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.

 

[jcsd]

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

I use the word 'curve' in a more general sense. I'm trying to reframe what has been said in terms of geometry (in this case the geometry of spacetime) to make you see why what you have said is not true.

Even in gallilean relativity accelerating frames are not equiavlent to non-accelerating ones as you get inertial forces in accelerating frames. A simple example of this is how you're pushed back into your seat when a car accelerates.

 

[Doc Al]

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

I can rewrite it in terms of equations, if you like. But to understand what they mean and where I get them from, you need to understand some relativity:

Time dilation: T_ship = T_ground/λ

Length contraction: D_ship = D_ground/λ

Couple that last with Distance = speed X time.

In this situation, λ = 1.5.

 

[Dale]

Wow, >180 posts in just under 7 hours. This must be some record.

By the fact that there are that many posts I gather that the OP is still making the same mistake?

 

[James S Saint]

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.

Something seems odd about that.

If I had said that they were going at .5c instead of .75c, the ground clock would read 2secs and that equation would make their clocks read 1.74secs.

So instead of them seeing that they traveled 2Ls in 1.33 secs, they would see that they traveled 2Ls in 1.73 secs. Or did I miss my math somewhere?

The end result would be that they resolved that they traversed 2Ls in 1.73secs = 1.15c.

 

[James S Saint]

Wow, >180 posts in just under 7 hours. This must be some record.

By the fact that there are that many posts I gather that the OP is still making the same mistake?

We might just now be getting around to the mistake.

 

[Dale]

(sorry if this has already been addressed)

James, one thing that you need to learn in relativity is that when you are talking about relative quantities (like speed or time or distance) you need to specify the reference frame in which they are relative.

This post above is a great example:

If I had said that they were going at .5c instead of .75c,

In which frame is their speed measured?

the ground clock would read 2secs and that equation would make their clocks read 1.74secs.

In which frame are these readings simultaneous?

So instead of them seeing that they traveled 2Ls

In which frame is the distance 2 ls?

 

[James S Saint]

(sorry if this has already been addressed)

James, one thing that you need to learn in relativity is that when you are talking about relative quantities (like speed or time or distance) you need to specify the reference frame in which they are relative.

This post above is a great example:In which frame is their speed measured?

In which frame are these readings simultaneous?

In which frame is the distance 2 ls?

All the same as before. I merely changed the speed to .5c instead of .75c.

the only frame involved is the initial or ground frame for these measurements... well until we started calculating the moving clocks with respect to each other.

 

[Dale]

All the same as before. I merely changed the speed to .5c instead of .75c.

the only frame involved is the initial or ground frame for these measurements.

Thanks for the clarification, I didn't want to read through all the posts to find out for sure, but it wasn't clear from the above post.

In the ground frame, if you are approaching the sign from the right at .75 c and your brother is approaching the sign from the left at .75 c then the distance between the two of you is reducing at 1.5 c in the ground frame.

In your frame, your velocity is (by definition) 0, and the sign's velocity is .75 c towards you, and your brother's velocity is .96 c towards you. The distance between your brother and the sign is decreasing at .21 c.

In all cases nothing is going faster than c even if some coordinate separation is > c.

 

[James S Saint]

Read it again.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.

But wait.

If their clocks read .88secs, that would mean that they observed that they traversed 2Ls in .88 secs rather than 1.33 secs. That is even worse, instead of 1.5c, they would calculate/measure 2.27c ..??

 

[James S Saint]

Thanks for the clarification, I didn't want to read through all the posts to find out for sure, but it wasn't clear from the above post.

In the ground frame, if you are approaching the sign from the right at .75 c and your brother is approaching the sign from the left at .75 c then the distance between the two of you is reducing at 1.5 c in the ground frame.

In your frame, your velocity is (by definition) 0, and the sign's velocity is .75 c towards you, and your brother's velocity is .96 c towards you. The distance between your brother and the sign is decreasing at .21 c.

In all cases nothing is going faster than c even if some coordinate separation is > c.

Well that is the result of an equation that is used to calculate how fast my brother would be leaving me if he was leaving the sign at .75c away from me.

If you don't think so, then calculate how fast I would observe him moving away from me IF he had been traveling away from me and the sign.

 

[Bussani]

You're still trying to work out speed using clocks in the moving ship and distances measured from at rest. Like PassionFlower said earlier, if you use that logic you can conclude that you traveled faster than light if you make a 10 lightyear long journey (measured from rest) in only 5 years (experienced on your ship), since you're not accounting for time dilation. Relativity can't help it if you come to broken results like this.

 

[Dale]

Well that is the result of an equation that is used to calculate how fast my brother would be leaving me if he was leaving the sign at .75c away from me.

I almost never use shortcut formulas because I inherently mistrust them (or rather, I mistrust my own ability to use them correctly in all situations). That was directly from the Lorentz transform, so it definitely applies.

If you don't think so, then calculate how fast I would observe him moving away from me IF he had been traveling away from me and the sign.

Without doing the math, if he were traveling at .75 c away from the sign in the ground frame then he is going at the same speed as you so he would also be at rest in your frame.

Well, see you again in another hundred posts or so.

 

[James S Saint]

You're still trying to work out speed using clocks in the moving ship and distances measured from at rest. Like PassionFlower said earlier, if you use that logic you can conclude that you traveled faster than light if you make a 10 lightyear long journey (measured from rest) in only 5 years (experienced on your ship), since you're not accounting for time dilation. Relativity can't help it if you come to broken results like this.

Except that I wasn't using their clocks until someone suggested that I would have to. But if I did, the speed problem gets worse, not better because their clocks slow down, making them think they went even faster.

My original calculation was using the "ground speed" frame which yields a 1.5c rather than a 2.27c speed by using their clocks.

 

[James S Saint]

Without doing the math, if he were traveling at .75 c away from the sign in the ground frame then he is going at the same speed as you so he would also be at rest in your frame.

Oh, that's right. I got my brain on backwards this morning.. sorry.

So now the issue is why that equation results in something that doesn't make since. The excuse "oh but it does, you are just wrong" doesn't help a bit.

 

[Sumo]

Oh, that's right. I got my brain on backwards this morning.. sorry.

So now the issue is why that equation results in something that doesn't make since. The excuse "oh but it does, you are just wrong" doesn't help a bit.

Because as Doc Al pointed out, you sitting in your ship would only see the distance to your brothers ship as being 0.66 LS. And this distance is real; as you said, there is no absolute frame. So the fact that you measured the distance between the ships to be 2ls is only 2ls from a particular frame.

So 0.66ls / 0.88s = 0.75c, which is what you stated. You would see your brother moving towards you at 0.75c. (at least I think I got that math right)

* sorry, you would see the sign approaching at that speed.

 

[Bussani]

Except that I wasn't using their clocks until someone suggested that I would have to. But if I did, the speed problem gets worse, not better because their clocks slow down, making them think they went even faster.

My original calculation was using the "ground speed" frame which yields a 1.5c rather than a 2.27c speed by using their clocks.

You can't mix and match measurements from different frames, so don't even worry about using the clock in the ship and the distance measured from the ground. You'll never get a real result from that. Basically, because of time dilation and everything, it's perfectly acceptable for you to get somewhere in a shorter proper amount of time than should be possible. That doesn't break any universal laws.

As people have said from the start, going by the ground frame's measurements, the closing speed is allowed to be 1.5c. Because of the lack of data (that is, we don't have someone in the ground frame during the movement to tell us that both ships were moving) it's impossible to come to a concrete conclusion about who was moving how fast, but assuming that only one ship moved at 1.5c would be the most illogical conclusion to come to. Anyone with the right knowledge would realize that both ships would have to have moved to close the gap that fast.

You can't apply the rule of being able to consider yourself stationary when measuring the speeds of things relative to yourself when all of your measurements are coming from another reference frame. You can only conclude that nothing was going faster than c according to the frame those measurements came from. And as I said in the previous paragraph, that should lead you to realize that one person moving at 1.5c can't be right.

 

[Doc Al]

But wait.

If their clocks read .88secs, that would mean that they observed that they traversed 2Ls in .88 secs rather than 1.33 secs. That is even worse, instead of 1.5c, they would calculate/measure 2.27c ..??

Nope. You have to stick to a single frame to get sensible results. According to the ship frame, they only traveled a distance of 0.66 Ly. So they measure the speed of the approaching sign to be 0.66/.88 = 0.75c. (Of course.)

From the ground frame, the ship moves 1 Ly in 1.333 s, again a speed of 0.75c.

The ships do not traverse a distance of 2 Lys according to any frame in this problem.

 

[espen180]

But wait.

If their clocks read .88secs, that would mean that they observed that they traversed 2Ls in .88 secs rather than 1.33 secs. That is even worse, instead of 1.5c, they would calculate/measure 2.27c ..??

No, because in addition to time dilation, you have length contaction to worry about. If the observers mounted rulers on their ships and used them to measure the length of the track (2 ls in the ground frame), they would measure it to be 1.32 ls. You might think that this still constitutes v > c, but there is an additional effect, relativity of simultaneity, which says that once they begin to race, each observer will measure the other to have jumped ahead, spesifically such that their separation is 0.85 ls, so each ship measures the other ship's velocity to be 0.85Ls/0.88s=0.96c.

 

[James S Saint]

Because as Doc Al pointed out, you sitting in your ship would only see the distance to your brothers ship as being 0.66 LS. And this distance is real; as you said, there is no absolute frame. So the fact that you measured the distance between the ships to be 2ls is only 2ls from a particular frame.

So 0.66ls / 0.88s = 0.75c, which is what you stated. You would see your brother moving towards you at 0.75c. (at least I think I got that math right)

* sorry, you would see the sign approaching at that speed.

Nope. That doesn't apply. The distance is measured standing still. The clocks we can argue about because the clocks move and thus read different than a still clock. But I am getting that both would read that 2Ls would be traversed in less than light speed time.

 

[James S Saint]

As people have said from the start, going by the ground frame's measurements, the closing speed is allowed to be 1.5c. Because of the lack of data (that is, we don't have someone in the ground frame during the movement to tell us that both ships were moving) it's impossible to come to a concrete conclusion about who was moving how fast, but assuming that only one ship moved at 1.5c would be the most illogical conclusion to come to. Anyone with the right knowledge would realize that both ships would have to have moved to close the gap that fast.

Nono.. not only would you be presuming the outcome of a measure so as to make it "correct", but the theory is about the ability to observe anything traveling faster than light. We know from the outside perspective how it was that the distance got traversed in such short order, but THEY do not. They cannot know who was moving nor how much. Thus they observe a movement, speculated to be one of them, that they measure to be faster than light. The theory says that they could never get in that situation. Their measurements should always be less than light speed.