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.