Perspective on Relativity and Length Contraction

1. Nov 8, 2013

ZZ Specs

Hello All,

I like to think that I have a decent background in physics and related material, but I confuse myself quite easily on, well, everything. I guess I think about it too much, really. So feel free to give complex answers and I will do my best to follow along accordingly. Anyways, here is my current dilemma at hand:

I picture an astronaut travelling home at near-light speeds to Earth from a distant star. Through relativistic principles, let's say that the astronaut experiences only 2 years of travel, while the Earth and its observers experience 4 years pass by. According to the theory of relativity, how exactly does his viewpoint compare to an observer on Earth watching him fly home? Does length contraction fit in anywhere to either perspective, since the viewpoints are from the start and finish points?

Will the astronaut witness:

a) time moving very fast on Earth, such that 4 Earth years of activity go by while he only experiences 2 years of aging (and thus Earth will witness the astronaut moving very slowly)

or

b) the length between him and Earth "shrink", such that he only travels 2 years' worth of distance (while an observer on Earth sees this length "elongate", thus appearing that the astronaut actually traveled 4 years' worth of distance)

Essentially, how does relativity play spatial contraction vs. time dilation? Or will the astronaut undergo some combination of both? Does length/spatial contraction only work from a removed, outside perspective that does not lie along the path of the relativistic-moving object (in this case, the astronaut)?

Please let me know if there's anything I can clear up to make the questions more clear. Thanks!

2. Nov 8, 2013

Simon Bridge

For both your questions - every observer is at rest in their own inertial frame.
This means, for everyone, it is the other guy who is moving.
The other guys clocks are slow, the other guys lengths are short.

Try the following FAQ:
http://www.physicsguy.com/ftl/html/FTL_intro.html
... you need the bits on space-time diagrams to help you make sense of the different perpectives.

3. Nov 9, 2013

ghwellsjr

Welcome to PF, ZZ Specs.

Some of your ideas are correct and some of them are mixed up so rather than answer your questions one by one, I'll just show you what's correct using spacetime diagrams as Simon Bridge suggested.

First, we have a spacetime diagram for the original rest frame of the Earth/Star/Astronaut. The Earth is depicted in blue, the Star in red and the Astronaut in black. The dots represent one-year increments of time for each body. Since the Astronaut is colocated with the Star at the beginning and with the Earth at the end, I simply show black dots on the other body's colored lines during those intervals:

There are several things to note:

1) The distance between the Earth and the Star is 3.46 light-years.

2) The speed of the Astronaut once he leaves the Star is 0.866c.

3) At the Coordinate Time of 0, the Astronaut leaves the star and arrives at the Earth at the Coordinate Time of 4 years.

4) During the trip, the Astronaut ages by 2 years.

5) The Time Dilation factor at 0.866c is 2.

6) The Doppler factor at 0.866c is 3.732.

Now we want to determine what each observer actually sees. We do this by drawing in light signals propagating along 45-degree diagonals starting at one body and ending at another one. First, we'll do this for what the Earth sees:

Note that since the Star is 3.46 light-years away from the Earth, the Earth observers will see the Star as it was 3.46 years ago. This means that even though the Astronaut leaves the Star at the Coordinate Time of 0, the Earth observers do not witness this fact until 3.46 years later.

Then they start seeing the Astronaut travel in high speed toward them. They also see the Astronaut's clock running 3.732 times faster than their own so that from their time of 3.46 years to 4 years, a difference of 0.54 years, they see the Astronaut's clock progressing through 2 years. This matches the Doppler factor if we divide 2 by 0.54.

Note also that during the travel interval, the Earth observers see the Astronaut's clock gain 2 years while theirs gain 0.54 years for a net gain of 1.46 years so now the Astronaut's clock is only 3.46-1.46 or 2 years earlier than theirs. However, they continue to see clocks on the Star at 3.46 years earlier than their own.

Next we want to determine what the Astronaut sees. Here's a spacetime diagram showing the light signals coming from the Earth to the Astronaut:

Prior to departure, the Astronaut sees the Earth's clocks 3.46 years earlier than his own (just like the Earth saw the Astronaut's clock). But while he's traveling, he sees the Earth's clocks running 3.732 faster than his own so that during his 2-year trip, he sees the Earth's clock go through 7.464 years. At the end of his trip, the Earth's clocks have gained 5.464 years compared to his own so that instead of being 3.46 years behind his, they are now 2 years ahead of his and they remain that way for the rest of his time on Earth. (The times for the Earth and the Star are the same as the Coordinate Times in their rest frame.)

Finally, we want to see what things look like in the rest frame of the Astronaut while he is taking his trip. Actually, since it's his rest frame, it's the Earth and Star that are traveling:

This diagram has the same numbers as the previous diagram but it shows that the distance between the Earth and the Star is contracted to 3.46/2 = 1.73 light-years as indicated at the bottom of the diagram.

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4. Nov 29, 2013

ZZ Specs

Mr. Simon Bridge, thank you for the link. It contained many helpful resources.

Mr. ghwellsjr, thank you for your detailed reply. That was extremely informative and cleared up a whole lot of info on the subject, I very much appreciate you taking the time to explain this all to me. It is very helpful and tremendously generous of you to take the time to explain all of this. Quite a community here eh?

I apologize for my late response; I'm an undergrad Mathematics student in my last semester and the work piles up like crazy, I totally lost track of keeping up with this forum after I posted. If you wouldn't mind, I would like to pitch another few questions to you guys:

You explain that the Earth would see the astronaut leave at time = 3.46 years, and then see him arrive at time = 4 years. So the Earth witnesses the astronaut travel 3.46 light-years over the course of .54 years; wouldn't the astronaut then, from the Earth's viewpoint, *appear* to move faster than the speed of light?

Also, by the phenomena stated here, is it reasonable to suggest that objects moving apart each see the other in slow motion, while objects moving closer each see the other in fast motion?

Lastly, you finish the explanation by stating that the astronaut sees the distance between the earth and the star contract to half its original length. Could you please explain just the most fundamental basics on why length contraction occurs, and in what cases it can be observed or factored in to a reference frame?

Thank you all so much for your time and help. I excitedly look forward to a further understanding of the wonderful universe around us.

5. Nov 30, 2013

Simon Bridge

The analysis there should have helped you with your questions below - but I think there is one detail extra you need:

Not when the time it took the light to reach the Earth is taken into account.

ghwellsjr was just being precise.

In order to "see" something, light has to come from that thing and arrive at your eyes - so you always see things after the event, as it were. When the rocket fires it's engines, the light from that event takes some time to travel to the Earth - meantime, the rocket is also travelling.

So when the Earth sees the astronaut start out, they realize that the astronaut set out 3.46 years previous to them seeing it happen.

It is a common convention to write about some observer seeing something to include the calculation about how long it took the light to travel. This is important when you read about simultaniety for example.

How do you mean "slow motion".
The relative rate that time flows does not depend on the direction of travel.

The FAQ in my link does that - most basically, length contraction (and the other transformations) a a form of perspective effect.

The main thing to hold on to is that there is no "true" length or time, and physics works like normal inside each inertial frame.

From the astronaught's POV the earth is travelling towards him at some speed v - it reaches him in some time t that he measures on his watch ... therefore, the Earth was d=vt away. That's just normal physics. It's just that the d he calculates is different from the one the Earth people calculate - using the same method: but timing with their clocks.

6. Nov 30, 2013

ZZ Specs

You are entirely correct, my apologies. I read the article when you first posted it but missed the session on length contraction this time around. This is what I've now gathered from it and from your explanation, at its most basic level:

The speed of light is constant. But different frames can experience time differently; so to preserve c = V = d/t for different values of t, d must also be variable, leading to the phenomenon of length contraction.

I understand this much, that the astronaut did not *actually* travel faster than light. But according simply to the human being watching the whole scenario through his telescope:

He sees the astronaut leave from the star 3.46 light-years away at time t = 3.46 years [call t = 0 the point at which the astronaut actually left; let us speak only from the earthbound observer's reference]. The astronaut takes 4 earth years to reach earth, arriving at t = 4 years. The observer on earth can obviously see the astronaut here on Earth at this time t = 4 years. So what does the observer see when the astronaut *appears* to cover 3.46 light years in only .54 years? Does the astronaut appear (visibly speaking; barring math) to move at faster-than-light speeds, or is this now a situation of length contraction due the variation of t?

This is in reference to ghwellsjr's second spacetime diagram, "what the Earth sees."

Again you are right; the terminology/conceptualization is tedious but absolutely crucial to the problem.

According to ghwellsjr, when the astronaut heads toward Earth, each frame's clock appears to operate faster to "account for the time lost to the limitation of the speed of light" (I use quotes because I am not quite sure this is the right way to represent this phenomenon; but ghwellsjr does demonstrate how they see each other's clock move faster).

So when the astronaut and the Earth move closer together, the Earth witnesses the astronaut's clock move faster (beginning at t = 3.46 years and continuing to t = 4). Say the astronaut departed from the Earth to the star at .866c; would the Earth witness his clock move faster or slower?

Thank you so much for your patience; I'm working on a 'thesis' of sorts (trying to expand on the interconnectedness of time and space; for kicks, mostly) and am really hoping that in the process I will straighten out a much stronger understanding of physics.

ZZ

7. Nov 30, 2013

ghwellsjr

Yes, it would appear that the astronaut is traveling faster than c, but you should also ask yourself the question: how fast does the light appear to be traveling? Think of a race between the astronaut and a flash of light leaving the distant star at the same time on their way to Earth. At time 3.46 years, the Earthlings suddenly see the flash of light marking the beginning of the race but they also see the flash at the same time so they "see" the light traveling towards them at an infinite speed so you can't really say that the astronaut appears to be traveling faster than the light appears to travel.

Yes, but think about how fast light appears to be traveling away from you. Suppose you turn on a laser pointing at a distant planet. You won't see the laser beam hit the planet until the light gets back to you. Therefore, this light appears to travel at c/2.

I didn't say the astronaut could see that distance as contracted, I said we could see it. No observer can see either length contraction or time dilation as these are different in different frames and different frames will always depict the same experiences for all observers.

Although no observer can see length contraction, they can use radar techniques to determine what it is according to their own rest frame. I explain this in a similar scenario in this thread. Have a look and see if it helps.

You're welcome.

8. Nov 30, 2013

Staff: Mentor

Clearly not. If the earth sees him leave before he arrives then he travelled slower than light. He would only travel faster than light if they saw him leave after he arrived.

Yes. This is Doppler shift. Note that this is what is directly seen, it does not account for the finite speed of light.

9. Nov 30, 2013

ZZ Specs

I see, and I am referring strictly to the appearance of the astronaut. Interesting phenomenon. When you say "they 'see' the light traveling towards them at an infinite speed", is this because as soon as they see the light being emitted, it immediately reaches their eyes (the light beam traveling at the same speed with the light they use to observe it), and thus it appears to move infinitely fast (relative to observation)?

So even without relative motion between two objects, you can experience a sort of time dilation based strictly on their separative distance? That's a very interesting conjecture, thank you for that.

Can you explain a little bit further what you mean by "no observer can see length contraction or time dilation"?
I get that the traveler would not see himself or his own frame's length contract, but he could (potentially, in my imagination) witness contraction of lengths in front or around him, no?

On a side note... that's quite a discussion over there... haha. Do you often encounter much stubbornness when addressing proposed questions or 'paradoxes'?

I am not implying that he actually traveled faster than any actual light; simply that, as ghwellsjr has shown, he *appears* to have traveled faster than 3E8 m/s by appearing to cover 3.46 years in a time of .54 years according to the principles of light's finite speed (speaking only in terms of what the observer visibly sees, not what mathematically holds up). As ghwellsjr also pointed out, any beam of light that left for earth along with the astronaut would still be observed to move faster than the astronaut, preserving the comparison.

Thanks for your post though; when you say "This is Doppler shift. Note that this is what is directly seen, it does not account for the finite speed of light" what do you mean it does not account for the finite speed of light?

10. Nov 30, 2013

Simon Bridge

You are getting a lot of excellent help so I'll only be brief.
That's the math - physically, the universe has to arrange matters so that this happens.

For everyone to agree about the speed that light travels, they must disagree about what distances and/or times this involves.

The astronaut appears, from the Earth POV, to be approaching faster than the speed that light travels in the lab.
It's an illusion - not to be confused with length contraction.

It helps to imagine the clock as making a small pulse of light each second (or so). As the moving clock gets closer, the inertial observer sees rapid pulses. As the clock retreats, the pulses come more slowly. As the clock passes, the pulse rate chages from fast to slow. This is the Doppler shift.
Time dilation is in addition to this effect.

The interconnectedness comes from when you consider where the missing lengths and times "went" so to speak.
The time dilation and length contractions make a nice symmetry when distance and time are treated as manifestations of the same object.

11. Dec 1, 2013

ghwellsjr

The way you word this implies that there could be a difference between "seeing the light" and the light "immediately reaches their eyes" when, in fact, they are one and the same thing. And someone who didn't know the difference might conclude that light travels infinitely fast.

No, that has nothing to do with Time Dilation. I'm simply taking a naive approach to observations that no scientist would actually do. I'm just pointing out that measuring the speed of light by starting a stop watch when you see the light start at one point and stopping it when you see the light arrive at another point can lead to all kinds of discrepancies, including a determination that light can propagate at anywhere between c/2 to ∞.

Time Dilation and Length Contraction are frame dependent. When you transform events marking time intervals or lengths, they can take on different intervals or lengths. But the observers cannot directly see these effects. Every frame must depict all observations of all observers in exactly the same way.

If you go back to post #3 and study the first and last diagrams, you will see that in the first one, only the black astronaut's time is dilated while in the last one only the blue earth's and red star's time is dilated. And in the first diagram the distance between the earth and the star is not length contracted but in the last diagram it is. And yet in all the diagrams, I show how everything all observers can see and all their measurements remain the same. In order for an observer to determine any Time Dilation or Length Contraction, they must make some Doppler observations, make some radar measurements, assume that the outgoing radar signals take the same amount of time is the returning echoes, and then do some calculations and from this, they can determine what the Time Dilations and Length Contractions of other observers/objects are in their own rest frame. Furthermore, then can then transform to any other frame moving with respect to their own and see what the Time Dilations and Length Contractions are in these other frames, including frames in which their own time is dilated and their own lengths are contracted.

I have no way of knowing whether any particular person is being stubborn or not but it still provides and opportunity for learning, if not by the person being directly addressed, but by many others watching the interchange. However, I am most satisfied when everyone learns.

Last edited: Dec 1, 2013
12. Dec 1, 2013

TheBC

I read a quite a few 'appears'/'appearance' vocabulary in this thread. Why should a moving train car 'appear' contracted (as if an optical illusion is involved), and not simply 'be' contracted? It does not make sense.
In the sketch I show how relative moving travellers experience reciprocally a moving contracted train car.
The moving train car does not 'appear' to fit between the passenger's fingertips (and does not 'appear' contracted to be able to fit between his fingertips). The train car simply does fit between his fingertips. The signals (information of the events) of the front and rear of the moving train at his fingertips are travelling simultaneously, from his fingertips through his arms and reaching his brain simultaneously. That's no illusion or arbitrarily chosen frame calculation. There is a shorter train between the fingertips at the end of the passenger's arms. One may call this 'measuring from the passenger's frame'; it simply means what the passenger's 3D space of simultaneous events (his 3D reality at one moment in time) is made of.

If you consider a train car at rest in front of you, you do not say the train car at rest 'appears' x meters long. You don't do this because you do not refer to any 'optical illusion'.
Stating that a moving train car 'appears' contracted insinuates there is some optical illusion involved. But there is no optical illusion invoved. Therefore there is also no reason to state that a moving train car 'appears' contracted. If the moving train car 'appears' contracted, then a train car at rest also only 'appears' measuring a certain length.

The reason why people often think the contraction only 'appears' as such is because the contraction is reciprocal. How can relative moving train cars be contracted reciprocally if its not an optical illusion effect?
Because the 3D worlds of simultaneous events of both passengers are different. Consider the train cars as 4D spacetime structures instead of evolving 3D objects. The different contracted train cars are different objects of simultanous events, cuts with different 'direction' through the 4D spacetime train car structures (like cutting a loaf of bread in different directions results in different 'objects'/slices of bread). Minkowski or Loedel diagrams visualize this very well.

Last edited by a moderator: Dec 1, 2013
13. Dec 1, 2013

Simon Bridge

Because a distinction is being made between the lawrentz contraction and optical effects due to the speed of light.

14. Dec 2, 2013

phyti

Knowing the image of the launch left simultaneously with the ship, when the image arrives, the ship should be close behind in this case.

The separation should be 3.46 yr * (1-.866) = .464 ly

Time to arrival is .464 ly/.866 = .536 yr.

The difference in travel time for light t1 and the ship t2, cannot equal t2.

t2 - t1 ≠ t2, since light has a finite speed. (ref: Ole Romer, James Bradley)

Since .536 is not the total trip time for the ship, it does not move faster than c.

Even if light speed was instantaneous, then the expression would be true, but earth would observe launch at t=0, and ship does not move faster than light.

15. Dec 3, 2013

universal_101

Length Contraction is observer dependent(i.e. it is not Lorentz invariant), same as Time Dilation is observer dependent. But because, Time Dilation exhibits measurable physical effects(differential aging-Twin Paradox-fast moving muons) which are ofcourse Lorentz invariant, whereas unlike Time Dilation, Length Contraction has no measurable physical effect, it is safe to say it is an apparent effect(or observer dependent).

16. Dec 3, 2013

WannabeNewton

It's not an "apparent" effect in the illusory sense: scenarios like the Ehrenfest paradox show that it can have consequential effects. Let's not get into this debate again because it's already been beaten to death here.

17. Dec 3, 2013

PAllen

Fast moving muons suggest length contraction is at least as real as time dilation unless you want to claim there is something less preferred about the the muon frame. In the muon frame, the only possible explanation for how it reaches the ground is that the atmosphere is extremely thin it its (the atmosphere's) direction of net motion. That muons reach the ground is an invariant fact. SR then states that explanation is frame dependent, but that time dilation and length contraction are on the same footing as explanations. If one is 'real', so is the other.

18. Dec 3, 2013

universal_101

Same here for the debate part, I can't afford to get red cards for this forum is very valuable(from time to time). But for the record, as long as the term illusory stands for non-real, non-physical effect, the only way to be sure that Length contraction is not illusory, is to have a measurement of a real, physical effect which directly measures Length Contraction(like the differential aging of fast moving muons for Time Dilation) and not the implied ones.

19. Dec 3, 2013

Staff: Mentor

I think it was ZapperZ who suggested some time ago in another thread that contraction of particle bunches in an accelerator could be considered as "direct" evidence of length contraction. I can't find the thread at the moment, though.

20. Dec 3, 2013

universal_101

I think you are correct, but only if SR is considered beyond any doubt the fundamental physics.
For example, if a theory is incorrect, then so is its implications, and that is why most scientists try to prove a theory(or make it more fundamental) by making physical measurements independently(i.e. no implied proofs).

And yes I can claim that instead of Length contraction there is something different about the Muon frame w.r.t the Earth frame, as long as there is no direct proof of Length Contraction.

the only possible explanation view is correct as I mentioned above, but only if SR is the only theory that can explain relativistic effects and already proven to be more fundamental than reality itself. Then only the implication, that there should be a real Length contraction, is legitimate.

21. Dec 3, 2013

PAllen

In this forum, we discuss SR and GR, not alternative theories. In effect, you admit that per SR, length contraction is not illusory. Note also, that one other theory at least consistent with SR predictions (LET, which has a preferred frame) also relies on non-illusory length contraction.

22. Dec 3, 2013

universal_101

Yes, it was for the Free Electron Laser electron bunch, and there are many examples like this, for example,

1.) Magnetic force on a moving charge from the frame of moving electrons in a current carrying wire.(Force being the physical effect due to Length Contraction in electrons frame)
2.) The fact that muons, despite having very short lifetime reaches the Earth surface(A physical effect/fact) means atmosphere is length contracted in muons frame.
3.) The fact that MMX experiment produces null result(a physical measurable effect) implies length contraction in the longitudinal direction w.r.t a moving frame.

23. Dec 3, 2013

universal_101

, You make it sound like as if I'm posting in the wrong section of the forums, but nonetheless i get the point, I should stick to the SR !
I never denied that, that per SR, length Contraction is real. And it is exactly this which is the whole point of discussion, i.e. if it were to be an apparent effect as per SR, there would have been NO objections.

There is almost NO difference between LET and SR mathematically and the effects thus produced, (i.e. they are virtually identical except for a preferred frame of anyone's choice, which by no means can be detected even in theory.)

Last edited: Dec 3, 2013
24. Dec 3, 2013

Staff: Mentor

This is complete nonsense. Do you have a mainstream reference for this criterion? I have never seen a reputable source demand something this extreme of any scientific theory.

25. Dec 3, 2013

PAllen

I wonder what people think of the following scenario (unfortunately not feasible, in practice):

Imagine two space beacons at mutual rest, separated by e.g. a million kilometers. A rocket passes them, turns, passes them again, turns, etc. On each passage the speed of the rocket relative to the beacons increases. The rocket can directly measure (theoretically) the relative speed of a beacon by measuring the time it takes to traverse the length of the rocket. The rocket can also measure the time it takes for both beacons to pass. [All acceleration occurs during the turnarounds, so no proper acceleration is measured in the rocket while the beacons are passing.]

On the first, slow, crossing the rocket measures D as the beacon separation by computing v (measured locally as described above) multiplied by the crossing time. However, following the identical procedure each time to measure v, and crossing time, D will be found to get smaller and smaller as v increases closer to c on each passing.

Note that since v is measured locally by the rocket on each passing, using its own clock, time dilation plays no role in this (at least for the rocket).

1) Do any of the participants believe that if the enormous engineering problems could be solved, that the observed result would be different from what is described above?

2) If not, is there any explanatory model other than distance contraction (at least for the rocket)?

Last edited: Dec 3, 2013