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

[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.

 

[Doc Al]

But I am getting that both would read that 2Ls would be traversed in less than light speed time.

You still don't get it. In the ground frame, the ships move 1 Ly in 1.333 s--a speed of 0.75c. You can calculate a closing speed of 1.5c, but nothing is actually moving at that speed.

 

[Sumo]

Nope. That doesn't apply. The distance is measured standing still.

But why do you insist on an absolute measure of distance? One of the premises that you agreed on was that this doesn't exist. Yet here you are talking as if the distance measured standing still is somehow the only real one.

 

[Doc Al]

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.

If the ships do not know anything about a ground frame or any measurements made in such a frame, then the best they can do is measure their speeds relative to each other. That speed, for the nth time, is 0.96c, not 1.5c. No one measures anything to move faster than light.

 

[James S Saint]

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.

No
No
No
You cannot change what they already measured while they were standing still. They did not make a measurement while they were moving such as to perceive a shorter distance.

They measured standing "still" 2Ls. They looked at both clocks and got that the distance vanished in faster than light time. How much faster is all we were calculating.

THE ONLY FRAME OF MEASUREMENT IS THE INITIAL FRAME. They make NO measurements while moving. Thus the distance, as they calculate at the end and from the beginning remains at 2Ls. Granted the clocks slow down. But that only makes it seem faster.

 

[Mentz114]

THE ONLY FRAME OF MEASUREMENT IS THE INITIAL FRAME. They make NO measurements while moving. Thus the distance, as they calculate at the end and from the beginning remains at 2Ls. Granted the clocks slow down. But that only makes it seem faster.

No, if you use their clocks, you must use the corresponding contracted distance to calculate the velocity.

 

[Doc Al]

OK, I think this thread has gone on long enough. I don't see any progress being made here. You've said the same thing many times, and you've been corrected many times. Time to stop this merry-go-round.