arkain101 Posted June 2, 2009 Report Posted June 2, 2009 Introduction: Just so you know a bit about my purpose, I'm not trying to prove anything in science wrong. Ever since I got interested in physics, I have taken an alternative route to learning it. Instead of only opening a book and learning physics only as an observer, I've wanted to also live it like the scientists of our past did. I've taken to a passion of discovering as much of the answers and proofs on my own as I can, using my own real experimentation, thought experimentation, and logical deductions. Of course I can read an equation and pump out the numbers (from time to time :lol:) but I also want to find a way to prove it hands on through my own discoveries. It has taken me a long way, and I must say it is incredibly rewarding to be able have the similar, if not the same eureka moments (with a little help) as the scientists that gave us these remarkable laws and theories. In this particular case, I am attempting to do just that. I don't know for certain how the age of the universe is calculated; all the reasons that suggest it is accelerating; nor all of the theories related to explaining the cause (ie, dark energy). In the sense that, I can not sit down and write out all the existing information you would find in a textbook on this manner. However, I've reached a point where a theory has exposed itself as a likely candidate to succeed, and I am going to attempt to produce an answer that is accurate to values that already exist. If it does succeed (If!), and produce accurate results, not only will it be a another exciting personal achievement, but I believe it would provide an accurate explanation for the observed size,expansion, and acceleration of space-time. Couldn't I just read a book or take a course and learn how other people did it (for some part I have). Sure, but this is a hobby, and I have no deadlines, and it just would not be nearly as much fun. :) Developing an equation. I've decided to take on a project to formulate an equation based on a theory. The equation is intended as well as expected to predict the precise age of the universe, based on existing measurements. The theory is quite simple, it considers the the expansion of space-time as well as the acceleration of space-time, to be caused by a deceleration of the universe. The Basic Theory To elaborate, it predicts that prior to the beginning of space and time was an evironment where all velocity was equal to C (or at least very, very close)and all of space and time was infinitely dilated relative to all locations (the lack of locations), and that in the moment of the big bang, this velocity decreased, and so did the spacial contraction, and time dilation. Almost instantaneously, space-time "grew" larger relative to each and every individual, chunk of energy and its neighboring chunk. Such that, as we look out into the cosmos into some distant past time, we observe a time where the expansion of space and time was occurring more quickly than it is here and now in the present. A premise of this theory is that time must be thought about in two respects. That is, the present (as in time) exists in all locations of the universe (where a frame of reference is placed), and because of this, the past affects the future, which implies the present can be produced also by the past. Time: We are familiar with the idea, that our past is formed by the future. For example: An event is expected to happen(future), then it happens (present), then it is gone (past). However, what this theory also takes into account is that, An event happens (present), the energy from that event, travels, and becomes an expected future event for another location, which in turn, creates another present event, which thus continues the process.But the theory is not the focus here. The focus here is on a specific equation based on the theory, which I need help to formulate Quick Checkup and Review It is likely that I have been misinformed and/or misinterpreted some the implications of special relativity theory, so the first step would be to clear that possibility out of the way. If my interpretation of SR is wrong, then it would not be worth pursuing. Likewise, if they are right then it will be worth pursuing. So there is a lot to cover. I will begin with my understanding of spacial contraction. To see if it is correct. So, lets put this into an example. Suppose a specially designed spaceship is going to make a relativistic trip. -The velocity [math]v[/math] is defined as constant 0.8c-The initial distance of the trip, measured at rest is 10 light years. -We give the velocity of the ship to be [math]0.8c[/math] or [math]2.3983396 \times 10^9m/s [/math], the ship travels the distance at constant v. [math]v = 299,792,458m/s \times 0.8 = 239,833,966m/s[/math] (I can't fit the full values on my calculator so I can only express it as following, step by step): [math]L_o = \frac {L}{\gamma} = \left( 1 - \frac {(2.3983396\times10^9)^2}{(2.9979245\times10^9)^2}\right)^{\frac {1}{2}}[/math] [math]\downarrow[/math] [math]\gamma = \left( 1 - \frac {5.7520328\times10^{18}}{8.9875513\times10^{18}}\right)^{\frac {1}{2}}[/math] Which we can also expressed as: [math]\gamma = \sqrt { 1 - \frac {5.7520328\times10^18}{8.9875513\times10^18}}[/math] [math]\downarrow[/math] [math]\gamma = \sqrt { 1 - 0.64}[/math] [math]\downarrow[/math] [math]\gamma = \sqrt {0.36}[/math] or [math]\gamma = (0.36)^{\frac{1}{2}}[/math] [math]\downarrow[/math] [math]L =\frac {L_o}{\gamma}[/math] [math]L =\frac {L_o}{0.6}[/math] Therefore we would calculate the [math]L_o[/math] (relativistic length) [math]L_o = L \times \gamma[/math] [math]L_o = 10 \times 0.6[/math] [math]L_o = 6[/math] In units of light years. (Edited for the purpose of cleaning up the Starting Post) Quote
arkain101 Posted June 2, 2009 Author Report Posted June 2, 2009 If the math works out, could anyone help me clean it up? I had difficulty find an official type of resource for this formula. Once I get this out of the way I can decide whether it is worth continuing with this, and get into the heart of the matter, a much more complex process. I realize so far this is quite simple, but I expect that the progression of this equation is going to require calculus and some expertise in these fields of mathematics and physics. Quote
Ben Posted June 2, 2009 Report Posted June 2, 2009 [math]factor = ( 1 - 0.64)^{\frac {1}{2}}[/math] [math]factor = 0.32[/math]Well, I have absolutely no idea what you trying to do here. First, what does the term "factor" on the LHS mean? As far as I am aware, a factor is an element that enters into a closed product, like in 6 = 2 x 3, then 2 and 3 factor 6. Second, unless there is some nuance I am missing (not at all unlikely!) you seem to be claiming that the square root of 1 - 0.64 = 0.36 is 0.32. Say what - it ain't. Even I, who owns no calculator, can see that Quote
arkain101 Posted June 2, 2009 Author Report Posted June 2, 2009 Well, I have absolutely no idea what you trying to do here. First, what does the term "factor" on the LHS mean? As far as I am aware, a factor is an element that enters into a closed product, like in 6 = 2 x 3, then 2 and 3 factor 6. Second, unless there is some nuance I am missing (not at all unlikely!) you seem to be claiming that the square root of 1 - 0.64 = 0.36 is 0.32. Say what - it ain't. Even I, who owns no calculator, can see that Thus Joe concludes the voyage will take (2.67 light years)/0.66 c = 4 years in each direction. In each of Jane's frames, the journey length is shorter by the factor 1/γ = (1 − v2/c2)^1/2 = 0.75, so for her the distance is only 0.75*2.67 = two light years, so at 0.66 c it takes her only 3 years in each direction. Twin Paradox (from Einstein Light: relativity in film clips and animations) Well, I am looking around for how to calculate distance or spacial contraction, and that above is what I've found. I looked over the math that I did late last night, and I got different numbers. [math]factor = ( 1 - 0.64)^{\frac {1}{2}}[/math] [math]factor = (0.356)^{\frac {1}{2}}[/math] [math]factor = 0.5973[/math]? I honestly don't know what to do with a decimal to the power of 1/2? Quote
Ben Posted June 2, 2009 Report Posted June 2, 2009 Then let me try to help you. Can you figure out what is 1 - 0.64? Do you really think it is 0.356? I don't, I think it is 0.36. Any number "decimal" or not, raised to the power 1/2 is the square root of that number. Want a proof? No of course you don't, but it is a simple application of the law of indices. So I ask again, what is the square root of 0.36? Is it unique? And again, what is the "factor" on the LHS of your equalities? Quote
modest Posted June 2, 2009 Report Posted June 2, 2009 And again, what is the "factor" on the LHS of your equalities? By glancing at Arkain's post (I didn't read it), it seems clear he's finding the Lorentz factor. Arkain, with v=.8 the Lorentz factor (gamma) is 1 and 2/3, the inverse (1/gamma) is three fifths or 0.6. ~modest ***** EDIT ****** -We give the velocity of the ship to be [math]0.8c[/math]... [math]D_i[/math] is the initial distance of 10 light years... [math]D_f = 5.9 \small{light years}[/math]... Good so far??? Yes, it'd be 6 ly. Length contraction is [math]L=Lo/\gamma[/math] or 10/(5/3) = 6 Quote
arkain101 Posted June 3, 2009 Author Report Posted June 3, 2009 Responding to ModestBy glancing at Arkain's post (I didn't read it), it seems clear he's finding the Lorentz factor. Yes, it'd be 6 ly. Length contraction is [math]L=Lo/\gamma or 10/(5/3) = 6[/math] Yes. I believe I was trying to find the lorentz factor ( I wasn't sure what this factor was, other than a percentage shorter distance for the relativistic traveller). However, I just have not applied the transformation equations to the contraction of distance, so this aspect I was most unfamiliar with. It seems the most common application for length contraction is to that of the moving body (relative to an observer at rest), as opposed to the distance contraction for the traveling observer. However, at a closer look, I realize that a time dilation is coupled with a length contraction, the two go together. Responding to BenThen let me try to help you. Can you figure out what is 1 - 0.64? Do you really think it is 0.356? I don't, I think it is 0.36. Any number "decimal" or not, raised to the power 1/2 is the square root of that number. Want a proof? No of course you don't, but it is a simple application of the law of indices. So I ask again, what is the square root of 0.36? Is it unique? And again, what is the "factor" on the LHS of your equalities? Hello Ben, It appears that the values I produced originally were incorrect. It appears we have a hang up on this rather simple example. I started using approximate values for the velocity and [math]c[/math], and ran the numbers twice, making an edit, which would likely be the cause for the different numbers. Although it is not that important to me (that I made these arithmetic errors), I will attempt to clear it up and get that out of the way. So that we can focus on the standardized equations. ( I am purely using latex (no paper), and I often make mistakes using it in this manner. I think what I will do in the future is construct and check the work on paper, before using latex.) [math]v = 299,792,458m/s \times 0.8 = 239,833,966m/s[/math] (I can't fit the full values on my calculator so I can only express it as following, step by step): [math]L_o = \frac {L}{\gamma} = \left( 1 - \frac {(2.3983396\times10^9)^2}{(2.9979245\times10^9)^2}\right)^{\frac {1}{2}}[/math] [math]\downarrow[/math] [math]\gamma = \left( 1 - \frac {5.7520328\times10^{18}}{8.9875513\times10^{18}}\right)^{\frac {1}{2}}[/math] Which we can also expressed as: [math]\gamma = \sqrt { 1 - \frac {5.7520328\times10^18}{8.9875513\times10^18}}[/math] [math]\downarrow[/math] [math]\gamma = \sqrt { 1 - 0.64}[/math] [math]\downarrow[/math] [math]\gamma = \sqrt {0.36}[/math] or [math]\gamma = (0.36)^{\frac{1}{2}}[/math] [math]\downarrow[/math] [math]L =\frac {L_o}{\gamma}[/math] [math]L =\frac {L_o}{0.6}[/math] Therefore we would calculate the [math]L_o[/math] (relativistic length) [math]L_o = L \times \gamma[/math] [math]L_o = 10 \times 0.6[/math] [math]L_o = 6[/math] In units of light years. I am a bit shaky on the use of [math] L, L_o, \gamma[/math] in those equations. Modest appears to have use [math] L[/math] as the relativistic length? Where as I was originally using [math] L_o[/math]. Need to figure that out. Any number "decimal" or not, raised to the power 1/2 is the square root of that number. Want a proof? No of course you don't, but it is a simple application of the law of indices. I recalled this some time last night. It has been awhile since I have done math and it slipped my mind. I remember posting a particular proof about that fact some time ago. Of course I want proof. That is why I am here asking these questions. I would have to ask you what you mean to imply by that statement. I will avoid assuming I know, for the greater good. ;) So I ask again, what is the square root of 0.36? Is it unique? I hope that is covered in the above. Moving ForwardOkay well, this is good news, I am on the right track as I suspected, but at least I know that for sure now. Because of this, I will bother to move forward; I am going to have to do some heavy review and study in order to get back into 'ye old mathematics'. :eek: What I am going to be attempting to do is produce an equation that will predict the redshift to distance ratio. That is, the equation will produce a value of a specific redshift for a given distance, or, a specific distance for a given redshift. Furthermore, if that succeeds, it should be able to be integrated into an integral form equation. (I am jumping a bit ahead of the gun here, but this is my assumptions) Quote
modest Posted June 3, 2009 Report Posted June 3, 2009 I can give you some pointers, if you like, Arkain. [math]v = 299,792,458m/s \times 0.8 = 239,833,966m/s[/math] <...> [math]L_o = 6[/math] In units of light years. You really don't want to convert from light units to SI units then back to light units. You can keep v=0.8, c=1, and d=10 throughout... [math]\gamma = \left( 1 - \frac {5.7520328\times10^{18}}{8.9875513\times10^{18}}\right)^{\frac {1}{2}}[/math] It's:[math]\gamma = \frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/math]set v=0.8 and c=1[math]\gamma = \frac{1}{\sqrt{1-\dfrac{.8^2}{1^2}}}[/math][math]\gamma = 5/3[/math] [math]L =\frac {L_o}{\gamma}[/math] [math]L =\frac {L_o}{0.6}[/math] Therefore we would calculate the [math]L_o[/math] (relativistic length) [math]L_o = L \times \gamma[/math] [math]L_o = 10 \times 0.6[/math] [math]L_o = 6[/math] You kind of got lost there because you had the wrong gamma. It's not 0.6, it's 5/3. Length contraction is:[math]L =\frac{L_0}{\gamma}[/math]with the correct gamma:[math]L =\frac{10}{5/3} = 6 \ Ly[/math] You can do this all in one step by writing length contraction:[math]L =L_0 \sqrt{1-\dfrac{v^2}{c^2}}[/math]Notice this is not Lo times the Lorentz factor, it's Lo times the inverse of the Lorentz factor (1/gamma). this gives you:[math]L =10 \sqrt{1-\dfrac{.8^2}{1^2}} = 6 Ly[/math]in one step. ~modest Quote
arkain101 Posted June 3, 2009 Author Report Posted June 3, 2009 Noted. I knew I was off to a strange start with the math from that twin paradox link, but it was the only source I could find that explained the contraction of space on a trip. That clears up a lot of confusion I had. I am much more familiar with the above, I think I'll stick with that. Quote
Ben Posted June 3, 2009 Report Posted June 3, 2009 Arkain: Let me give you some friendly (and hopefully useful) advice. Don't play with your calculator and try to plug in specific numbers. Otherwise, you will never see the full picture. This applies globally, I believe. Regarding the law of indices. Since [math]x^1 = x[/math] by definition, we will have that [math]x^1 \times x^1 = x \times x = x^2 = x^{1+1}[/math]. Generalizing, one will have that [math]x^m \times x^n = x^{m+n}[/math] So that [math] x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{\frac{1}{2} + {\frac{1}{2}}} = x^1 = x[/math], thus that [math] x^{\frac{1}{2}} = \sqrt{x}[/math] Quote
arkain101 Posted June 4, 2009 Author Report Posted June 4, 2009 Ben, thoroughly understood. First Steps: Important Elements After some work, I've concluded that in order for the theory to agree with astronomical measurements, there must be included an cosmological [math]a[/math], that is deceleration, occurring at the present [math]t[/math]. However, the [math]a[/math] implies a force [math]F[/math] to be responsible. It appears that force has a relation ship with gravity. [math]a = \frac{M}{F}[/math] Which implies a force:[math]F = \left( \, \sqrt {\frac {E}{C}}\, \right) \, {a}[/math] I assume it has some relationship with gravity. [math] \left( \, \sqrt {\frac {E}{C}}\, \right) \, {a} = G \frac{m_1 m_2}{r^2}[/math] The force and acceleration are proportional to a particular time point and distance measurement. However, the force is energy dependent, which implies that it is not an external accelerating force, but rather an internal emission force, specific to the present, and may have relativistic variable values, on observed distance. I can't explain to much more clearly at this point. I am including the acknowledgments the theory must make to abide to observed measurements, in the broad scale of things of space-time. Ie, the exponential dependent redshift and acceleration of distant bodies. Quote
arkain101 Posted June 4, 2009 Author Report Posted June 4, 2009 Represented in one dimension of space [math]X[/math], and one dimension of time [math]t[/math] Relative to an observer located at [math]PTL[/math] (present time location), taking measurements on [math]-PTL[/math] (past time locations) or, [math]-t[/math] location, we can produce the following values based on the observers (thought experiment) type of movement in [math]t[/math] direction: (For example): For Velocity, moving along the [math]X[/math] axis in the [math]-t[/math] direction velocity (of an object or -PTL) increases "[math]+v[/math]" and vice versa. The [math]+t[/math] direction moves closer to the [math]PTL[/math] in the [math]X[/math] axis, and [math]-t[/math] direction moves away from the [math]PTL[/math] in the [math]X[/math] axis. [math]Velocity = (+v, -t)(-v, +t)[/math]That is, increases in the negative time direction, and velocity decreases in the positive time direction. [math]acceleration = (+a, -t)(-a, +t)[/math] [math]distance = (+x, -t)(-x, +t)[/math] [math] (+t, -v, -a, -x)(-t, +v, +a, +x)[/math]That is, in the positive time direction: velocity, acceleration, and distance decrease. Where as, in the negative time direction: velocity, acceleration, and distance increase Elaborating on Acceleration:To be clear on the acceleration aspect. Acceleration does not refer to the object itself accelerating through space. It refers to the increase of space-time expansion, as an observer see's an object (located in the [math]-PTL[/math] ,or can be said as [math]-t[/math] location) move in the [math]+t[/math] direction. So it should be said clearly that, it does not suggest the universe IS accelerating locally, but that it can be observed that space-time WAS accelerating. Furthermore, the idea acknowledges that the value of acceleration does not need to remain constant, but can be different rates at different times. For example, due to the nature of the relativistic equations. [math]L' = L \, \sqrt{1-v^2/c^2}[/math] At a very distant past time, where v was very close to the value c, any change in v, would produce much more rapid change in; acceleration of space, expansion of space, observed velocity of expansion; as opposed to a lower value of v attributed to a more present time. (where v can be treated as an average energy value, based on the [math]t[/math] location), as a kind of energy to density relationship. This produces conclusions: a)The universe is expanding (space) more slowly in the near the present, than it is relative to the past. Where the present is thought of as locality. That is, most local location, both spacial and time. (more local space-time zone) For example: As we observe a standard candle, we observe it moving in the [math]+t[/math] direction at a [math]-t[/math] location relative to our [math]PTL[/math]. That is, as standard candle, is observed to redshift more and more (relative to our PTL) as it moves in the [math]+t[/math] direction. We see this as an acceleration(from our PTL). It is seen as an acceleration because the distance (or space) that light travels through from the -PTL to the PTL, is still expanding while it makes its journey, on a seemingly growing path. This growing path grows exponentially as the Observer imagines moving in the [math]-t[/math] direction. Which is to say, all locations, all times of (space-time) are undergoing anti-spacial-retraction, from a cosmic (present t location) decreasing velocity v. Quote
arkain101 Posted June 4, 2009 Author Report Posted June 4, 2009 I've realized this is starting to require a lot of development of the theory itself prior to the intended equation. Until another time I am going to shift this theoretical part to another topic and return to this topic at a possible later time. I will link the new topic discussing the theory aspect here later. I've got some exciting ideas to share and work through, see you there. Quote
arkain101 Posted June 5, 2009 Author Report Posted June 5, 2009 Find the continuation of the theory here: A Cosmological Theory - Extending on the Big Bang Quote
arkain101 Posted January 6, 2013 Author Report Posted January 6, 2013 (edited) I can't even look at my old posts without getting a headache anymore.. what drove my ambition back then, wow. HAHA Edited January 6, 2013 by arkain101 Quote
Pmb Posted January 19, 2013 Report Posted January 19, 2013 Introduction: Just so you know a bit about my purpose, I'm not trying to prove anything in science wrong. Ever since I got interested in physics, I have taken an alternative route to learning it. Instead of only opening a book and learning physics only as an observer, I've wanted to also live it like the scientists of our past did. I've taken to a passion of discovering as much of the answers and proofs on my own as I can, using my own real experimentation, thought experimentation, and logical deductions. Of course I can read an equation and pump out the numbers (from time to time :lol:) but I also want to find a way to prove it hands on through my own discoveries. Let me rmind you of a famous quote by Isaac Newton - If I have seen further than most it has been because I have stood on the shoulders of giants. There's nothing wrong with cracking a physics book and learning like all the past scientists have. You can still do what you aim to do. You'll just be able to do it more efficiently. A great deal of these scientists never did experiments but relied on the experimental data of others. Johannes Kepler studied the motion of planets and kept incredinly accurate data of what he observed. Newton took tha data and made discoveries from that data such as the inverse square law of gravity. Einstein did something similar with Maxwell'sequations. What are these experiments that you've been doing? I'm curious as to how you're doing these experiments, inerpreting the data and recording the results unless you already know at least the basics of physics and at the most rudimentary of mathematics such as algebra, trigonometry, calculus and vector calculus. Did you already learn this mathematics? Quote
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