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TimeLord

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Everything posted by TimeLord

  1. 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.
  2. 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?
  3. 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.
  4. A strange thing just happened. I was playing Civ 2 & thinking about various random things when I started thinking of the word "moratorium". I thought I heard it on SG1 or something. So I looked it up to see what it meant. The dictionary wasn't meaningful to me, so i checked wikipedia. Turns out that on this day in 1969 there was the "Moratorium to End the War in Vietnam". Weird how stuff like that happens. http://en.wikipedia.org/wiki/Moratorium_to_End_the_War_in_Vietnam
  5. Pi is still a constant. That's the way it's defined. pi = circumference / diameter of a circle.
  6. 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.
  7. 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... 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. :)
  8. Re: [censored] yeah, I'm a time traveller. Lots of sand. A few cactii.
  9. Re: [censored] yeah, I'm a time traveller. Lots of sand. A few cactii.
  10. I thought anon's post was rather amusing. I'm not aware of anyone who takes these forums or claims seriously. If one travelled back in time, the technology would be so outdated to them, it would be ludicrous. Consider giving up your 3GHz processor for an AppleII and a BBS. On the other hand, a traveller from the past would surely find such machines confusing and get along better without them. One could, of course, claim to originate from the present era and travel to and from other times. However, there is no benefit to believing or disbelieving such a story, as it will get one no closer to their own goals. Anyway, there's no real point to this post either, so we both just wasted a little more time. ;-)
  11. Re: Question regarding the paradox of time travel *DELETED* Post deleted by TimeLord
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