Is Faster-Than-Light Travel Possible Relative to Other Celestial Bodies?

[GoodPR]

As for you guys arguing about sound, I don't know about in a mathmatical way, but in a thought experiment way sound is identical to light.
As someone moving very quickly towards you emits a sound, the sound waves compress giving a higher pitched sound with more energy, and opposite when they move away from you.
Light does the same, the wavelength compresses and it gives you more energy,
And in both cases sound or light shone at you from perpendicular to the object is unchanged.


my question remains however,
Can you travel with enough velocity towards Earth, that from Earths perspective, you would appear as a black hole, (because of your incredible mass).

 

[GoodPR]

Also, in closer relation to the title of the thread,
If I am accelerating at a rate that feels like 1 g constantly it would provide me with a constant acceleration regardless of how fast I am going,

It is known that the faster you travel the more mass you accumulate, however we know that the rocket itself will never feel itself to have more mass, because it is always in a rest frame.
Therefor, traveling closer and closer to the speed of light, will not require more rocketfuel to maintain the 1g acceleration, from the perspective of the rocket itself. It is only the time that rocketfuel burns for, that changes relative to the rocket and Earth. From the rockets perspective it burns 1 kg of fuel persecond.
From Earth perspective it burns 1kg of fuel per .1 second, or whatever.

Therefor, if accelerating at 1g took 2 years to reach light speed relativity excluded, if we gave the rocket 3 years of fuel, the rocket would feel as though it is traveling faster than light compared to Earth, even though from the rockets perspective light is still traveling light speed away from the rocket, and from the Earths perspective we have not attained light speed.

And the method which allows all of this to be true at the same time without violating relativity, is that the relative mass would play an increasing role in balancing the passage of time, so that all parties can experience the same things, without any laws being violated.

In the case where the rocket thinks its going faster than light based on its acceleration over time, from Earths prospective the rocket would appear to be a black hole moving towards earth, at less than light speed. From the rockets perspective, Earth would appear to be a black hole moving towards him at less than light speed.
Due to the warping of space time of the black holes nothing is violated.

 

[russ_watters]

If a rocket had an inertial navigation system that was programmed to calculate velocity via v=a*t, then yes, it could calculate a velocity faster than C. But so what? That just means it was programmed wrong.

 

[GoodPR]

Not entirely true.
Rest mass does not change in any frame regardless of your velocity, and time always ticks by at 1 second per second.
So for the sake of the rocket, that would be an accurate measurement of its true velocity.

It is only when you want to compare your velocity with something else that you need to account for relativity.(which obviously for the case of a rocket is always true)
Since we know that the rocket's rest mass never changed, it is time that is different between the rocket and the observer. The rocket perceives itself buring 1 kg of fuel per second. However on Earth it seems to burn 1kg per 0.1 seconds, or whatever.

Since Light speed is a constant regardless of your velocity or gravity, I believe you could relate your speed directly to that, when you reach the speed of light based on your simple formula, I believe any photon you are emitting directly in front of you would either double in Energy or become two photons as observed by the stationary object your traveling towards.
Keep in mind that despite the simple calculation showing your going faster than light, I'm not claiming light doesn't move light speed away from you in all directions.

 

[russ_watters]

Not entirely true.
Rest mass does not change in any frame regardless of your velocity, and time always ticks by at 1 second per second.
So for the sake of the rocket, that would be an accurate measurement of its true velocity.

No, it wouldn't - otherwise, you could plug that number into appropriate equations and get nonsensical responses. Velocity is something you measure between two objects.

What you would be measuring is rapidity: http://en.wikipedia.org/wiki/Rapidity

 

[GoodPR]

Did you read what you quoted and stop? Because I immediately said the same thing as you did after the quote.

 

[eekf]

I am going to comment on quite a few submissions made before.

1. There will be no Doppler shift between two objects moving at the same velocity whether you are using sound in a medium propagating sound or a medium (or non-medium) propagating light. There will be a Doppler shift on the transmit side, and the inverse at the receive side.

2. You would be able to set up an experiment such as the Michelson-Morley experiment for both environments. When you do this (remembering that we actually know that the speed of sound can be exceeded) you can now check to see if you can measure a speed relative to an "absolute rest" environment, which in the case of sound would be the medium.

However, what becomes suddenly clear if you use the sound environment, is that the sound going 90 degrees to your velocity, will not reach you when reflected. In order for the sound you send sideways to reach you, you have to send it slightly ahead of you, depending on your speed. If you travel close to the speed of sound in the medium, it will only reach you if you send it out close to straight ahead.

Apart from the distance increase of the path in the moving frame, you will therefore also have a Doppler shift in the 90 degree arm which will increase in relation to your velocity (wrt the medium), corresponding (either totally or to a large degree) with the shift in phase you are expecting because of the longer path length the light has to travel in an "absolute frame" when your velocity relative to this frame increases.

Interference will be in the "absolute rest" frame, before you measure your intervals. You may also have to consider the phase shifts and possibly Doppler effects at the "mirrors".

I have not checked these calculations, but have been lead to believe that you get a Null result for all velocities. Please do not quote me on this. I have the document where this is described by an 80 year old scientist, but have not worked through it. Do the calculations.

3. Assuming the constancy of the speed of light through the vacuum (or sound in a sound medium) is a fact (and we have very little if anything to doubt this), the question becomes if the speed of light in each reference frame is aconstant in both directions. The Lorentz transformation assumes this to be true and then just works with the speed of light c.

Now it is important to consider this: I can either assume the speed of light in both directions is c constant in any moving frame, and use the average of the time to an object and back to determine the distance, or I can assume that they are different, but that by assuming it constant I can have a model which describes mathematics for any frame without knowledge of my velocity relative to my "absolute rest" frame, bearing in mind that I might get some discrepancies.

The part I struggle with most when assuming that the actual relative speed of light is the same in both directions, is the Sagnac effect. The Sagnac effect is used in (amongst others) laser Gyro's and is also needed in keeping the GPS system up to date. In gyro's it is used to send light in two directions to come back to the starting point, where the difference in travel time between the two pulses are measured, when the gyro is rotating.

I have read that SRT explains this effect just as well as other models because the "extra" length causes the time difference. If I evaluate a small straight section of the gyro, I can however not find how to explain this if the assumption is that the constant c is the same in both directions. The velocity of the environment is constant. There should therefore not be any path length difference between the directions the light is traveling in according to SRT.

4. I want you to understand that I am not claiming that it is possible to move faster than light. If atoms are actually held together through EM fields (as we believe currently), and there is a "preferred absolute reference frame" (which most mainstream modern scientists do not believe), the atoms may actually fly apart when approaching or exceeding light speed.

I am saying that the Lorentz transformation prevents traveling faster than the speed of light, but that the mathematical models we use to represent the world has restrictions and assumptions (believes) built into them. We have to try and understand those restrictions.

We humans have a tendency to restrict ourselves with our own knowledge or beliefs. When we "know" it is impossible to go faster than light or for an object to be heated by a colder object or for energy to be created, we tend to overlook opportunities.

The more knowledge we accumulate, the more we tend to restrict ourselves by what we know. Most of the times we do not even realize that we have made a subtle change to what we assume (believe), and that that change is restricting us.

5. I am not trying to start a new belief or theory.

 

[jreelawg]

My take is that relativity only makes sense mathematically not intuitively.

 

[Razzor7]

My take is that relativity only makes sense mathematically not intuitively.

It makes as much intuitive sense as any theory. It's just wha' happens.

 

[GoodPR]

EEkf
I don't even see what your trying to say, everything you said appeared to be in agreeance with all modern theories, the only thing you seemed to make a statement on was that the more knowledge we seem to gain, the more we close our minds to other knowledge. On that I'd have to agree, I think any GUT out there is going to be a simple equation, at face value.

 

[eekf]

My take is that relativity only makes sense mathematically not intuitively.

The only way I could understand it was to plot a spacecraft traveling at 0.8c and emmitting a "spherical" light pulse every second. After 4 seconds (as seen from a stationary frame), I drew the position of the light pulses with compasses. I ended up with a drawing looking exactly like the drawings in high school physics books representing the sound waves of an aircraft approaching the speed of sound.

The problem comes in if you consider the positions of the light pulses from the spacecraft 's point of view. If you use the same clock synchronisation (not rate) in the spacecraft 's reference frame, you end up with the first light pulse being much further behind the spacecraft than in front of it. In addition, the light pulse behind would seem to have traveled much faster than the speed of light (actually 1.8c).

If you use this mathematical model, you end up with calculations which have the speed of light dependent on the direction you are traveling in and not isotropic. By adjusting your clocks, you can however rescale to have the light in both directions move at c. You end up that you have to change the scaling of your x-axis as well as your clocks if you want the transformation to be consistent through all reference frames.

If you consider the same setup for sound in air, you realize that the ony reason you can (or have to) do this is because the speed of sound in the medium is a constant, independent of your relative movement. If the air was traveling with you (i.e. in a closed van), you can use the same transformations, but the speed of sound (as viewed from outside the van) would not be isotropic.

 

[eekf]

EEkf
I don't even see what your trying to say, everything you said appeared to be in agreeance with all modern theories, the only thing you seemed to make a statement on was that the more knowledge we seem to gain, the more we close our minds to other knowledge. On that I'd have to agree, I think any GUT out there is going to be a simple equation, at face value.

Actually what I am saying is not in complete agreeance with modern theories, or should I say better: the interpretations given to modern theories.

Consider a spacecraft 1 lightsecond long with two independent atomic clocks (one in the front and one in the back) which is synchronised when it is in "a stationary Earth" reference frame. It then accelerates to a speed of 0.9c. Once stabilised at this speed, a light pulse is sent from the front to the rear at t=0 and reflected back to the front from the rear.

I am saying that the time difference measured between transmitting from the front till reflection at the rear will be much lower than the time difference measured from the reflection to the reception at the front.

As such I am saying that clocks have to be re-synchronised after reaching 0.9c for c to be measured constant in both directions.

I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame.

 

[matheinste]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

 

[eekf]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

Possibly. I believe experiments like this have been done. See http://xxx.lanl.gov/abs/astro-ph/0608223.

 

[eekf]

eekf,

But your interpretation would allow you to detect absolute motion.

Matheinste

Another recent paper: http://blog.hasslberger.com/docs/Cahill_Experiment.pdf.

The conclusions of this paper (amongst others):
1. Speed of light is anisotropic (eight different experiments).
2. Fitzgerald-Lorentz contraction is real effect in interferometers.

It might be that if the Doppler shift because of the fact that the 90 degree light in the arm in interferometer equipment can not come back to interfere (interference can only be with light that is not at 90 degrees and angle is dependent on v) might even indicate no contraction needed.

 

[Ich]

I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame.

I believe that
1. Interpretations must not disagree on facts, therefore, if your "interpretation" does not agree with mainstream results, it is a disguised ATM theory.
2. If ATM proponents would spend only a tenth of their effort disproving the mainstream on actually learning what it says, there'd be no more ATM proponents.

Please calculate your example in SR, and present the result here. If you can't, you're surely not in a position to challenge anything. If you can, your "belief" is moot.

 

[vin300]

Here's the values:If 0.9c is its velocity, the timr taken from front to back in the frame of reference of the ship is 1/1.9s which is 0.52s and the time taken for the reverse is
1/0.1s,i.e. 10s

 

[eekf]

I believe that
1. Interpretations must not disagree on facts, therefore, if your "interpretation" does not agree with mainstream results, it is a disguised ATM theory.
2. If ATM proponents would spend only a tenth of their effort disproving the mainstream on actually learning what it says, there'd be no more ATM proponents.

Please calculate your example in SR, and present the result here. If you can't, you're surely not in a position to challenge anything. If you can, your "belief" is moot.

I agree with vin300's calculation.

I would like to know what you believe the whole purpose of clock (re-)synchronisation in each reference frame is, if it is not to get c isotropic?

Please tell me what ATM theory is about. I do not know it. I am not trying to disguise any theory. I am trying to make sense of what I see - especially in the mathematics.

 

[matheinste]

I agree with vin300's calculation.

I would like to know what you believe the whole purpose of clock (re-)synchronisation in each reference frame is, if it is not to get c isotropic?

Isn't this reasonong back to front. Clocks in a common inertial reference frame are synchronized to make them show the same time,not to make light speed isotropic. The standard synchronization process depends on the average of the two way directional speed of light being constant which it assumed to be. If clocks are not first synchronized how can you measure speed.

vin300's calculation assumes that the light inside the spaceship travels different distances. It does not. Light emitted from the centre of the ship travels the same distance from emitter to front and back to emitter as it does from emitter to rear and back to emitter.

Matheinste

 

[vin300]

vin300's calculation assumes that the light inside the spaceship travels different distances. It does not. Light emitted from the centre of the ship travels the same distance from emitter to front and back to emitter as it does from emitter to rear and back to emitter.

Matheinste

Er a little misconception Light inside the spaceship does go different distances in the f of stationary observer because the spaceship itself is in motion.If you'd like to know how derive it's simple t(back to front) is [L/c-v ]t(f to b) is [L/c+v]

 

[matheinste]

Saying the spaceship is in motion is meaningless.

Matheinste.

 

[eekf]

Isn't this reasonong back to front. Clocks in a common inertial reference frame are synchronized to make them show the same time,not to make light speed isotropic. The standard synchronization process depends on the average of the two way directional speed of light being constant which it assumed to be. If clocks are not first synchronized how can you measure speed.
Matheinste

Actually we are trying to measure elapsed time between events at different points in the same reference frame, not speed. To do that we have to synchronise our clocks in the reference frame at the speed it is moving.

Please look at the following quoted from Einstein 1905:
"We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' ."

From your last posting I understand that you actually agree that you have to resynchronise the clocks at v=0.9c before you can measure or calculate anything sensical in SRT, especially if you are comparing the times of different events at different positions.

Do I understand you correctly?

 

[matheinste]

------From your last posting I understand that you actually agree that you have to resynchronise the clocks at v=0.9c before you can measure or calculate anything sensical in SRT, especially if you are comparing the times of different events at different positions.------

If the clocks,during acceleration remained at rest relative to each other then they remain in synch. But I believe whether they remain at rest relative to each other depends upon the acceleration being of a certain type.

Yes clocks do need to be synchronized, or resynchronized IF required, to make time measurements. To calculate speed you need to know both time and distance.

All events take place in every reference frame. We usually measure the time between two events from the perspective of a single inertial reference frame at our convenience.

Matheinste.

 

[vin300]

Eh, now visualise. Light does not, like a person in the spaceship move at the velocity of the ship.
The standard velocity addition formula is
vector(R) = vector(u)+vector(v) [Sorry I don't know using tex]
where R is the resultant velocity, u is the velocity of the object in the f of the ship and v is the velocity of the ship w.r.t. a stationary frame

Relativistic vel addition is
vector(R)=[vector(u)+vector(v)]/1+ dot(uv)/c^2

If you use this formula, you get the velocity of light as c in all frames

Whatever be the velocity of the ship, velocity of light in the ship remains c.

In the frame of reference of the observer on the earth, the length of the ship contracts, he sees the length of the ship to be L/(gamma)
The time for the light from the back to reach the front is
t =L/c+v(gamma)
The time to reach the back is L/c-v(gamma)

I had not used the idea that the length of the ship is contracted for the stationary observer that's a discrepancy in the calculation

 

[Ich]

I would like to know what you believe the whole purpose of clock (re-)synchronisation in each reference frame is, if it is not to get c isotropic?

Hmm... it's about synchronizing clocks?
Anyway, that doesn't matter. Either you synchronize clocks to make c isotropic, or you synchronize clocks assuming that c is isotropic.
If you agree that, following SR, the clocks are out of synch (matheinste is wrong on that point), why do you state "I believe most modern interpretations are that the clocks will stay synchronised because they are in the same reference frame."?
How could this be a question of interpretation, as long as we agree on the terminology (e.g. what "synchronization" means)?

Please tell me what ATM theory is about.

You will tell us. Sooner or later.

 

[matheinste]

If you use this formula, you get the velocity of light as c in all frames

Whatever be the velocity of the ship, velocity of light in the ship remains c.

In the frame of reference of the observer on the earth, the length of the ship contracts, he sees the length of the ship to be L/(gamma)
The time for the light from the back to reach the front is
t =L/c+v(gamma)
The time to reach the back is L/c-v(gamma)

I apologise. I was in the ship's frame.

Matheinste.

 

[matheinste]

Hello Ich,

Rergarding accelerated clocks,perhaps this answer I received a long time ago in another thread confused me. I cannot remember who it was from but it was certainly one of the regular, knowledgeable responders. In answer to the question "will the previously synchronized clocks be in synch,as viewed by acomoving observer, after the acceleration the reply was-----
--------Assuming that the world lines of the two clocks and the world line of this observer guy all look the same in the original rest frame (except for their starting position), the answer is definitely yes. This is an immediate consequence of the definition of proper time and the postulate that clocks measure proper time.

You may have read about the scenario where the clocks are attached to opposite ends of a rigid rod. (It's been discussed in this forum a few times). The rod will be getting shorter (Lorentz contracted) when its speed increases (in the original rest frame). So the rear must be accelerating faster than the front, and later it must be decelerating faster than the front. In this case, the world lines will not be identical.

They might still have the same proper time though, e.g. if the deceleration profile is the "opposite" of the acceleration profile, so that the first half of the world line of the rear is a mirror image of the second half of the world line of the front and vice versa. But they may not have the same proper time in general. --------

This is the reason I mentioned a certain kind of acceleration was needed for clocks to stay in synch. I am however not sure of my ground and would welcome any clarification.

Matheinste.

 

[Ich]

Hello matheinste,

--------Assuming that the world lines of the two clocks and the world line of this observer guy all look the same in the original rest frame (except for their starting position), the answer is definitely yes. This is an immediate consequence of the definition of proper time and the postulate that clocks measure proper time.

I think I remember this discussion.
The clocks stay in synch wrt their initial rest frame.
One can see that "immediately" (as is claimed) because, starting from two simultaneous events in the initial frame, after "running through" identical world line segments, the clocks will show the same time. And the respective events will still have an equal time coordinate in the initial frame.
That means that they are out of synch in their new common rest frame.

 

[matheinste]

Thanks Ich,

That seems to make sense. So the clocks are in synch when viewed from their initial frame and so MUST be out of synch when viewed by someone in their new inertial frame as they cannot be in synch in two frames in relative inertial motion with respect to each other.

Matheinste.

 

[A.T.]

There will be no Doppler shift between two objects moving at the same velocity whether you are using sound in a medium propagating sound or a medium (or non-medium) propagating light.

(...)

Apart from the distance increase of the path in the moving frame, you will therefore also have a Doppler shift in the 90 degree arm which will increase in relation to your velocity

You are contradicting yourself.

The MM-setup doesn't rely on clock synchronization or exact distance measurement, and would not give a null result in water with sound. I cannot follow your argument about the Doppler shift in the 90°-arm, that miraculously cancels out the expected phase shift. If I move trough a sound medium parallel to a wall and make a loud sound, I receive the echo without any Doppler shift. It doesn't even matter if the wall also moves with me or rests in the medium. This sound travels exactly the same way, it would in the 90°-arm of a sound-MM-setup.

I have not checked these calculations, but have been lead to believe that you get a Null result for all velocities.

(...)

I am not trying to start a new belief or theory.