# Lorentz factor equal to infinity + lots of questions

1. Aug 12, 2010

### 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!

2. Aug 12, 2010

### bcrowell

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

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

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.

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.

3. Aug 12, 2010

### fluidistic

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
?

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.

4. Aug 12, 2010

### bcrowell

Staff Emeritus
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.

5. Aug 12, 2010

### fluidistic

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

6. Aug 13, 2010

### Passionflower

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.

7. Aug 13, 2010

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.

8. Aug 13, 2010

### Fredrik

Staff Emeritus
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.

9. Aug 13, 2010

### fluidistic

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

10. Aug 13, 2010

### George Jones

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

Last edited: Aug 13, 2010
11. Aug 13, 2010

### Passionflower

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.

Last edited: Aug 13, 2010
12. Aug 13, 2010

### bcrowell

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

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?"

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

13. Aug 13, 2010

### bcrowell

Staff Emeritus
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.

14. Aug 13, 2010

### Passionflower

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?

15. Aug 14, 2010

### Fredrik

Staff Emeritus
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.

Last edited: Aug 14, 2010
16. Aug 14, 2010

### Passionflower

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.

17. Aug 14, 2010

### Fredrik

Staff Emeritus
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.

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?

Last edited: Aug 14, 2010
18. Aug 14, 2010

### Passionflower

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.

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:
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.

Last edited: Aug 14, 2010
19. Aug 14, 2010

### Fredrik

Staff Emeritus
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.

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.

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).

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:
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.

Last edited: Aug 14, 2010
20. Aug 14, 2010

### Passionflower

So lets 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.

Last edited: Aug 14, 2010