Is the Speed of Light Truly Unattainable in Relativity?

[DonB]

...Almost always, when we speak of velocity measured by an observer, we mean as measured in the observer's rest frame, not somebody else's frame...

DrGreg, your response brings up a tangent question I hope you (or others) can answer -- a question I don't recall thinking of before. Accepting that I, as an at-rest observer (but "at rest" relative to what?), see a light beam approaching me at c, if I were to towards or away from that beam, would I still see it at c, or at c plus/minus my own velocity?

 

[DonB]

The latter is experimentally verified and consistent with Maxwell's Equations, which have been verified to ridiculous accuracy. If you assumed the former, nothing would make sense; high energy physics could not be explained at all.

Without experiment, there's nothing wrong with either idea. In fact, the former makes much more sense logically. Why should light be so special? But physics is an experimental science and every experiment we have ever done, contrary to what we expect and what feels "right", shows that the latter explanation as light being constant to all reference frames is the correct one.

Thank Pengwuino. I understand and accept (at least to the degree that I grasp any of this) your explanation. My main point was to explain the reason behind having the question in the first place..., but certainly not to argue its experimental validity.

 

[Pengwuino]

DrGreg, your response brings up a tangent question I hope you (or others) can answer -- a question I don't recall thinking of before. Accepting that I, as an at-rest observer (but "at rest" relative to what?), see a light beam approaching me at c, if I were to towards or away from that beam, would I still see it at c, or at c plus/minus my own velocity?

You define yourself to be at rest. All your measurements must be taken in your frame of reference.

If you moved away or towards the beam with a constant velocity (ie while continuing to be an inertial observer), the speed of light would remain 'c'.

 

[DonB]

You define yourself to be at rest. All your measurements must be taken in your frame of reference.

If you moved away or towards the beam with a constant velocity (ie while continuing to be an inertial observer), the speed of light would remain 'c'.

Okay..., that's what I had assumed the answer would be. But it brings up the question, why specific "at rest"? From the sounds of your response, whether I'm walking or not walking, it's all the same.

 

[Pengwuino]

Okay..., that's what I had assumed the answer would be. But it brings up the question, why specific "at rest"? From the sounds of your response, whether I'm walking or not walking, it's all the same.

It's usually the simplest. Walking at constant velocity works as well perfectly fine (which is at the core of special relativity). However, if you accelerate, you no longer are an inertial observer and everything mentioned already no longer applies. When you talk about accelerations, you can no longer freely move between frames of reference and see the same physics occurring.

 

[nitsuj]

Okay..., that's what I had assumed the answer would be. But it brings up the question, why specific "at rest"? From the sounds of your response, whether I'm walking or not walking, it's all the same.

Im gunna throw a guess out there and say it's because..., again, c is constant. As you accelerate your measurements change as your velocity changes relative to what they once were.

Again, not from math, I think the graphical representation is 4 dimensions, you either move through time (rest frame) or space & time. So as you continue to accelerate, your frame of spacetime properties are changing from what they once were, c is staying the same. If you stay still, you will have constant whatever properties (momentum, and other junk) and will again have the opportunity to measure c on a level playing field.

Is that right'ish Pengwuino? or anyone

 

[WannabeNewton]

DrGreg, your response brings up a tangent question I hope you (or others) can answer -- a question I don't recall thinking of before. Accepting that I, as an at-rest observer (but "at rest" relative to what?), see a light beam approaching me at c, if I were to towards or away from that beam, would I still see it at c, or at c plus/minus my own velocity?

Vector addition is different in SR than the one in euclidean space: http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity

 

[Pengwuino]

Im gunna throw a guess out there and say it's because..., again, c is constant. As you accelerate your measurements change as your velocity changes relative to what they once were.

Again, not from math, I think the graphical representation is 4 dimensions, you either move through time (rest frame) or space & time. So as you continue to accelerate, your frame of spacetime properties are changing from what they once were, c is staying the same. If you stay still, you will have constant whatever properties (momentum, and other junk) and will again have the opportunity to measure c on a level playing field.

Is that right'ish Pengwuino? or anyone

I can't exactly understand what you're trying to say.

Vector addition is different in SR than the one in euclidean space: http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity

This isn't really what he's asking since velocity addition in that sense talks about moving between different frames of reference and measuring massive particles movement. If you're looking at how the speed of light changes as you move at constant velocities, it doesn't.

 

[nitsuj]

Well the speed of light is measured as c in inertial frames, or locally flat space i.e. where the general metric reduces to the minkowski metric, so global measurements wouldn't give c because you can't even define velocity properly globally due to the path dependence of parallel transport.

If you call that a measurement so be it.

 

[nitsuj]

I can't exactly understand what you're trying to say.

To say it differently, as you accelerate that Lorentz transformation thing happens. I imagine as you "continuously" accelerate the transformations occur congruently and that is what would spoil the measurements in calculating c is constant. (at rest you are in the same "frame" as c, proper time)

 

[WannabeNewton]

This isn't really what he's asking since velocity addition in that sense talks about moving between different frames of reference and measuring massive particles movement. If you're looking at how the speed of light changes as you move at constant velocities, it doesn't.

He asked if he would measure a speed of light greater or less than c if he were to move along with or away from a beam of light and the reason it remains c no matter what is easily seen through the velocity - addition formula of SR.

 

[ghwellsjr]

DrGreg, your response brings up a tangent question I hope you (or others) can answer -- a question I don't recall thinking of before. Accepting that I, as an at-rest observer (but "at rest" relative to what?), see a light beam approaching me at c, if I were to towards or away from that beam, would I still see it at c, or at c plus/minus my own velocity?

Have you ever thought about what it means to "see a light beam approaching" you and how you would measure its speed? Has it occurred to you that you cannot see the beam, all you can see is the portion of it when it reaches you? If you're thinking about seeing a beam from a search light, for example, the only reason why you can see it as a beam is because there are dust particles in the air that are illuminated by the beam itself and then scattered in all directions and you're seeing the ones that are scattered in your direction. But if you were in the vacuum of space, you would not see any beam; you would just see the spot where it hits your eye or your detector. It may look like it is coming from a source far away, but you cannot see the image of that source until the light from it reaches you, correct?

So now, how do you measure its speed since you cannot see it until it reaches you? First off, it doesn't matter if you are "at rest" relative to anything, just that you are not accelerating. Secondly, you don't have to worry about any frame of reference or any theory about relativity or ether or anything else. Thirdly, you don't have to worry about the source of light, just that it is not accelerating, which just means that it is not changing.

Now in order to actually measure the speed of light coming from a fixed distant source, you need to have some equipment. Since speed is length divided time, you need to have a rigid measuring stick of a known fixed length and a timing device. Since the beam is coming at you all the time, you also need a shutter so that you can start and stop the beam to give you something to observe and know when the light has traveled a certain distance. One last thing you need is a mirror. So you fix your shutter at one end of your measuring stick along with your timer which you start together. The pulse of light now travels to the other end of your measuring stick where it reflects off the mirror and back to you where you have a light detector that stops the timer. Then you calculate the speed of light as twice the distance between your shutter/detector and your mirror, divided by the time interval. The value that you get will be c.

Now you fire your rockets and head toward the light source until you reach whatever speed you want. You turn off your thrusters and you repeat your measurement exactly as you did before and you get the exact same value for the speed of light.

Now you turn your rocket around and head away from the light source until you are going in the other direction from your first measurement and go as fast as you want. You stop, turn around and repeat the measurement. You get the same answer.

Do you accept this as a factual statement of what would really happen if you could carry out this experiment?

 

[Pengwuino]

He asked if he would measure a speed of light greater or less than c if he were to move along with or away from a beam of light and the reason it remains c no matter what is easily seen through the velocity - addition formula of SR.

Hmm, I seem to enjoy complicating things beyond what's needed and possibly making them nonsensical. You're right.

 

[nitsuj]

Accepting that I, as an at-rest observer (but "at rest" relative to what?), see a light beam approaching me at c, if I were to towards or away from that beam, would I still see it at c, or at c plus/minus my own velocity?

I think it would be at rest to time itself. c. rate/flow,ect At one with the continuum

Qm seems to say that it's impossible to predict velocity/position and some other things simultaneously. Almost like it is physically impossible to pin down the point of "now" (as we know it).

 

[DonB]

It's usually the simplest. Walking at constant velocity works as well perfectly fine (which is at the core of special relativity). However, if you accelerate, you no longer are an inertial observer and everything mentioned already no longer applies. When you talk about accelerations, you can no longer freely move between frames of reference and see the same physics occurring.

Okay, so the only real distinction that need to be made is accelerating or not accelerating (which differentiates using SR or GR). Other distinctions do not matter. I can go with that.

 

[DonB]

Have you ever thought about what it means to "see a light beam approaching" you and how you would measure its speed? Has it occurred to you that you cannot see the beam, all you can see is the portion of it when it reaches you? ...

Forgive me, for I have taken the lazy route and not stated myself as specifically as I suppose I should have. I was not particularly interested in all the mechanics. Rather, I simply wanted to know that as light approached me at c (relative to me) while I am "at rest", would that relative-to-me speed change if I were no longer at rest (e.g., if I were walking towards or away from the beam). What I really saw, or did not see, was really a mute point. My apologies if my casual description conveyed the wrong idea.

 

[Dale]

I simply wanted to know that as light approached me at c (relative to me) while I am "at rest", would that relative-to-me speed change if I were no longer at rest (e.g., if I were walking towards or away from the beam).

The phrase "relative to me" means "in the frame where I am at rest". In that frame it makes no sense to speak of walking towards or away from the beam. You are stationary in that frame by definition.

 

[jtbell]

as light approached me at c (relative to me) while I am "at rest", would that relative-to-me speed change if I were no longer at rest (e.g., if I were walking towards or away from the beam).

No. Regardless of whether you are standing still, or walking towards or away from the source of the light, or running towards or away from it, or flying in a spaceship at 0.9c towards or away from it, the light always moves at c towards you.

 

[DonB]

No. Regardless of whether you are standing still, or walking towards or away from the source of the light, or running towards or away from it, or flying in a spaceship at 0.9c towards or away from it, the light always moves at c towards you.

Thanks. Like I mentioned to an earlier reply, this is what I thought relativity held to; but the regular mention of an object being "at rest" made me question this, and so I just needed verification.

 

[ghwellsjr]

Has your original question been answered?

 

[DonB]

Has your original question been answered?

Thanks for asking. I would have to answer your question is both yes and no. The "yes" part is that, I'm told, that light has not ref. frame, and thus my movement relative to it and can not be determined. The "no" part is this: at any micro-fraction of a nano second the distance between the beam and me is a definite, measurable distance. Thus, at both T1 and T2 (Time 1 and Time 2) that distance can be measured, and with delta D factored with delta T, a speed can be determined at which, relative to the beam, I am moving (or, conversely, it is moving relative to me). Maybe this is something that I just have to accept for now.

 

[PAllen]

Thanks for asking. I would have to answer your question is both yes and no. The "yes" part is that, I'm told, that light has not ref. frame, and thus my movement relative to it and can not be determined. The "no" part is this: at any micro-fraction of a nano second the distance between the beam and me is a definite, measurable distance. Thus, at both T1 and T2 (Time 1 and Time 2) that distance can be measured, and with delta D factored with delta T, a speed can be determined at which, relative to the beam, I am moving (or, conversely, it is moving relative to me). Maybe this is something that I just have to accept for now.

But these are your distances and times, and lead to you computing the light moves at c. You need to ask about the light's measurements of distance and time, which you've been told are undefinable. In the limit as v->c, you get 0/0.

 

[ghwellsjr]

Thanks for asking. I would have to answer your question is both yes and no. The "yes" part is that, I'm told, that light has not ref. frame, and thus my movement relative to it and can not be determined. The "no" part is this: at any micro-fraction of a nano second the distance between the beam and me is a definite, measurable distance. Thus, at both T1 and T2 (Time 1 and Time 2) that distance can be measured, and with delta D factored with delta T, a speed can be determined at which, relative to the beam, I am moving (or, conversely, it is moving relative to me). Maybe this is something that I just have to accept for now.

If you're going back to my description of how to measure the speed of a light beam from post #42 and you are concerned about the time it takes for the shutter/detector to transmit the start (at T1) and stop (at T2) signals to the timer, as long as it is the same for both times, then it won't have any bearing on delta T, the difference beween between T2 and T1.

If you're not going back to post #42, then I have no idea what you're talking about so could you elaborate?

 

[DonB]

But these are your distances and times, and lead to you computing the light moves at c. You need to ask about the light's measurements of distance and time, which you've been told are undefinable. In the limit as v->c, you get 0/0.

So, you're saying that due to the warp in time/space from the light's traveling at c, the distance from me to the light beam is different from my perspective than it is from the light's perspective? Guess I hadn't grasped that part of relativity before.

I don't know if you've noticed, but I ask a question about this 0/0 (based upon DH's FAQ article) in a separate thread.

 

[DonB]

If you're not going back to post #42, then I have no idea what you're talking about so could you elaborate?

It seems that PAllen identified my problem area (post #52).

 

[kevinfrankly]

I'm a relative newbie to relativity (no pun intended -- I know you've heard that one too many times), as well as to this forum, so forgive me if this is a dumb question...

As I understand it, there is a significant percentage of those that believe that people, spaceships, etc. will never travel at the speed of light (c), right?

If that is so, where is my logic (below) faulty:

1. For any two given things (A and B), the speed of A relative to B will always be the same as the speed of B relative to A.

2. Relativity insists that light always travels at c relative to all things.

Thus, if light is moving at c relative to me, then am I not moving at c relative to that particular beam of light (...and, in fact, all beams of light at any given moment in time)? And thus, the speed of light is not only something that is not unattainable, but has rather never been unattained?

What am I missing?

CMIIW
isnt one of einstein postulate said that the speed of light is constant independent of the observer, or in other words the speed of light is just the same for any observer.

 

[Dale]

1. For any two given things (A and B), the speed of A relative to B will always be the same as the speed of B relative to A.

This is because the inverse of a boost is a boost of the same speed (and opposite direction).

2. Relativity insists that light always travels at c relative to all things.

Thus, if light is moving at c relative to me, then am I not moving at c relative to that particular beam of light

There is no boost to c.

Another way to think of it is that there is a symmetry between two massive observers. There is no sense in which one is "at rest" that doesn't apply equally well to the other, i.e. each can be boosted into a frame where the spacelike components of its four-velocity are 0. There is no such symmetry between a massive observer and light, the massive observer is timelike and the light is lightlike. Light cannot be boosted into a frame where the spacelike components are non-zero.

 

[DonB]

But these are your distances and times, and lead to you computing the light moves at c. You need to ask about the light's measurements of distance and time, which you've been told are undefinable. In the limit as v->c, you get 0/0.

After sleeping on this, I suppose that ultimately what you are saying is that Statement 1 of my original post in this thread (1. For any two given things (A and B), the speed of A relative to B will always be the same as the speed of B relative to A) -- a statement that I think I picked up for Einstein's own material -- is not always true. Specifically, for currently attainable speeds it is true, but as an object approaches the speed of light this becomes less and less true. Would that be a fair conclusion..., and thus point out the flaw in the original logic?

 

[PAllen]

After sleeping on this, I suppose that ultimately what you are saying is that Statement 1 of my original post in this thread (1. For any two given things (A and B), the speed of A relative to B will always be the same as the speed of B relative to A) -- a statement that I think I picked up for Einstein's own material -- is not always true. Specifically, for currently attainable speeds it is true, but as an object approaches the speed of light this becomes less and less true. Would that be a fair conclusion..., and thus point out the flaw in the original logic?

No, it is exactly true for any speed less than c. It is meaningless for c. The relativity statement you are mis-quoting was between material bodies which always move less than c relative to each other. You have the following properties, actually:

1) If B is moving relative to A at less than c, then A's motion relative to B is the same (in the opposite direction).

2) If B is moving at c relative to A, it is moving at c relative to all matter, and there is no such thing as motion relative to B. Further, B has no rest mass and no rest frame.