Is the Speed of Light Truly Unattainable in Relativity?

[DonB]

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?

 

[Pengwuino]

It does not make sense to speak of "what does a photon see?" and thus, "what velocity does a photon see". The idea of something observing a velocity depends on two things; the ability to measure distances and the ability to measure times. Neither of these things can be done for photons.

 

[I like Serena]

Welcome to PF, DrDon!

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)?

This very reasoning had been confusing physicists.
Then Einstein and others came with the revolutionary idea to say, what if c is constant for any observer?
Through what hoops do we have to wring the mathematics of it (Lorentz transformations), to make it work?

There! Relativity theory!

 

[jtbell]

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

When we say that A is moving at some velocity relative to B, we mean that A is moving at that velocity in a reference frame in which B is at rest. However, if B is a light beam or photon, there is no such (inertial) reference frame! Light travels at speed c in any inertial reference frame. So the general consensus is that it's meaningless to talk about the velocity of something relative to a photon or light beam.

Note also the following FAQ in the FAQ section at the top of this forum:

Rest frame of a photon

 

[DonB]

It does not make sense to speak of "what does a photon see?" and thus, "what velocity does a photon see". The idea of something observing a velocity depends on two things; the ability to measure distances and the ability to measure times. Neither of these things can be done for photons.

If I may interact, without intending (or appearing) to argue...

Your response seems to make relativity equal to perception -- what does something "see". But Einstein's arguments seem to indicate other he didn't feel this way, for he often referenced movement relative to the ground, a train, etc. If his references are valid, then why is it not equally valid to ref. to another inanimate object like light?

 

[DonB]

Welcome to PF, DrDon!

Then Einstein and others came with the revolutionary idea to say, what if c is constant for any observer?...

There! Relativity theory!

Thank you for the welcome.

That's what I'm saying..., if c is constant to any observer, then isn't each observer moving at that constant c relative to the light?

 

[ZapperZ]

Please start by reading the FAQ subforum in the Relativity forum.

Zz.

 

[WannabeNewton]

If I may interact, without intending (or appearing) to argue...

Your response seems to make relativity equal to perception -- what does something "see". But Einstein's arguments seem to indicate other he didn't feel this way, for he often referenced movement relative to the ground, a train, etc. If his references are valid, then why is it not equally valid to ref. to another inanimate object like light?

You cannot lorentz boost to a photon's reference frame so you cannot say you are moving at 'x' relative to the photon because for that you would need to have lorentz boosted to the photon's rest frame which doesn't exist and the lorentz transformations fail in the process.

 

[DonB]

Thanks for the response. If I may interact with your ideas, without intending (or appearing) to argue that I'm right (just trying to understand)...

When we say that A is moving at some velocity relative to B, we mean that A is moving at that velocity in a reference frame in which B is at rest.

This doesn't seem to match what I've read by Einstein. He gives examples of referencing movement of men who are walking or moving on trains (e.g., walking 3 MPH on a train moving 60 MPH).

Furthermore (I am asking in ignorance), do we really know that light is not at rest relative to its own frame of ref.?

However, if B is a light beam or photon, there is no such (inertial) reference frame!

Interesting. How do we know that? Is it just that it's a "theoretical" impossibility (which assumes all the thought supporting the theory is absolutely sound), or is there experimental evidence from which this is drawn?

Light travels at speed c in any inertial reference frame.

Interesting thought (just thinking out loud)... If one accepts this, and yet one has yet to prove by experimentation that light has no ref. frame, then arguably he has to allow the possibility that relativity requires that light must travel at c relative to itself. Interesting indeed!

Note also the following FAQ in the FAQ section at the top of this forum:

Rest frame of a photon

I thought I had read all the FAQ's before, but I may have missed this one. I'll check it out.

Thanks!

 

[DonB]

Please start by reading the FAQ subforum in the Relativity forum.

Zz.

Thanks Zz. As I explained above, I thought I had. Apparently I missed a section of it. My apologies.

 

[nitsuj]

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?

Who would agree that you are moving at c? It is c that is constant, your velocity varies. But again, even though your velocity changes you will always measure c as constant.

If that is your point, what is the consequence of saying as far as EM waves/photons go, you have always moved at c and they (photons) NEVER move? How long does that thought hold up for you? What movement do you perform when lights shine on the left and right of you, from the photons perspective ofcourse

 

[DonB]

Note also the following FAQ in the FAQ section at the top of this forum:

Rest frame of a photon

Thanks for pointing this link out. DH does a great job explaining this fascinating concept.

I know this will give me major points on the crackpot scale, but that's not where I'm coming from. I'm nothing more than a newbie genuinely trying to understand (and who can't find anyone locally who knows beans about this to help me out). DH concludes that trying to grasp the concept of a rest ref. frame for a photon is "undefined" and "indeterminate" (which I understand and do not question). But is it legit to resign ourselves to the conclusion that undefined and indeterminate means non-existent? Can I legitimately say that because I do not understand X=3/0 -- which is undefined -- that there is no expression "X=3/0"? In my naivety I understand "undefined" and "indeterminate" to speak of my personal inability (specifically, to wrap my mind around something), but not necessarily the inability of the item in question to exist. Am I missing something?

Thanks again for the link.

 

[DonB]

Who would agree that you are moving at c?

I don't know; that's why I'm asking in the first place.

It is c that is constant, your velocity varies.

But that first requires that we determine what the velocity is referenced to, right? Does my velocity vary in ref. to a beam of light, which as you say, is always constant?

But again, even though your velocity changes you will always measure c as constant.

If that is your point...

No, it isn't my point.

...what is the consequence of saying as far as EM waves/photons go, you have always moved at c and they (photons) NEVER move?

I have moved, and they have not moved..., relative to what? I probably misunderstand, but your question seems to argue for some absolute point of ref. that (in my understanding) relativity argues against. I thought that relativity argues that for two entities that are moving in ref. to each other, it is impossible to determine which is moving and which is at rest -- or, if in fact both are moving.

...What movement do you perform when lights shine on the left and right of you, from the photons perspective

None; but then again, there is no movement as I pass a friend as we both float through space. Am I moving? Or is he moving? Relativity seems to argue that answering that question is impossible without first determining the point of ref. I'm simply apply the same relativity principles to me passing a beam of light out in space..., and asking if there is some reason that that is different.

 

[nitsuj]

I don't know; that's why I'm asking in the first place.


But that first requires that we determine what the velocity is referenced to, right? Does my velocity vary in ref. to a beam of light, which as you say, is always constant?


No, it isn't my point.


I have moved, and they have not moved..., relative to what? I probably misunderstand, but your question seems to argue for some absolute point of ref. that (in my understanding) relativity argues against. I thought that relativity argues that for two entities that are moving in ref. to each other, it is impossible to determine which is moving and which is at rest -- or, if in fact both are moving.


None; but then again, there is no movement as I pass a friend as we both float through space. Am I moving? Or is he moving? Relativity seems to argue that answering that question is impossible without first determining the point of ref. I'm simply apply the same relativity principles to me passing a beam of light out in space..., and asking if there is some reason that that is different.

1.) it is for you to think about, if no one would agree you are moving at c, then you are not moving at c

2.) Yes in "reference" to the beam of light. You are suggesting that is what determind you move at c all the time. I am suggesting that isn't true, since you can move at different velocities and c doesn't.

3.)okay

4.) Relative to each other, you suggested a photon can be used as a frame of reference, which can't be done. I offered the thought of having a light shine on the left and right of you means you traveled towards the photons at c, while the light source traveled away at c, and that this happened on both sides of you at the same time.

5.)"I'm simply apply the same relativity principles to me passing a beam of light out in space..., and asking if there is some reason that that is different." Ah okay, so you now know you can't. That photon reference frame FAQ explains pretty clearly why it doesn't makes sense to use a photon as a rest frame.

I recently learned the difference between coordinate time and proper time. That helps me understand why a photon is not remotely avaible in coordinate time as a frame of reference for something that is subject to proper time. Turining the idea on it's head means nothing and despite that being a great "tool" for conceptualizing SR/GR, it can't be used for the laws of physics itself.

Laslty. EM is massless and HAS TO move at c, you are not massless and have to not move at c.

Lastly lastly, I'm a relative, relativity "newbie" too. It's fun stuff for sure!

 

[DonB]

1.) it is for you to think about, if no one would agree you are moving at c, then you are not moving at c

2.) Yes in "reference" to the beam of light. You are suggesting that is what determind you move at c all the time. I am suggesting that isn't true, since you can move at different velocities and c doesn't.

3.)okay

4.) Relative to each other, you suggested a photon can be used as a frame of reference, which can't be done. I offered the thought of having a light shine on the left and right of you means you traveled towards the photons at c, while the light source traveled away at c, and that this happened on both sides of you at the same time.

5.)"I'm simply apply the same relativity principles to me passing a beam of light out in space..., and asking if there is some reason that that is different." Ah okay, so you now know you can't. That photon reference frame FAQ explains pretty clearly why it doesn't makes sense to use a photon as a rest frame.

I recently learned the difference between coordinate time and proper time. That helps me understand why a photon is not remotely avaible in coordinate time as a frame of reference for something that is subject to proper time. Turining the idea on it's head means nothing and despite that being a great "tool" for conceptualizing SR/GR, it can't be used for the laws of physics itself.

Laslty. EM is massless and HAS TO move at c, you are not massless and have to not move at c.

1) So, you're saying that the reality of my speed is totally dependent upon whether others will agree that I am moving? Sounds to me like reasoning that I can't know how fast I'm traveling in my Jeep without getting someone else to tell me the speed.

2) No, that is not what I'm suggesting. Two rocks traveling through space are going a given speed relative to each other, even though neither one "determined" the speed of the other. As long as the two rocks exists, they are going a given speed relative to each other, and that speed is the same for A relative to B, or B relative to A. So, if we can validly claim that a beam of light is going at c relative to me, then my question is why can we not similarly say that I am moving at c relative to the light? As the photon passes by me, does it not "see" me passing by it at c?

4.a) I'm not arguing that your statement is wrong; but I am wanting to know how we know that it is right -- the 'why'.

4.b) Sorry, I just don't follow you.

5) Begging to differ in my ignorance, but as I spelled out in an earlier post, I'm missing how we can legitimately jump from something being "undefined" to conclude that it is "non-existent."

 

[WannabeNewton]

As long as the two rocks exists, they are going a given speed relative to each other, and that speed is the same for A relative to B, or B relative to A.

The reason you can say that is because you can freely lorentz boost from B to A and A to B so that you can evaluate relative speeds from the rest frames of A and B. You cannot lorentz boost to the frame of a photon so how do you suppose the same statement carries over?

 

[pervect]

You can assume that a photon has a reference frame like you're used to, or you can assume that Maxwell's equations work.

You can't do both.

Maxwell's equations are not consistent with having an unchanging electric field and an unchanging magnetic field just propagating through space. Which is what you'd get if you assumed a photon, somehow, had a reference frame.

People looked for along time at how to fix up Maxwell's equations without success, but there didn't appear to be anything wrong with them.

So, Einstein realized that the problem was ssuming (based on no particular necessity to do so) that a photon had a reference frame in the first place. The problem was solved.

Once it's been pointed out that it leads to a contradiction to assume that a photon has a reference frame, it seems to me to be not particularly productive to ask "why". It's a classic reducto ad absurdum proof - assume A, show that you get a contradiction, then you realize that A must be false.

It's only slightly more complicated here, because the proof that Maxwell's equations do in fact work is experimental.

 

[ghwellsjr]

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?

If you are A and a particular beam of light is moving at c relative to you and I am B and that same beam of light is moving at c relative to me and you and I are moving relative to each other, then how can we both be moving at c relative to that beam of light?

 

[DonB]

If you are A and a particular beam of light is moving at c relative to you and I am B and that same beam of light is moving at c relative to me and you and I are moving relative to each other, then how can we both be moving at c relative to that beam of light?

Good question. But the same thing can be equally said about turning the question around and asking how can light be moving at a constant c relative to both of us when we are moving relative to each other..., yet it is accepted nonetheless. So my question becomes, why accept the latter as a given while thinking the former impossible and unworthy of consideration?

 

[DonB]

You can assume that a photon has a reference frame like you're used to, or you can assume that Maxwell's equations work.

You can't do both.

Maxwell's equations are not consistent with having an unchanging electric field and an unchanging magnetic field just propagating through space. Which is what you'd get if you assumed a photon, somehow, had a reference frame...

Thank Pervect. This is the kind of "why" I was looking for. Now off to research what all Maxwell says.

 

[ghwellsjr]

Good question. But the same thing can be equally said about turning the question around and asking how can light be moving at a constant c relative to both of us when we are moving relative to each other..., yet it is accepted nonetheless. So my question becomes, why accept the latter as a given while thinking the former impossible and unworthy of consideration?

Are you aware that in the Theory of Special Relativity, the speed of light is only c for observers at rest within a Frame of Reference and that for moving observers the speed of light is not c?

 

[PAllen]

Are you aware that in the Theory of Special Relativity, the speed of light is only c for observers at rest within a Frame of Reference and that for moving observers the speed of light is not c?

What? The speed of light is c for all inertial observers (in vaccuum, etc). For accelerating observers (e.g. inside a rapidly accelerating rocket), it is still c as a local measurement. Nonlocal measurements for an accelerated observer run into the same issues of coordinate choice and conventions as GR.

What is 'at rest in a frame of reference'?

 

[nitsuj]

Are you aware that in the Theory of Special Relativity, the speed of light is only c for observers at rest within a Frame of Reference and that for moving observers the speed of light is not c?

What!??

ghwellsjr You have a unique ability to point out these blantant details in SR/GR that I miss.

I have always thought c is constant, and signed off on that as being...well...constant when perceiving things sr/gr.

What is a scenario that a moving observer wouldn't measure c to be constant?

 

[nitsuj]

5) Begging to differ in my ignorance, but as I spelled out in an earlier post, I'm missing how we can legitimately jump from something being "undefined" to conclude that it is "non-existent."

I won't be able to even remotely define for you why this is the way physics explains it i.e. conclude it's undefined/non-existant ect. But I will say from my understanding there is little difference between c and proper time. Given that, you suggest it is you who exists purely in the "time dimension" and that EM is in 3D.

see how physics conlcudes one of those two perspectives you offer is right.

 

[WannabeNewton]

What!??
What is a scenario that a moving observer wouldn't measure c to be constant?

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.

 

[ghwellsjr]

Are you aware that in the Theory of Special Relativity, the speed of light is only c for observers at rest within a Frame of Reference and that for moving observers the speed of light is not c?

What? The speed of light is c for all inertial observers (in vaccuum, etc). For accelerating observers (e.g. inside a rapidly accelerating rocket), it is still c as a local measurement. Nonlocal measurements for an accelerated observer run into the same issues of coordinate choice and conventions as GR.

What is 'at rest in a frame of reference'?

What!??

ghwellsjr You have a unique ability to point out these blantant details in SR/GR that I miss.

I have always thought c is constant, and signed off on that as being...well...constant when perceiving things sr/gr.

What is a scenario that a moving observer wouldn't measure c to be constant?

I wasn't talking about what any observer would measure as the speed of light, that will always be c because they are making a round trip measurement and this has nothing to do with the Theory of Special Relativity. It's what happens in nature. In SR, we pick a Frame of Reference and assign the one-way speed of light to be c and all clocks stationary in that frame are synchronized, running at the same rate and having the same time on them. This results in an observer at rest in that frame measuring the round trip speed of light to be c, but also the two parts of the round trip take the same time, meaning that the one-way speed of light for both halves of the trip is also c.

But for an observer in motion in that frame of reference, although he will also measure the round-trip speed of light to be c, the times that it take for the two parts of the trip are not equal and so the one-way speed of light is not c.

 

[PAllen]

I wasn't talking about what any observer would measure as the speed of light, that will always be c because they are making a round trip measurement and this has nothing to do with the Theory of Special Relativity. It's what happens in nature. In SR, we pick a Frame of Reference and assign the one-way speed of light to be c and all clocks stationary in that frame are synchronized, running at the same rate and having the same time on them. This results in an observer at rest in that frame measuring the round trip speed of light to be c, but also the two parts of the round trip take the same time, meaning that the one-way speed of light for both halves of the trip is also c.

But for an observer in motion in that frame of reference, although he will also measure the round-trip speed of light to be c, the times that it take for the two parts of the trip are not equal and so the one-way speed of light is not c.

That's a strange way of putting it. The observer in motion would use the same process to define their own frame, and come up with speed of light as c in all directions. Using reasonable measurement methods, no one measures speed of light different from c (in vaccuum), one way or two way. Are you trying to say that if B is in motion relative to A, A interprets B's measurement devices as distorted and de-synchronized relative to A's? (That, of course, is trivially true and clear).

 

[DrGreg]

But for an observer in motion in that frame of reference, although he will also measure the round-trip speed of light to be c, the times that it take for the two parts of the trip are not equal and so the one-way speed of light is not c.

This is a somewhat bizarre use of language, and as you can see by the reaction of other posters, potentially misleading. 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. It's an understood convention when we refer to an "observer" rather than some other description of an object.

The "speed" you are referring to is often referred to as "closing speed" between two objects (neither described as "observer"), to distinguish between the standard use of the term "speed".

And in fact on further thought, your claim is wrong anyway. In the circumstances you refer to, not only is the 1-way closing speed not c, the 2-way closing speed isn't c either. So I think I may have misunderstood you, and I've no idea what you are talking about (unless you are talking about a variant of Lorentz Ether Theory and not relativity at all).

 

[Pengwuino]

Good question. But the same thing can be equally said about turning the question around and asking how can light be moving at a constant c relative to both of us when we are moving relative to each other..., yet it is accepted nonetheless. So my question becomes, why accept the latter as a given while thinking the former impossible and unworthy of consideration?

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.

 

[ghwellsjr]

This is a somewhat bizarre use of language, and as you can see by the reaction of other posters, potentially misleading. 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. It's an understood convention when we refer to an "observer" rather than some other description of an object.

The "speed" you are referring to is often referred to as "closing speed" between two objects (neither described as "observer"), to distinguish between the standard use of the term "speed".

And in fact on further thought, your claim is wrong anyway. In the circumstances you refer to, not only is the 1-way closing speed not c, the 2-way closing speed isn't c either. So I think I may have misunderstood you, and I've no idea what you are talking about (unless you are talking about a variant of Lorentz Ether Theory and not relativity at all).

I think you haven't read carefully what I wrote and that is why you have misunderstood me.

I'm trying to answer DrDon's questions using his language and terminology. Sorry if it's bizarre.

 

[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?