practice and the notes written here: https://www.thenakedscientists.com/forum/index.php?topic=45718.msg397740#msg397740 you can insert these pictures directly into the right places in a forum post.
I'm in the UK, about ten miles from Cambridge. There's a 'Relativity and Gravitation Group' in Cambridge, I need to somehow engage with them ideally. Thanks for the feedback, if you want to know anything else or have further advice I'm here.QuoteI'd welcome any suggestions on how to proceed (with publishing a paper)The advice I would always give is to take your ideas to an establishment like a University. You'll have to do the usual stuff, like convince them that the ideas have some potential. Depending on which country you're in, this may also cost you some money. For example, you may need to register as a research student.
To the best of my knowledge, The Naked Scientists (TNS) Forum is not registered with the Arxiv print server. So TNS could not even get your paper up on the Arxiv website let alone recommend it for publication in some recognised journal.
I AM NOT A MODERATOR OR TNS STAFF, that is merely how it is to the best of my knowledge.
There's only a small number of regular users but some of them would probably be willing to discuss your work. That's pretty much all the forum can do. However, my post is already quite long so I'm going to end here.
Best Wishes.
Post by: Eternal Student on 23/05/2025 15:16:08
....If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c. x/t is the speed of light c in the observer's frame....
I think that I do now see what you're trying to say. An observer who knew nothing about SR and assumed Galilean transformations apply, would think that the moving body should record different speeds for the speed of light.
You really would need a sentence or two explaining this. Otherwise your audience will throw the work straight out. Indeed it's a violation of the way SR is usually presented and the properties that light seems to have.
I can now appreciate the abstract you've written for your paper.....
...This paper reexamines relativity from somewhat altered first principles to build an alternative perspective...
So, in conclusion, it's not merely just a faster derivation of some formulae, it's a genuine "new theory" that might also re-create some formulae corresponding to existing results in SR along the way. I'm not a moderator but this does seem to qualify for re-location to the "new Theories" section of the forum. That isn't something you should regard as being a negative thing. "New Theories" isn't where junk is always put, it's just where new theories should be put.
....If you have a new theory please put it in New Theories. This not intended to stifle discussion on a topic, neither is it aimed at you personally suggesting you are wrong, or that your theory is wrong. It is just how the forum is organised. Neither is it discrimination aimed at you, everyone is treated the same. If someone posts a question about marine life in the "Non Life Sciences" board it will be moved to "Life Sciences".....
[ Taken from "How do we moderate discussions... ", https://www.thenakedscientists.com/forum/index.php?topic=66954.0 ]
I don't intend to conceal my intentions and have tried to be honest and open in my appraisal. I would support @paul cotter 's suggestion to move this forum thread.
- - - - - - - - - - - -
What I like is that you have an interest in Physics and have made a serious attempt to re-examine some existing theory. I hope this interest continues. You've also been extremely courteous and polite in making replies and discussion with anyone else who had replied to you and that is appreciated.
Finally, I'm just a Mathematician with an interest in Physics and I don't claim that my views and opinions are of any great importance.
- - - - - - - - - - - -
You asked for general advice and I would still suggest getting involved with a University and on to one of their undergraduate or graduate programs. It's their business to know how to get a research paper published and exactly what may need adjusting. Getting involved with the group you described at Cambridge also sounds sensible. It may be that you end up learning other bits of Physics along the way before they consider supporting your research paper - but I don't see that as such a bad thing, you get to learn some more physics.
What's the benefit of learning about the history, old theories and old experiments?
If you're going to make an adjustment in the assumptions about the behaviour of light then you need to consider all the experiments that have already been done. The Michelson - Morley experiment ( https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment ) and similar work suggests that light travels at the same speed regardless of how an inertial reference frame is assumed to be moving. You can create a flash of light while at rest in one frame and then shift to being at rest in another frame of reference with off-set velocity +0.5c (perform a "boost") to measure the speed of that light but that speed still seems to be 1c and not 1c +/- 0.5c as suggested by a Galilean transformation.
What's the benefit of learning about other stuff besides SR?
You should aim to teach people "how to juggle" and not just how to keep 2 balls in the air. You can keep 2 balls moving with some of them in the air by passing one ball low between one hand and the other while only one of the balls is ever going high in the air. I could teach that to anyone in 5 minutes. However, you can't develop that method to juggle 3 balls or 4 balls properly. Even if you only want to juggle 2 balls, you should teach people how to do that properly, both balls should be made to pass high in the air and never passed low between the hands. That is a method that can be developed and will be useful for juggling 3 or more balls.
(https://content.artofmanliness.com/uploads/2020/11/juggle.jpg)
There may be some other way to think about and derive some formulae for SR - and it may be faster and easier but if it only works for SR then it's of limited value. It should be something that can be developed and is still useful and consistent with things like General Relativity and hopefully many other theories, notions and experiments done in other areas of Physics.
I would say that the notion of a spacetime interval and its invariance is the single most important property we should emphasise in SR. That extends to notions of a metric and leads the way to GR. I wouldn't want to have any other assumptions for SR because that will make more work or problems further on. We would then need some other motivation and some new assumptions for GR and other theories in Physics that do seem to work well.
Knowing about other bits of Physics is essential if we are going to identify how the universe works and what the most fundamental laws of nature may actually be. Tweaking and adjusting an assumption we previously had in SR will have knock-on effects.
It is quite possible to obtain a valid final result, such as a velocity addition formula, from assumptions that were actually incorrect but it would be of no value beyond being a memory-aid for those forumulae in an examination (e.g. my thumb and fingers have nothing to do with the way a force results from a current in a magnetic field but it's still the way we all remember that force law while holding our thumb and fingers out).
Best Wishes.
Post by: A-wal on 23/05/2025 15:59:19
This doesn't reproduce the velocity addition formula among other things by chance, it's gets to them by the most direct rout. It does apply to GR as well but I wanted to keep it in this section of the forum so I didn't mention that directly.
I'm trying to keep this out of the new theories section because if this does seem to conflict with standard relativity in any way I want to know, I really don't think it does. I think you still may have misunderstood certain aspects of the derivation (my fault for not being clear enough), I never meant to suggest that the speed of light relative to any observer can possibly be anything other than c, it's the single true postulate (the same rules in every frame should be a given) that absolutely everything else is built on.
When I said "If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c. x/t is the speed of light c in the observer's frame" I was talking about the speed of light relative to a body that's in motion relative to the observer. If a body is moving away from you at half the speed of light then in your frame the two way speed of light relative to that body as measured by you the observer is 0.5c, if the body is moving towards you then the two way speed of light relative to that body as measured by you the observer is 1.5c.
I do use the spacetime interval, "(x'/t')^2+(t'/t)^2 = c^2 is a body's overall speed through space and time = c.".
I appreciate the feedback but I think what I actually need to do is not so much learn how it's usually handled but simply stop assuming people will know what I'm talking about if I don't explicitly tell them, so as to present it in a way that makes sense to people how are used to seeing it formulated differently.
I'll show you the most important parts of the derivation with formatted equations and clear descriptions of what they're doing.
Post by: Halc on 23/05/2025 17:03:30
The main sections of the forum are about questions on the consensus views, and this is neither a question nor consensus, but rather a proposal of a new methodology that may or may not produce similar empirical measurements.
I wanted to point out some very confusing wording which borders on wrongness sometimes.
For the speed of light v = x/t = c where x is space, t is time and v is velocity.velocity (not speed) is indeed space over time. Speed of light is a specific speed and needs a specific distance x and a specific time t. Else all you're saying is that the speed of light is a speed like any other, which it isn't.
Quote
For c to have the same value in both frames the value of x/t varies between them.No. Any specific distance x in one frame does not translate at all to another (F'). The distance in F' would be x'.
Likewise with time. Neither speed nor velocity nor distance is frame invariant. Only intervals are frame invariant.
The speed of light is indeed invariant, but the velocity of light is not, and x/t actually gets you velocity of light, not speed of light.
In you linked paper, there are 3 principles listed up front.
The second is "All uniformly accelerated frames of reference are locally inertial reference frames"
That's nonsense. In an inertial frame, an object at rest sans force tends to remain at rest. In an accelerated frame (uniform or not), it requires a force to remain at rest, as evidence by dropping a rock, which comes to rest only after a force continuously gets applied to it.
I agree: Use time 'dilation' instead of contraction. Don't invent new terms for existing concepts. Time dilation isn't always contraction. Sometimes a remote clock runs faster than yours.
Understand the difference between proper time (which is frame invariant) from coordinate time (which dilates, and thus is frame dependent). Use the correct terms when referencing time.
Likewise, understand the difference between proper acceleration (which can continue indefinitely) from coordinate accleration (which cannot). Use the correct terms when referencing acceleration, and don't just use the ambiguous 'acceleration'.
The speed of light is v = x/t = c ...This is a contradiction. If x/t=c in all frames, and c is constant, then x/t is the same in all frames, but you say it isn't. This would be clarified if you just say that in frame F', x'/t' = c, which is still wrong for the reasons I gave above, but it conveys that we're taking the ration of two different values than we were in the first frame. 'space' and 'time' are not values, so x/t is pretty meaningless as something that equals some specific value. x/t is simply velocity, not a specific velocity.
for c to have the same x/t value in all frames the value of x/t must vary between them
Quote
When I said "If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c.The rate of change in separation of a light pulse and a body moving in F has nothing to do with where the observer (stationary in F) is positioned.
So for that rate of separation to be 0.5c, the body and the light need to be moving in the same direction relative to F, and for 1.5c, in opposite directions. Whether that body or light pulse is moving away from or towards the observer is utterly irrelevant.
Meanwhile, c = (x/t)−v = 0.5 basically says that light speed is half light speed, a mathematical contradiction. Put some rigor into your posts and into your paper. It will really help. Do this before presenting it to anybody.
Quote
a = (x'+x''+x''')/(((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))(square root of(c^2/(c^2−((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))^2))))The loss of Latex here is unfortunate, but even text can be formatted for readability.
Line up the parens with long reach, just like you line up the braces in C code.
Use sqrt(X) or √(X) instead of 'square root of' which is needlessly wordy.
c? or c2 instead of c^2 which is admittedly only a minor improvement
Edit: I noticed that the square root symbol works, but the 'squared' symbol does not (showing up as a question mark), so only the superscript version came out right. What's the use of those symbols if only some of them work?
Post by: Eternal Student on 23/05/2025 19:28:48
...What's the use of those symbols if only some of them work?
Thery're a monument to the old days, they did once work.
- - - - - - - - - - -
Is it a good time to remind everyone that there's a donation page and a forum does cost some money in maintenance?
If you ( @A-wal ) don't continue with some physics then it may not have been a worthwhile use of anyone's time or any sponsor's money. So come on, @A-wal , consider the replies from everyone, notice that it isn't all negative and when you're in a position to encourage others to study physics - do that.
Mis-judging an idea presented is a mistake I/we could make, that's not too terrible. What would be terrible is if we put you off doing more physics.
Best Wishes.
Post by: A-wal on 23/05/2025 22:48:12
I am moving this to new theories because yes, that's what it is, and not because it's wrong or anything.It produces identical empirical measurements! It's the same model, it seems you're just unable to recognise it when it's written in form that deviates from the one that you were taught, even if it's the simplest (although not well written to be fair) possible form of the model.
The main sections of the forum are about questions on the consensus views, and this is neither a question nor consensus, but rather a proposal of a new methodology that may or may not produce similar empirical measurements.
This is ridiculous, there was not a single thing that in any way contradicts standard derivations of relativity, not one! And yet you moved it anyway.
I wanted to point out some very confusing wording which borders on wrongness sometimes.That doesn't even make sense. The two way distance over time (x/t) of light in any given reference frame is always c. "For the speed of light v = x/t = c where x is space, t is time and v is velocity." is an entirely valid statement.For the speed of light v = x/t = c where x is space, t is time and v is velocity.velocity (not speed) is indeed space over time. Speed of light is a specific speed and needs a specific distance x and a specific time t. Else all you're saying is that the speed of light is a speed like any other, which it isn't.
This also makes very little sense. If the speed of light (its x/t value) in any given frame is always the same then because of the x/t values that define the motion of observers relative to each other when they're in different frames the value of x and/or t MUST vary between frames. This is just saying that if observers that are in motion relative to each other agree that x/t = c for light's speed relative to themselves then they must disagree on the value of x and/or t of each other, commonly (and somewhat inaccurately) referred to as length contraction and time dilation.QuoteFor c to have the same value in both frames the value of x/t varies between them.No. Any specific distance x in one frame does not translate at all to another (F'). The distance in F' would be x'.
Likewise with time. Neither speed nor velocity nor distance is frame invariant. Only intervals are frame invariant.
The speed of light is indeed invariant, but the velocity of light is not, and x/t actually gets you velocity of light, not speed of light.
In you linked paper, there are 3 principles listed up front.No, that right there is utter nonsense. By using the example of a falling rock you've already chosen your arbitrary base frame from which you measure acceleration. You've also proved my point for me, in general relativity the rock is at rest while falling and in constant accelerated motion once the ground is continuously applying a force on it.
The second is "All uniformly accelerated frames of reference are locally inertial reference frames"
That's nonsense. In an inertial frame, an object at rest sans force tends to remain at rest. In an accelerated frame (uniform or not), it requires a force to remain at rest, as evidence by dropping a rock, which comes to rest only after a force continuously gets applied to it.
There is nothing that defines a uniformly accelerating frame as anything other than an inertial frame, other than a less accelerated frame by which you can measure an acceleration. But you can always insert a new less accelerated 'inertial' frame in order to redefine the previous one as accelerated frame. All observers agree (unlike velocity which is reciprocal) on which bodies are accelerating relative to other bodies but there is no ground state of zero acceleration just as there's no ground state of zero velocity.
Acceleration is relative, get over it. I didn't mention any of this until you brought it up after moving the discussion here.
I agree: Use time 'dilation' instead of contraction. Don't invent new terms for existing concepts. Time dilation isn't always contraction. Sometimes a remote clock runs faster than yours.I will not, because it's stupid. If you can give me one actual example of observable time dilation I'll use it. I can think of one, when an observer has a large enough acceleration relative to a body that the non reciprocal time contraction that they're undergoing with respect to the less accelerated body will overpower the time contraction caused by their relative velocity and cause them see that body as time dilated. That (and only that) is how the term should be used.
I can't use it in the derivation, it would stop it working. I have to describe what actually occurs.
Understand the difference between proper time (which is frame invariant) from coordinate time (which dilates, and thus is frame dependent). Use the correct terms when referencing time.I know the difference between proper time and coordinate time and the difference between proper and coordinate acceleration thank you very much. I know exactly what comments like this are supposed to do. If I spoke to you like that I'd expect to be banned. I'll be very careful how I word this, when that tactic of implying error by making a correction of something that doesn't need correcting is employed it's not the target of the comment that it makes look stupid.
Likewise, understand the difference between proper acceleration (which can continue indefinitely) from coordinate accleration (which cannot). Use the correct terms when referencing acceleration, and don't just use the ambiguous 'acceleration'.
Your intention is clear, to drag the discussion down to a level where the intended topic is longer being discussed.
Seriously?The speed of light is v = x/t = c ...This is a contradiction. If x/t=c in all frames, and c is constant, then x/t is the same in all frames, but you say it isn't. This would be clarified if you just say that in frame F', x'/t' = c, which is still wrong for the reasons I gave above, but it conveys that we're taking the ration of two different values than we were in the first frame. 'space' and 'time' are not values, so x/t is pretty meaningless as something that equals some specific value. x/t is simply velocity, not a specific velocity.
for c to have the same x/t value in all frames the value of x/t must vary between them
Don't lose my temper, don't lose my temper. And where's the rest of that quote that you intentially left out? You know, the part that would would put it into the right context?QuoteWhen I said "If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c.The rate of change in separation of a light pulse and a body moving in F has nothing to do with where the observer (stationary in F) is positioned.
So for that rate of separation to be 0.5c, the body and the light need to be moving in the same direction relative to F, and for 1.5c, in opposite directions. Whether that body or light pulse is moving away from or towards the observer is utterly irrelevant.
Meanwhile, c = (x/t)−v = 0.5 basically says that light speed is half light speed, a mathematical contradiction. Put some rigor into your posts and into your paper. It will really help. Do this before presenting it to anybody.
When I said "If a body is moving away from the observer at half the speed of light then light's speed relative to that body as measured by the observer is c = (x/t)−v = 0.5c, if they're moving towards you then light's speed relative to that body as measured by the observer is c = (x/t)+v = 1.5c. x/t is the speed of light c in the observer's frame" I was talking about the speed of light relative to a body that's in motion relative to the observer. If a body is moving away from you at half the speed of light then in your frame the two way speed of light relative to that body as measured by you the observer is 0.5c, if the body is moving towards you then the two way speed of light relative to that body as measured by you the observer is 1.5c.Maybe this is news to you but light doesn't have a measurable one way speed so I used it's two way speed. If something is moving away from me at 0.5c and I measure the two way speed of light to be 1c then light (when the light is moving is moving away from me towards the object) at 0.5c relative to that body as measured by me, if that object is moving towards me then light (when the light is moving is moving away from me towards the object) is moving at 1.5c relative to that body as measured by me.
Okay. Writing it as code is actually a good idea. I only just found out that ∑ is a for loop, I thought it was some kind of tensor.Quotea = (x'+x''+x''')/(((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))(square root of(c^2/(c^2−((((((t'+t''+t''')square root of((c^2−((x'+x''+x''')/(t'+t''+t'''))^2)/c^2)))−(((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))))+((((t'+t'')square root of((c^2−((x'+x'')/(t'+t''))^2)/c^2)))−(((t')square root of((c^2−(x'/t')^2)/c^2)))))^2))))The loss of Latex here is unfortunate, but even text can be formatted for readability.
Line up the parens with long reach, just like you line up the braces in C code.
Use sqrt(X) or √(X) instead of 'square root of' which is needlessly wordy.
c? or c2 instead of c^2 which is admittedly only a minor improvement
Post by: A-wal on 23/05/2025 22:49:20
Velocity addition. All relative velocities are below the speed of light, which defines the scaling factor when combining velocities.
v = v'+v'' where v' is the initial velocity and v'' is the velocity to be added on.
All measurable velocities already have the scaling factor included so we need to get rid of that first.
F1.png (20.66 kB . 979x449 - viewed 924 times)This gives us the 'raw' velocities ready to be summed, x'/y added to x''/z, this obviously gives us what the combined velocity would be when not taking relativity into account because we've removed its scaling. Then we simply apply the scaling factor to the combined raw velocities.
F2.png (27.32 kB . 1095x312 - viewed 917 times)Acceleration addition, same thing.
a = a'+a'' where a' is the 1st acceleration and a'' is the 2nd acceleration.
The correcting factor applies once to a so a' and a'' must first be uncontracted.
a' = v''−v' where v' is the initial velocity, v'' is the velocity after 1st acceleration.
This gives us the difference in relative velocity between the initial velocity and the velocity after the first acceleration. The rate of acceleration makes no difference so just the velocity difference is fine.
a'' = v'''−v'' where v''' is the final velocity after the 2nd acceleration is applied.
This gives us the second velocity increase after the second acceleration.
Now we uncontract them to get the raw accelerations to be combined.
F5.png (37.77 kB . 1178x438 - viewed 942 times)Then we recontract the total after summing them.
F6.png (27.3 kB . 1098x299 - viewed 870 times)We can always introduce a new inertial base frame if we want and redefine the previous one as accelerating. Simply add the new frame and make all the previous inertial bodies inertial again within this new frame, then just uncontract whatever acceleration (as a velocity difference now) that you just added, uncontract what their velocity was in the previous now accelerated frame and then add them together and contract the total.
You can do this from within all frames that were previously considered inertial before you redefined them as accelerating, and to all previously accelerating frames by combining the accelerations in the same way.
You can keep on adding new inertial frames to redefine previous inertial bodies as having any amount of proper acceleration you like because you're using a scaling factor that doesn't give a crap. It will always keep relative velocities below c and you can include infinite acceleration at a point in space and it will become unreachable because of non reciprocal length (space) contraction or infinite acceleration at a point in time and it will become unreachable because of non reciprocal time CONTRACTION.
Relative (but still non reciprocal) acceleration makes it singularity proof.
Post by: Eternal Student on 24/05/2025 00:42:40
What tends to happen in most forums is that the regulars will close ranks when one of them comes under attack. We're all just human so that behaviour is just as likely to happen here as in any other forum.
Fortunately, @Halc is probably big enough and confident enough not to worry too much about some comments, so we don't have to rush to defend him. As you can see clearly written in post #7, I have made comments that may have prompted moving the forum thread and @Halc may have just been a moderator on duty at the time.
As mentioned before, I / we could have mis-judged something, I am / we are only human.
Best Wishes.
Post by: A-wal on 24/05/2025 01:11:35
I showed a derivation that reproduces the velocity addition formula precisely in a far simply way by simply removing the scaling factor that c imposes before combining the velocities and reapplying the scaling factor, you would think that by itself would be enough to show that this has merit.
Post by: paul cotter on 24/05/2025 09:18:09
Post by: A-wal on 24/05/2025 13:39:59
Acceleration is not relative, it is absolute as it can easily be measured with an accelerometer without any external reference. The definition of an inertial frame of reference is one without acceleration. If you were to include accelerated frames of reference in the definition of inertial frames then what would be a non-inertial frame?Great question! I wanted to get into this when you asked about it before but I didn't want the thread to be moved into here and also at the time I was talking about how acceleration is measured from within an inertial frame in the conventional sense anyway.
An inertial frame is defined as the least accelerated frame (or set of frames I suppose but I think of acceleration as defining a frame and relative velocities being subsets within each frame) within a given system. It's the ground state from which acceleration is measured relative to that frame.
Accelerometers do not measure acceleration, they measure a push. If a body is being pushed from a specific direction an accelerometer will feel a force pushing it in the opposite direction. If on the other hand a body is under uniform acceleration no force will be felt and in the absence of a less accelerated frame to compare it to can be considered as inertial.
This works really well with gravitation because it tends to be quite evenly distributed throughout each body. In the case of gravity the difference in acceleration over a body is called tidal force, this is what an accelerometer measures. It's not just gravity that can be defined as inertial, acceleration is acceleration.
Post by: A-wal on 02/07/2025 16:52:17
There's two frames, from the less accelerated frame's perspective the time contraction of clocks in the frame that's accelerating relative to them is purely a function of relative velocity, T = t(√((c^2−v^2)/c^2)), we want the perspective of the frame that's accelerating relative to the other frame and how they measure clocks in the less accelerated frame to show how the relative acceleration creates the time difference.
There must be a simple formula to calculate the time dilation by the acceleration. This seems to be a way to know who is correct.Of course there is, T = (2(√(c^2/(c^2−v^2))))((vx)/c^2) where x is space applies a non‐reciprocal time dilation in the form of a time jump on the less accelerated frame from the perspective of the frame that's accelerating relative to it, that's proportional to distance. It simplifies to this because we're using constant relative velocities and instantaneous relative accelerations.
This overcooks the time difference so that despite the Earth twin's clocks slowing during the coasting phases of the journey from the perspective of the traveling twin both twins will agree on the time difference once the traveling twin returns because the time jump is proportional to distance, this is a simultaneity shift.
A simpler way to look at it, if the turnaround point is at rest relative to Earth then the distance is shortened by the square root of ((c^2−v^2)/c^2) for an observer that is in motion relative to that frame, so the time that it takes to travel that shortened distance is reduced by the same amount, so the shortened time for the traveling twin is simply T = t(√((c^2−v^2)/c^2)).
Post by: hamdani yusuf on 04/08/2025 13:20:22
Acceleration is not relative, it is absolute as it can easily be measured with an accelerometer without any external reference. The definition of an inertial frame of reference is one without acceleration. If you were to include accelerated frames of reference in the definition of inertial frames then what would be a non-inertial frame?If an accelerometer is brought in ISS, will it show its absolute acceleration?
If an accelerometer is free falling from an airplane, will it show its absolute acceleration?
Post by: paul cotter on 04/08/2025 15:54:22
Post by: hamdani yusuf on 05/08/2025 03:08:23
Both of these examples are cases of free fall, so the accelerometer reading will be zero.Does it mean that they are absolutely not accelerating?
Post by: paul cotter on 05/08/2025 07:36:31