Can Negative Energy Theoretically Enable Faster-Than-Light Travel in Sci-Fi?

In summary: and time-travel means causality and theorems about the nature of reality. If you change reference frames such that causality would be violated that the negative energy wouldn't necessarily look as negative, lessening the exceedence of the speed of light and allowing causality to be restored. I'm wondering if that even makes sense in the concepts of the paradox, even though negative energy can't exist in the equations.There is no one answer to this. It would depend on the specific scenario and the rules that were in place for the time-travel.
  • #36
Right, so in this case light does not necessarily follow the least-time curse? In other words it ain't that smart.

For the people inside the shaft, since potential is not changing in the hollow space, can they tell the difference between being in the hollow with no acceleration, and being in free space with no acceleration? In other words it acts just like any intertial reference frame in which SR is valid.

I hope you can see where I'm going. SR says that you can't travel faster than light, and we apply this to spaceship inside the hollow of the planet. However, we have clarified that there is a faster path outside of the hollow.

Now the thing that I've been working toward is seeing the equivelence to the effects of the hypothetical cylinder planet, and the effects of a scifi ftl machine. It seems to me that if you use the shortcut outside the hollow of the planet, it has the *same effect* as using an ftl machine. Space is basically flat inside the hollow, correct? If a ship was here it is equivilant to a ship in free space, and is bound by c from SR. He can either use some ftl drive, or take a shortcut through the outside of the planet. Do they not have the same effect?
 
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  • #37
Longstreet said:
I hope you can see where I'm going. SR says that you can't travel faster than light, and we apply this to spaceship inside the hollow of the planet. However, we have clarified that there is a faster path outside of the hollow.

Now the thing that I've been working toward is seeing the equivelence to the effects of the hypothetical cylinder planet, and the effects of a scifi ftl machine. It seems to me that if you use the shortcut outside the hollow of the planet, it has the *same effect* as using an ftl machine. Space is basically flat inside the hollow, correct? If a ship was here it is equivilant to a ship in free space, and is bound by c from SR. He can either use some ftl drive, or take a shortcut through the outside of the planet. Do they not have the same effect?
Well, are you familiar with the idea of "traversable wormholes" which are allowable solutions to the equations of general relativity? In a sense they're similar, in that by going through a wormhole you can get to a distant location faster than a light beam which takes the long route outside of the wormhole, but you never locally exceed the speed of light while inside the wormhole, a beam of light shone through the wormhole would beat you to your destination. However, it's known that traversable wormholes in GR lead to the possibility of "closed timelike curves", paths through spacetime which allow your worldline to curl around and intersect an earlier point on the worldline--time travel into your own past, in other words. I'm sure the cylinder situation wouldn't allow closed timelike curves, so that's an important difference. I would guess that any form of apparent FTL involving a finite region of weirdly curved spacetime that gives you a shortcut through an otherwise flat spacetime would probably lead to the possibility of CTC's, just because of simultaneity issues.

Also, to get back to the title of this thread, it's known that traversable wormholes require negative energy to hold them open, and I think Hawking proved that any solution to GR that allows for CTCs (or maybe just any solution involving CTCs in a finite region, the theorem might not apply to something like Godel's http://www.daviddarling.info/encyclopedia/G/Godel_universe.html idea) requires negative energy.
 
  • #38
Yes I've seen that. I'm talking about "inhouse" ftl; not ripping open the universe. There are no loops in this example. The ultimate goal of this line of questions does actually relate to the origional post. The only difference between the flat space in the hollow of the cylinder and normal free flat space is there are areas of higher potential, and hence shorter time/paths. If negative energy has higher potential than the background, as opposed to lower potential as with gravity, then a path of negative energy should provide a shorter path between two points than free space.
 
  • #39
I happened to think of another question. What are the implications of such a cylinder mass/planet moving parallel to it's length. Obveously my conclusion that the result of ftl in this case for people inside the planet could be wronge. But I haven't figured out why yet. However, time travel proplems and causality problems arise when there is another frame involved, and the conclusion from the principle of relativity that no frame is special says that all ftl would be the same in every frame if there was such a thing (allowing the preceeding problems).

However, any other inertial frame here involves the planet moveing at massive velocities along with the inhabitants. I am wondering if the other frame is even inertial anymore, even in the exact center of the planet, when the planet is moveing past it (something like space-time dragging?). In other words, the two frames are no longer equal.
 
  • #40
Longstreet said:
I happened to think of another question. What are the implications of such a cylinder mass/planet moving parallel to it's length.

We can rephrase the question: what does the "gravity" of the planet look like to an observer moving with respect to it?

The answer in English would include the following effects:

1) The apparent gravity of the planet would increase by a factor of [itex]\frac{1}{1-\beta^2}[/itex], where [itex]\beta[/itex] is the relativistic velocity in geometric units, i.e [itex]\beta = v/c[/itex]

2) There would be a strong "gravitomagnetic field" from the viewpoint of the moving observer. This would cause an apparent change in the "weight" (vertical component of the "gravitational force") of the observer depending on his velocity. This effect would produce a force that apparently decreases the weight of an observer that was co-moving with the planet. There are Christoffel symbols associated with both a linear effect and a quadratic effect with respect to velocity.

3) Gravitomagnetic forces also cause effects that translate into an x-component of force associated with a z-component of velocity (z being "up", x being the "direction of motion of the planet").

4) Some non-linear effects which I won't attempt to describe in English.

The answer in math would look like this:

The metric of an infinite plane gravitational source for z>0 is the rindler metric, which can be expresses as an orthonormal basis of one-forms

[tex]
d\hat{t}=(1+gz)dt: d\hat{x}=dx: d\hat{y}=dy: d\hat{z}=dz
[/tex]

The "gravitational field" of this observer is described by his Christoffel symbols

[tex]
\Gamma^{\hat{z}}{}_{\hat{t}\hat{t}} = -\frac{g}{1+gz}
[/tex]

This is the mathematical representation of the 'z' component of the "gravitational force".

The only other non-zero Christoffel symbol for this metric (two symbols which are equal by symmetry) is not related to any sort of force, because its raised index is time, not space:
[tex]
\Gamma^{\hat{t}}{}_{\hat{z}\hat{t}} = \Gamma^{\hat{t}}{}_{\hat{t}\hat{z}} = -\frac{g}{1+gz}
[/tex]We can "boost" these basis one-forms to ask what the "gravitational field" (Christoffel symbols) for the moving observer are.

Because the metric is locally Minkowskian for the orthonormal basis of one forms (by defintion, this is why the one-forms are "orthonormal") this "boost" is the familiar boost from special relativity, and not anything more complicated.

Thus we have[tex]
d\hat{t}=\frac{(1+gz)dt-\beta dx}{\sqrt{1-\beta^2}}: d\hat{x}=\frac{dx-\beta(1+gz)dt}{\sqrt{1-\beta^2}}:
d\hat{y}=dy:
d\hat{z}=dz
[/tex]

From this we can calculate the Christoffel symbols in the boosted basis for the moving observer:

[tex]
\Gamma^{\hat{z}}{}_{\hat{t}\hat{t}} = -\frac{g}{(1+gz)(1-\beta^2)}
[/tex]

This represents the increased "gravity" the moving observer feels. The conversion to latex is a bit tedious and there are a lot of Christoffel symbols, so I'll just paste the computer output.

[tex]CC = \Gamma: (1) = \hat{t}: (2)=\hat{x}: (3) = \hat(y): (4) = \hat(z)[/tex]

Code:
                (4)                          g
             CC      [ (1)  (1)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )               (4)                          beta g
            CC      [ (2)  (1)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )                (1)                          g
             CC      [ (4)  (1)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )                (2)                        beta g
             CC      [ (4)  (1)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )               (4)                          beta g
            CC      [ (1)  (2)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )                                              2
                (4)                       beta  g
             CC      [ (2)  (2)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )               (1)                          beta g
            CC      [ (4)  (2)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )                                               2
               (2)                         beta  g
            CC      [ (4)  (2)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )                (1)                          g
             CC      [ (1)  (4)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )                (2)                        beta g
             CC      [ (1)  (4)] = ----------------------
                                                       2
                                   (1 + g z) (-1 + beta )               (1)                          beta g
            CC      [ (2)  (4)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )                                               2
               (2)                         beta  g
            CC      [ (2)  (4)] = - ----------------------
                                                        2
                                    (1 + g z) (-1 + beta )

The equations of motion with respect to the local "frame" coordiantes [itex] q^0 = \hat{t}: q^1 = \hat{x} : q^2 = \hat{y} : q^3 = \hat{z}[/itex] are given by the geodesic equations.

[tex]
\frac{d^2 q^i}{d\tau^2} + \Gamma^i{}_{jk} \frac{d q^j}{d\tau} \frac{d q^k}{d\tau} = 0
[/tex]

Summation over the repeated indicies j and k is implied: the above expresion gives 4 equations, one for each value of i.

It's instructive to take the Newtonian limit of the above equations: then dt/dtau = 1, and dx/dtau = vx, dy/dtau = vy, etc. This makes the interpretation of the Christoffel symbols in terms of "forces" somewhat intuitive by comparing the equations of motion to the Newtonian equations of motion.

We then get

[tex]
\ddot{\hat{z}} - g' (1 - 2 \beta \, \hat{vx} + \beta^2 \, \hat{vx}^2) = 0
[/tex]
[tex]
\ddot{\hat{y}} = 0
[/tex]
[tex]
\ddot{\hat{x}} - g' (2\beta \, \hat{vz} + 2 \beta^2 \, \hat{vx} \, \hat{vz}) = 0
[/tex]

Here [tex]g' = \frac{g}{(1+gz)(1-\beta^2)}[/tex]

Note that the [itex]\hat{vx}^2[/itex] term in the first equation will be negligible and can be dropped (in the Newtonian approximation) as can the [itex] \hat{vx}\hat{vz}[/itex] term in the second equation, because [itex]\hat{vx}[/itex] << 1 and [itex]\hat{vz}[/itex] << 1.

Thus, in the low velocity realm, we will see an increase in weight of an object (change in g value), along with the gravitomagnetic effects that cause moving objects in the x direction to experience acceleration in the z direction proportional to their x-velocity, and a similar effect that causes motion the z direction to cause acceleration in the x-direction.

Note by "low velocity realm" I allow [itex]\beta[/itex] to be large, it is only vx, vy, and vz that must be small.
 
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  • #41
Longstreet said:
Yes I've seen that. I'm talking about "inhouse" ftl; not ripping open the universe. There are no loops in this example. The ultimate goal of this line of questions does actually relate to the origional post. The only difference between the flat space in the hollow of the cylinder and normal free flat space is there are areas of higher potential, and hence shorter time/paths. If negative energy has higher potential than the background, as opposed to lower potential as with gravity, then a path of negative energy should provide a shorter path between two points than free space.
When you say "a path of negative energy" here, are you suggesting something like a cylinder composed of negative energy with a path through it? I think I remember pervect did say in some previous thread that you would experience "reverse time dilation" near a large clump of negative energy, ie instead of your clock running slower than that of distant observers, it would run faster...so is your idea that a light-clock at the center of such a cylinder would run faster, therefore some sort of "FTL" would be possible by traveling through it? Like if we positioned one end of such a cylinder near the sun and another near alpha centauri, light could get from one end to the other in less than 4 years? This is an interesting idea, I'd like to hear pervect's comments on it...

Also, in your most recent post #39, were you still talking about an ordinary cylinder or were you talking about a negative-energy cylinder?
 
  • #42
pervect, I am interested now inside the cylinder since you said time is slow there as well. My thinking being that it gets close to an inertial frame (free space) if the mass isn't having a net acceleration on the observers. Here is a picture to illustrate. However, to equate the causality problems with ftl in special relativity I'd have to assume it is still such an inertial frame if the observer is moving inside. I had figured that the mass density and total energy of the planet would apear to increase, causing the increase in weight like you mention. However, that effect wouldn't seem impact people who feel weightless to begin with inside the cylinder. The gravito-"magnet" effect I am not familiar with at all.

Yes. That is the idea JesseM. I am using the cylinder as a test case of seeing what the effect would be, although not exactly usefull in itself. My post #39 is a massive (positive mass) cylinder.
 

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  • #43
I'm going to have to think more to figure out what happens inside the planet. So take the following to refer to what happens outside the planet.

Imagine a moving charge in a magnetic field. The charge experiences a Lorentz force F = q x B.

Suppose the charge is moving in the x direction, and the magnetic field is in the y direction. Then the charge will be deflected in the 'z' direction by a magnetic force.

Gravity acts approximately the same way with weak fields, but there are various differences in the numerical factors.

There is some more detail in the wikipedia article:
http://en.wikipedia.org/wiki/Gravitoelectromagnetism

Note that some people in fringe physics have attempted to hijack the term :-(. I'll quote the relevant section of the Wikipedia on this directly, in case this valuable section gets edited out in some future version of Wiki:

Despite the electromagnetism in gravitoelectromagnetism, and despite the similarity of the GEM force law to the Lorentz force law, gravitomagnetism should not be confused with any of the following:

* claims to have constructed anti-gravity devices,
* Eugene Podkletnov's claims to have constructed gravity-shielding devices and gravitational reflection beams,
* the so-called Electric Universe "theory", which claims to identify gravity as a form of electromagnetism.

These and other claims are considered pseudophysics by mainstream science.

Gravitoelectromagnetism, on the other hand, is firmly part of our gold standard theory of gravitation, general relativity, has testable predictions, and is in the final stages of being directly tested.
Moving on to our example - let 'z' be up, and 'x' be the direction of motion of the observer with respect to the planet.

We see from the detailed analysis in terms of the Christoffel symbols that we have a gravitomagnetic force that causes particles moving in the 'x' direction to experience a 'z' component of "force" (or acceleration with respect to our coordinate system), and it causes particles moving in the 'z' direction to experieince an 'x' component of force.

Thus it is essentially a gravitomagnetic force in the 'y' direction. (I haven't worried about sign, except to note that particles co-moving with respect to the planet appear lighter).

People on the planet itself see no gravitomagnetic force, just as someone near a stationary charge sees no magnetic field.

People moving with respect to the planet see a gravitomagnetic force, just like somone near a moving charge will see a magnetic field.
 
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  • #44
Just an interestsing note:

[tex]
d\hat{t}=(1+gz)dt: d\hat{x}=dx: d\hat{y}=dy: d\hat{z}=dz
[/tex]

and the associated line element

ds^2 = -(1+gz)^2 dt^2 + dx^2 + dy^2 + dz^2

is one of the more convenient forms of the metric for an accelerated observer, but it's not unique. I recently ran across

[tex]
d\hat{t}=e^{az} dt : d\hat{x} = dx: d\hat{y} = y : d\hat{z} = e^{az} dz
[/tex]

with the associated line element

ds^2 = -e^{2az} dt^2 + dx^2 + dy^2 + e^{2az} dz^2

This is the same physical metric in a differing coordinate system. Computation of the Christoffel symbols (in the ONB, not the coordinate version) yields the same results as one would expect with

[tex](1 + g z1) = e^{a z2}[/tex]

with one interesting exception

[tex]\Gamma^\hat{z}{}_{\hat{z}\hat{z}}[/tex] is not the same. This purely spatial Christoffel symbol is thus an artifact of the coordinate system being used, and has no physical significance.
 
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