smkngstarr Posted October 10, 2007 Share Posted October 10, 2007 Does time actually exist, if so prove it. now before you start telling me that i dont know what Im talking about. Prove to me that time does actually exist. [/color] Link to comment Share on other sites More sharing options...
bogz Posted October 10, 2007 Share Posted October 10, 2007 Well if time did not exist you wouldn't be able to count the number of occurrences of an event within a fixed interval. But we can do that, so time does exist? Link to comment Share on other sites More sharing options...
smkngstarr Posted October 10, 2007 Author Share Posted October 10, 2007 so by your reckoning counting to ten proves the existence of time Link to comment Share on other sites More sharing options...
bogz Posted October 10, 2007 Share Posted October 10, 2007 Best I can do, hopefully some better attempts are on their way from other folks (which I look forward to reading but yea, counting to 10 doesn't prove the existence of time, but doesn't it disprove the non existance of time hehe Link to comment Share on other sites More sharing options...
Darby Posted October 10, 2007 Share Posted October 10, 2007 smkngstarr, Does time actually exist, if so prove it. now before you start telling me that i dont know what Im talking about. Prove to me that time does actually exist. I'm not sure what you want here. You can't proove to me that you exist let alone me prooving to you that time does or doesn't exist. No matter what your response is to that statement I can always fallback on the existential belief that "it's all in my mind", i.e., nothing but I exists and everything else is just my alone-in-the-void mind's eye hunger for company.It's not actually necessary to proove that time exists. In physics time is simply a metric that we use to measure change. To that extent I'm perfectly satisfied. It works, it follows specific rules and it makes consistently correct predictions about the universe, within the limits of the domain under observation, that are verified by experiment and observation. Within the postulates and limits of the domain covered by Special Relativity I know that the speed of light is a fixed constant for all inertial observers. Therefore 1 second of time is ~300,000 kilometers of length and 1 meter of length is ~1/300,000,000ths of a second of time. Using the Lorentz Transformation I also know how an inertial observer moving at a velocity different than mine will perceive my clock and yardstick. No problem. What's your point?. Link to comment Share on other sites More sharing options...
satown Posted October 10, 2007 Share Posted October 10, 2007 Scalar Vector Link to comment Share on other sites More sharing options...
RainmanTime Posted October 10, 2007 Share Posted October 10, 2007 Interesting. I may agree, or not, with your diagrams. I need more info. Can you express your vector idea in standard unit vector notation? Or can you show how a cross product would be formulated between a spatial vector and a time vector? RMT Link to comment Share on other sites More sharing options...
smkngstarr Posted October 11, 2007 Author Share Posted October 11, 2007 Not quite sure what the second reply is trying to impart. I guess I just dont understand it. My original point was that . I believe that time travel is impossible. because if there is not an actual force reffered to as time . How can it be manipulated to allow travel within it. After reading alot of the posts in this forum I realize that the majority of the people who post here are significantly more educated than myself. Link to comment Share on other sites More sharing options...
Darby Posted October 11, 2007 Share Posted October 11, 2007 Or can you show how a cross product would be formulated between a spatial vector and a time vector? Ray,Refer to his 3D cube above for a graphic. If we follow the identity of a cross product, use the bottom right hand corner of his cube and label one horizontal leg S "space", the other horizontal leg T "time". The vertical leg in that corner of the cube is the cross product "pseudovector" SxT= "spacetime". The problem that I see is that you can't arbitrarily rotate your system of coordinates (at least mathematically) but in the real world we shouldn't be constrained to a preferred orientation for your coordinate system. If you arbitrarily rotate the coordinates the sign of both S and T flip (to negative for instance) but their product is still positive...which shouldn't be the case. BTW: I don't want to get too technical here because smkgstarr has stated that we're getting beyond his knowledge base. I think that his original question was seriously posed and I don't want to leave him behind...on his own thread. Link to comment Share on other sites More sharing options...
TimeLord Posted October 11, 2007 Share Posted October 11, 2007 Can you express your vector idea in standard unit vector notation? It's a little difficult using only text, but I'll try. A 4D vector could be represented as:a = where I've used the .x notation from programming instead of a subscript. Anyway, moving on... Or can you show how a cross product would be formulated between a spatial vector and a time vector? The vector product in n dimensions requires (n-1) vectors and returns a vector perpendicular to all of them. In 3D, we only need 2 vectors to get a vector normal to the plane of the first two. Note that in all cases, the vector parameters must be nonparallel to return a nonzero vector. In 4D, suppose we have 3 vectors:a = b = c = We then define a function p(a,b,c) = | i j k t | | a.x a.y a.z a.t | | b.x b.y b.z b.t | | c.x c.y c.z c.t | which is just a 4x4 determinant of the same style used to solve 3D cross products, with an additional dimension. Upon solving this determinant, we get the following result: type vector4d x as single y as single z as single t as single end type function product4d(a as vector4d, b as vector4d, c as vector4d) as vector4d dim d as vector4d d.x = a.y*(b.z*c.t-b.t*c.z)+a.z*(b.t*c.y-b.y*c.t)+a.t*(b.y*c.z-b.z*c.y) d.y = a.z*(b.t*c.x-b.x*c.t)+a.t*(b.x*c.z-b.z*c.x)+a.x*(b.z*c.t-b.t*c.z) d.z = a.t*(b.x*c.y-b.y*c.x)+a.x*(b.y*c.t-b.t*c.y)+a.y*(b.t*c.x-b.x*c.t) d.t = a.x*(b.y*c.z-b.z*c.y)+a.y*(b.z*c.x-b.x*c.z)+a.z*(b.x*c.y-b.y*c.x) product4d = d end function In the case that a and b are 3D vectors and c = <0,0,0,1>, the result of p(a,b,c) simplifies to the 3D cross product of axb. Link to comment Share on other sites More sharing options...
Darby Posted October 11, 2007 Share Posted October 11, 2007 In 3D, we only need 2 vectors to get a vector normal to the plane of the first two. TimeLord,Good follow-up post. You can see that I "cheated" a bit because I reduced 3D space to 1D with time as the second dimension. As you discovered, moving out to a true 4D cross product leaves you with a nice array but not much of a graphic for our 2D/3D minds to grasp. I suppose that that's what the extra 6-8 years of post-grad and post-doc work for a physicist is really all about - making sense of otherwise "senseless" arrays and then finding something useful to apply them to. Try the cross product as RMT would have it with time as a 3 vector. Link to comment Share on other sites More sharing options...
RainmanTime Posted October 12, 2007 Share Posted October 12, 2007 Thanks Guys, I was hoping satown would expound on what I was asking for. Seeing as he was the one who posted the graphics, I was hoping he could provide the mathematical basis for the difference he is suggesting with these graphics. Timelord: Yes, I agree with your formulation of the standard cross product for a 4-vector. But it seems to me that this would be the mathematical formulation for satown's first graphic (i.e. the classical view that time is just a 4th dimension extension beyond 3-D space). In looking at his tesseract-like graphical construction, it seems he is implying something more, and as such it would seem he would be implying a different vector/tensor formulation. Try the cross product as RMT would have it with time as a 3 vector. Yes, Darby, you seem to understand what my theory is. The classical view has space-time as a 4-vector, and the math is expressed as TimeLord has reviewed it. But what if space-time is actually a 6-D manifold space? Yes, as you have pointed out, it is difficult to envision this from a 3-D perspective, and indeed it is even difficult for many people to see how a tesseract is trying to depict a 4-D space. But I tend to believe that a 3-D orthogonal formulation of time could go far in explaining the difference between "extant reality" (that which is sensed by an observer) as a subset of "potential realities" (the locus of all possible events, speaking from a quantum theory perspective). And I believe that it is the metric of information that would be the bridge between these two "versions" of reality.RMT Link to comment Share on other sites More sharing options...
RainmanTime Posted October 12, 2007 Share Posted October 12, 2007 Hi smkngstarr, My original point was that . I believe that time travel is impossible. I believe we think along similar lines, but let me expound on my beliefs. I believe that the "classical" conception of time travel (as our bodies, comprised of mass, being able to move backward or forward along our worldlines) is "impossible". And the reason I believe it is "impossible" is because we currently have limited view (and physical definition) of what Time really is. I believe that once we can more fully describe the reality of Time, we will come to find that a "version" of time travel IS possible, but that it would "look" nothing like the classical view as espoused in so many science fiction stories.For the longest time (no pun intended!), science believed that Mass, Space, and Time were individual "real" measures of reality. Einstein (along with Lorentz, Mach, Maxwell, and others) were the first to quantify that the ideas of "Space" and "Time" as distinct measures is really just an approximation to the full nature of what we now know as the "space-time manifold". In other words, claiming that some body is "at this point in space" and at "this point in time" is artificial (because of relativity), and a more accurate way to describe a body is by specifying the body as being "at this point in space-time". This also perfectly aligns with the physical fact that VELOCITY (a measure of space per unit time, or space-time) is the primary physical "contraction metric" for how space-time can be warped. I propose that science will eventually take one step further, and explain to us that Mass must also be fully integrated with space-time to understand the physical (and aphysical) aspects of the universe in their totality. I call this the Integrated Matrix of Massive SpaceTime. ©2004 Tree-O-Life.org. How can it be manipulated to allow travel within it. As I mention above, I believe step 1 is getting a better handle on how Time actually relates to both Space AND Mass. But let me use a comparitive analogy to help you understand how "time travel" MIGHT look (quite different from the classical view):Currently, we can say with certainty that in order to move a body of Mass through SPACE, it requires Energy conversion. All forms of propulsion that we know of essentially rely on a change in Mass over some Time period in order to propel the body through Space. Now, using the same 3 variables (Mass, Space, and Time), let us say we wish to "move through Time" (instead of Space). If Mass-per-Time needs to be modified to move through Space, then would it not make sense that Mass-per-Space (i.e. density) would need to be modified to move through Time? Such would be the case if you accept (as I do) that the Integrated Matrix of Massive SpaceTime must follow conservation laws (as outlined by Emily Noether). RMT Link to comment Share on other sites More sharing options...
bogz Posted October 12, 2007 Share Posted October 12, 2007 If Mass-per-Time needs to be modified to move through Space, then would it not make sense that Mass-per-Space (i.e. density) would need to be modified to move through Time? Such would be the case if you accept (as I do) that the Integrated Matrix of Massive SpaceTime must follow conservation laws (as outlined by Emily Noether). Do you have any links that expand on modifying mass-per-space (density) to move through time? Does that mean we are getting less dense as we get older? hehe ;-) Link to comment Share on other sites More sharing options...
RainmanTime Posted October 12, 2007 Share Posted October 12, 2007 Hi bogz, Do you have any links that expand on modifying mass-per-space (density) to move through time? Unfortunately, I don't... this is just part of my Massive SpaceTime theory. HOWEVER, when you think about what Einstein told us, we are aware that the greater the mass density of an object, the greater its gravitational force. And Einstein told us that gravity warps space-time. Hence, the higher the mass density of an object, the larger the local warpage of space-time. So that is at least some evidence that density is a key to manipulating space-time (of which time is just an approximate measure of the whole space-time manifold). Does that mean we are getting less dense as we get older? hehe ;-) Actually, when you think on a cosmological scale, that is EXACTLY what the universe is doing as time progresses. We know the universe is expanding (and accelerating) so by definition it is becoming less dense. :)RMT Link to comment Share on other sites More sharing options...
TimeLord Posted October 15, 2007 Share Posted October 15, 2007 You say that density changes as time progresses. This is correct, but not the typical line of thinking I've encountered. Assuming mass remains constant, it is the volume which changes. This is because we only perceive the 3 spatial dimensions of a 4D vector. So if an object is temporally displaced relative to an observer, it will appear scaled up or down in size. For example, consider a sphere with radius r located at (x0,y0,z0,t0): (x-x0)^2 + (y-y0)^2 + (z-z0)^2 + (t-t0)^2 = r^2 r is its 4D radius, but we only see the 3D projection of the 4D object at this instant, which has radius R in 3 dimensions: (x-x0)^2 + (y-y0)^2 + (z-z0)^2 = R^2 Clearly, this will change its apparent volume, since V = (4/3)*pi*R^3. An object moving in time at a rate different than an observer will appear to scale in size. Funny nobody mentions this when talking about mass-energy conversions. Link to comment Share on other sites More sharing options...
Darby Posted October 16, 2007 Share Posted October 16, 2007 TimeLord, Clearly, this will change its apparent volume, since V = (4/3)*pi*R^3. An object moving in time at a rate different than an observer will appear to scale in size. Funny nobody mentions this when talking about mass-energy conversions. Good post. Thank you.The ratio that you suggest is, obviously, considered in general relativity. The ratio that suggests both "radius" and "pi" are variables, not constants. When you add into the mix the gravitational constant (which might be a constant - who knows) then "pi" itself is a variable. The curvature of spacetime will affect the value of "pi". Just how much "pi" varies describes the value of the spacetime curvature. "Pi", 3.1416..., only applies to plane geometry - an area of spacetime that is astymptocically flat (locally "flat"). Given a large enough area, even on the surface of the Earth, "pi" will not have the same value as Euclid gave it in plane geometry. Link to comment Share on other sites More sharing options...
TimeLord Posted October 16, 2007 Share Posted October 16, 2007 Pi is still a constant. That's the way it's defined. pi = circumference / diameter of a circle. Link to comment Share on other sites More sharing options...
Einstein Posted October 16, 2007 Share Posted October 16, 2007 Timelord Assuming mass remains constant The latest I heard is that the sun looses 4 billion tons of mass per second. So it is very reasonable to assume that the mass of the entire universe is constantly decreasing. Since all the suns in the universe loose mass during the nuclear fusion process. Since we know that mass is declining, and spatial volume is increasing, it might be reasonable to connect the two quantities. Maybe mass is just changing state into an increased spatial volume. The change seems to be occuring in one direction only. Time also seems to be moving in one direction only. I've been speculating for some time now that the flow of time is related to the change in mass of the universe. Link to comment Share on other sites More sharing options...
TimeLord Posted October 16, 2007 Share Posted October 16, 2007 Stars do lose mass over time (due to nuclear reactions), but not all massive objects do. Furthermore, no star loses mass forever. At some point, it runs out of "fuel" to fuse. I think time travel can and should be specified on a per-particle basis, rather than saying a star gives off energy because it moves through time. I presently don't know if or how time dilation as described by relativity relates to actual translational motion through time, but time dilation is not limited to accelerated reference frames. Nuclear reactions can be exoergic, endoergic, or neither. Thus it is possible to cause accelerated or decelerated time dilation with such reactions, instead of the one-way time dilation caused by high velocity. Consider the following: dt = dt0 / sqr(1 - (v/c)^2) This implies that dt > dt0 in all cases (unless v is imaginary). However, dt / dt0 = m / m0 dt = dt0 * m / m0 = dt0 * (m / m0) * (c^2 / c^2) = t0 * E / E0 Q = E - E0 Q / E0 = E / E0 - 1 E / E0 = 1 + Q / E0 dt = dt0 * (1 + Q / E0) Q may take on positive or negative values, so dt may be greater than or less than dt0. Of course, this has the obvious disadvantage of requiring nuclear reactions to slow down or speed up time. And then, I'm not sure it moves the particle(s) through time at all. Link to comment Share on other sites More sharing options...
Einstein Posted October 16, 2007 Share Posted October 16, 2007 Timelord Q may take on positive or negative values, so dt may be greater than or less than dt0. Of course, this has the obvious disadvantage of requiring nuclear reactions to slow down or speed up time. And then, I'm not sure it moves the particle(s) through time at all. The answers wont be found in the math. Once the answer is found though, it may be described with math. I'm just pointing out that the direction mass is taking in the universe appears to be predominately downhill. If the change in mass is the primary cause for the flow of time, then it becomes evident that being able to control this flow would require the science to be able to manipulate mass. Link to comment Share on other sites More sharing options...
RainmanTime Posted October 16, 2007 Share Posted October 16, 2007 This is a great thread, BTW, and I should also say this is the kind of thread that is intended for this particular forum. Well done, gents! Let's see if my tiny grey matter can add something to the mix to keep it going. Assuming mass remains constant, it is the volume which changes. True, but it is the assumption that mass remains constant (with time) that is, IMO, part of the "problem" we face in seeing mass as somehow segregated from spacetime. And when taken with regard to this statement: An object moving in time at a rate different than an observer will appear to scale in size. Funny nobody mentions this when talking about mass-energy conversions. Yes, it is quite funny. And again, IMO this disjoint assumption of "constant mass" (or ignoring the internal mass-energy fluctuations of a body) are what contribute to ignorance of how time, or more specifically spacetime, relates to mass on cosmological and quantum scales. One can grasp the distinction in two different ways that we teach classical mechanics of energy conservation:When I teach elementary fluid dynamics, I draw the comparison between the classical mechanical energy equation (m*g*h1 + 0.5*m*v1^2 = m*g*h2 + 0.5*m*v2^2) and Bernoulli's equation for pressure energy (P1 + rho*g*h1 + 0.5*rho*v1^2 = P2 + rho*g*h2 + 0.5*rho*v2^2). The equations are virtually identical but that one holds for solid matter and the other is applicable to fluids. Yet while they are VIRTUALLY identical, students always ask why the pressure form has P1 and P2 (static pressures) and the matter form has no equivalent. The answer is that in the mechanical/matter form we are assuming mass remains constant and therefore we ignore the internal mass-energy of the solid object that is moving from state 1 to state 2. However, the fluid dynamic treatment cannot ignore the "internal energy" represented by static pressure since it CAN and DOES change from point to point in a flowfield. This is precisely why Einstein's "energy density" model follows (and looks an awful lot like) the fluid dynamic paradigm, because at both cosmological and quantum scales we cannot afford to "assume constant mass" nor ignore internal mass-energy. I believe it also provides a clue to how relativity and quantum theory will eventually be united. We now know that "mass/matter" (I believe they are distinctly different entities, and have defined them as such, but most use the terms interchangeable) is only a VERY small portion of total cosmological energy (~4%), with Dark Matter (DM) and Dark Energy (DE) rounding out the remaining 96% of energy. It is my belief that normal baryonic matter (the ~4%) is not and cannot be isolated from DM and DE. They interact. In much the same way that static pressure, head pressure, and dynamic pressure interact with one another in the fluid model. And this is also part of my hunch that, at some point, we will no longer be able to treat mass/matter as a scalar quantity. To understand how baryonic matter, DM, and DE interact, we will need to adopt a vector (tensor) model for mass.... and that will eventually lead to what I call the Massive SpaceTime 3x3 Matrix. RMT Link to comment Share on other sites More sharing options...
TimeLord Posted October 16, 2007 Share Posted October 16, 2007 The assumption of constant mass was an example of a special case in which mass does not change while an object moves through time. The purpose was simply to show that the volume of that object perceived by a temporally stationary observer would change. With constant mass, it is easy to see that density would change. Of course, mass can and does change, depending on kinetic energy, binding energy, and maybe something else too. By the way, has anyone else noticed that (v/c)^2 = .5*m*v^2 / .5*m*c^2 = 2*U / BE, where U = kinetic energy, and BE = total mass-energy? Link to comment Share on other sites More sharing options...
RainmanTime Posted October 16, 2007 Share Posted October 16, 2007 The assumption of constant mass was an example of a special case in which mass does not change while an object moves through time. The purpose was simply to show that the volume of that object perceived by a temporally stationary observer would change. Understood and agreed, TimeLord. I was not disputing you, only taking the discussion further. and maybe something else too. That's the part I am interested in. I am thoroughly convinced that mass is only perceived as a scalar, but its true nature is a vector. Another clue that tells me this is true is that there is NO conservation of mass law...only conservation of energy. Thus, mass is a degenerate (i.e. approximate...incomplete) form and measure of energy. By the way, has anyone else noticed that (v/c)^2 = .5*m*v^2 / .5*m*c^2 = 2*U / BE, where U = kinetic energy, and BE = total mass-energy? Interesting indeed. I've never looked at it like that. Would you say that this fact supports the idea that we live in (on!) a Brane Universe? I like to make the analogy that our limited perceptive existence (that 4% portion of witnessed energy) lives on the surface (hence the reason for squaring the Lorentz Contraction ratio) of a Hubble Bubble.I'd appreciate your thoughts... RMT Link to comment Share on other sites More sharing options...
TimeLord Posted October 17, 2007 Share Posted October 17, 2007 Quite alright. I don't know about branes. My previous comment (regarding (v/c)^2) was no more than a restatement of the lorentz factor as 1/sqr(1-2U/BE). It's probably of little consequence. As for "dark" energy and matter, I think they're only a by-product of incomplete physical models. I suspect that, in time, all shall be accounted for by "normal" matter. Long ago, I came to the conclusion that a charged particle is defined in a plane & rotates its spin axis to be parallel to its acceleration vector. Thus, the electric field does not point equally in all directions (radially), but the alignment is so fast that we can treat such a particle as a point charge. Mass may well be similar, but I don't yet know for sure. I have considered the slight EM attraction caused by 2 charged particles of opposite charge, such as a proton & electron, placed in very close proximity to one another (such as in an atom). Though the charges nearly cancel, there is a slight difference in the coulomb force exerted on a distant charged particle because the electron & proton are at different distances. This, combined with electrostatic induction could well induce "gravity". However, this might be yet another effect, aside from gravity. Consider the following over-simplified example: Q = charge on distant particle qe = electron charge qp = proton charge = -qe Fe = coulomb force from electron Fp = coulomb force from proton re = distance between distant particle & electron rp = distance between distant particle & proton e0 = permittivity of free space Ra = atomic radius = rp - re (assuming the proton is further than the electron from the distant particle) Ftot = net force exerted on atom & distant particle Fe = qe*Q/(4*pi*e0*re^2) = (Q*qe/(4*pi*e0))*(1/re^2) Fp = qp*Q/(4*pi*e0*rp^2) = -(Q*qe/(4*pi*e0))*(1/rp^2) Ftot = F1 + F2 = (Q*qe/(4*pi*e0))*(1/re^2 - 1/rp^2) The value of (1/re^2 - 1/rp^2) is incredibly small, so the force exerted is very small, but nonzero. Furthermore, Ftot will increase with increasing radius and/or total charges. The above example is for hydrogen, since it's "simple". Anyway, it's possible that I made mistakes here or overlooked something. Let me know if so. Link to comment Share on other sites More sharing options...
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