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*To*: Steve VanDevender <stevev@efn.org>*Subject*: Re: C-ship*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Thu, 05 Jan 1995 10:55:40 +0100

>I have a 486 Linux system with a small DOS partition. If >Circumspace isn't too big I could probably squeeze it on there. OK, it's about 180 Kb, you can download it with next URL: http://www.cpedu.rug.nl/~N0642983/circum.zip >Velocity in spacetime physics is not quite the concept you're >used to. An object "at rest" has a "vertical" worldline. An >object in motion has a "tilted" worldline. ("Vertical" and >"tilted" with respect to the t-axis, of course.) Lorentz >transformations modify the tilt of worldlines, except when the >worldline is that of a light ray; then it only shortens or >lengthens displacements along that worldline (which is a funny >thing to say, because the interval between any pair of points >along any light ray is zero). I understood this, it's just a graph with time along the vertical axis and a space-dimension on the horizontal axis (x-axis). Most diagrams are scaled so that a worldline of lightray has a 45 degree angle. And when you LT this, the non-light worldlines get a bit steeper, am I right? But is it possible to see in a such a diagram from what direction the light-ray comes? (I know the diagram has only 1 dimension, but you can extend it to 2 dimension in a horizontal plane and the time in the vertical direction) What I thus don't know is how you can visualize abberation in such diagrams. > > I realized the next morning that it indeed could not be true, but still > > wasn't sure where I went wrong. Now I think I know why: If the observer is > > "at rest" and the objects move, then the curvature is caused by the finite > > travel of light which causes to show objects at places where they WERE. > > But when the observer moves and the objects are "at rest", then the > > curvature is caused by abberation. > > Do you understand this subtle difference? > >Well, no. Curvature (caused by finite light travel time) and >aberration (caused by Lorentz transformation of light rays) are >two different things. No I think not, if an object is at rest in a certain frame, then the lightrays of all points of the object are all over the place. So if you are a some place you know for sure where the object is (we assumed it doesn't move). When the observer moves, it still gets the lightrays, from the original direction, but then abberation deforms the view and there is your curvature. Now the objects move in the restframe of the observer and the observer is at rest. There is no abberation because the observer doesn't move, but now the finite light travel time causes the curvature. >I think where the confusion is arising is that to get what the >camera sees, you must be in the frame of the camera. You can do >the calculations from any frame, including the ones where the >camera is at rest and the one where the lattice is at rest. But >to figure out what the camera sees, you need to transform all >your results into the frame where the camera is at rest. >Otherwise you get what other observers see the camera seeing. If you put the observer at rest in his rest frame, there is no abberation, since he doesn't move. The two situations I mentioned differ in the fact that the one time the observer is at rest and the objects move and the other time the observer is moving and the objects are at rest. If both the observer and the objects are moving then both curvature due to "finite light travel time" and abberation. Of course you can always find a frame where only one of the two situations is present. > > I tried checking it, but I don't get the same answer. At the non-substituted > > result I already have a different answer. How sure are you about the > > calculation? Did you do it by hand or did you use a mathematical program? > > > > To be short I also have some negative exponent terms: ...exp(-a * t')^2 and > > ....exp(-a * t') > >Because the equation is of the form > >(big mess) = 0 > >where (big mess) contains terms of exp(-a * t'), I multiply >through by exp(a * t') to cancel those, producing terms of exp(a >* t')^2 and exp(a * t'). I'm not sure how you got terms of >exp(-a * t')^2 but if you multiply through at the right stage you >shouldn't. (t- a/a^2 Sinh[a T])^2 - (x-ax/a^2(Cosh[a T]-1))^2 -(y-ay/a^2(Cosh[a T]-1))^2 -(z-az/a^2(Cosh[a T]-1))^2 = 0 The addition of these 4 terms is is what I feed my calculation program. Substituting the Cosh and Sinh with the Exponential functions gives me quite a bunch of terms: (T=t') - 1/(2*a^2) - (3*ax^2)/(2*a^4) - (3*ay^2)/(2*a^4) - (3*az^2)/(2*a^4) + 1/(4*a^2*E^(2*a*T)) - ax^2/(4*a^4*E^(2*a*T)) - ay^2/(4*a^4*E^(2*a*T)) - az^2/(4*a^4*E^(2*a*T)) + ax^2/(a^4*E^(a*T)) + ay^2/(a^4*E^(a*T)) + az^2/(a^4*E^(a*T)) + (ax^2*E^(a*T))/a^4 + (ay^2*E^(a*T))/a^4 + (az^2*E^(a*T))/a^4 + E^(2*a*T)/(4*a^2) - (ax^2*E^(2*a*T))/(4*a^4) - (ay^2*E^(2*a*T))/(4*a^4) - (az^2*E^(2*a*T))/(4*a^4) + t/(a*E^(a*T)) - (E^(a*T)*t)/a + t^2 - (2*ax*x)/a^2 + (ax*x)/(a^2*E^(a*T)) + (ax*E^(a*T)*x)/a^2 - x^2 - (2*ay*y)/a^2 + (ay*y)/(a^2*E^(a*T)) + (ay*E^(a*T)*y)/a^2 - y^2 - (2*az*z)/a^2 + (az*z)/(a^2*E^(a*T)) + (az*E^(a*T)*z)/a^2 - z^2 = 0 As you can see there are 1/Exp[2aT], 1/Exp[aT], Exp[aT], Exp[2aT] terms, these terms do not cancel out. Here is the same stuff but now with Cosh and Sinh: - (ax^2/a^4) - ay^2/a^4 - az^2/a^4 + t^2 - (2*ax*x)/a^2 - x^2 - (2*ay*y)/a^2 - y^2 - (2*az*z)/a^2 - z^2 + (2*ax^2*Cosh[a*T])/a^4 + (2*ay^2*Cosh[a*T])/a^4 + (2*az^2*Cosh[a*T])/a^4 + (2*ax*x*Cosh[a*T])/a^2 + (2*ay*y*Cosh[a*T])/a^2 + (2*az*z*Cosh[a*T])/a^2 - (ax^2*Cosh[a*T]^2)/a^4 - (ay^2*Cosh[a*T]^2)/a^4 - (az^2*Cosh[a*T]^2)/a^4 - (2*t*Sinh[a*T])/a + Sinh[a*T]^2/a^2 = 0 So again I ask you if you are using a mathematical program, I do, and am quite confident about its accuracy. I've tried a lot of things but I simply don't see how I could simplify the formula more. (cosh(u)^2 - sinh(u)^2 = 1 won't work...) Timothy

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