Lorentz factor equal to infinity + lots of questions

[fluidistic]

I've had my second class of Modern Physics and I have some "obvious" doubts that many of you certainly had, although I didn't find anything answering the first doubt on a quick google search.
1)Say I am in an inertial reference frame. I can see that any photon going around me do it at a constant speed equal to c. Therefore, I'm tempted to say that if I choose a reference frame on a moving photon, it would be inertial. But the problem would be in calculating anything that involve distances and time with respect to the photon's frame of reference. Some Lorentz transformation would involve the Lorentz factor $$\gamma$$ which would be equal to $$\infty$$ which is impossible. So it seems that the reference frame of any photon cannot ever be inertial. Is this right? If so, why?!
2)Another question is: If I'm accelerating in a laser's beam (toward the source of the laser), would I see the photons hitting me with a speed faster than c? I know that I would observe a Doppler effect and thus the momentum of the photons would probably be greater than if I wasn't accelerating, but I'm not sure since I've never studied those things yet.

3)How can I know if I'm an inertial reference frame of reference? I'm guessing by looking at any photons hitting my system and measure their speed.

4) In the twin paradox, one man stays on Earth while the other travels with respect to Earth at great speeds. I think that by using Lorentz transformations, one can see that an interval of time of the man in space correspond to a lesser interval of time than the one of the man on Earth. That's why when the man in space comes back to Earth they have aged differently and that the space man is less old than the Earth's man. My question is: Looking only at the 2 men, we can't say that one is moving while the other doesn't. Maybe the space man can say that he's in an inertial reference frame? (or not due to changes in acceleration because he's getting away from Earth and thus the gravitational force is less strong.)


I hope these are not too many questions, I'm a newbie in Modern Physics but I'd appreciate any kind of answers (even complicated ones) and also if you answer "only" one question.
Thanks!

 

[bcrowell]

2)Another question is: If I'm accelerating in a laser's beam (toward the source of the laser), would I see the photons hitting me with a speed faster than c? I know that I would observe a Doppler effect and thus the momentum of the photons would probably be greater than if I wasn't accelerating, but I'm not sure since I've never studied those things yet.

Relativistic combination of velocities doesn't happen by straight addition. Any velocity combined with c equals c.

3)How can I know if I'm an inertial reference frame of reference? I'm guessing by looking at any photons hitting my system and measure their speed.

Great question! The answer is that you can't. This is one of the considerations that led Einstein from special relativity to general relativity.

4) In the twin paradox, one man stays on Earth while the other travels with respect to Earth at great speeds. I think that by using Lorentz transformations, one can see that an interval of time of the man in space correspond to a lesser interval of time than the one of the man on Earth. That's why when the man in space comes back to Earth they have aged differently and that the space man is less old than the Earth's man. My question is: Looking only at the 2 men, we can't say that one is moving while the other doesn't. Maybe the space man can say that he's in an inertial reference frame? (or not due to changes in acceleration because he's getting away from Earth and thus the gravitational force is less strong.)

Even though there is no fundamental way to tell the difference between an inertial frame and a noninertial frame, there is still a way to tell the difference between a geodesic (world-line of something that's free-falling) and a non-geodesic. The stay-at-home twin has a world-line that's a geodesic. The traveling twin doesn't, and he can tell this by looking at an accelerometer.

Say I am in an inertial reference frame. I can see that any photon going around me do it at a constant speed equal to c. Therefore, I'm tempted to say that if I choose a reference frame on a moving photon, it would be inertial. But the problem would be in calculating anything that involve distances and time with respect to the photon's frame of reference. Some Lorentz transformation would involve the Lorentz factor $$\gamma$$ which would be equal to $$\infty$$ which is impossible. So it seems that the reference frame of any photon cannot ever be inertial. Is this right? If so, why?!

FAQ: What does the world look like in a frame of reference moving at the speed of light?

This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.)

What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.

 

[fluidistic]

Thanks a lot bcrowell for your detailed reply.
I'm still confused about the second question that you answered by "Relativistic combination of velocities doesn't happen by straight addition. Any velocity combined with c equals c.". Ok, I've no problem with this, but then why is Einstein's second postulates of Special Relativity

As measured in any inertial frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

instead of

As measured in any frame of reference, light is always propagated in empty space with a definite velocity c that is independent of the state of motion of the emitting body.

?

Correct me if I'm wrong, but it makes no sense to think about choosing a frame of reference as being "over" a photon since no observer could ever be there so it's thinking about an impossibility.
Ok for the geodesic, seems interesting.

 

[bcrowell]

Yeah, that's a good question. Special relativity (SR) is the special case of relativity where there's no gravity. In SR there is a clear distinction between inertial and noninertial frames, just as there is in Newtonian mechanics.

But in general relativity (GR) we have the equivalence principle. As an example of the e.p., imagine that you're in an elevator, and you feel pressure between your feet and the floor. There is no way for you to tell whether (i) you're in a nonaccelerating elevator immersed in a gravitational field, or (ii) you're in an accelerating elevator with no ambient field. Therefore you can't tell whether you're in an inertial frame or not. Possibility i doesn't arise in SR because gravity isn't considered as a possibility.

 

[fluidistic]

Ok thanks a lot for all. My doubts vanished for now.

 

[Passionflower]

3)How can I know if I'm an inertial reference frame of reference? I'm guessing by looking at any photons hitting my system and measure their speed.

Great question! The answer is that you can't. This is one of the considerations that led Einstein from special relativity to general relativity.

That is not correct.

The speed of light measured by an inertial frame is constant but it is not constant if the frame is not inertial. If we know the speed of light is c and we measure it and it is not c we must conclude we are in a non-inertial frame.

 

[Dead Boss]

The stay-at-home twin has a world-line that's a geodesic.

Really? I thought that only free-falling observer is following a geodesic. If observer stands on ground, he has a non-zero acceleration. Please, correct me if I'm wrong.

 

[Fredrik]

Really? I thought that only free-falling observer is following a geodesic. If observer stands on ground, he has a non-zero acceleration. Please, correct me if I'm wrong.

What you're saying is correct in GR, but the twin paradox is a scenario in SR, where massive objects have no effect on spacetime geometry.

 

[fluidistic]

That is not correct.

The speed of light measured by an inertial frame is constant but it is not constant if the frame is not inertial. If we know the speed of light is c and we measure it and it is not c we must conclude we are in a non-inertial frame.

Can someone confirm this? That was a guess of mine, but I've been showed wrong.

 

[George Jones]

Yeah, that's a good question. Special relativity (SR) is the special case of relativity where there's no gravity. In SR there is a clear distinction between inertial and noninertial frames, just as there is in Newtonian mechanics.

But in general relativity (GR) we have the equivalence principle. As an example of the e.p., imagine that you're in an elevator, and you feel pressure between your feet and the floor. There is no way for you to tell whether (i) you're in a nonaccelerating elevator immersed in a gravitational field, or (ii) you're in an accelerating elevator with no ambient field. Therefore you can't tell whether you're in an inertial frame or not. Possibility i doesn't arise in SR because gravity isn't considered as a possibility.

I am not going to talk about frames, but I will talk about observers. An observer in an accelerating rocket in special relativity and an observer hovering in a static gravitational field (general relativity) are both non-inertial observers; both observers have non-zero proper acceleration.

Non-zero 4-acceleration is indicated by an accelerometer that consists of two main parts, a hollow sphere like a basketball inside of which is a slightly smaller sphere. Initially, the centres of the spheres coincide, so that there is a small, uniform (vacuum) gap between the spheres. When acceleration is non-zero, the gap will be closed, contact between the spheres will be made, and an alarm that indicates non-inertial motion will sound. If the ship is not accelerating, no alarm will sound.

Place the accelerometer in an accelerating rocket and the alarm sounds. Place the accelerometer near the surface of the Earth, and assume that the accelerometer is small enough that tidal forces can be neglected. When the accelerometer is held at rest (in space), the alarm sounds, but if the accelerometer falls freely, no alarm sounds.

In both special and general relativity, the accelerometer sounds the alarm when 4-acceleration is non-zero, i.e., when an observer is non-inertial.

That is not correct.

The speed of light measured by an inertial frame is constant but it is not constant if the frame is not inertial. If we know the speed of light is c and we measure it and it is not c we must conclude we are in a non-inertial frame.

The local speed of light as measured by all observers, inertial and non-inertial, is c. The coodinate speed of light can be anything.

 

[Passionflower]

The local speed of light as measured by all observers, inertial and non-inertial, is c.

True, although one cannot measure the speed of light locally one needs at least some spatial distance.

In an accelerating frame an increasing radar distance between the measurement positions increases (or decreases if measured in the opposite direction) the measured speed of light. Obviously when this distance approaches zero the measured speed will also approach c.

For each accelerating (with constant proper acceleration) frame there exists a region where the measured speed of light approaches zero.

 

[bcrowell]

Really? I thought that only free-falling observer is following a geodesic. If observer stands on ground, he has a non-zero acceleration. Please, correct me if I'm wrong.

Sure, you're technically correct. To get rid of that issue, have the stay-at-home twin live on the International Space Station.

In both special and general relativity, the accelerometer sounds the alarm when 4-acceleration is non-zero, i.e., when an observer is non-inertial.

Maybe we just have different definitions of "inertial" in mind. There are three theories running around here, so let's consider what the word would mean in each theory.

In Newtonian mechanics, the standard definition of "inertial" says that a person standing on the ground is in an inertial frame, while a person in an accelerating elevator is not. This contradicts the criterion you've used in the quote above.

In SR, there are no gravitational interactions, so the contradiction between your definition and the standard definition in the Newtonian context becomes irrelevant.

In GR, I think what you're calling "inertial" is what I would refer to as "free-falling" or "geodesic."

So it looks to me like there may just be some confusion here because when I use "inertial," I mean the traditional Newtonian definition of the term. There are clearly two different notions here, which we could call "Newtonian-inertial" and "geodesic-inertial." Is it standard in GR to use the bare term "inertial" to mean "geodesic-inertial," counting on context to imply that it doesn't mean "Newtonian-inertial?"

My #2 read like this:

Even though there is no fundamental way to tell the difference between an inertial frame and a noninertial frame, there is still a way to tell the difference between a geodesic (world-line of something that's free-falling) and a non-geodesic. The stay-at-home twin has a world-line that's a geodesic. The traveling twin doesn't, and he can tell this by looking at an accelerometer.

Maybe it would be more consistent with standard usage to phrase it like this:

Even though there is no fundamental way to tell the difference between a frame that Newton would consider to be accelerating and one that he would consider to be nonaccelerating, there is still a way to tell the difference between a geodesic (world-line of something that's free-falling) and a non-geodesic. The stay-at-home twin has a world-line that's a geodesic. The traveling twin doesn't, and he can tell this by looking at an accelerometer.

 

[bcrowell]

3)How can I know if I'm an inertial reference frame of reference? I'm guessing by looking at any photons hitting my system and measure their speed.

Great question! The answer is that you can't. This is one of the considerations that led Einstein from special relativity to general relativity.

That is not correct.

The speed of light measured by an inertial frame is constant but it is not constant if the frame is not inertial. If we know the speed of light is c and we measure it and it is not c we must conclude we are in a non-inertial frame.

I've probably caused a huge amount of confusion here by using "inertial" in a way that may not be standard. Actually there is also the question of what fluidistic meant by the term: the Newtonian definition, or the relativistic notion of geodesic motion.

Here's my shot at writing a clearer answer.

You definitely can't tell anything by measuring the speed of photons, because it's always c. It doesn't matter what your frame of reference is doing; no matter what, you'll always measure c as the speed of light. (That doesn't refer to coordinate velocities, but velocities measured by local experiments, in which we essentially measure the speed of light relative to our apparatus, as it passes directly by or through our apparatus.)

There is no local experiment that will distinguish between what Newton considered to be an inertial frame and what Newton considered to be a noninertial frame. This is the equivalence principle.

Local experiments can distinguish between free-falling frames and non-free-falling frames. For example, you can use an accelerometer.

 

[Passionflower]

You definitely can't tell anything by measuring the speed of photons, because it's always c. It doesn't matter what your frame of reference is doing; no matter what, you'll always measure c as the speed of light. (That doesn't refer to coordinate velocities, but velocities measured by local experiments, in which we essentially measure the speed of light relative to our apparatus, as it passes directly by or through our apparatus.)

First, please define what you mean by local.

How do you explain that the ruler distance and radar distance of a Born rigid object undergoing a constant proper acceleration is different if you keep assuming that the measured speed of light from one edge to the other is equal for both inertial and accelerating motion?

 

[Fredrik]

I would define a local measurement as a measurement performed in a region of spacetime that's so small that the effects of curvature and acceleration are negligible. I would also say that GR is defined in terms of local measurements. (I don't know if any of the books cover this well. These are my own thoughts on the subject). A theory is defined by a set of axioms that tells us how to interpret the mathematics as predictions about results of experiments. Einstein's equation can't define GR on its own (just as Minkowski spacetime can't define SR on its own). We also need to specify how to perform measurements of length and time.

Time isn't a problem. In both SR and GR, we can take the time axiom to be "A clock measures the proper time of the curve in spacetime that represents its motion". Length is more complicated, because of how curvature and acceleration can deform a measuring device in a region of spacetime where those things aren't negligible.

The only way I know to deal with the problem is to avoid it, by only defining measurements of very short distances using radar: "The distance between the reflection event and the midpoint between the emission and detection events is equal to half the time between emission and detection". This is of course only approximately true, but the approximation becomes exact if we let that time go to zero.

So by my definitions, the "measurement" you (Passionflower) were talking about (measuring the speed of light using a device that's being deformed by its own acceleration) is just an incorrectly performed measurement. This doesn't necessarily mean that what you said was wrong, but it does mean that we need a more general definition of "length measurement" before we can even discuss it.

 

[Passionflower]

So by my definitions, the "measurement" you (Passionflower) were talking about (measuring the speed of light using a device that's being deformed by its own acceleration) is just an incorrectly performed measurement.

That is not what I wrote at all, it has absolutely nothing to do with some 'deformation' of the measuring apparatus!

Do you perhaps disagree with the clock hypothesis?

The explanation why the ruler and radar distance is different for a constantly accelerating object (with a length > 0) is very simple: the speed of light is simply different in an accelerating frame than in an inertial frame. And this is fully consistent with SR and GR.

 

[Fredrik]

That is not what I wrote at all, it has absolutely nothing to do with some 'deformation' of the measuring apparatus!
...
the speed of light is simply different in an accelerating frame than in an inertial frame.

Perhaps you should define what you mean by an "accelerating frame". I would define it using the radar notion of simultaneity (even when the reflection event is far away). This is what MTW calls a "proper reference frame" of an object, and what Dolby & Gull used in that article that often gets mentioned in the twin paradox threads. In that "accelerating coordinate system", the world line of a ray that passes through the origin has slope 1 at the origin, so there is no way that a measurement of the speed of light by a small enough measuring device can get a value that is significantly different from 1.

To get a different value, you need either an unorthodox definition of "accelerating coordinate system", or a larger measuring device.

There are of course coordinate systems where the speed of light has any value you want it to, but you spoke specifically about accelerating coordinate systems, not about non-inertial coordinate systems in general.

You may be right that deformation of the measuring device is irrelevant in this specific scenario. I don't have time to think about that now. But we clearly have to account for deformations in some situations, if we insist on using a more general notion of "measurement" than the local measurements I described.

Do you perhaps disagree with the clock hypothesis?

I have no idea why you're asking me that. I'm talking about special relativity and general relativity, not some "alternative" theory. Did you see what I wrote about time measurements in those theories?

 

[Passionflower]

Perhaps you should define what you mean by an "accelerating frame".

You could simply replace 'accelerating frame' with 'accelerating object' in that sentence if that makes it any clearer for you.

With regards to my question to you about the clock hypothesis: You mentioned something about acceleration 'deforming' the measurement apparatus. The measurement apparatus to measure the two way speed of light is simply a clock (and a mirror). If you assume acceleration 'deforms' such a clock you must clearly have a problem with the clock hypothesis.

I am truly at a loss what the objections to my writings are here. I simply corrected an inaccuracy someone made and I get replies that the inaccuracy is accurate in the limit or your reply about some deformation of the measurement apparatus which is in my mind not even in question here.

Added:

so there is no way that a measurement of the speed of light by a small enough measuring device can get a value that is significantly different from 1.

So it seems that you understand what I am talking about.
Am I correct that you actually agree with me that, at least in principle, one can determine if one travels non-inertially by comparing the measured speed of light with the speed in a non-inertial frame?

I wrote:

True, although one cannot measure the speed of light locally one needs at least some spatial distance.

In an accelerating frame an increasing radar distance between the measurement positions increases (or decreases if measured in the opposite direction) the measured speed of light. Obviously when this distance approaches zero the measured speed will also approach c.

Which explains the issue: as this distance approaches zero, the measured speed of light will approach the measured speed of light in an inertial frame.

 

[Fredrik]

You could simply replace 'accelerating frame' with 'accelerating object' in that sentence if that makes it any clearer for you.

Sounds like you mean "measuring device that was designed for local measurements but is now accelerating". In that case, accelerating coordinate systems aren't involved at all.

The measurement apparatus to measure the two way speed of light is simply a clock (and a mirror).

This method is also used to define which set of events is the x-axis of the accelerating coordinate system, and to determine the assignment of x coordinates to points on the x axis. That assignment is such that the speed of light is always 1. That's how the accelerating coordinate system is defined.

If you assume acceleration 'deforms' such a clock you must clearly have a problem with the clock hypothesis.

Violations of the clock hypothesis contradict both SR and GR. Nothing I said does that. I also stated exactly what SR and GR (as I define those theories) say that a clock measures, and my axiom clearly doesn't leave any room for violations of the clock hypothesis. So it makes no sense to say that I "must" have a problem with it. To say that I do is to claim that I'm not even talking about SR or GR.

Also, as you said yourself, there measuring device has other parts than just a clock. If you e.g. attach two detectors that light can pass through to opposite ends of a ruler, the deformation of that ruler by inertial forces will be a major concern. If you intend to reflect light off a mirror, you need a solid object to hold it in place. That object can be deformed. If you intend to have the mirror floating freely instead, you need some other way to keep track of how far away it is. The obvious way (and probably the only way) is to use light, but that doesn't make sense if you intend to measure the speed of light. (If the speed of light would change, you would use a different estimate of the distance to the mirror and end up with the same result for the calculated speed of light anyway).

I am truly at a loss what the objections to my writings are here.

Since the slope of the world line of a ray of light at the origin is 1, the speed of light is 1 (=c) in the accelerating coordinate system too. It's wrong to say that it isn't, and you did say that "it is not constant if the frame is not inertial".

It's true that if we take a measuring device that was designed for local measurements, and use it when it's accelerating, it might give us a result that's different from 1 for the speed of light, but that doesn't mean that the speed of light isn't 1 in the accelerating coordinate system. It just means that a device designed for local measurements doesn't work when it's accelerating.

Edit: I wrote all of the above before I saw this addition to your post:

Am I correct that you actually agree with me that, at least in principle, one can determine if one travels non-inertially by comparing the measured speed of light with the speed in a non-inertial frame?

I agree that we can use a measuring device designed for local measurements to find out if we're doing inertial motion or not, but I would say that it's because such a device doesn't work properly when it's doing non-inertial motion. I do not agree that the result is the speed of light in the accelerating frame.

Edit 2: I haven't actually worked out what the result will be if we use a local measuring device that's doing non-inertial motion, so at the moment I can only say that I expect the result to be ≠1 in general, but perhaps it will be =1 in some special cases, like constant proper acceleration that's small enough to keep component parts in approximately Born rigid motion.

 

[Passionflower]

So let's get down to brass tacks: In your opinion if we have a spaceship with on the floor a clock and at the roof a mirror to measure the roundtrip time of light do we, yes or no, get exactly the same result when the spaceship is traveling inertially or traveling Born rigid with a constant acceleration?

You know my answer: it is no, not exactly.

 

[Fredrik]

So let's get down to brass tacks: In your opinion if we have a spaceship with on the floor a clock and at the roof a mirror to measure the roundtrip time of light do we, yes or no, get exactly the same result when I am traveling inertially or traveling Born rigid with a constant acceleration?

I discussed this specific problem with Hurkyl two years ago. (Weird...feels more like 5 years ago). Link. He chose to consider a rocket of proper length 1, so when it's doing inertial motion, the roundtrip time (tail-head-tail or head-tail-head) is 2. He then calculated the roundtrip times for Born rigid motion with the rear of the rocket having constant proper acceleration 1, and got the results tail-head-tail=1.39 and head-tail-head 2.77. So if his calculations are correct, and I think they are, the roundtrip times when the rocket is accelerating are clearly different from the roundtrip times when the rocket is doing inertial motion.

 

[Dale]

Great question! The answer is that you can't. This is one of the considerations that led Einstein from special relativity to general relativity.

I disagree. An inertial frame in SR is, by definition, a frame where Newton's laws hold good meaning that inertial objects (attached accelerometer reads 0) move in straight lines at constant speed (worldline has the form (t,vt)) and also where Maxwell's equations hold good (light moves at a coordinate speed of c). Both of these conditions can be determined through measurements.

EDIT: I see Passionflower already mentioned that.

 

[DrGreg]

Passionflower, when relativists talk about a "local" measurement, the rigorous interpretation of this is that you make measurements and calculation within a small region of spacetime and consider the calculus limit as the size of region shrinks to zero. Practically it means that if the region is small enough, the difference between the measured value and the limiting value can be considered negligible. It is in this "local" sense we can say all observers (free-falling or not) measure the same local speed of light.

Bear in mind that the very definition of (instantaneous) speed is a calculus limit of an average speed anyway, so to say the local (two-way) speed of light is constant isn't being loose with words, it is technically correct.

 

[Austin0]

Passionflower, when relativists talk about a "local" measurement, the rigorous interpretation of this is that you make measurements and calculation within a small region of spacetime and consider the calculus limit as the size of region shrinks to zero. Practically it means that if the region is small enough, the difference between the measured value and the limiting value can be considered negligible. It is in this "local" sense we can say all observers (free-falling or not) measure the same local speed of light.

Bear in mind that the very definition of (instantaneous) speed is a calculus limit of an average speed anyway, so to say the local (two-way) speed of light is constant isn't being loose with words, it is technically correct.

How does this equate with measurements taken at significantly different R distnaces in a Born accelerating system if the clocks are running at non-negligably different rates?

 

[DrGreg]

How does this equate with measurements taken at significantly different R distnaces in a Born accelerating system if the clocks are running at non-negligably different rates?

What I said assumes that local measurements are made using "locally correct rulers" (measuring proper distance) and "locally correct clocks" (measuring proper time). The Rindler coordinate clocks (in general) run fast or slow compared with proper time, so the coordinate speed of light is generally not equal to c in Rindler coordinates. But the local speed (using local proper rulers and local proper clocks) is always c.

 

[fluidistic]

Ok so if I'm not misunderstanding, according to PassionFlower the answer to my question 2 would be a yes. Is this right? I would indeed measure a speed greater than c for the photons hitting me since I'm accelerating toward them.
And my answer to my question 3 would also be a yes? If I measure that the speed of light in vacuum is not exactly c, then it means that I'm not in an inertial frame (and thus I'm accelerating. I don't know with respect to what I'm accelerating, but I would detect an acceleration in an accelerometer which seems what matters).

 

[bcrowell]

Great question! The answer is that you can't. This is one of the considerations that led Einstein from special relativity to general relativity.

I disagree. An inertial frame in SR is, by definition, a frame where Newton's laws hold good meaning that inertial objects (attached accelerometer reads 0) move in straight lines at constant speed (worldline has the form (t,vt)) and also where Maxwell's equations hold good (light moves at a coordinate speed of c). Both of these conditions can be determined through measurements.

EDIT: I see Passionflower already mentioned that.

See my #12 and #13. I created confusion by using "inertial" to refer to the Newtonian notion of an inertial frame.

 

[bcrowell]

Ok so if I'm not misunderstanding, according to PassionFlower the answer to my question 2 would be a yes. Is this right? I would indeed measure a speed greater than c for the photons hitting me since I'm accelerating toward them.

If that's what passionflower is saying, then passionflower is incorrect. All observers measure the speed of light to be c when they make local measurements. It doesn't matter whether the observers are accelerating.

 

[bcrowell]

First, please define what you mean by local.

See DrGreg's #23.

How do you explain that the ruler distance and radar distance of a Born rigid object undergoing a constant proper acceleration is different if you keep assuming that the measured speed of light from one edge to the other is equal for both inertial and accelerating motion?

It's not an assumption, it's an observed fact.

The reason your counterexample isn't a counterexample is that your experiment isn't local, in the sense of DrGreg's #23. The effect on the value of c that you get in your example is proportional to the length of the apparatus, L. When we talk about measuring c locally, we mean $c=\lim_{L\rightarrow0} c_L$, where $c_L$ is the experimental result found with an apparatus of length L.

[EDIT] By the way, I don't think the use of the Born-rigid object is necessary in order to come up with examples of this type, and it leads to some logical problems.

First the logical problem. A Born-rigid object is not just an object made out of a strong material, it's an object being acted on by a set of prearranged external forces that are chosen so as to prevent the object from distorting as seen in a particular frame. In order to prearrange those forces, you have to establish a notion of simultaneity, as in, "OK, everybody push *now*." But to establish simultaneity you probably need to do the kind of radar measurements involved in Einstein synchronization, which take the constancy of the speed of light as a matter of definition.

To see that it's not necessary to invoke Born-rigidity to get an example of this kind, just consider the Sagnac effect. It is possible for light to take a different amount of time to go around a triangle in the clockwise direction than it takes for it to go around in the counterclockwise direction. This happens whenever the apparatus is rotating. This establishes that the speed of light can't possibly constant in the nonlocal sense. But the effect is proportional to the area of the triangle, so the speed of light is still constant in the local sense, i.e., in the limit as the size of the apparatus approaches zero.

 

[Fredrik]

Ok so if I'm not misunderstanding, according to PassionFlower the answer to my question 2 would be a yes. Is this right? I would indeed measure a speed greater than c for the photons hitting me since I'm accelerating toward them.

If that's what passionflower is saying, then passionflower is incorrect. All observers measure the speed of light to be c when they make local measurements. It doesn't matter whether the observers are accelerating.

I think what Passionflower had in mind was to take a measuring device that correctly measures the speed of light to be c when it's not accelerating, and use it while it is accelerating. This would in general not qualify as a local measurement, and the result would be >c. (He didn't say that the result of a local measurement would be ≠c).

This doesn't mean that the speed of light in "the accelerating coordinate system" is >c.

 

[Passionflower]

Ok so if I'm not misunderstanding, according to PassionFlower the answer to my question 2 would be a yes.

Yes, using a clock, emitter, and a mirror you would, at least in principle, be able to detect a different speed of light. Some may have a problem with this measurement, it is not 'local' enough for them, so it is 'not valid' they insist that to 'really' measure the speed of light one has to put the mirror against the emitter so that they can proclaim: 'see acceleration makes no difference'.

If we have a Born rigid (why Born rigid? Just to keep it from being needlessly complicated) spaceship and we measure the two way speed of light from the floor to the ceiling and back the result depends on two factors:

1. The rate (and direction) of proper acceleration
2. The height of the spaceship

In addition to a change in the measured speed of light we can also, at least in principle, determine that the radar distance and ruler distance between the floor and the ceiling are no longer identical if the spaceship is accelerating.

 

[Austin0]

What I said assumes that local measurements are made using "locally correct rulers" (measuring proper distance) and "locally correct clocks" (measuring proper time). The Rindler coordinate clocks (in general) run fast or slow compared with proper time, so the coordinate speed of light is generally not equal to c in Rindler coordinates. But the local speed (using local proper rulers and local proper clocks) is always c.

I don't quite follow you here. What is the defnition of locally proper clocks and rulers in this context?
I can see that extreme localization of measurement would minimize the effects of the acceleration differential between the front and the back but don't follow how this would eliminate the effects of the difference in clock rate which would maintain the same proportionality independant of the length of the measurement.
By proper are you meaning a clock in the CMIF for that location??

 

[Fredrik]

I don't quite follow you here. What is the defnition of locally proper clocks and rulers in this context?
I can see that extreme localization of measurement would minimize the effects of the acceleration differential between the front and the back but don't follow how this would eliminate the effects of the difference in clock rate which would maintain the same proportionality independant of the length of the measurement.
By proper are you meaning a clock in the CMIF for that location??

He didn't say "locally proper", he said locally "correct". Think of it as clocks that are small enough for curvature and acceleration to be irrelevant, and rulers with distance markings that on small scales agree with radar measurements. Proper time is defined here. A clock measures the proper time of the curve in spacetime that represents its motion.

 

[DrGreg]

I don't quite follow you here. What is the defnition of locally proper clocks and rulers in this context?
I can see that extreme localization of measurement would minimize the effects of the acceleration differential between the front and the back but don't follow how this would eliminate the effects of the difference in clock rate which would maintain the same proportionality independant of the length of the measurement.
By proper are you meaning a clock in the CMIF for that location??

I'm just saying you use your own clock measuring your own personal time, not a "Rindler coordinate clock" that deliberately runs at the wrong speed to keep itself synced to the observer's "master clock". Time how long it takes for light to reflected back to you from a mirror and calculate the average round-trip speed, and take the limit as the distance to the mirror drops to zero. You'll always get c.

 

[Passionflower]

Time how long it takes for light to reflected back to you from a mirror and calculate the average round-trip speed, and take the limit as the distance to the mirror drops to zero. You'll always get c.

Yes but the limit is not the only thing that is interesting, when we increase the distance to the mirror we get different values and the direction of measurement is also a factor.

For instance if we have a Born rigid spaceship with proper constant acceleration and we want to measure the roundtrip time between the front and the back then we find that the roundtrip time measured from the front is longer than measured from the back of the spaceship.

We can even, in principle, test this with massive objects as well (were we assume a fully elastic reflection), then if the initial proper velocity is $> 1/\sqrt{2}$ the result if similar as for light but if the velocity is $< 1/\sqrt{2}$ the roundtrip time measured from the front is shorter than measured from the back of the spaceship. When the velocity is exactly $1/\sqrt{2}$ the two directions are of equal duration, this speed seems to 'ignore' gravitation.

All this is independent of the actual rate of acceleration. And of course, by the equivalence principle, the same would happen if we measure this in a tower.