Is it Possible to Travel Faster than the Speed of Light?

[James S Saint]

Wrong. Look at the diagram. The time elapsed on the worldline clock is simple to calculate.

The worldline clock?

They each have a clock of their own. Both clocks measure 1.333secs. They have to use their own clocks as it is an issue of what they each observe for themselves.

 

[James S Saint]

There are 3 frames involved here, the ground frame and the frame of each observer. Don't let the symmetry of the situation fool you these are 3 very distinct frames.

The only frames that matter are the ones used to measure. Only one is used to measure anything.. and everything.

 

[jcsd]

There is no measuring going on as they move. They take ALL measurements together at the beginning and end while they were not moving with respect to each other. Thus there is no frame changing involved in any measuring. But from the PERSPECTIVE of each, the other seems to have come to them at 1.5c.

If both observers start from not moving then they must've accelerated, even if we assume that acceleration is instantaneous they are not inertial observers. In this case it is the travellers themselves who are changing between inertial frames and what has been said so far only applies for inertial frames.

 

[jcsd]

The only frames that matter are the ones used to measure. Only one is used to measure anything.. and everything.

But see above, this is not an inertial frame. Rather it is two inertial frames 'stictched' together.

 

[Doc Al]

You seemed to have missed a point somewhere. The only measurements made are in the "initial/ground" frame.

The measurement is made of 2Ls at t(0), travel is made. When they meet, they stop and look at their clocks. Both clocks read that they traveled for 1.333secs.

All measurements are from the initial frame.

When they meet, they stop and look at the clocks that were at rest in the ground frame all along. Not the clocks that were moving along with the ships. Their ship clocks would not agree with those measurements.

 

[James S Saint]

Okay, but in the ground frame, neither of the travellers is traveling at greater than c. We started by specifying that each was traveling at 0.75 C in opposite directions in that frame.

We cannot apply measurements made in one frame to another, so in order to work out what was going on in the other frames we'd either need to take new measurements in those frames or alternatively use the Lorentz transformation to convert our previous measurements into the other frames. And lo and behold when we do this we find that neither traveller obserevrs the other traveling at greater than c.

There
Are
No
Other
Frames that are used.

 

[James S Saint]

If both observers start from not moving then they must've accelerated, even if we assume that acceleration is instantaneous they are not inertial observers. In this case it is the travellers themselves who are changing between inertial frames and what has been said so far only applies for inertial frames.

But they make no measurement as they move. Frames only matter to the measurements involved. Both measurements are made from the same initial frame.

 

[James S Saint]

But see above, this is not an inertial frame. Rather it is two inertial frames 'stictched' together.

Can you explain how it is not an "inertial frame" without presuming an absolute frame?

 

[espen180]

The time dilation involved at merely .75c is minuscule to the 1.5c measurement realized. Neither travelers clock would slow by 25%.

It is not. Here, I'll do the calculation for you!

Distance to sign in ship frame: 1Ls/(γ(.75c))=1Ls/1.5=0.66Ls
Time until arrival at sign: 0.66Ls/0.75c=0.88s
From a spacetime diagram you can see that A will observe B to be 0.19Ls away from the sigs when the race starts.
Since they arrive at the same time, B is traveling at 0.19Ls/0.88s=0.21c wrt the sign, giving a relative speed of 0.96c to A. The same result is found by Lorentz velocity addition.

 

[Mentz114]

They each have a clock of their own. Both clocks measure 1.333secs. They have to use their own clocks as it is an issue of what they each observe for themselves.

And they each have a worldline, so we're saying the same thing. I can't explain any better why you're wrong. If we start with the velocites defined in the ground ( signpost ) frame, then you will see your twin approaching at 0.96c, not 1.5c. As has been said many times - 1.5c is not the velocity of anything in this scenario. The signpost observer is adding two velocites, but it's not what is observed by you or your twin or from any frame.

I'll leave you to it now.

 

[jcsd]

But they make no measurement as they move. Frames only matter to the measurements involved. Both measurements are made from the same initial frame.

In which case allmeasurements are made in the gorund frame, but as I said earlier we started by stating that in this frame both observers we're traveling at 0.75 c (in opposite directions).

What you've done is construct an eccentric frame from an obserevrr who undergoes two bouts of instaneous acceleration. The coordinate time in this frame (as measured by you) does not even correspond to the proper time experinced by the observer, it's the time experinced in the ground frame.

I wouldn't even call that a non-inertial frame, I'd just call it some sort of parameterization.

 

[James S Saint]

When they meet, they stop and look at the clocks that were at rest in the ground frame all along. Not the clocks that were moving along with the ships. Their ship clocks would not agree with those measurements.

Ok, so what would their ship clocks read if we secretly knew that they were traveling at .75c with respect to their initial frame?

 

[Tomsk]

If you see your brother off in the distance 2ls away, and you want to meet up at a clock half way between you, you can get there in a time less than 2s (according to that clock) no problem, because it is only 1ls away. The only way to BOTH meet up there is to BOTH accelerate towards the clock. Any problem with that? The only measurements made are made in the initial ground frame.

 

[jcsd]

Can you explain how it is not an "inertial frame" without presuming an absolute frame?

In special relativity velocities are relative, but accelerations are not. The value of an accleartion can change from frame to frame, but an object is either accelerating or it isn't.

 

[James S Saint]

The clock on the ground would read 1.333secs. And they could easily just look over at that one rather than their own, but I'm curious.

 

[jcsd]

Consider Gallilean relativity in that velocities are also relative, whilst accelerations are not.

 

[James S Saint]

If you see your brother off in the distance 2ls away, and you want to meet up at a clock half way between you, you can get there in a time less than 2s (according to that clock) no problem, because it is only 1ls away. The only way to BOTH meet up there is to BOTH accelerate towards the clock. Any problem with that? The only measurements made are made in the initial ground frame.

The problem with using a centered clock is that it gives clue to the travelers. If perhaps we strung 10,000 clocks across the flight path (the ground), they would read the 1.333secs, no matter which clock they stopped at and not know that they had each traveled the same distance.

But even if they use their own clocks, I think you will find that their time dilation doesn't make up the difference.

 

[Bussani]

Even if you use a ground clock, so all of your measurements really did come from one frame, you'd still come to the conclusion everyone keeps telling you: according to the ground measurements, the closing speed between the two ships was 1.5c. If you come to the incorrect conclusion that someone must have moved at 1.5c from that, then it's your own fault for not having enough data and for not thinking, "Hm, I guess both moved at .75c."

Relativity only says that you won't see someone coming at you at >c. That's it. It doesn't say that you can't take measurements in the ground frame, move toward each other, and then conclude that the closing speed was >c. If all the measurements are from the ground frame, then we get a result according to the ground frame. This result isn't anything special.

 

[James S Saint]

In special relativity velocities are relative, but accelerations are not. The value of an accleartion can change from frame to frame, but an object is either accelerating or it isn't.

Emm.. that doesn't seem to ring true. An acceleration is merely the rate of change of a velocity thus if the velocity was changing relative to two parties, the acceleration would have the opportunity to change. It wouldn't have to be different.

 

[James S Saint]

Even if you use a ground clock, so all of your measurements really did come from one frame, you'd still come to the conclusion everyone keeps telling you: according to the ground measurements, the closing speed between the two ships was 1.5c. If you come to the incorrect conclusion that someone must have moved at 1.5c from that, then it's your own fault for not having enough data and for not thinking, "Hm, I guess both moved at .75c."

Relativity only says that you won't see someone coming at you at >c. That's it. It doesn't say that you can't take measurements in the ground frame, move toward each other, and then conclude that the closing speed was >c. If all the measurements are from the ground frame, then we get a result according to the ground frame. This result isn't anything special.

"Seeing" IS taking measurements. No one ever actually sees light travel at all. When we say "observe travel" we MEAN "by the measurements". The only measurements are saying that a distance of 2Ls vanished in 1.333 secs. Now it is speculated that the travelers clocks will read different than a ground clock. So what will their clocks read.

We ONLY care about what is apparent to the travelers at the end of the travel, not during.

 

[Doc Al]

Ok, so what would their ship clocks read if we secretly knew that they were traveling at .75c with respect to their initial frame?

That was already answered multiple times. For example, post #24: The moving ship would measure the time to reach the sign to be about 0.88 s.

And that speed is hardly a secret; you used it to set up the problem.

 

[Mentz114]

Now it is speculated that the travelers clocks will read different than the ground clock. So what will their clocks read.

170 posts in and you finally realize that the travellers' clocks will disagree with the ground clock. Landmark. You've been told how to calculate the time on the travellers clocks.

Reading off the first graph I get tau = 0.937 s.

 

[Bussani]

"Seeing" IS taking measurements. No one ever actually sees light travel at all. When we say "observe travel" we MEAN "by the measurements". The only measurements are saying that a distance of 2Ls vanished in 1.333 secs. Now it is speculated that the travelers clocks will read different than a ground clock. So what will their clocks read.

We ONLY care about what is apparent to the travelers at the end of the travel, not during.

See, what relativity forbids is for you to measure your brother as coming towards you faster than light while you're traveling towards each other. You're not taking measurements when this happens, thus relativity isn't being violated. All of your measurements are coming from the ground frame, so you get a ground frame result. According to that frame, the closing speed is allowed to be 1.5c. There is no problem. You seem to think this goes against relativity, but it doesn't.

 

[jcsd]

Emm.. that doesn't seem to ring true. An acceleration is merely the rate of change of a velocity thus if the velocity was changing relative to two parties, the acceleration would have the opportunity to change. It wouldn't have to be different.

Draw a straiggtht line (assume length doesn't enter into it as you can always extend a straight line), you can rotate and translate that straight line into any other startight line. Now draw a curve that isn't a straight line, you cannot rotate and translate that curve into any other curve without deforming it. You cannot roate and translate that curve into any otherb curve.

The straight lines and curves we're talking about here are the worldines of observers on a spacetime diagram, the staright lines are inertial obserevers, the curves that are not staright lines are the non-inertial observers.

 

[James S Saint]

That was already answered multiple times. For example, post #24: The moving ship would measure the time to reach the sign to be about 0.88 s.

Post #24 doesn't say anything about the time of the travelers clock. I have been watching for that.

Now, what equation did you use to claim that the time dilation caused a 1.333 time to be reduced to .88? That is like 33% reduction.

And that speed is hardly a secret; you used it to set up the problem.

Secret from the travelers, NN.

 

[jcsd]

Also to add to my last post, try stitching two non-paralell staright lines together so they join at a point.You cannot rotate or translate the resulting curve into a straight line.

 

[espen180]

Post #24 doesn't say anything about the time of the travelers clock. I have been watching for that.

Now, what equation did you use to claim that the time dilation caused a 1.333 time to be reduced to .88? That is like 33% reduction.

Secret from the travelers, NN.

The time dilation constant AKA the Lorentz factor, which you assumed was negligible.

t'=γt where \gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}

For v=.75c, γ=1.5.

 

[Doc Al]

Post #24 doesn't say anything about the time of the travelers clock. I have been watching for that.

Read it again.

Now, what equation did you use to claim that the time dilation caused a 1.333 time to be reduced to .88? That is like 33% reduction.

For a speed of 0.75c, the time dilation factor is about 1.51. So if the ground observers measure the travel time to be 1.333 s, the ship clocks will measure 1.333/1.51 = 0.88 s.

You could also view it in terms of length contraction. According to ground measurements, the distance traveled is 1 Ly. From the ship frame, that distance is only 1/1.51 = .66 Ly. So the time works out to be about 0.88 s.

 

[James S Saint]

Also to add to my last post, try stitching two non-paralell staright lines together so they join at a point.You cannot rotate or translate the resulting curve into a straight line.

?? I have no idea what that meant. Two straight lines that somehow result in a curve??

 

[James S Saint]

For a speed of 0.75c, the time dilation factor is about 1.51. So if the ground observers measure the travel time to be 1.333 s, the ship clocks will measure 1.333/1.51 = 0.88 s.

I asked by what equation you got that, not merely the result.