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Negative Energy and a SciFi ftl |
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| Jan17-06, 09:55 PM | #1 |
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Negative Energy and a SciFi ftl
I know crazy unsupported theories aren't supposed to be posted, but I have been working on some fiction which requires faster than light travel. I have read many other guides to ftl in SciFi, and I had origionally fitted in a form of hyperdrive, which would simply avoid the issue by ignoring the relevent dimensions of space. But, I don't like it either. Anyway, if it's ok here is what I made up so far. Admin I don't mind to delete if it really shouldn't be here:
Also it seems that the ship would have to be isolated from the effects of negative energy for the ftl to be usefull for a number of reasons; the ship falling apart and the crew aging and dieing from the opposite of the twin paradox. So, I guess my question now is does this make any more sense than the other SciFi forms of ftl, or is it just as "magical"? What really would happen if you had negative energy, assuming of course it could exist appart from normal energy. |
| Jan24-06, 01:47 AM | #2 |
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Fermions (i.e. half-spin particles: electrons and the like) don't propogate at c. Photons are actually bosons, they're spin 1, although not all bosons propogate at c either.
One thing that's always bothered me about FTL in Sci-Fi is that there's an implicit assumption that you can define simultaneaty across large distances in a curved space, which I do not think is true. |
| Jan24-06, 03:12 AM | #3 |
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In general, its not. Simultaneity is not universal, not even in flat minkowksi space. |
| Jan24-06, 03:49 PM | #4 |
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Negative Energy and a SciFi ftl
Your approach is of course "magical" because it is not currently believed anything can go faster than light. But when some way is found to circumvent that all ideas about FTL will come under new scrutiny.
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| Jan24-06, 04:08 PM | #5 |
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I don't know why I'm so concerned over the details. I think that even if ftl was plausable theoretically it wouldn't be very practical. After I posted I though, you know, I wish I didn't post that since there really isn't any way to qualify anything. I have read in other forum about Heim theory, and of course there was an article proposing that alternate geometries would allow alternate speeds of light; using dark energy particles I think, which is similar to the concept of negative energy.
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| Jan24-06, 04:35 PM | #6 |
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But be really aware of what dicerandom said:
When I read that Honor Harrington zips hither and yon and when she gets back her elapsed time exactly adds up to that of the people who stayed home, I can only smile, and concentrate on the adventures and personalities. |
| Jan24-06, 07:06 PM | #7 |
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Here is a question. Say there is a cylinder planet where the diameter is << than the length. Bob and Alice live millions of miles apart. There are also satelites in orbit as in the picture I attached. If they send a laser beam direclty from one house to the other, and also a beam that is relayed through the salelites, which signal arives at the other house first?
Edit: I guess the question should be, can the laser relayed through the satelites ever get there before the direct one. |
| Jan24-06, 07:25 PM | #8 |
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| Jan24-06, 07:45 PM | #9 |
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Yes. Clues are planet, millions of miles, and the general construct is a question reguarding general relativity. You don't even need SR to answer the question in flat space.
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| Jan25-06, 01:50 AM | #10 |
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Note: this is very much off the cuff, basically I'm assuming that the metric for your cylinder world will have roughly the same characteristics as the Kerr metric for a spherical rotating mass.
With the spinning cylinder you're going to end up with two major curvature effects (I'm thinking of Christoffel symbols here): 1) There's the standard radial 'pull' [itex]\Gamma^r_{tt}[/itex] due to the mass of the cylinder. 2) There's also a contribution like [itex]\Gamma^\phi_{tt}[/itex] due to the rotation of the cylinder. Now your light from the laser will follow a geodesic, so you should be looking at the geodesic equation: [tex]\frac{d^2 x^\alpha}{d\tau^2} + \Gamma^\alpha_{\mu \nu} \frac{d x^\mu}{d\tau} \frac{d x^\nu}{d\tau} = 0[/tex] So initially your photon will have velocities in t and z, the t component of the velocity will result in "accelerations" in the r and [itex]\phi[/itex] directions, so basically your laser will end up dropping towards the surface of the cylinder and picking up rotation along with the cylinder as well. Now, unless the cylinder is rotating extremely quickly I can't imagine this effect (namely the spin) being nearly large enough to justify taking the light beam out to a larger radius. I think that the magnitude of the rotation will also be limited like it is with the case of the Kerr black hole, so you'll have a definite upper limit imposed on how quickly you can spin this thing. |
| Jan25-06, 01:59 AM | #11 |
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I'm more interested in time and distance differences between the two houses, and between the "satelites". Will the light beam traveling from one house to the other take any longer than from one satellite to the other.
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| Jan25-06, 07:09 AM | #12 |
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While you could work out the geodesic equation in the curved coordinates, the easiest solution is to use Minkowski coordinates (i.e. non-rotating coordinates), and then you know that light follows a straight-line path.
Your diagram does not show the curvature of the cylinder. the ideal path will be tilted slightly "up" because the cylinder is curved. |
| Jan26-06, 01:58 AM | #13 |
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Maybe I need to be more explicit. Time passes more slowly in the houses than in the satelites, right? This is my only assumption so far from what I have gathered from other websites, sense I don't know the math to find out myself.
Say you have a light clock manufactured by AB corp. One clock is in Alice's house on the surface and one is in the satellite orbiting above her house several thousand miles away (this is a really big planet btw.). If alice looks up at the satelite will she see the clock running faster than her own? I'm guess the answer is yes, but that the clock also looks smaller so the speed of light is not violated here? Now, the two houses (Alice's and Bob's) can act like one giant light clock, and the two satellites act like a giant light clock, and lets say that AB corp. built them too (ie they are idential in their reference frame). If my previous assumptions are true, then Alice and Bob will, from the surface of the planet, see the satellite light clock run faster than their house light clock. But that the satellites are somehow closer together than the houses preserving speed of light. But now how can Alice and Bob still independently verify that they see their satellite orbiting directly above their house? Ok, I'll eagerly await replies. :) |
| Jan26-06, 03:48 PM | #14 |
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The "speed" of light, as definied by the rate of change of the distance coordinate with respect to the time coordinate, is not constant. What is constant is the speed of light as measured by a local observer, using local clocks and rulers. An example might help. Let us suppose you are in a spaceship accelerating at 1g, using acceleration to create "artificial gravity" rather than rotation as in your example. Using standard coordinates, clocks "above" you will tend to tick faster. The metric of space will, however, not change at all. You can find the "speed" of light to be as fast as you like by looking at a light beam sufficiently far "above" you. At 1g of acceleration, roughly 1 light year above you light will be travelling twice as fast (in coordinate terms) than it will be at your location. It's not directly relevant, perhaps, but it's interesting to note that a person 1 light year above you will accelerate at only 1/2 g to "hold station" a constant distance above you (while you accelerate at 1g). This is related to the "Bell spaceship pardox". |
| Jan26-06, 04:25 PM | #15 |
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It may be a minor point, but the planet is solid, and Alice and Bob live on the surface like on earth. But the matter has been stretched into a cylinder (instead of a sphere). The cylinder isn't necessarily rotating.
Ok, so what I am hearing is that Alice can indeed observe what looks like faster than light travel between the two satellites, as long as she doesn't try to measure it herself. By that I mean that Alice knows Bob's house is 3 million km away by surface travel. Ignoring the fact that you can't have a prefectly straight laser beam over that distance because of gravity (ie they have some kind of relay system that make practically a straight path operating at c), they should be able to send emails directly between the houses in 10 seconds (3x10^9m/3x10^8m.s = 10s). The two satellites are directly above each house, and they can fire a signal directly up to hit the satellite. The satellite can then send a signal to the other and then back down to the other house. I don't know how to calculate the time difference (so I leave out exact satellite distances), but lets just say for argument sake it takes 9 seconds to relay the emails. pt1: From the point of view of Alice and Bob is this not a form of ftl communication? In other words, the fastest they can signal each other on the surface is in 10 seconds with a beam traveling at c. But an alternate path, shorter I guess, only takes 9 seconds. Is this causing any "time travel" problems? pt2: If it is possible to have negative energy, would the argument not be reversed that a cylinder planet made of negative energy could relay a signal faster than the satellites, instead of slower? pt3: Is this a form of ftl communication, and do "time travel" problems occur? variations: Since the length of the cylinder planet is still undefined, it could stretch from here to alpha centari. Also, it needen't really be a planet, but simply a small path of energy/negative energy. The planet analogy is simply to relate to what we already know from GR. |
| Jan26-06, 05:28 PM | #16 |
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| Jan26-06, 06:39 PM | #17 |
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Why make your planet a cylinder? If you want to make life simple, make it a very large "Alderson disk".
[add] I assume you are familiar with that idea from your SF background, if not the wikipedia has an article on it at http://en.wikipedia.org/wiki/Alderson_Disc I discussed the metric for an infinite Alderson disk (i.e. a 2-d plane) in another thread recently - if we have the plane being the x-y plane, the metric is just ds^2 = (1+g|z|)dt^2 - dx^2 - dy^2 - dz^2 Light in such a metric does not follow a straight line path. Furthermore, a lightbeam following the optimum path will cover more distance than it will if constrained to move on the z=0 plane, if that's what your eariler remarks were about. This is not really going "faster than light", though, it's just an observation that the optimum path for light is not a path that's restricted to the z=0 plane. In coordinate terms it might look like FTL, but in local terms it's just that light, following the optimum path, gets there before light following a non-optimum path. There is a very simple way to compute the optimum path. Imagine that, instead of being on a planet with gravity, you are on a big accelerating spaceship with the apparent gravity being due to your acceleration. Then you can adopt an inertial coordinate system, and light travels in straight lines in the inertial coordinate system, while the spaceship accelerates (and thus does not travel in a straight line). The light that will get between two points the fastest is light that travels a straight line in an inertial coordinate system. From the POV of the passenger on the rocket, the light will follow a curved trajectory, a lot like a thrown softaball, as if it were being "bent" by gravity. Light that is forced to co-move with the rocket (z=same z as rocket) will travel a longer distance, and hence take a longer time. You can see this from the fact that this path is "curved" in the inertial frame. Thus it is possible to cover more than 1 light-year of distance on the z=0 plane in 1 year by shooting light upwards at the right angle and letting it follow its "natural" path. All this means is that the real distance (the Lorentz interval) between two points is shorter than the coordinates would make you think. [add] Here's an anology. If you are on the Earth, and you want to take the shortest distance between two points at the same lattitude, you do not follow a curve of constant lattitude. Instead, you follow a "great circle" path. I believe that this problem (trajectory of light in an accelerating metric) has been worked out in detail by another poster BTW - if you're interested, I'll try and look up the details (when I have more time, right now I have to get going). |
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