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

[James S Saint]

Okay, you experts, please explain this one to me.. LOGICALLY.

I am traveling in my UFO along highway 40 at .75 times the speed of light. I am 1 light second (Ls) away from a speed limit sign. I know that I am traveling at that speed because I can see the speed limit sign coming at me at what appears to be .75 times the speed of light as measured by my clock. Of course, it appears that the sign is approaching me at .75c rather than me moving, but that isn't anything new to me.

But then I see my brother in his UFO rental coming in the opposite direction toward me and he is also 1 Ls away from the sign. As I observe, I can see that the distance between him and that same sign post is reducing at the same rate as the distance between me and the sign post. Thus, I conclude that he must have his foot on the accelerator with the same enthusiasm as I. He must be also traveling at .75 c.

We reach the sign at the same time because we were an equal distance from it and traveling at the same speed. It took me 1.333 seconds to make the 1 Ls distance at .75c and it took him that same amount of time ( = 1.333 sec).

So in 1.333 secs, he and I reached the sign.

But now there is where things seem to get a little confusing. The distance of 2 Ls between us got reduced to 0 in only 1.333 secs. That means that he traveled a 2Ls distance toward me in only 1.333 secs. That is 1.5 times the speed of light.

I can see and measure that he is approaching me at 1.5c... ? (the signpost has become irrelevant)

 

[Passionflower]

I can see and measure that he is approaching me at 1.5c... ?

You won't.

Velocity addition is not linear and, except in the simplest cases, neither commutative nor associative.

 

[HallsofIvy]

You cannot just assert "I can see and measure that he is approaching me at 1.5c". HOW would you "see and measure" that?

I would see and measure his speed as
\frac{.75c+ .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= \frac{1.5c}{1+ 0.5625}= \frac{1.5}{1.5625}c which is 96% the speed of light.

 

[James S Saint]

You won't.

Velocity addition is not linear and, except in the simplest cases, neither commutative nor associative.

Nonono.. wait.

I measured the distance that reduced concerning both he and I. It reduced the same amount in the same amount of time. I can validly observe that.

I can also observe that I am approaching the sign at .75c.

It doesn't matter what I would calculate because at the moment we reached each other, only a certain amount of time on my clock would have passed. It takes no mathematics except to see that in only one second, he, who was twice as far away as the sign, has reached me. That is NOT an issue of math. It is simple logic.

So again, LOGICALLY explain how he could each me in less time than a theoretical photon would have traveled the same distance?

 

[James S Saint]

You cannot just assert "I can see and measure that he is approaching me at 1.5c". HOW would you "see and measure" that?

I would see and measure his speed as
\frac{.75c+ .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= \frac{1.5c}{1+ 0.5625}= \frac{1.5}{1.5625}c which is 96% the speed of light.

I did not just assert it. I explained that I measured the distance being reduced at a rate. That is how anyone measures speed.

His distance reduced at the same rate as mine. That is observable. He reached me in less time than it would have taken a photon to travel that same distance. That is also measurable. Everything is validly measurable and yet the result is that he got to me faster than light would have if i were standing still like the sign post.

 

[Passionflower]

He reached me in less time than it would have taken a photon to travel that same distance.

An object with mass cannot reach an observer quicker than light, it would have to travel faster than light.

 

[James S Saint]

An object with mass cannot reach an observer quicker than light, it would have to travel faster than light.

Yes, I heard that rumor too. Now explain the scenario I presented.

 

[Passionflower]

that rumor

I see.
Are you here to learn or debunk relativity?

 

[James S Saint]

You cannot just assert "I can see and measure that he is approaching me at 1.5c". HOW would you "see and measure" that?

I would see and measure his speed as
\frac{.75c+ .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= \frac{1.5c}{1+ 0.5625}= \frac{1.5}{1.5625}c which is 96% the speed of light.

And btw, that math reflects me traveling to the sign and him traveling from the sign in the same direction. That is not my scenario.

 

[James S Saint]

I see.
Are you here to learn or debunk relativity?

I never deny logic, regardless of the fame of the professor, else I would be highly religious.

Explain the LOGIC, as requested. (please)

 

[alxm]

I never deny logic, regardless of the fame of the professor, else I would be highly religious.

Explain the LOGIC, as requested. (please)

There's no "LOGIC" here. If you are moving at 0.75c towards a sign in one direction, and the other guy is moving at 0.75c towards the sign in the other direction, then it does not follow 'logically' that your speed relative each other is 1.5c. That is an assumption you made on how relative velocities work. A faulty assumption.

 

[Dale]

I can see that the distance between him and that same sign post is reducing at the same rate as the distance between me and the sign post. Thus, I conclude that he must have his foot on the accelerator with the same enthusiasm as I. He must be also traveling at .75 c.

Here is the problem. You are starting with a flawed premise. In this arrangement the fastest your brother can possibly be closing with the sign is <0.25c.

In the sign's frame you could each be closing with the sign at 0.75c, but not in your frame.

 

[espen180]

But then I see my brother in his UFO rental coming in the opposite direction toward me. As I observe, I can see that the distance between him and that same sign post is reducing at the same rate as the distance between me and the sign post. Thus, I conclude that he must have his foot on the accelerator with the same enthusiasm as I. He must be also traveling at .75 c.

This is an impossible scenario.

If your brother was also traveling at .75c relative to the ground, you would measure his speed to .96c relative to you, while an observer stationary on the ground would measure his speed relative to you to be 1.5c.

In a nutshell. your argument is a bare assumption. You first assume that you measure his speed relaive to you to be 1.5c, then ask for an explanation. You also seem to assume that velocities can be linearly added together, which is wrong according to the Lorentz transformations.

You have to be careful about in which frame you report measurements, and not mix measurements from different frames.

 

[Doc Al]

I did not just assert it. I explained that I measured the distance being reduced at a rate. That is how anyone measures speed.

No, you merely asserted it based on how you think things work. But they don't work that way.

His distance reduced at the same rate as mine. That is observable. He reached me in less time than it would have taken a photon to travel that same distance. That is also measurable. Everything is validly measurable and yet the result is that he got to me faster than light would have if i were standing still like the sign post.

Nonsense. If you both were moving towards the sign at 0.75c with respect to the ground, then you'd measure that your brother is coming towards you at 0.96c with respect to you and thus you observe him getting closer to the sign at a rate of only 0.21c.

Explain the LOGIC, as requested. (please)

It's not LOGIC that's problem, it's lack of knowledge.

 

[James S Saint]

Okay guys, I can see that I need to add some detailed numbers to the scenario to make the point more clear. I have edited it with some details. Please reread it.

It doesn't require any sophisticated math

 

[Borek]

I can be completely wrong, I am far from my area of expertise and I even think I should shut up, but...

Is there anything wrong with the fact that the distance between two objects gets smaller faster than c? Neither of the objects moves faster than c, and my understanding is when they observe each other they won't see the other moving faster then c, and when observed from the outside neither moves faster than c. Just the distance changes faster then c, but that's something different.

Assuming that just because one of them moves at 0.75c means the other can get faster than 0.25c seems wrong to me. Following this line of thinking, what happens to the light emitted by our Sun? If it goes in the direction of the Earth with c, everything emitted in the opposite direction has to stop?

 

[espen180]

Okay guys, I can see that I need to add some detailed numbers to the scenario to make the point more clear. I have edited it with some details. Please reread it.

It doesn't require any sophisticated math

It still has the same problems.

 

[James S Saint]

I can be completely wrong, I am far from my area of expertise and I even think I should shut up, but...

Is there anything wrong with the fact that the distance between two objects gets smaller faster than c? Neither of the objects moves faster than c, and my understanding is when they observe each other they won't see the other moving faster then c, and when observed from the outside neither moves faster than c. Just the distance changes faster then c, but that's something different.

The distance getting smaller is the only measure of speed. So it does matter that a distance gets smaller faster than c.

 

[James S Saint]

It still has the same problems.

Please stop merely asserting that it must be wrong, and explain WHY. The logic seems perfect to me. The tiny bit of math is trivial. We are dealing with 2Ls, 1.333 secs, and 0 Ls.

 

[espen180]

I can be completely wrong, I am far from my area of expertise and I even think I should shut up, but...

Is there anything wrong with the fact that the distance between two objects gets smaller faster than c? Neither of the objects moves faster than c, and my understanding is when they observe each other they won't see the other moving faster then c, and when observed from the outside neither moves faster than c. Just the distance changes faster then c, but that's something different.

Assuming that just because one of them moves at 0.75c means the other can get faster than 0.25c seems wrong to me. Following this line of thinking, what happens to the light emitted by our Sun? If it goes in the direction of the Earth with c, everything emitted in the opposite direction has to stop?

Well, as you point out, different observers report different results. An observer standing by the signpost will observe both traveling at .75c and their relative speed to be 1.5c. No problem. James' statement is that the observers in the ships will measure their relative speed to be greater than c, which is erronous.

As for the light question, a photon doesn't have an intertial reference frame, you we cannot ask what the world looks like for a photon.

 

[espen180]

Please stop merely asserting that it must be wrong, and explain WHY. The logic seems perfect to me. The tiny bit of math is trivial. We are dealing with 2Ls, 1.333 secs, and 0 Ls.

Here is your logic.

1. You state "Observer A measures their relative speed to be >c.
2. You ask for an explanation.

There is none to give. You started with a physically impossible scenario. A will NOT observe B's speed to be .75c relative to the signpost!

 

[James S Saint]

If you try to use this famed formula;

\frac{.75c+ .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= \frac{1.5c}{1+ 0.5625}= \frac{1.5}{1.5625}c which is 96% the speed of light.

You actually have in MY scenario;

\frac{.75c+ -.75c}{1+ \frac{(.75c)(-.75c)}{c^2}}= \frac{0.0c}{1- 0.5625}= \frac{0.0}{0.4345}c

The formula doesn't really apply

 

[alxm]

Please stop merely asserting that it must be wrong, and explain WHY. The logic seems perfect to me.

You already got an explanation, repeatedly. There's no logic whatsoever in what you're saying, you're just blindly asserting that 'logically' speeds add up linearly, i.e. using simple addition.

If I get 5% bank interest one year and 10% the year after, is it 'logic' to assume your money has increased by 15%? Compound interest is not linear, either.

 

[Doc Al]

Okay guys, I can see that I need to add some detailed numbers to the scenario to make the point more clear. I have edited it with some details. Please reread it.

The numbers don't change anything.

It doesn't require any sophisticated math

More sophisticated than you think!

I am traveling in my UFO along highway 40 at .75 times the speed of light. I am 1 light second (Ls) away from a speed limit sign. I know that I am traveling at that speed because I can see the speed limit sign coming at me at what appears to be .75 times the speed of light as measured by my clock. Of course, it appears that the sign is approaching me at .75c rather than me moving, but that isn't anything new to me.

Nothing wrong here.

But then I see my brother in his UFO rental coming in the opposite direction toward me and he is also 1 Ls away from the sign.

Careful here: You are claiming that both you and your brother are at the same distance from the sign at the same instant. That's different than what you think it means. You are probably thinking of a scenario in which you and your brother are at the same distance and speed at a given instant with respect to the ground--but according to your measurements, you would not see him at the same distance from the sign at the same time according to you.

As I observe, I can see that the distance between him and that same sign post is reducing at the same rate as the distance between me and the sign post.

Sorry, no can do!

Thus, I conclude that he must have his foot on the accelerator with the same enthusiasm as I. He must be also traveling at .75 c.

Nope.

We reach the sign at the same time because we were an equal distance from it and traveling at the same speed. It took me 1.333 seconds to make the 1 Ls distance at .75c and it took him that same amount of time ( = 1.333 sec).

So in 1.333 secs, he and I reached the sign.

If you both were 1 Ls away from the sign as measured by ground observers (not you!) and you both traveled at 0.75c with respect to the ground, then it would take 1.333 seconds according to ground observers's clocks for you both to reach the sign. But according to you, the distance is only about 0.66 Ls (length contraction) and thus it only takes about 0.88 s to cover that distance.

But now there is where things seem to get a little confusing. The distance of 2 Ls between us got reduced to 0 in only 1.333 secs. That means that he traveled a 2Ls distance toward me in only 1.333 secs. That is 1.5 times the speed of light.

Not really. Again, you see the distances length contracted. But more important than that, you do not agree that your brother was at an equal distance from the sign at the same moment. According to you, at the moment that you were a certain distance from the sign, your brother was a lot closer. (This is the relativity of simultaneity at work. Trickier than you think!)

I can see and measure that he is approaching me at 1.5c... ? (the signpost has become irrelevant)

Nope. As already explained, if you both move towards the sign at 0.75c with respect to the ground, your speed with respect to each other is only 0.96c.

 

[espen180]

If you try to use this famed formula;


You actually have in MY scenario;

\frac{.75c+ -.75c}{1+ \frac{(.75c)(-.75c)}{c^2}}= \frac{0.0c}{1- 0.5625}= \frac{0.0}{0.4345}c

The formula doesn't really apply

You clearly have not taken the time to learn the theory properly.
Here's a read: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/veltran.html

 

[James S Saint]

You already got an explanation, repeatedly. There's no logic whatsoever in what you're saying, you're just blindly asserting that 'logically' speeds add up linearly, i.e. using simple addition.

If I get 5% bank interest one year and 10% the year after, is it 'logic' to assume your money has increased by 15%? Compound interest is not linear, either.

I have debunked the only attempts at explanation. The math was inappropriate in one post and the others merely proclaim that my story is wrong. No explanations are being given, only assertions.

If you cannot follow the simple logic requested and correct at what point an error was made, please don't bother to post.

 

[Doc Al]

If you try to use this famed formula;

I would see and measure his speed as
\frac{.75c+ .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= \frac{1.5c}{1+ 0.5625}= \frac{1.5}{1.5625}c which is 96% the speed of light.

You actually have in MY scenario;

\frac{.75c+ -.75c}{1+ \frac{(.75c)(-.75c)}{c^2}}= \frac{0.0c}{1- 0.5625}= \frac{0.0}{0.4345}c

The formula doesn't really apply

You just don't understand how to use that formula! That second speed is not the speed of your brother with respect to the ground (which would be -.75c), but the speed of the ground with respect to your brother, which is +.75c. Halls' use of the formula is correct.

 

[Doc Al]

I have debunked the only attempts at explanation. The math was inappropriate in one post and the others merely proclaim that my story is wrong. No explanations are being given, only assertions.

If you cannot follow the simple logic requested and correct at what point an error was made, please don't bother to post.

You've 'debunked' nothing. Your understanding is incorrect and your errors have been pointed out several times.

You actually have to learn a bit of physics! "Simple logic" is not enough.

 

[James S Saint]

Careful here: You are claiming that both you and your brother are at the same distance from the sign at the same instant. That's different than what you think it means.

Do I have to insert two more signposts, one a Ls further away where he is and another where I am at the time? The signposts would be exactly 2Ls apart and he and I are at those posts respectively at the same moment. That is the "setup" I do not need to measure that fact. That is the story itself.

but according to your measurements, you would not see him at the same distance from the sign at the same time according to you.

I do not need to "see" him at that sign to know that he is there. Theoretically we could have prearranged to startup and get at those points by merely accelerating at equal rates. By the same method, I can know that he is traveling at the .75c. I don't really have to "see" him.

Sorry, no can do!

? Why not??

 

[James S Saint]

You've 'debunked' nothing. Your understanding is incorrect and your errors have been pointed out several times.

You actually have to learn a bit of physics! "Simple logic" is not enough.

Sorry, but logic trumps physics. Without logic, there would be no physics.

 

[espen180]

I do not need to "see" him at that sign to know that he is there. Theoretically we could have prearranged to startup and get at those points by merely accelerating at equal rates. By the same method, I can know that he is traveling at the .75c. I don't really have to "see" him.

You don't seem to understand how this works. Both you and your brother will observe the other to be closer to the signpost. Also the .75 speed in wrt. the ground, you wrt. you.

Sorry, but logic trumps physics. Without logic, there would be no physics.

Sorry, this is wrong. Experiment trumps everything, and experiment doesn't agree with you.

 

[espen180]

Do I have to insert two more signposts, one a Ls further away where he is and another where I am at the time? The signposts would be exactly 2Ls apart and he and I are at those posts respectively at the same moment. That is the "setup" I do not need to measure that fact. That is the story itself.

The number of signposts doesn't change anything. You will still observe that when you get up to speed, the distance between the signposts has shrinked.

 

[Doc Al]

Do I have to insert two more signposts, one a Ls further away where he is and another where I am at the time? The signposts would be exactly 2Ls apart and he and I are at those posts respectively at the same moment. That is the "setup" I do not need to measure that fact. That is the story itself.

Apparently you've never heard of the 'relativity of simultaneity', one of the key features of special relativity. Frames in relative motion will not agree that two events took place at the same time! If someone on the ground measures that you were equidistant from the signpost at the same time, then you would not agree.

I do not need to "see" him at that sign to know that he is there. Theoretically we could have prearrange to startup and get at those points by merely accelerating at equal rates. By the same method, I can know that he is traveling at the .75c. I don't really have to "see" him.

You both are traveling at 0.75c with respect to the ground. In order to determine what you would measure from your moving reference frame, you must apply relativity.

 

[James S Saint]

You don't seem to understand how this works. Both you and your brother will observe the other to be closer to the signpost. Also the .75 speed in wrt. the ground, you wrt. you.

As stated before, I really don't actually have to see or measure him. I can know he is there at that speed merely by knowing that he accelerated at the same rate as I. I know the actual distance never changed because I can see that there is nothing to change it, so the idea that the distance would LOOK shorter is irrelevant even if true. The fact of the matter, regardless of what I see, is that we actually approach each other in less time than light would have traveled that same distance.

Sorry, this is wrong. Experiment trumps everything, and experiment doesn't agree with you.

That is absurdly incorrect but is a different thread.

The number of signposts doesn't change anything. You will still observe that when you get up to speed, the distance between the signposts has shrinked.

Observation of the distance is irrelevant to the fact that I know where he is even without observation. All you are claiming is that my perceptions will be skewed. The problem is that faster than light travel was achieved whether I observed the event taking place accurately or not. The end result is what matters.

 

[alxm]

Sorry, but logic trumps physics. Without logic, there would be no physics.

You haven't used any logic though. All you've done is assert that velocities add linearly. That is an assumption based on the fact that it's true (to a good approximation) at the everyday speeds you move at. It was also the assumption of physics up until 1905.

Let's make a different assumption: The speed of light in vacuum is measured to be the same by all observers, independently of their inertial frame of reference. It does follow logically from this assumption (together with the assumption that the laws of physics are the same regardless of inertia) that velocities do not add linearly. Either you have a length contraction (Lorentz's solution) or time dilation (Einstein's).

These are three different, but logically consistent viewpoints, starting from three different assumptions. The first assumption is false, it been repeatedly shown experimentally countless times since the Michelson-Morely experiment, that the speed of light is constant, regardless of the observer's speed. The other two assumptions lead to different predictions, and it turns out Einstein's assumption was the correct one. This has also been verified countless times, and has lead to predictions ranging from relativistic mass, to the existence of antiparticles, to the fact that gold is yellow.

The theory is completely logically consistent. If you have a problem with it, it's not the logic, it's the assumptions behind it. If you don't believe c is constant, then you better come up with an experiment to prove it. And you'll also need to find a theory that explains all the other results as well, because classical-mechanical velocity addition does not.

 

[espen180]

You are completely flawed in both your logic and physics skills and seem to be unwilling to learn.

That is absurdly incorrect but is a different thread.

If you are not willing to learn the accepted theory, there is hardly anything left to discuss here.

 

[James S Saint]

Apparently you've never heard of the 'relativity of simultaneity', one of the key features of special relativity. Frames in relative motion will not agree that two events took place at the same time! If someone on the ground measures that you were equidistant from the signpost at the same time, then you would not agree.


You both are traveling at 0.75c with respect to the ground. In order to determine what you would measure from your moving reference frame, you must apply relativity.

Like espen180, you are confusing the matter by claiming that my observation will be skewed. It doesn't matter what I observe other than the fact that we were doing the exact same thing in opposite directions and we got to a destination which required that the ACTUAL distance between us got reduced faster than light travels.

I could have said that we started our clocks at 4LS away and started up to travel the distance. We accelerated fast enough to ensure that we each reached the right posts when we were going .75c.

No observing is really necessary other than the beginning point and the end point, thus any skew in observation is actually irrelevant. Relativity of simultaneity doesn't apply because we would not be traveling with respect to each other at the beginning points and end point.

So what you are saying, is that to make the story clear, I have to leave out all observations and measurements and go merely by the start and end points, which I can do.

So what are you going to say then?

 

[Doc Al]

Observation of the distance is irrelevant to the fact that I know where he is even without observation.

You think you know, but you're just basing this on your pre-relativistic understanding of how things work.

All you are claiming is that my perceptions will be skewed.

This has nothing to do with 'skewed perceptions'; we're talking about measurements.

The problem is that faster than light travel was achieved whether I observed the event taking place accurately or not.

Uh, no.

 

[James S Saint]

You haven't used any logic though. All you've done is assert that velocities add linearly.

No, I haven't. You are apparently not reading carefully.

 

[James S Saint]

You think you know, but you're just basing this on your pre-relativistic understanding of how things work.

WHAT??

Come on now. He and I agree to behave exactly identical. That is how I know. That has NOTHING to do with relativity.

This has nothing to do with 'skewed perceptions'; we're talking about measurements.

Perceptions ARE measurements.

 

[Borg]

I don't really have to "see" him.

You need to think about it though. That is why you're missing the point. If you and your brother each shine a light when your cross the '1 second marks', where will you be when you see his? In the reference frame of an observer at the signpost, where does he think the ships are when he sees the lights? Once you understand those questions, you will begin to understand the answer to your question. Your logic is jumping between observers and you can't approach the problem that way.

But then I see my brother in his UFO rental coming in the opposite direction toward me and he is also 1 Ls away from the sign.

When you 'see' him cross his one second mark, you will have passed yours therefore the event that you think that you 'see' (you passing your one second mark when you think that he passes his) will never occur.

 

[Doc Al]

Like espen180, you are confusing the matter by claiming that my observation will be skewed. It doesn't matter what I observe other than the fact that we were doing the exact same thing in opposite directions and we got to a destination which required that the ACTUAL distance between us got reduced faster than light travels.

Careful with claims about ACTUAL distances--you do realize that distance is frame-dependent, right?

In any case, you are partially correct: The closing speed of the two ships according to ground observers is 1.5c. So what? This has nothing to do with anything moving at faster than light speeds in any frame, right?

That does not mean that you'll see your brother coming toward you at 1.5c according to your measurements. Of course, if all you are interested in is measurements made in the ground frame, then you can dispense with relativity.

 

[espen180]

WHAT??

Come on now. He and I agree to behave exactly identical. That is how I know. That has NOTHING to do with relativity.

It has everything to do with relativity. You cannot assume relativity is wrong and use that as an argument against relativity. That is called the http://en.wikipedia.org/wiki/Bare_assertion_fallacy" .

 

[Doc Al]

WHAT??

Come on now. He and I agree to behave exactly identical. That is how I know. That has NOTHING to do with relativity.

Everything is completely identical (except for your directions, of course) in the frame of the ground observers. But not necessarily in other frames.

Perceptions ARE measurements.

By 'measurements' I mean observations after correction for light travel time.

But it seems like you are just thinking about the closing speed of the two ships, which can certainly be greater than light speed. See my last post.

 

[James S Saint]

Careful with claims about ACTUAL distances--you do realize that distance is frame-dependent, right?

"Actual" in this case, now that I am talking about us both starting from a stand still, is whatever we first measured when standing on the ground. I can also merely say that we both stopped when we met, so the end reference would be the same as the beginning. IF we had started our clocks together and did everything exactly the same way, but in opposite directions, our CLOCKS would be identical and BOTH read that we narrowed the distance between us at faster than light speed.

So what? This has nothing to do with anything moving at faster than light speeds in any frame, right?

What it means is that if I measure the distances as stated, before and after, I have to conclude that the 2Ls distance between us reduced faster than light and thus, ignoring the ground (which is merely a different frame) the measure of our relative speed to each other ends up being faster than light.

So yes, it DOES mean that I "see" my brother get to me faster than light would have.

 

[Borg]

He must be also traveling at .75 c.

So yes, it DOES mean that I "see" my brother get to me faster than light would have.

You stated in the first post that your brother is traveling less than the speed of light. How does he shine a light and manage to get to you before the light? Or are you going to say that the speed of light is 1.75 times the speed of light...

 

[Austin0]

WHAT??

Come on now. He and I agree to behave exactly identical. That is how I know. That has NOTHING to do with relativity.


Perceptions ARE measurements.

AS Doc Al just pointed out what you are thinking is fine as applied from the signpost frame 1.5c no problem. But given your scenario if your brother passes you and you actually measure his velocity by clocks at the front and back of your ship you will find the velocity to be as stated by others previously.
You are just mentally putting yourself on the ground and then assuming the same perceptions would apply in your ship. Logically true perhaps but not in compliance with what you would actually measure.

 

[Doc Al]

"Actual" in this case, now that I am talking about us both starting from a stand still, is whatever we first measured when standing on the ground.

You are talking about distances as measured by the ground frame.

I can also merely say that we both stopped when we met, so the end reference would be the same as the beginning. IF we had started our clocks together and did everything exactly the same way, but in opposite directions, our CLOCKS would be identical and BOTH read that we narrowed the distance between us at faster than light speed.

The fact that the distance between ships as measured by the ground frame will change at faster than light speed does not mean that you, using your own measuring rods and clocks, would measure the speed of the other ship with respect to you at greater than light speed.

What it means is that if I measure the distances as stated, before and after, I have to conclude that the 2Ls distance between us reduced faster than light and thus, ignoring the ground (which is merely a different frame) the measure of our relative speed to each other ends up being faster than light.

Again, when you speak about distances and times you must specify a reference frame.

So yes, it DOES mean that I "see" my brother get to me faster than light would have.

No it doesn't, except in the sense already discussed: The closing speed as measured by ground observers (not you!) will be 1.5c. The relative speed of the ships--the speed of one ship in the frame of the other--will be 0.96c.

These distinctions are crucial to understanding what relativity is saying.

 

[Bussani]

Like Doc Al's pointed out, relativity doesn't forbid the distance between the two ships from receding at a faster than light rate according to a ground observer. That's nothing special.

Side question, if I may ask: who would reach the sign first according to all 3 observers ("you" in the vehicle, your brother in the oncoming vehicle, and a person standing beside the sign)? I assume if the person by the sign sees them reach it simultaneously, the other two won't agree with him.

 

[James S Saint]

You stated in the first post that your brother is traveling less than the speed of light. How does he shine a light and manage to get to you before the light?

I wasn't referring to HIM shining the light. I was referring to the fact that there was 2Ls between us and 1.333 secs later, there was no distance between us. Light couldn't do that.Guys...

Do we agree that there is no absolute frame?

The ground merely serves as an initial means to measure the distance between us both before and after. It is NOT an absolute frame, right? It merely let's us have the same reference when we start and end.

The fact is that we were 2Ls apart and merely 1.333 secs later, we were together by BOTH clocks. We can ignore the ground. Forget the ground.

Speed is measured by distance divided by time => 2Ls/1.333secs = 1.5c