Debunking the Existence and Duration of Virtual Particles

[kith]

I'm not sure I understand your question. But, with interacting fields you'll still get creation and annihilation operators for those fields, which will correspond to real particles.

But are the corresponding states still eigenstates? I picture it to be somehow like this: I have the Dirac field for electrons and the electromagnetic field for photons. I have free Hamiltonians $H_D$ and $H_{EM}$ with eigenstates corresponding to certain numbers of electrons and photons. My complete Hamiltonian reads $H_D + H_{EM} + H_{int}$. Are the eigenstates of the free Hamiltonians still eigenstates of the complete Hamiltonian or where do I get the "old" real electrons and photons when I consider the complete system?

I found this a great next step:
http://www.damtp.cam.ac.uk/user/tong/qft.html

I'm thinking about working through Schweber or Weinberg, because they draw more connections to the familiar non-relativistic QM I already know. For example, I want to read in deatail about second quantization. /edit: I've just read Tong's nice part about "recovering quantum mechanics". I think, I'll have some use for this text, thank you!

 

[krickea]

On the contrary one could argue that nonvirtual particles do not exist. Every particle is virtual since it is always en route from one interaction to the next.

That's seems good.

A Photon is emitted high up in the atmosphere, an Auroral photon, and travels to
an observers Eye. Is it a virtual photon? How will a good
Quatum description of the whole procedure look like?
Krickea

 

[tiny-tim]

this discussion about whether eg the photons we see are real particles or virtual particles is missing the whole point of this thread …

the photons we see certainly exist in the maths …

the maths describes them as existing in particular numbers, as being at (or very near) a particular position at a particular time, and as having creation and annihilation operators

if you want to argue that those photons must be slightly off-shell, therefore they are off-shell particles, fine

but going further and say that if they're off-shell particles they must be virtual particles is redefining "virtual particle" to include two entirely different things …

i] off-shell particles with particular numbers positions and operators (see above)

ii] off-shell things which the maths does not even purport to describe as having particular numbers positions and operators

proving that something normally regarded as real actually has all the attributes of reality except one does not even begin to prove that something with none of those attributes should be regarded as real!

(or indeed as existing in any sense)​

 

[muppet]

Thanks for your comment, muppet.

Yet, I still don't understand why real particles are favoured over virtual ones. I try to summarize my current understanding.

Real particles are plane wave initial states. Interactions are described by the S-Matrix which produces plane wave final states again. Griffiths argues, that these idealized states do exist only approximatively, because every photon detected in an experiment has to have been emitted somewhere a finite time ago. So in this view, the long lived real photons mediate the interaction between far away objects (like emitter and detector) and virtual ones between near objects (like particle-particle scattering); there's no fundamental difference between real and virtual particles. What's wrong with this viewpoint?

Where do real particles arise in non-perturbative QFT if I include interactions (which are necessary for emission+absorption processes)? Maybe someone can sketch the typical approach and/or provide me a summary article or something like that because I'm not familiar with non-perturbative QFT at all.

[I hope there are enough actual questions in this post. ;)]

Nothing's really wrong with that viewpoint. It may be helpful to distinguish between two distinct questions here:

1)Why are the particles that hang around in the real world (to which we usually suppose that the external lines in Feynman diagrams correspond) always on-shell?
2)If a "real" particle leaves one scattering event and flies off to participate in another one, it may be regarded as mediating an interaction between the other particles involved in the two scattering processes; doesn't this blur the distinction between real and virtual particles?


My above comment about cluster decomposition and all that was intended as an answer to 2), although it was perhaps both too indirect and too technical; the point was that QFT automatically imposes that any 'virtual' particle that hangs around for any length of time will appear to be on-shell.

As for 1) remember that in the S-matrix approach the asymptotic 'in' and 'out' states, eigenstates of the free Hamiltonian, coincide with the eigenstates of the real Hamiltonian in the limit of plus or minus infinite time. What this means (from the mathematical definition of a limit!) is that for any finite experimental resolution there exists a time T such that for all times later than T (for the infinite future, or before -T, for the infinite past) the asymptotic states are indistinguishable from the real eigenstates by any experiment. What makes the S-matrix work is that this time is really really short. I'd strongly recommend the book by John R. Taylor on nonrelativistic scattering theory for a good discussion of this point. For a bit more on why this might be, keep reading.

I can picture this in the free case. But if I have two interacting fields, can I still define meaningful creation/annihilation operators for the states of one field alone? Are the eigenstates of the free Hamiltonians still eigenstates of the complete Hamiltonian? So do in fact the "real" particles emerge from such a consideration? From what I know about QM I would guess no.

[Btw: I agree with your view on Griffiths' book. I really like it and I learned a lot for practical purposes, but still have many fundamental questions. But to answer these, I need to dig into much heavier stuff, it seems. ;)]

But are the corresponding states still eigenstates? I picture it to be somehow like this: I have the Dirac field for electrons and the electromagnetic field for photons. I have free Hamiltonians $H_D$ and $H_{EM}$ with eigenstates corresponding to certain numbers of electrons and photons. My complete Hamiltonian reads $H_D + H_{EM} + H_{int}$. Are the eigenstates of the free Hamiltonians still eigenstates of the complete Hamiltonian or where do I get the "old" real electrons and photons when I consider the complete system?


I'm thinking about working through Schweber or Weinberg, because they draw more connections to the familiar non-relativistic QM I already know. For example, I want to read in deatail about second quantization. /edit: I've just read Tong's nice part about "recovering quantum mechanics". I think, I'll have some use for this text, thank you!

As you're only just beginning to learn field theory, and I'm only just beginning to understand some of it (after about two years of trying to...) I'm not sure how intelligible what I'm about to say will be, but I'll give it a go. Hopefully what follows will also start to address some of Lapidus' questions.

The only exactly solvable QFTs of which I know that have anything at all to do with the real world are free theories. (I think there are mathematically interesting results in certain special theories with loads of symmetry, but don't know of any that relate to known particle physics.) We describe interacting theories by perturbing around free theories. This turns out to change both everything and nothing. For example, one can calculate the mass and the coupling in QFT as functions of the parameters m and lambda that you write down in the Lagrangian; just adding interactions means that the parameter "m" is no longer the mass of the particles! And no, the eigenstates of the interacting theory are not those of the free theory. However, you can argue that the spectra of the free and interacting theories (i.e. the set of energy eigenvalues, and hence particle masses) should be the same, so long as you ignore bound states (for more on this, search for a thread 'isometric operators- spectrum preserving?' or something like that). So you know what some of the answers should be from experiments, and hence you can relate the physical mass to the parameter m (this is the business of 'renormalisation', that you may have heard of).

Non-perturbative results in QFT are few and far between, and I'm afraid I don't know a great deal about them. One can (formally) construct, from the classical action that describes your theory, the so-called effective action- which is basically a description of the system that takes all quantum fluctuations into account, right from the outset. In practice, however, one can only usually compute an approximation to it in some power of hbar. (The good news, incidentally, is that the leading order term in this expansion is just the classical action, from which we get the ordinary equations of motion.)

As an aside, Weinberg's book says that "the expression 'second quantization' is misleading, and it would be a good thing if it were retired permanently".

 

[kith]

but going further and say that if they're off-shell particles they must be virtual particles is redefining "virtual particle" to include two entirely different things

Griffiths doesn't regard them as virtual particles because they are off-shell, but because they appear as internal line in the Feynman diagram describing the combined process emission+absorption! Isn't this the common definition of "virtual particle"?

Proving that something normally regarded as real actually has all the attributes of reality except one does not even begin to prove that something with none of those attributes should be regarded as real!

I didn't attempt to do that. [Actually, I'm not claiming things here at all (since I have only basic knowledge in QFT), but I'm asking questions and seek wisdom. ;)]

My initial question was how to distinguish between real photons and virtual ones. The answer has been, that real particles appear in the eigenstates of free fields. So to me, in a description including interactions, real particles seem to be an approximation for long-lived virtual particles and can be mathematically described in a way, which is not possible for short-lived ones, because there the approximation doesn't hold.

So my current understanding is this: If I use non-perturbative QFT, I get only 'real' particles but I can't describe interactions and therefore experiments. To describe them, I use perturbation theory (Feynman diagrams) from where I also get 'virtual' particles. They are hard to distinguish from 'real' ones because of the cluster decomposition principle. The exact description of particles does neither correspond to 'real' particles nor to 'virtual' ones. 'Real' particles are distinguished due to the fact, that this approximation is valid at almost all times.

Is this correct?

 

[kith]

Thanks for your extensive comment, muppet.

So you seem to agree, that there is no clear distinction between real and virtual particles.

I don't really get the bottom line from the second part of your text. Maybe, I'll come back to it, if I have some more basic knowledge. I'll put Taylor on my list, but a 500-pages on nonrelativistic scattering theory alone, may exceed my reading capabilities. ;)

 

[muppet]

Thanks for your extensive comment, muppet.

So you seem to agree, that there is no clear distinction between real and virtual particles.

I don't really get the bottom line from the second part of your text. Maybe, I'll come back to it, if I have some more basic knowledge. I'll put Taylor on my list, but a 500-pages on nonrelativistic scattering theory alone, may exceed my reading capabilities. ;)

You mean the part about the leading order term in an expansion of the effective action in powers of hbar? You'll definitely come across that eventually. Read it as "quantum field theory reproduces classical field theory when Planck's constant goes to zero"

Happily, the discussion in Taylor I'm referring to is in chapter two; chapter one is mathematical preliminaries that I skipped and referred back to as and when needed. The full book is too much for my reading capabilities too- I picked it up as the treatment of scattering in the quantum field theory text by Peskin and Schroeder is based on 3 chapters of that book.

I suppose the best way to summarise my opinion on virtual particles is by analogy with ordinary NRQM.
If you have a simple harmonic oscillator potential, you expand your state in oscillators of different frequencies, and you'll be working with states that correspond exactly to the eigenstates of the system.
If you have a system "close to" a SHO, you can perturb around your idealised SHO; it's a good calculational scheme, and it gives you an intuition for the physics, but you're no longer talking about an exact correspondence.
If you have an arbitrary potential, you can still expand states in the SHO basis, as it's a perfectly good basis. But the calculations will be intractable and you'll have no understanding of the physical picture of the situation at all.

Free particles are the idealisation, analogous to the SHO. The virtual particles exchanged in scattering processes are exactly like perturbations around the SHO; they're an intutive way of understanding the result of poking a system you understand well, but you shouldn't do any rigorous philosophy treating them as entities in themselves. The confusion arises because what we think of as "states" and "potentials" in QFT are created using the same operators. But the "particle" states behave differently in the presence of interactions to our conception of what a particle should be, as they're no longer eigenstates of the Hamiltonian that actually describes the system. Any circumstance in which a state created by the action of a field behaves like we imagine a real particle should will be one in which we can neglect the effect of interactions upon the time evolution of that state. And situations exist where perturbation theory breaks down and the particle picture conveys almost no useful information, because we're no longer basing our approximation on a picture that describes the physics well.

As a totally unrelated aside, proofreading this post reminded me of an observation one of the permanent staff here in my university recently made to me:

I like the way you speak, in these closed sentences like something out of 'Yes Minister' that somehow eventually manage to end.

(For those unfamiliar with the reference: )
I hope it's not too verbose to be of some help.

 

[byron178]

The opening query of this post is the claim that virtual particles 'flat-out don't exist'. Hawking is a highly imaginative theorist, but his exploitation of flat-out non-existent particles is hardly comparable to that of exploiting the aether, or phlogiston.

The non-observation, incidentally, is not a falsification of Hawking evaporation proper. We need a positive observation of a radiation-signature of some process antithetical to Hawking evaporation: if x is happening, y cannot be happening.
_______

On, the other hand, I may have the cart before the horse: Hawking radiation would be as much evidence of virtual particles as Hawking BH-evaporation.

So its safe to say virtual particles don't exist in physical reality?also wouldn't they violate causality since they can travel faster than light?

 

[byron178]

I know virtual particles travel faster than light,and faster than light travel has to also deal with time travel backwards in time because from what i understand if a particle were to travel faster than light it will in one frame travel to the past,now do virtual particles travel backwards to the past from them traveling faster than light?

 

[Vanadium 50]

I know virtual particles travel faster than light

Well, they don't.

"It AIN'T so much the things we don't know that get us into trouble. It's the things we know that just ain't so."

 

[tiny-tim]

I know virtual particles travel faster than light …

no, https://www.physicsforums.com/library.php?do=view_item&itemid=287" in the coordinate-space representation (of a Feynman diagram) are on-mass-shell, so virtual photons travel at the speed of light, while virtual electrons travel at any and every speed slower than light

however, in the momentum-space representation, the virtual particles are both on- and off-mass-shell: both virtual photons and virtual electrons travel at all possible speeds

(of course, this is all just maths … virtual particles aren't real … the clue's in the name! )

and faster than light travel has to also deal with time travel backwards in time because from what i understand if a particle were to travel faster than light it will in one frame travel to the past

no, traveling is always forward in time

but a faster than light object traveling forward in time from A to B in one frame may be traveling forward in time from B to A in another frame

now do virtual particles travel backwards to the past from them traveling faster than light?

virtual particles don't actually exist, they don't have an A or B to travel between

i suspect you're thinking of the rule that an anti-particle (real or virtual) traveling forward in time can be thought of as a particle traveling backward in time

 

[ThomasT]

... a faster than light object traveling forward in time from A to B in one frame may be traveling forward in time from B to A in another frame

So whether an object is moving from A to B or from B to A depends solely on the observational frame of reference? I don't think so.

 

[yoron]

On the contrary one could argue that nonvirtual particles do not exist. Every particle is virtual since it is always en route from one interaction to the next.

Beautifully put. We put a lot of trust in 'propagation', don't we? And sometimes even seem to ignore indeterminacy (HUP), as being a principle of its own, instead referring to 'virtual particles' as if they were limited by 'time' to be unmeasurable. I used to like the idea of 'virtual particles' but?

I think I like indeterminacy more.

 

[tiny-tim]

So whether an object is moving from A to B or from B to A depends solely on the observational frame of reference? I don't think so.

think again! …

if we regard something as moving north at speed v faster than light, then an observer moving at speed slower than light but faster than c²/v will regard the same thing as moving south at a speed faster than light …

do the maths, and you'll agree ​

 

[byron178]

no, https://www.physicsforums.com/library.php?do=view_item&itemid=287" in the coordinate-space representation (of a Feynman diagram) are on-mass-shell, so virtual photons travel at the speed of light, while virtual electrons travel at any and every speed slower than light

however, in the momentum-space representation, the virtual particles are both on- and off-mass-shell: both virtual photons and virtual electrons travel at all possible speeds

(of course, this is all just maths … virtual particles aren't real … the clue's in the name! )no, traveling is always forward in time

but a faster than light object traveling forward in time from A to B in one frame may be traveling forward in time from B to A in another framevirtual particles don't actually exist, they don't have an A or B to travel between

i suspect you're thinking of the rule that an anti-particle (real or virtual) traveling forward in time can be thought of as a particle traveling backward in time

Doesn't relativity say that if matter or a particle were to travel faster than light it will in one frame of reference travel backwards in time? i also found this article maybe i misread it that says virtual particles travel faster than light. http://www.desy.de/user/projects/Physics/Quantum/virtual_particles.html

 

[tiny-tim]

Doesn't relativity say that if matter or a particle were to travel faster than light it will in one frame of reference travel backwards in time?

i don't really understand the concept of traveling backwards in time …

if an observer sees something moving, he sees it moving forward in time …

that's the way we see things …

why would an observer think that anything was traveling backward in time?

but yes, one observer may see an object traveling forward in time faster than light from A to B while another observer may see it traveling forward in time from B to A

(in other words: causality is not invariant )​

i also found this article maybe i misread it that says virtual particles travel faster than light. http://www.desy.de/user/projects/Physics/Quantum/virtual_particles.html

this usenet physics FAQ on https://www.physicsforums.com/library.php?do=view_item&itemid=287" is very misleading (it doesn't even consider whether they are simply a mathematical device) …

even the introductory section "What are virtual particles?" fails to answer its own question: after an initially promising overview, it (appropriately!) creates virtual particles out of nowhere without defining or describing them

as to faster-than-light, it says …

… the virtual photon's plane wave is seemingly created everywhere in space at once, and destroyed all at once. Therefore, the interaction can happen no matter how far the interacting particles are from each other.​

… from which it somehow gets to …

… the virtual photon can go from one interacting particle to the other faster than light!​

i'm not certain i understand what the first section means, and i am certain that the second section doesn't follow from it

as meoremuk says …

In my opinion, virtual particles are "made up" concepts.… Usually, the S-matrix is calculated from the Hamiltonian by using perturbation theory. These calculations involve a large number of rather complicated integrals.

In 1949 Feynman invented an ingenious technique of representing these integrals by diagrams. Each line and vertex in the diagram corresponded to a certain factor in the integrand. This technique enormously simplified manipulations with integrals of the perturbation theory.

The diagrams looked so nice that many people (including Feynman) started to use them to "explain" in layman terms what occurs in scattering events. Lines were interpreted as virtual particles that "move" between vertices, etc. etc. These explanations became so "sticky" that many people now believe that at large magnification a collision of electrons really looks like a web of virtual particles jumping back and forth. In my opinion, these beliefs have nothing to do with reality.

 

[sheaf]

In case it hasn't been posted before, http://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/" is quite informative on the subject of virtual particles.

 

[byron178]

i don't really understand the concept of traveling backwards in time …

if an observer sees something moving, he sees it moving forward in time …

that's the way we see things …

why would an observer think that anything was traveling backward in time?

but yes, one observer may see an object traveling forward in time faster than light from A to B while another observer may see it traveling forward in time from B to A

(in other words: causality is not invariant )​

this usenet physics FAQ on https://www.physicsforums.com/library.php?do=view_item&itemid=287" is very misleading (it doesn't even consider whether they are simply a mathematical device) …

even the introductory section "What are virtual particles?" fails to answer its own question: after an initially promising overview, it (appropriately!) creates virtual particles out of nowhere without defining or describing them

as to faster-than-light, it says …

… the virtual photon's plane wave is seemingly created everywhere in space at once, and destroyed all at once. Therefore, the interaction can happen no matter how far the interacting particles are from each other.​

… from which it somehow gets to …

… the virtual photon can go from one interacting particle to the other faster than light!​

i'm not certain i understand what the first section means, and i am certain that the second section doesn't follow from it

as meoremuk says …

So virtual particles travel faster than light but don't exist?

 

[Vanadium 50]

Byron, a number of people have noticed you keep posting the same question over and over and over and over and over...

Do you not understand the replies? If not, it would help us to know what you don't understand, or at least what your background in physics is.

 

[Constantin]

think again! …

if we regard something as moving north at speed v faster than light, then an observer moving at speed slower than light but faster than c²/v will regard the same thing as moving south at a speed faster than light …

do the maths, and you'll agree ​

I've been trying really hard to imagine this, but it seems I'm running out of brainpower.
I'm sure the maths are right, but can you help me imagine how this works ?

And if one observer sees the object moving north and another sees it moving south, then there should exist a third observer who sees the object as stationary, right ? But I still can't imagine it.

 

[byron178]

Byron, a number of people have noticed you keep posting the same question over and over and over and over and over...

Do you not understand the replies? If not, it would help us to know what you don't understand, or at least what your background in physics is.

I don't understand how virtual particles don't travel backwards in time when they travel faster than light.

 

[tiny-tim]

I'm sure the maths are right, but can you help me imagine how this works ?

See http://en.wikipedia.org/wiki/Velocity-addition_formula"

And if one observer sees the object moving north and another sees it moving south, then there should exist a third observer who sees the object as stationary, right ?

No, to see it as stationary, an observer would need to have the same velocity, v, ie also faster than light, wouldn't he?

But there would exist a third observer (with speed c²/v, of course, slower than light) who sees the object as moving infinitely fast!

So virtual particles travel faster than light …

byron, that's not true … read my https://www.physicsforums.com/showpost.php?p=3584038&postcount=256" again

 

[juanrga]

ive been reading on this forum that virtual particles flat out don't exist?then why is it said they exist for a certain amount of time?

Virtual particles are not detected as rest of particles. However, virtual particles appear in the equations of QFT and some people speak about their existence in this narrow sense.

In rigor, QFT does not say that «they exist for a certain amount of time», because «spacetime» in QFT has only a formal meaning.

 

[DevilsAvocado]

think again! …

if we regard something as moving north at speed v faster than light, then an observer moving at speed slower than light but faster than c²/v will regard the same thing as moving south at a speed faster than light …

This is probably one of the most astonishing claims I’ve seen in a while... are you sure this is not a 'mix-up' with RoS...??

do the maths, and you'll agree ​

Eh well, maybe, but let’s skip the advanced math for awhile, and use our common sense and pictures to begin with:

• We stand on the South Pole together with Alice and her brand new superluminal spaceship.
  
  
• Alice has decided to set a new speed record, traveling to the North Pole.
  
  
• Bob is flying his old "c²/v wreck" to inspect the event at the equator.
  
  
• Alice takes off at superluminal speed in direction towards the North Pole.
  
  
• Bob is dropping his jaw, because according to you; he will see Alice taking off at the North Pole, in direction towards the South Pole!

Now the question I have for you, my friend: – If Bob now travels with his old wreck to the South Pole to talk to Alice about what went wrong, where is Alice actually located, the North Pole or the South Pole?

...

 

[Fredrik]

This is probably one of the most astonishing claims I’ve seen in a while... are you sure this is not a 'mix-up' with RoS...??
...
Eh well, maybe, but let’s skip the advanced math for awhile, and use our common sense and pictures to begin with:

I'll use pictures, but I won't use common sense. Think about it in terms of spacetime diagrams. Let's make north the positive x direction. Imagine a tachyon gun that's fired at event A with x coordinate 0. The beam moving at speed v>1 hits the target at event B with x coordinate L>0. Now imagine the simultaneity lines of an observer moving at speed u in the positive x direction. (It has to be in the positive x direction). The larger the u, the more the simultaneity lines get "tilted" toward the line x=t representing light speed. The slope of the simultaneity lines can get arbitrarily close to the slope of the line x=t. So clearly, if u is large enough, there will be a simultaneity line that is "below" A and "above" B, meaning that in that guy's (comoving inertial) coordinate system, B is the earlier event.

Since the time axis is drawn in the "up" direction in a spacetime diagram, the world line of an object with speed u have slope dt/dx=1/u. If that object is the spaceship containing the observer, then its speed is <1 and its simultaneity lines make the same angle with the x-axis as its world line makes with the t axis. So the slope of the simultaneity lines is 1/u.

• Bob is dropping his jaw, because according to you; he will see Alice taking off at the North Pole, in direction towards the South Pole!

Now the question I have for you, my friend: – If Bob now travels with his old wreck to the South Pole to talk to Alice about what went wrong, where is Alice actually located, the North Pole or the South Pole?

That's an interesting question. The simplest possibility is that Alice doesn't exist anymore. In the south pole's comoving inertial frame, Alice (the tachyon) is created at A and destroyed at B. In Bob's (original) comoving inertial frame, Alice is created at B and destroyed at A. So A is an emission event in some coordinate systems and an absorption event in some coordinate systems. The same goes for B of course. When Bob has landed at the south pole, there is no Alice, at no point in "space" as defined by his new comoving inertial system, which is of course the one comoving with the south pole.

Suppose that the people at the south pole create Alice at an earlier time, and have "her" bounce back and forth between two tachyon mirrors until one of the mirrors is removed at event A. Suppose also that the people at the north pole are equipped to "catch" Alice at event B, and then have her bounce back and forth between two tachyon mirrors. In that case, the answer to your question must be that Alice is at the north pole.

On the other hand, if that scenario is possible, then it must also be possible to set things up so that from the point of view of Bob's original comoving frame, Alice is kept bouncing back and forth between tachyon mirrors at the north pole for some time before event B, where she is released, and later caught at event A, where she ends up bouncing back and forth between tachyon mirrors at the south pole. If this is how things were set up, then the answer to your question must be that Alice is at the south pole.

 

[DevilsAvocado]

I'll use pictures, but I won't use common sense.

Phew, that’s a big relief!

()

Think about it in terms of spacetime diagrams. Let's make north the positive x direction. Imagine a tachyon gun that's fired at event A with x coordinate 0. The beam moving at speed v>1 hits the target at event B with x coordinate L>0. Now imagine the simultaneity lines of an observer moving at speed u in the positive x direction. (It has to be in the positive x direction). The larger the u, the more the simultaneity lines get "tilted" toward the line x=t representing light speed. The slope of the simultaneity lines can get arbitrarily close to the slope of the line x=t. So clearly, if u is large enough, there will be a simultaneity line that is "below" A and "above" B, meaning that in that guy's (comoving inertial) coordinate system, B is the earlier event.

Since the time axis is drawn in the "up" direction in a spacetime diagram, the world line of an object with speed u have slope dt/dx=1/u. If that object is the spaceship containing the observer, then its speed is <1 and its simultaneity lines make the same angle with the x-axis as its world line makes with the t axis. So the slope of the simultaneity lines is 1/u.

Phew²... I think the 'magic' word here is tachyon, a hypothetical subatomic particle... I have absolutely no problem with that, and we can easily use 'common sense' to understand that there is nothing strange at all seeing a superluminal tachyon arriving at the end point, before we see it leaving the starting point. A no-brainer!

Also we can thank "the old one" for that macroscopic objects like Alice and her new spaceship is not allowed to travel faster than light, according to Einstein... Bob would have to spend all his free time at the dentist, repairing his jaw...

If I may become a little bit 'picky'; I would like to add that we would probably be able to distinguish the "real Alice" from her "mirage-redshifted-image", and by this we would be able to conclude that she is actually moving in the direction from the South Pole to the North Pole...

http://en.wikipedia.org/wiki/Tachyon"

Because a http://en.wikipedia.org/wiki/Tachyon" always moves faster than light, we cannot see it approaching. After a tachyon has passed nearby, we would be able to see two images of it, appearing and departing in opposite directions. The black line is the shock wave of Cherenkov radiation, shown only in one moment of time. This double image effect is most prominent for an observer located directly in the path of a superluminal object (in this example a sphere, shown in grey). The right hand bluish shape is the image formed by the blue-doppler shifted light arriving at the observer—who is located at the apex of the black Cherenkov lines—from the sphere as it approaches. The left-hand reddish image is formed from redshifted light that leaves the sphere after it passes the observer. Because the object arrives before the light, the observer sees nothing until the sphere starts to pass the observer, after which the image-as-seen-by-the-observer splits into two—one of the arriving sphere (to the right) and one of the departing sphere (to the left).

That's an interesting question. The simplest possibility is that Alice doesn't exist anymore.

I hear you; my brain has visited this "place" several times, reading thru this thread...

The serious (and perhaps simplest) answer is that Alice would cease to exist long before going > c. She have to accumulate infinite energy and would in this process acquire infinite mass = squashed in the heaviest 'black hole' of all times...

the answer to your question must be that Alice is at the north pole.
...
the answer to your question must be that Alice is at the south pole.

Now we’re talking! Thanks buddy! :grumpy:

()


P.S. If 'anyone' wishes to ask the question; "Does this mean that we can travel backwards in time?" the answer is as always NO.

 

[tiny-tim]

… let’s skip the advanced math for awhile…

it's not advanced, it's very elementary math … see the link

…, and use our common sense …

ok, and your common sense should tell you that Alice cannot go faster than light …

any argument based on the "common sense" concept of tachyonic Alice ageing is founded on nonsense, not common sense

(how would you apply the same "common sense" argument to a tachyonic particle without a concept of ageing? )

 

[DevilsAvocado]

it's not advanced, it's very elementary math … see the link


ok, and your common sense should tell you that Alice cannot go faster than light …

any argument based on the "common sense" concept of tachyonic Alice ageing is founded on nonsense, not common sense

(how would you apply the same "common sense" argument to a tachyonic particle without a concept of ageing? )

With all due respect, I think you missed the main point:

where is Alice actually located, the North Pole or the South Pole?

Please feel free to exchange "Alice" to anything that pleases you; "object", "something", tachyon, etc. The important question here is; how one "FTL object" can be detected/observed at two different end points, depending on the frame of reference?

I have never heard anything like it, and I would appreciate it very much if you could (at least in principle) describe how this works (i.e. more than "do the maths").

If you still refers to "do the maths" and this http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity", it looks to me that you may not have the complete answer for this very simple question.

(Don’t worry about the conversation between me & Fredrik, there are some 'silly jokes' there, just for fun nothing more.)

 

[tiny-tim]

… where is Alice actually located, the North Pole or the South Pole?

Please feel free to exchange "Alice" to anything that pleases you; "object", "something", tachyon, etc. The important question here is; how one "FTL object" can be detected/observed at two different end points, depending on the frame of reference?

i'm sorry, i don't understand what the question is, nor what the difficulty is …

can you spell it out, please? ​

 

[tiny-tim]

hmm … how about i ask you a question instead …

i shine a laser beam at the the moon, so that the spot travels across the moon's surface faster than light (that's very easy! ) …

i say it travels from A to B faster than light, another observer says it travels from B to A faster than light …

where's the difficulty? ​

 

[DevilsAvocado]

i'm sorry, i don't understand what the question is, nor what the difficulty is …

can you spell it out, please? ​

Okay, your own words, my bolding:

if we regard something as moving north at speed v faster than light, then an observer moving at speed slower than light but faster than c²/v will regard the same thing as moving south at a speed faster than light …

Problem: I’m aware of Relativity of Simultaneity, but I have never heard of "Relativity of Direction".

Question: If "something" is moving north, let’s say from the South Pole to the North Pole, and another observer is experiencing this "something" as if moving south, i.e. from the North Pole to the South Pole. Where will this "something" finally end up, the South Pole or the North Pole?

Or to keep it simple: One object can’t be in two places at once.

 

[Constantin]

Problem: I’m aware of Relativity of Simultaneity, but I have never heard of "Relativity of Direction".

If you start moving faster than a certain object, it will appear reversing direction.
I'm not saying it's about moving faster than Alice in this case though, just that your problem is very simple.

One object can’t be in two places at once.

"Two places at once" is wrong. It will be in two different places for two different observers. Or for the same observer after he changes his motion. That's normal and very simple as well.

 

[tiny-tim]

Question: If "something" is moving north, let’s say from the South Pole to the North Pole, and another observer is experiencing this "something" as if moving south, i.e. from the North Pole to the South Pole. Where will this "something" finally end up, the South Pole or the North Pole?

yes!

one observer says it ends up at the South Pole, another observer says it ends up at the North Pole …

where's the difficulty? ​

Or to keep it simple: One object can’t be in two places at once.

nobody's saying it is

 

[DevilsAvocado]

one observer says it ends up at the South Pole, another observer says it ends up at the North Pole

Great, now we’re almost there. Instead of "saying" that it ends up at the South Pole and the North Pole, both observers travels to "their Pole" to check their measurement apparatus.

What will they find?

 

[tiny-tim]

Great, now we’re almost there. Instead of "saying" that it ends up at the South Pole and the North Pole, both observers travels to "their Pole" to check their measurement apparatus.

What will they find?

you mean one observer (with the given velocity) times his start so as to reach A when the spot of light does, and the other observer (with the other given velocity) times his start so as to reach B when the spot of light does?

then the first one will find he's just in time to catch the spotlight before it turns off, and so will the second one!