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

[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

 

[Borg]

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.

Please explain what you think you see if he does shine a light when he crosses that point. Do you think that you would see his light traveling at 1.75 times the speed of light?

 

[Doc Al]

Do we agree that there is no absolute frame?

Right!

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.

Nothing absolute about the ground frame, that's for sure!

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.

What do you mean by 'forget the ground'? You gave distances and times that only make sense in the frame of the ground! Specifying a distance without mentioning the frame you are using is meaningless, right? (Unless you secretly do believe in some absolute frame!)

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

Again, that's the closing speed according to the ground frame. You can't 'forget the ground'.

 

[Bussani]

Forget the ground.

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

You can't say, "Forget the ground," and then use distances according to the ground.

 

[DrGreg]

Speed is measured by distance divided by time

Speed relative to a frame is measured by distance in the same frame divided by time in the same frame.

You are mixing frames.

 

[James S Saint]

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.

Yes, but I'm not talking about what a ground observer sees. I see the distance of 2Ls get reduced to 0 in only 1.333 secs. That is what matters. Perhaps I cannot actually see the ground at all or even know it exists. As stated just prior, the ground only serves as an equal frame for us to begin and end. Both of us would end up seeing that 2Ls of distance vanished in only 1.333 secs.

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.

All observers would see us meet at the sign. The ground person would see each of us traveling .75c toward the sign. We would each see ourselves approaching the sign at .75c. But regardless of what we might observe of each other, our end measurements would be that we reached each other in only 1.333 secs causing a 2Ls distance to be traversed.

 

[espen180]

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

You are - whether you ralize it or not - working in the ground frame, in which the relative speed of the ships is only constricted to be smaller than 2c, and in which your result is correct. You cannot, however, extrapolate the measurement of the ground frame to the ship frame without using the Lorentz transformation equations, which will tell you that according to an observer on either of the ships, the other ship is approaching at .96c.

 

[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.

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.

No, it doesn't. If you and he were traveling in the same direction, at the same .7t c, then it would be
\frac{.75c- .75c}{1+ \frac{(.75c)(.75c)}{c^2}}= 0, of course.

 

[James S Saint]

Again, that's the closing speed according to the ground frame. You can't 'forget the ground'.

No it is according to US.

We measure 2Ls of distance. In merely 1.333 secs, we measure NO distance between us. How is that not 1.5c?

 

[jcsd]

Yes, but I'm not talking about what a ground observer sees. I see the distance of 2Ls get reduced to 0 in only 1.333 secs. That is what matters. Perhaps I cannot actually see the ground at all or even know it exists. As stated just prior, the ground only serves as an equal frame for us to begin and end. Both of us would end up seeing that 2Ls of distance vanished in only 1.333 secs.


All observers would see us meet at the sign. The ground person would see each of us traveling .75c toward the sign. We would each see ourselves approaching the sign at .75c. But regardless of what we might observe of each other, our end measurements would be that we reached each other in only 1.333 secs causing a 2Ls distance to be traversed.

But think what happens in the ground observers frame when one of you shines a light at the other, the light, traveling at the speed of light, reaches the other before the pair of you meet.

Is the light therefore traveling at greater than 1.5 times the speed of light? That would be totally illogical as we started by assuming it travels at the speed of light.

 

[HallsofIvy]

Wow, 57 posts in less than 3 hours! That may be a record. James S Saint, you are doing all of your computations as if Newtonian physics applied. Naturally, you are going to get a contradiction to relativity.