Is Faster than Light travel impossible?

[Doc Al]

Now what is the equation the track uses to measure the length of the rod ?

How would you like them to measure it? I suspect you'd like them to use a beam of light, else there wouldn't be much to talk about. So if they send a beam of light from one end (A) to the other (B), and they measure the travel time to be Δt, then the length of the rod in the track frame will be: L = cΔt.

 

[jmallett]

How would you like them to measure it? I suspect you'd like them to use a beam of light, else there wouldn't be much to talk about. So if they send a beam of light from one end (A) to the other (B), and they measure the travel time to be Δt, then the length of the rod in the track frame will be: L = cΔt.

Well, that's what Einstein said, so in terms of understanding Einstein's Special Theory of relativity this is going well.

But, the "stationary observer", the track observer we are talking about, sees something quite different from that, and relatively more complex, because he knows the speed v, and because he knows the speed of light c, therefore he needs to make 2 calculations, one to the front end, and one to the back end, and he has to take account of the fact that the rod is moving and that the speed of light is not instantaneous.

 

[Doc Al]

But, the "stationary observer", the track observer we are talking about, sees something quite different from that, and relatively more complex, because he knows the speed v, and because he knows the speed of light c, therefore he needs to make 2 calculations, one to the front end, and one to the back end, and he has to take account of the fact that the rod is moving and that the speed of light is not instantaneous.

And so?

 

[jmallett]

And so?

you don't see it ?

 

[Doc Al]

you don't see it ?

See what?

Depending on what the track frame measures, they can calculate the length of the rod with just one equation. If the train and the light beam are both moving to the right, the length of the moving rod according to the track frame will be L_track = (c - v)Δt_track.

See if you can derive that equation.

 

[jmallett]

See what?

Depending on what the track frame measures, they can calculate the length of the rod with just one equation. If the train and the light beam are both moving to the right, the length of the moving rod according to the track frame will be L_track = (c - v)Δt_track.

See if you can derive that equation.

That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.

 

[Doc Al]

That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.

Nope. There is only one beam of light that travels from one end of the rod (A) to the other (B). It doesn't 'belong' to the train or the track. Both frames can analyze the motion of that same beam of light.

 

[cfrogue]

That particular equation only works if the track can use the train's light, so it can't be valid in this useage. That's a simplification of what occurs. The track must use the track's light.

You can only say this if you have another theory of light.

DA presented a valid equation.

What is your theory?

 

[jmallett]

You can only say this if you have another theory of light.

DA presented a valid equation.

What is your theory?

That particular theory was the basis of the ether. Mitchelson Morely and a number of others disproved it. Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light.

I don't yet have a full theory, just a whole lot of questions which I can't answer, but I am putting together some math on the subject.

Certainly in the situation Doc Al proposes we need to add in the distance deltaL which the rod passes through during the duration of the measurement. That's simple math and I will write it up for you. I hope soon to be able to share those equations, and I hope that someone will point out any errors I might have made. In addition to that I am working on an a demonstration with two bodies moving without reference to anything, i.e. truly relative, that use their own electromagnetic radiation (maybe light) to make a common measurement.

 

[Doc Al]

Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light.

Nope--Einstein makes no such assumption. As I've already stated, light does not 'belong' to any reference frame. What Einstein does assume is that all light moves at the same speed with respect to any reference frame.

Certainly in the situation Doc Al proposes we need to add in the distance deltaL which the rod passes through during the duration of the measurement.

The equation I provided already takes into consideration the fact that the rod moves a distance vΔt_track during the time it takes for the light to travel from one end to the other.

(Note: Discussion of personal theories is not permitted.)

 

[jmallett]

Nope--Einstein makes no such assumption. As I've already stated, light does not 'belong' to any reference frame. What Einstein does assume is that all light moves at the same speed with respect to any reference frame.


The equation I provided already takes into consideration the fact that the rod moves a distance vΔt_track during the time it takes for the light to travel from one end to the other.

(Note: Discussion of personal theories is not permitted.)

You are using the same light to move at the same speed relative to two objects that have a relative motion.

Ahh, now we have it. The high priests statements are to believed, regardless of whether they correctly interpret Einstein or not and all other readings of his work are not permitted.

Seems I am in the wrong temple here. If no discussion of what Einstein says is permitted then I simply genuflect to your excellence and leave.
Adieu

 

[darkhorror]

Why don't you first try to understand relativity and what we are saying. As all this stuff has been tested and is shown to work. If I am moving close to the speed of light with respect to you and shine a beam of light. We both shine lights. With respect to me both beams of light move at the speed of light with respect to me. Also to you both beams of light moving at the speed of light with respect to you.

Do you understand that?

 

[cfrogue]

That particular theory was the basis of the ether. Mitchelson Morely and a number of others disproved it. Einstein went forward with the assumption that light is relative to the frame of reference in which it exists, i.e. the train's light is not the track's light.

I don't yet have a full theory, just a whole lot of questions which I can't answer, but I am putting together some math on the subject.

Certainly in the situation Doc Al proposes we need to add in the distance deltaL which the rod passes through during the duration of the measurement. That's simple math and I will write it up for you. I hope soon to be able to share those equations, and I hope that someone will point out any errors I might have made. In addition to that I am working on an a demonstration with two bodies moving without reference to anything, i.e. truly relative, that use their own electromagnetic radiation (maybe light) to make a common measurement.

If you would like to PM your views to me, feel free. It is just this forum is not a place to present ATM ideas.

You certainly can ask questions here though.

Also, one should respect the mainstream ideas since most are backed by experimental data.

a demonstration with two bodies moving without reference to anything, i.e. truly relative, that use their own electromagnetic radiation (maybe light) to make a common measurement.

Before digging into this too hard, look at the work of Dingle first. Frame to frame clock synchronization is not logically decidable using relative frames of motion.

 

[ExecNight]

Hello i said

Ergosphere..

Why is it deleted? Who deleted it without notice?

Actually it adds a lot to this discussion considering its an area around a rotating black hole where space time itself travels FTL, and actually is outside event horizon so how about answering instead of deleting =)

 

[Doc Al]

I deleted your one word post. Now that you've expanded your thought to a full sentence, I still don't see the relevance to the particular discussion in this thread.

 

[JesseM]

Hello i said

Ergosphere..

Why is it deleted? Who deleted it without notice?

Actually it adds a lot to this discussion considering its an area around a rotating black hole where space time itself travels FTL, and actually is outside event horizon so how about answering instead of deleting =)

Nothing travels FTL in the ergosphere in a true physical sense--the worldlines remain timelike, so if you pick any event on an object's worldline, then at later times the worldline will always remain within the future light cone of that event. Looking at the bottom section of http://home.pacbell.net/bbowen/swirl.htm it sounds like the notion of "space traveling FTL" refers to how stationary coordinates behave in a certain commonly-used coordinate system:

Between the static limit and the event horizon we have a region called the "ergosphere". As mentioned above, one cannot remain at rest with respect to the fixed stars while in the ergosphere. One can however, remain outside the event horizon and even return back into space if one desires. Inside the ergosphere, one sees stationary coordinate points whiz by him at FASTER than the speed of light.

On this subject, note that the idea that nothing can have a speed greater than light only applies when we are talking about coordinate speed in an inertial coordinate system, even in special relativity it is quite possible to come up with non-inertial coordinate systems where the coordinate speed of an object at one position may be greater than the coordinate speed of light at a different position. And all coordinate systems covering non-infinitesimal regions of curved spacetime (like any coordinate system used to describe a rotating black hole) are non-inertial ones.

 

[cfrogue]

Nothing travels FTL in the ergosphere in a true physical sense--the worldlines remain timelike, so if you pick any event on an object's worldline, then at later times the worldline will always remain within the future light cone of that event. Looking at the bottom section of http://home.pacbell.net/bbowen/swirl.htm it sounds like the notion of "space traveling FTL" refers to how stationary coordinates behave in a certain commonly-used coordinate system:

On this subject, note that the idea that nothing can have a speed greater than light only applies when we are talking about coordinate speed in an inertial coordinate system, even in special relativity it is quite possible to come up with non-inertial coordinate systems where the coordinate speed of an object at one position may be greater than the coordinate speed of light at a different position. And all coordinate systems covering non-infinitesimal regions of curved spacetime (like any coordinate system used to describe a rotating black hole) are non-inertial ones.

are you claiming in GR light can travel > c?

I think you are taking a straight line path through hyperbolic space to conclude this and light cannot follow.

Please prove that light can travel > c.

 

[JesseM]

are you claiming in GR light can travel > c?

Not in a locally inertial frame. But in non-inertial frames (and all frames covering a non-infinitesimal region of curved spacetime in GR are non-inertial), light can definitely have a coordinate speed other than c...this is just as true in SR as in GR (for instance, light wouldn't move at c everywhere in accelerating Rindler coordinates in flat spacetime). Note for example that this page says, in the "General Relativity" section:

The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.

In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected).

 

[ExecNight]

So if i want a certain answer, i can use "a" coordiante system.
And if i want another answer i can use "the" coordinate system for my liking =)

 

[cfrogue]

Not in a locally inertial frame. But in non-inertial frames (and all frames covering a non-infinitesimal region of curved spacetime in GR are non-inertial), light can definitely have a coordinate speed other than c...this is just as true in SR as in GR (for instance, light wouldn't move at c everywhere in accelerating Rindler coordinates in flat spacetime). Note for example that this page says, in the "General Relativity" section:

None of this proves light can travel > c.

Prove it.

 

[JesseM]

None of this proves light can travel > c.

Prove it.

OK, suppose x,t represent the coordinates of some inertial frame. Here is a coordinate transformation which gives a non-inertial coordinate system:

x' = x + (10 light years/year^2)*t^2
t' = t

Suppose in the inertial frame a light beam was sent out at x=0 light years, t=0 years and received at x=4 light years, t=4 years. Then in the non-inertial frame, it was sent out at x'=0 light years, t'=0 years and was received at x'=164 light years, t'=4 years, meaning its average coordinate speed between being sent and received was (164 - 0)/(4 - 0) = 41 light-years/year = 41c.

 

[cfrogue]

OK, suppose x,t represent the coordinates of some inertial frame. Here is a coordinate transformation which gives a non-inertial coordinate system:

x' = x + (10 light years/year^2)*t^2
t' = t

Suppose in the inertial frame a light beam was sent out at x=0 light years, t=0 years and received at x=4 light years, t=4 years. Then in the non-inertial frame, it was sent out at x'=0 light years, t'=0 years and was received at x'=164 light years, t'=4 years, meaning its average coordinate speed between being sent and received was (164 - 0)/(4 - 0) = 41 light-years/year = 41c.

So you are saying we can communicate > c?

Do you understand LT?


You put gamma into the complex number system.

How do you resolve this?

 

[JesseM]

So you are saying we can communicate > c?

In a non-inertial coordinate system, yes. You can define a non-inertial coordinate system such that a snail moves faster than c if you want. That's the funny thing about non-inertial coordinate systems, distances and times can be defined in any arbitrary way you want. For example, I can define a non-inertial coordinate system where the distance from New York to Los Angeles is 0.00000000001 nanometers while the distance from Los Angeles to San Francisco is 30 trillion light years. Coordinate systems are just ways of labeling points in spacetime, and in non-inertial coordinate systems the labels are totally arbitrary.

Do you understand LT?

I sure do. Do you? In particular, do you understand that they only deal with inertial coordinate systems?

You put gamma into the complex number system.

No idea what you are saying here.

 

[cfrogue]

In a non-inertial coordinate system, yes. You can define a non-inertial coordinate system such that a snail moves faster than c if you want. That's the funny thing about non-inertial coordinate systems, distances and times can be defined in any arbitrary way you want. For example, I can define a non-inertial coordinate system where the distance from New York to Los Angeles is 0.00000000001 nanometers while the distance from Los Angeles to San Francisco is 30 trillion light years. Coordinate systems are just ways of labeling points in spacetime, and in non-inertial coordinate systems the labels are totally arbitrary.

I sure do. Do you? In particular, do you understand that they only deal with inertial coordinate systems?

No idea what you are saying here.

forget it

 

[yuiop]

In a non-inertial coordinate system, yes. You can define a non-inertial coordinate system such that a snail moves faster than c if you want. That's the funny thing about non-inertial coordinate systems, distances and times can be defined in any arbitrary way you want. For example, I can define a non-inertial coordinate system where the distance from New York to Los Angeles is 0.00000000001 nanometers while the distance from Los Angeles to San Francisco is 30 trillion light years. Coordinate systems are just ways of labeling points in spacetime, and in non-inertial coordinate systems the labels are totally arbitrary.

Even in such a arbitrary coordinate system, you can not transmit information (or travel) from A to B, faster than a photon can travel from A to B.

(Just thought I would mention that, just in case some beginner drops in here and gets really really confused :)

 

[JesseM]

Even in such a arbitrary coordinate system, you can not transmit information (or travel) from A to B, faster than a photon can travel from A to B.

Yes, I agree, and you're right that this should be added in case anyone else gets confused. But cfrogue was asking for a demonstration that light itself can travel faster than c, so clearly his question about traveling "faster than c" was about coordinate speeds, not about anything traveling faster than a physical light signal.

 

[Sylventhe]

After taking the time to write this, I realized once I got to the end, I really didn't have an observable point. However, I spent a good thirty or so minutes typing and formatting it, so I felt it a shame to waste, and maybe it will spark good discussion ... yeah, that sounds good...

This post may or may not contain a point. You have been warned.

I'm not entirely sure how I happened upon this discussion, but I felt I had to join in.

Let me first state for the record, I am NOT a physicist. I write software. Unfortunately, with the ability to write software also comes just enough knowledge of mathematics to hurt onesself. Most of the time I can keep this under control, but for some reason I feel compelled to use all the mathematical firepower I can bring to bear so I can shoot myself in the foot.

My apologies for this being overly-simplistic with these statements.

1. Assume a perfectly spherical cow of unit radius and mass.
2. Assume I fix myself to an arbitrary location in one-dimensional euclidian space. We shall designate to be x=0. (I further assume the reader is okay with this designation)
3. A theoretical object, we will simply designate "the sphere ... OF DOOM" (epic cinematic music in background played by a band lost somewhere on the positive x-axis) is traveling at me at $$\delta$$x = +2c. Let the distance between x = n and x = (n+1) be the distance "ts ... OD" travels in 1 unit of time.
4. "the sphere ... OF DOOM" emits light uniformly in all directions, and the light traveling down the x-axis does so at 1c (relative to me).
5. Let t=0 designate the moment in time when "ts ... OD" passes my location.

Let f(t) be the location of the sphere at time t.
Let l_n(t) be the location of light emitted by the sphere at time t=n.

At t=-3, f(t) = -3, l_-3(t) = -3.
At t=-2, f(t) = -2, l_-3(t) = -2.5, l_-2(t)=-2.
At t=-1, f(t) = -1, l_-3(t) = -2, l_-2(t)=-1.5, l_-1(t)=-1.
At t=0, f(t)=0, l_-3(t)=-1.5, l_-2(t)=-1, l_-1(t)=-.5.

So, at t=0, I observe light from the sphere emitted when the sphere was at x=0. I see the sphere at my location, x=0.
At t=1, I observe light from the sphere emitted when the sphere was at x=-1. I see the sphere at x=-1.
At t=2, I observe light from the sphere emitted when the sphere was at x=-2. I see the sphere at x=-2.
At t=3, I observe light from the sphere emitted when the sphere was at x=-3. I see the sphere at x=-3.

Using this (admittedly poor-excuse-for-) logic, from my point of view, what I see is absolutely nothing until the object reaches me. From that point forward, I see the object's light in reverse order, so even though the sphere is now behind me (wreaking havoc and mayhem somewhere in the positive x-axis), what I see is the sphere moving away from me down the negative x-axis.

Of course, what gets confusing is the whole "sphere emitting light" part. If the sphere travels at 2c, and the light at 1c, then the sphere is "dropping off" light along the way. If the sphere emits light, it would seem to follow the light leaves the sphere at 1c, which I would observe to be 3c.

So, other than proving that I have just enough knowledge of mathematics to hurt myself, but hopefully only myself, have I really said anything worthy of note here? Was it wrong to assume the perfectly spherical cow of unit radius and mass?

 

[matheinste]

Hello Sylventhe,

Perfectly spherical cows are OK for thought experiments but Spheres of Doom, or any other material particles are limited to subluminal speeds in Special Relativity.

Matheinste.