B Can an observer perceive he is traveling faster than light?

[GrantSB]

I have seen thought problems with an observer on a train or in a station, etc., but I have not seen ones with the observer traveling at relativistic speeds. It seems to me that at sufficient speed he would observe himself exceeding the speed of light due to the slowing of time. This seems like a contradiction.

Can anyone point me to a discussion on this topic?

 

[jbriggs444]

I have seen thought problems with an observer on a train or in a station, etc., but I have not seen ones with the observer traveling at relativistic speeds. It seems to me that at sufficient speed he would observe himself exceeding the speed of light due to the slowing of time. This seems like a contradiction.

Can anyone point me to a discussion on this topic?

Many discussions here. A key point is the principle of relativity. No matter how fast someone else thinks you are moving, you are always at rest in your own rest frame. The laws of physics provide no way to tell whether that frame is "really" moving or or is "really" at rest. It follows that an observer never perceives his own time to be slowed. An [inertial] observer only measures clocks that are moving relative to his rest frame to tick slowly.

 

[Pencilvester]

...travelling at relativistic speeds. It seems to me that at sufficient speed he would observe himself exceeding the speed of light

Also, just a simple correction, when we say “relativistic speeds”, we mean speeds at which special relativity is no longer negligible, i.e. a significant fraction of the speed of light, but never exceeding it.

 

[Dale]

It seems to me that at sufficient speed he would observe himself exceeding the speed of light due to the slowing of time.

Usually in relativity discussing what an “observer” sees is a sort of short hand for describing things in that observer’s rest frame. In an observer’s rest frame the observer is always at rest (by definition).

 

[GrantSB]

I agree with all the comments made, and used relativistic in the way it is described. My intereest is what the observer sees if he passes close to a "stationary" object, or at least stationary in our frame of reference. If the observer is travellingat .95 C (say) then there will be a significant slowing of time in his frame of reference (I get that he won't percieve this) and he will cover a significant distance in our frame of reference, with very little elapsed time in his. If he passes close to a staionary object then to him it would appear to be approaching him (and passing by) faster than the speed of light (or so it seems to me).

 

[Ibix]

In his frame of reference, he is stationary and it's the other object doing 0.95c, and its clocks that run slow and its rulers that are shortened. Where would any oddity in measurement come from?

 

[PeterDonis]

he will cover a significant distance in our frame of reference, with very little elapsed time in his

You can't mix quantities from different frames of reference. Dividing distance in our frame, by time in his frame, gives a meaningless number.

 

[GrantSB]

Ok, let me present this as a thought problem. Let's assume the distance to Alpha Centauri is 4 light years, and we know the path that a spaceship will take. In advance of the trip we place one beacon every light-day along the path, some 1400 in total. A ship traveling at .01 C would pass one every 100 days (or so). Now, assume the spacecraft is traveling at a speed such that in his frame of reference one month passes to complete the trip to Alpha Centauri. He will pass by all the beacons, but they will pass by every half hour (if he is to pass all 1400 in the month). He knows the spacing of the beacons, therefore could he not infer his speed by watching them go by, just as one might on a highway by timing vs. the roadside markers? Similarly, wouldn't the beacons appear to be approaching at a rate greater than C?

 

[PeterDonis]

Let's assume the distance to Alpha Centauri is 4 light years

Distance is frame-dependent. The distance is 4 light years in the frame in which the Earth and Alpha Centauri are at rest (we're assuming here that they are in fact at rest relative to each other, which is not actually true, but for this discussion it works as an approximately true assumption).

In advance of the trip we place one beacon every light-day along the path, some 1400 in total.

The beacons are one light-day apart in the Earth-Alpha rest frame. They are not one light-day apart in other frames. (Also, they are moving in other frames.)

could he not infer his speed by watching them go by, just as one might on a highway by timing vs. the roadside markers?

No. As I've already said, you can't divide distances in one frame by times in another frame. Such a calculation is physically meaningless.

What the spaceship observer will observe is that, in his frame, the beacons are moving past him, and are much less far apart than one light-day. The speed he will infer from the distances in his frame, divided by the times in his frame, is whatever speed gives a time dilation factor corresponding to one month "trip time" (which in his frame is the time it takes Alpha Centauri to reach him, assuming he starts out with Earth co-located with him, and moving away).

 

[Ibix]

He knows the spacing of the beacons,

You seem to be assuming that there's something special about the spacing of the beacons in their rest frame. There isn't. His own measurement of the distance between the beacons, which will be about half a light hour if I haven't messed up the mental arithmetic, is just as valid, and is the correct distance to combine with his own time measurements. Combining one frame's time measurement with another's distance measurement is a bit like multiplying the width of a rectangle by its diagonal and hoping to get the area of the rectangle.

 

[GrantSB]

Please ignore the ignorance of my questions. I am trying to understand.

So what you are saying is that when the spaceship leaves Earth, at the speed I described earlier, then the 1400 beacons will appear to be much closer together than 1 light day apart. I guess similarly, Alpha Centauri will appear much closer than 4 light years, as it is at the end of the 1400 beacons. I was going to say that it appears one light-month away, but this can't be right as he gets there in a month, and at that distance he would infer he was traveling at C.

Assume he has a triangulation device (or something else) on his ship and is constantly monitoring the (apparent) distance to Alpha Centauri. What happens if he decelerates? Does it appear to recede?

 

[GrantSB]

And what would he percieve the distance to Alpha Centauri to be at the outset of his trip?

 

[Ibix]

So what you are saying is that when the spaceship leaves Earth, at the speed I described earlier, then the 1400 beacons will appear to be much closer together than 1 light day apart. I guess similarly, Alpha Centauri will appear much closer than 4 light years, as it is at the end of the 1400 beacons.

Yes.

I was going to say that it appears one light-month away, but this can't be right as he gets there in a month, and at that distance he would infer he was traveling at C.

He's traveling at about .9998c (edit: oops! In this frame, Alpha Centauri and the beacons are traveling at .9998c) - a month is close enough.

Assume he has a triangulation device (or something else) on his ship and is constantly monitoring the (apparent) distance to Alpha Centauri. What happens if he decelerates? Does it appear to recede?

That's complicated. The rule of thumb you are using to think this way only works in an inertial frame (Peter and I have been implicitly assuming a rocket that passes Earth at constant speed on its way to Alpha Centauri). The maths to describe what is seen (or rather, how measurements are interpreted) by an accelerating observer is rather complex, and depends on how you define "distance", which is a rather flexible beast in a non-inertial frame. So the short answer is "it depends". Under reasonable assumptions, the answer is no it does not recede, largely because your definition of "now" in the phrase "where is Alpha Centauri mow?" changes, so naive assumptions about it being nearby "now" and then far away "now" break.

 

[Nugatory]

And what would he percieve the distance to Alpha Centauri to be at the outset of his trip?

Assume that he is already moving at his cruising speed as the Earth passes him and the string of buoys and alpha centauri are moving towards him (he's at rest and everything is moving, remember). The distance to alpha Centauri at the moment that the Earth and him pass one another will be one month times the speed - this is just a simple application of "distance equals speed times travel time"

Both the distance and the time will be less than in the frame in which the Earth is at rest. "Distance equals speed times travel time" will be true in both frames.

 

[Dale]

He knows the spacing of the beacons,

The spacing is not one light day in his frame. They are spaced much more closely, hence the fast rate of passing.

then the 1400 beacons will appear to be much closer together than 1 light day apart.

It is not just a matter of appearance. His most careful measurements of the distance will show that they are much closer together.

What happens if he decelerates? Does it appear to recede?

I don’t think this question can be answered at a “B” level. Here is a paper that describes one way to answer the question, but I would say it is “I” level material:

http://arxiv.org/abs/gr-qc/0104077

 

[PeterDonis]

Assume he has a triangulation device (or something else) on his ship and is constantly monitoring the (apparent) distance to Alpha Centauri. What happens if he decelerates?

Instantly? Nothing. Light travels at a finite speed, so the triangulated distance to Alpha Centauri cannot "jump" instantaneously just because the ship decelerates.

 

[Nugatory]

Assume he has a triangulation device (or something else) on his ship and is constantly monitoring the (apparent) distance to Alpha Centauri. What happens if he decelerates? Does it appear to recede?

It will be easier to reason about this problem if you assume that he is using a radar to measure the distance: send a flash of light towards alpha Centauri; it is reflected from alpha Centauri and returns to him. Assume (so we don't need to use any calculus and to avoid a bunch of unnecessary complications involving non-inertial frames) that he only changes speed between measurements - constant speed between the emission of a flash and receiving its reflection.

It would be a good exercise for you to try working out for yourself what happens to the distance.

Spoiler

The distance (in the frame in which he is at rest) between him and alpha Centauri at the time (again, in the frame in which he is at rest) is the one-half the round trip time for the radar signal. It will be greater as he slows.

 

[GrantSB]

Thank you all for interpreting the spaceship as already being at speed. This is just a thought problem, so I chose to ignore the crushing G forces of getting to this velocity in under a year. As for 'instantly', I never said instantly, I said constantly (i.e. he is looking forward). As he decelerates he sees something, ignoring issues of "now" and such, which I get can be thorny.

btw it is clearly beyond my abilities to work this out for myself :).

I must confess, what has been described by the group of you seems paradoxical to me. When the spaceship is traveling fast, Alpha Centauri appears to be quite close (1 light month) and approaching at .9998 C. OK, I'll accept that. Now assume there isn't a person in the spaceship, but some solid-state device that can survive extreme deceleration. In that case, in the frame of reference of the spaceship (as it slows), Alpha Centauri both appears to be approaching more slowly, and is further away. If it decelerates to rest in our frame of reference then Alpha Centauri will once again be 4 light years away. Ignoring the admittedly complex math, with sufficient deceleration couldn't it appear to recede at > C?

 

[Dale]

In that case, in the frame of reference of the spaceship (as it slows),

There is no unique standard definition for the frame of reference for a non inertial spaceship. Please see the paper I posted earlier for an example of how to rigorously define such a concept.

Again, this is simply not B-level material

Ignoring the admittedly complex math, with sufficient deceleration couldn't it appear to recede at > C?

You cannot ignore it and answer the question. Depending on the arbitrary choices you make in the mathematical definition of the non-inertial frame you can get any answer you want.

 

[PAllen]

Thank you all for interpreting the spaceship as already being at speed. This is just a thought problem, so I chose to ignore the crushing G forces of getting to this velocity in under a year. As for 'instantly', I never said instantly, I said constantly (i.e. he is looking forward). As he decelerates he sees something, ignoring issues of "now" and such, which I get can be thorny.

btw it is clearly beyond my abilities to work this out for myself :).

I must confess, what has been described by the group of you seems paradoxical to me. When the spaceship is traveling fast, Alpha Centauri appears to be quite close (1 light month) and approaching at .9998 C. OK, I'll accept that. Now assume there isn't a person in the spaceship, but some solid-state device that can survive extreme deceleration. In that case, in the frame of reference of the spaceship (as it slows), Alpha Centauri both appears to be approaching more slowly, and is further away. If it decelerates to rest in our frame of reference then Alpha Centauri will once again be 4 light years away. Ignoring the admittedly complex math, with sufficient deceleration couldn't it appear to recede at > C?

In a sense, yes. This is a coordinate velocity associated with coordinates adapted to continuous velocity change. Typically, such coordinates are described via hyperbolic rotation. Just as turning your head quickly can give a faster than c coordinate speed to the moon in a frame turning with your head, a hyperbolically rotating frame will have FTL coordinate speed for distant objects.

However, these are completely different from the notion of relative velocities. In SR, a relative velocity may be taken as a velocity of some body in an inertial frame in which the other body has a momentary velocity of zero. This will always be less than c, with no exceptions or caveats whatsoever.

 

[GrantSB]

Thank you all. I will read the referenced paper (on the long shot that I can make some sense of it :) ) and scratch my chin a bit. One final question: Is the forshortening of distance you have described the Lorentz contraction? I had always interpreted this as a characteristic of just the object itself, not it's entire frame of reference. I guess if you consider the object to be at rest then from the point of view of observer one might consider everything else to be moving rather quickly, and experiencing the effect.

If the answer is no, then that's probably good enough, I'm not trying to extend the conversation, just integrate the small bits I know.

I do have a separate question. It is still about time and speed, but different enough that it deserves a separate thread. I'll post this later.

 

[PeterDonis]

Is the forshortening of distance you have described the Lorentz contraction?

Yes.

I had always interpreted this as a characteristic of just the object itself, not it's entire frame of reference.

As should be evident from the answer just above, this is not correct.

 

[PeterDonis]

Is the forshortening of distance you have described the Lorentz contraction? I had always interpreted this as a characteristic of just the object itself, not it's entire frame of reference.

To elaborate a bit more on this: in the spaceship frame of reference, the Earth, Alpha Centauri, and all of the space beacons in between are moving, all at the same speed and in the same direction. So in the spaceship frame of reference, distances between all of these objects are Lorentz contracted. There is no fundamental difference between "objects" being contracted and "distances" being contracted. The key point is that you have a bunch of objects that, in the spaceship frame, are all moving in the same direction at the same speed.

 

[Dale]

One recommendation from my side. I have given you an “I” level reference for your “I” level question. However, you will most likely be better served to “table” it until you have mastered the “B” level stuff first.

Start purely with questions about inertial objects analyzed in inertial frames. Then move on to non inertial objects analyzed in inertial frames. Learn a bit about four vectors, manifolds, coordinate charts, and the metric. Only then revisit the topic of non inertial reference frames.

 

[Freixas]

I have a few tangential comments to offer.

First, I've been following PhysicsForums through the emails that get sent periodically. I have to say that I am impressed with the mentors and other frequent posters for their patience in responding to the a plethora of totally confused beginners (such as myself). When I created my thread, I didn't realize how frequent this this type of posting was, nor how often beginners (again, including myself) keep thinking the experienced physicists have it wrong. :-) So a tip of the hat to you all...

Second, I think there's a bias in the way a lot of thought experiments are posed which contributes to the confusion of beginners. Often, one observer is given in a context that we tend to assume is a "master" reference. For example, the Earth vs. spaceship or train vs. train platform. One appears to be "stationary" in some sort of absolute sense and the other moving.

This bias can be hard to shake as one is not always aware of having it. Even if one thinks that he recognizes that the references are arbitrary, there will be some unconscious assumption that gives a preference to one over the other.

I like to work things out by thinking of a totally empty universe and then putting one observer (A) in it. It's pretty clear that this observer can't tell if he is at rest or moving at a high fraction of c. It make no difference. Then I add a second observer (B) in a different inertial frame. Without the bias of Earth vs. ship, it is a lot easier to see that whatever is true for A observing B must also be true for B observing A.

To make this more fun, I like to add a whole set of A observers, spaced equidistant and all at rest relative to each other. Similarly, I add a set of B observers with the same characteristics. To work out what everything looks like to whomever, you wind up having to correctly apply time dilation, length contraction and the relativity of simultaneity. You cannot get the correct descriptions without considering all three.

As a side observation, beginners often hear about time dilation and length contraction. They don't hear about the relativity of simultaneity. In my own thread, it was my main point of confusion. You can't get the right answer by ignoring it.

Sorry, GrantSB, I haven't actually addressed your questions other than to suggest you avoid using labels like "ship", "Earth", "Alpha Centauri". Also, I skipped out on non-inertial frames. I was pretty happy to follow what was going using only inertial frames. Good luck!

 

[m4r35n357]

Second, I think there's a bias in the way a lot of thought experiments are posed which contributes to the confusion of beginners. Often, one observer is given in a context that we tend to assume is a "master" reference. For example, the Earth vs. spaceship or train vs. train platform. One appears to be "stationary" in some sort of absolute sense and the other moving.

In SR, all inertial frames can be considered on an equal footing. This doesn't not mean they are all identical or indistinguishable; in practice, we always select one frame to do the analysis, usually our own rest frame (this is a B-level thread so please don't shoot me mods!).

Labeling the frames makes explaining and understanding the problem simpler, and for example when there are just two frames in steady relative motion the problem is symmetrical, and we can look at it from a choice of two frames (train or station and so on). BUT, as soon as there are more than two frames in the problem, the symmetry is broken, so we tend to choose to pick one frame (an inertial one) and use that as a "master" (yes, like the Twin Paradox!). We do this to preserve our sanity.

Anyway, just wanted to check that you were aware of this; all inertial frames are on an equal footing, but that doesn't mean we can't single one out to make our lives easier.

 

[Freixas]

Labeling the frames makes explaining and understanding the problem simpler

Actually, what I was suggesting was that certain labels for inertial frames can make things more confusing for beginners as it leads to unconscious biases. For those who understand how it all works, the labels aren't a problem, of course.

 

[Herman Trivilino]

Is the forshortening of distance you have described the Lorentz contraction? I had always interpreted this as a characteristic of just the object itself, not it's entire frame of reference.

You mean two points on the object, like the two end points of a rod. The same is true of any two points that lie along the line of motion, whether they lie on the rod or not, they just have to be at rest relative to the rod. By the way, instead of saying "its entire frame of reference" say "its entire rest frame". That clarifies which of the reference frames you're talking about, because if the rod is in motion in some frame of reference, it's "in" that frame of reference, too.