Why is the speed of light constant for all observers?

AI Thread Summary
The discussion addresses the concept of the constant speed of light and its implications in special relativity, particularly focusing on length contraction. It explains that while the speed of light remains constant for all observers, the length of objects can appear different depending on their relative velocities. Specifically, a 5-meter car can fit into a 3-meter garage if the car is traveling at four-fifths the speed of light, due to the effects of length contraction. The conversation also touches on the behavior of clocks in different frames of reference, emphasizing that time dilation occurs during relative motion and that the effects of acceleration are crucial in determining which observer experiences time loss. Overall, the principles of special relativity clarify how measurements of time and length vary between different observers.
vidovnan
[SOLVED] variable speed of light

OK people, I'm sure you'll find this one easy. For myself, I have been working on it for a while now and am not getting far enough.

My work colleague has posed me this problem:

how can you fit a 5m(eter) car into a 3m garage?

Now, I know this has something to do with the variable speed of light. I know that the speed of light varies in a gravitational field. And I know that the physical dimensions of bodies vary when these are plunged into a gravitational field.

However, that's about all I know so far.

And my work colleague is soooo smug.

Then I found you...

Please help me challenge his smugness.
 
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Light actually does not vary in speed. All observers will always measure light as moving through their local frame at c.

What your friend's question involves is actually a phenomenon called length contraction. You see, the length of an object (like the car) is not a fixed quantity. Different observers moving at different relative speeds to the car will actually measure its length differently.

Let's say you're standing still beside the 3 m garage. A physicist would say you are using the garage as your 'rest frame,' and you are at rest in it. The garage's length, 3 m, is called its 'proper length.' The car's proper length is 5 m -- this means that a person standing still beside it would measure it as being 5 m in length.

If you propel the car to a high enough velocity, it is possible that you, standing beside your garage, will measure the car's length as being shorter than its proper length of 5 m.

The equation governing this effect (called length contraction) is this:

l = l0 / γ

where l0 is the proper length of the car, and l is the measured length of the car.

γ (lowercase Greek gamma) is a unitless number that is always greater than or equal to one. When you are at rest with respect to the object you're measuring, γ = 1, and you will measure its length as its proper length.

γ can be found by plugging in v into the following definition:

γ = 1 / sqrt(1 - v2 / c2)

To find the velocity the car must have relative to the garage to cause an observer standing beside the garage to measure its length to be 3 m, you must solve the equation:

3 = 5 / γ

3 = 5 * sqrt(1 - v2 / c2)

v = (4/5) c, or four-fifths the speed of light.

If you drive your 5 m (proper) car at four-fifths the speed of light with respect to your 3 m (proper) garage, it will fit inside it -- before promptly crashing through the back wall!

- Warren
 
Nothing to do with VSL theories (which are a far from conventional) or gravitational fields even, it's special relativity and length contraction, which is given by the following equation:

L = L0√(1 - v2/c2)

Solving for L = 3 and L0 = 5, we find that v = 0.8c. So in other words: to fit a 5m car into a 3m garage the car must be traveling at 4/5 the speed of light in a vacuum (relative to the garage).

edited to add: looks like chroot beat me to it!
 
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Hey, Thanx people! Now that's what I call service.

Remind me to visit this board more often.
 
Warren,

If you get the car going fast enough to measure its length as 3m instead of 5m,and don't cra**** into the garage, how long is the car after you slow it back down to a halt in your rest frame?

-zoob
 
5m of course, length contraction doesn't deform the object. Interestingly, due to the failure of simultaneity at distance, though an observber in the rest frame of the garage will see both the front and back end of the car in the garage at the same time, an observer in the rest frame of the car will not.
 
Originally posted by jcsd
5m of course, length contraction doesn't deform the object. Interestingly, due to the failure of simultaneity at distance, though an observber in the rest frame of the garage will see both the front and back end of the car in the garage at the same time, an observer in the rest frame of the car will not.
JCSD,

This being the case (which makes perfect sense to me) why is it that people claim a clock that is measured to be running slow when moving past an observer, will actually be found to have lost time when slowed down and returned to the observer's rest frame (which doesn't make sense to me)?
 
Originally posted by zoobyshoe
JCSD,

This being the case (which makes perfect sense to me) why is it that people claim a clock that is measured to be running slow when moving past an observer, will actually be found to have lost time when slowed down and returned to the observer's rest frame (which doesn't make sense to me)?

Hey Zooby, I thought I would jump in since I'm here.
I suspect that you don't really believe the statements above. From the frame of the garage, the car really is shorter. Next, if observers in both frames of reference must observe light to travel with speed C, and since we know the length contraction is given by

L = L0√(1 - v2/c2)

and we know that for any observer, C = L0/t0 = L/t,

it can easily be shown that

t = t0√(1 - v2/c2)

Therefore just as length contracts, observers in the frame of the garage will see clocks running slowly in the car. Again, this is not just an illusion; this is real. Now, it gets interesting since for an observer in the car, the garage is in motion and the car is at rest. So, an observer in the car sees clocks in the length contracted garage running slowly. Whose clocks are right? Both. When someone accelerates, in this case when the car comes to a stop and comes back, we have chosen a preferred observer - the garage - and we find that the clocks in the car have lost time. It we speed up to catch the car, assuming we started out in motion with the car and synchronized our clocks, we would find that the clocks in the frame of the garage have lost time. Again, the one who changes their state of motion is the one whose clocks have lost time. Note that in reality, the presence of gravity complicates this situation.
 
If the clock in the car comes back into the rest frame of the garage having lost time, why hasn't the car lost length?
 
  • #10
Originally posted by zoobyshoe
If the clock in the car comes back into the rest frame of the garage having lost time, why hasn't the car lost length?

When the car is in motion, it is shorter in the frame of the garage. Likewise, from the frame of the garage, the car's clocks are running more slowly. When the car stops, that is, when the frame of the car coincides with the frame of the garage, the two lengths L and L0 agree. Likewise, it we compare the ticks of the clocks, we find that again they agree – they occur at the same rate. However, and this was a key test of relativity, we find that while the frames of the car and garage did not coincide, ie, while the car is in motion as viewed from the garage, the clocks in the car really were running more slowly...just as observed and predicted.

This was finally verified I think in the early sixties using two atomic clocks; one on the ground, and one in a jet. After flying one of the clocks around for a while, and after accounting for the effects of gravity, the clock on the plane had indeed lost time as predicted to within the accepted margins of error. This has since be replicated in many other ways. Also, we see the lifespan of subatomic particles increase according to Relativity and their relative speed – since their clocks run more slowly, we see them live longer. This is seen in particle accelerators as well as in nature.
 
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  • #11
Originally posted by Ivan Seeking
Again, the one who changes their state of motion is the one whose clocks have lost time.
Both parties see the other as in motion according to their own frame. Both parties see the other's clock as running slow.
When both parites come to be in the same frame again neither can say if it was he or the other who decelerated. Both must find the the other's clock has lost some time, in which case they will both find both clocks to agree both as to the time and the rate of timekeeping. The illusion of slow clocks only exists while the relative motion is occurring. When the relative motion stops the illusion stops.
 
  • #12
Originally posted by zoobyshoe
Both parties see the other as in motion according to their own frame. Both parties see the other's clock as running slow.
When both parites come to be in the same frame again neither can say if it was he or the other who decelerated. Both must find the the other's clock has lost some time, in which case they will both find both clocks to agree both as to the time and the rate of timekeeping. The illusion of slow clocks only exists while the relative motion is occurring. When the relative motion stops the illusion stops.

We can tell who accelerates [decelerates] - the one who experiences a force. If the relative motion of observer B changes wrt observer A, and if observer A experiences no forces, then A knows that B has changed his state of motion. Likewise, B feels a force and is also aware of whose frame of reference has changed. This indicates who is at rest -the preferred observer. This is no illusion; it is a 98 year old, well tested theory. The clocks do not agree when we compare the results in the same frame of reference; and the two clocks vary by the amount predicted by Special Relativity [or General Relativity if required].

Edit: A key concept here is that until someone experiences a force, indicating a change in their state of motion, there is no preferred observer. We can define either [or any] frame of reference to be at rest as long as the state of motion remains constant.
 
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  • #13
So, it seems that acceleration is the critical point.

If we accelerate the car away from the garage, what does each observe about the other's clock?
 
  • #14
Originally posted by zoobyshoe
So, it seems that acceleration is the critical point.

If we accelerate the car away from the garage, what does each observe about the other's clock?

The clocks in the car would be seen to run slowly as before; and due to the acceleration.

From General Relativity: Clocks run more slowly in gravity fields.
 
  • #15
Originally posted by Ambitwistor All that is required for different elapsed times is that, by whatever means (acceleration or not), the two twins take spacetime paths of different lengths.
In which case the one who takes the longer path ages less?

This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.
 
  • #16
Originally posted by Ambitwistor
or a gravitational slingshot back home

How do we do this without any acceleration?
 
  • #17
Originally posted by zoobyshoe
In which case the one who takes the longer path ages less?


More

This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.

SR is only valid in the inertial frame - no forces.
We need GR to calculate the effects of acceleration [gravity].
 
  • #18
Originally posted by zoobyshoe
In which case the one who takes the longer path ages less?

A path in spacetime is called an interval.

The length of an interval is called its proper time. If you have a clock follow some path, the clock will measure that much time having been elapsed when moved along that path.
This is in General Relativity, right? I'm not finding a discussion that goes further than to say: "As a consequence of its motion the clock goes more slowly than when at rest." in SR.
This is not really true -- a distant observer will measure a clock as running slowly when it is moving at high relative velocity to the observer. According to the clock, however, everything is just fine.

Imagine Picard is flying along in the Enterprise at 0.9c with respect to the Earth. An observer on the Earth will measure Picard's clock as running slow compared to an identical Earth-bound clock. Picard, however, will see everything on the bridge of the Enterprise as running completely normally, but will measure the Earth-bound clock as running slowly.

If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!

- Warren
 
  • #19
In order to declare one frame valid and the other not, i.e. if we are to determine whose clocks have lost time, someone has to accelerate. Until that happens, boths frames of reference are valid.
 
  • #20
Originally posted by Ivan Seeking
In order to declare one frame valid and the other not, i.e. if we are to determine whose clocks have lost time, someone has to accelerate.
No. As Ambitwistor just demonstrated with some killer graphics :wink:, all you need to do is calculate the proper times along the paths of both twins, and compare.

Both frames are perfectly "valid," as are all frames. There's no such thing as an invalid frame.

- Warren
 
  • #21
Zooby, do you feel spoiled?
 
  • #22
Originally posted by chroot
No. As Ambitwistor just demonstrated with some killer graphics :wink:, all you need to do is calculate the proper times along the paths of both twins, and compare.

We still have no preferred observers in the inertial frame.


Both frames are perfectly "valid," as are all frames. There's no such thing as an invalid frame.

- Warren [/B]

If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.
 
  • #23
Originally posted by Ivan Seeking
We still have no preferred observers in the inertial frame.
I have no idea what this means.
If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.
This is also incorrect. Special relativity is all that's necessary to understand the twin paradox; Ambitwistor explained it nicely. The only thing you need general relativity to explain is gravitation.

- Warren
 
  • #24
Originally posted by chroot
I have no idea what this means.

There is no absolute reference frame. There is no absolute state of rest.

This is also incorrect. Special relativity is all that's necessary to understand the twin paradox; Ambitwistor explained it nicely. The only thing you need general relativity to explain is gravitation.

- Warren [/B]

How do we determine which twin is younger? One of them has to accelerate in order to leave earth; unless he was born in a state of relative motion as compared to his twin.
 
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  • #25
One of them has to accelerate in order to leave earth; unless he was born in a relative state of motion as compared to his twin.
 
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  • #26
Well, Ambitwistor's is the best explanation of the twin paradox I have run into. It's the first time I even understood the importance of the "Spacetime Interval". (And, as Chroot said, the graphics were killer.)

This gives me a much better sense of what people are saying is the case. I am fairly certain I don't grasp it yet, but the "spacetime interval" must surely have been the missing link I needed to start putting this together in my mind.

The problem for me has always been that if the two people in relative motion measure each others clocks as slow it strikes me as proof positive both clocks are fine and would agree on the total time elapsed if compared later in the same frame. The difference in length of the spacetime interval finally introduces the asymetry that accounts for the differences in the time elapsed in the two different frames.
 
  • #28
Originally posted by zoobyshoe
It's the first time I even understood the importance of the "Spacetime Interval".
The interval is a very important quantity in relativistic physics because it is invariant. No matter what coordinate system you use, or which observers you consider to be at rest, the interval they will measure for some path \Gamma is always the same. The interval is independent of observers and is a fixed quantity for any particular path through spacetime.

- Warren
 
  • #29
I have no idea why anyone has objected to anything I have said here. I will read the link and pick this up later.
 
  • #30
Originally posted by chroot If you think about it, it has to be that way... if it weren't, then some cosmic ray particle moving at 0.9c with respect to you somewhere in the depths of space would somehow affect YOUR clock!
This part, I don't get. I thought cosmic rays were photons, and as such, could never be observed going less than C.

In other words, I have been under the impression that even if I am traveling at 0.9c all photons whose speed I measure going in any direction relative to mine will be clocked going at C. Is this not the case?
 
  • #31
Originally posted by zoobyshoe
This part, I don't get. I thought cosmic rays were photons
No, sorry for the confusion -- some cosmic rays are photons -- but many are massive particles like protons and electrons. The photons, of course, will always be observed traveling at c, while the massive particles will always be < c.

- Warren
 
  • #32
Originally posted by chroot some cosmic rays are photons -- but many are massive particles like protons and electrons.
Interesting. How did they all get the same name, being such different things?
 
  • #33
In Six Easy Pieces he mentions cosmic rays as the highest energy photons we are aware of. Is there a term that can be used to differentiate these from the "riff raff" cosmic rays?
 
  • #34
Originally posted by Ambitwistor
It hasn't been clear to me what your points have been, but it has seemed at times that you have been implying that acceleration is necessary for the twins to experience different elapsed proper times, and/or general relativity is required to resolve the twin paradox in the presence of acceleration, neither of which is true.

Well, somehow we got off track I think...perhaps you are expecting Ivan the Terrible?

First, I was explaining the mechanics of the paradox; that's all.

Next, I addressed the issue of preferred observers; and I still think correctly so. Perhaps this language is out of favor, but specifically I meant that no absolute state of rest or motion exists. This is a still significant concept of SR; no?

Finally, I keep addressing the issue that if we wish to describe one frame of reference as preferred, this in response to Zooby's question about whose clocks run slowly - meaning to prefer one system over the other - a frame of rest must be defined. I meant this all within the context of SR. In all cases we must still define a frame of rest by which we determine who will age less quickly; true? With our twins, we know who is in motion - the one who leaves earth...and this requires acceleration. The page linked makes this assumption immediately
 
  • #35
Originally posted by Ivan Seeking
When the car is in motion, it is shorter in the frame of the garage. Likewise, from the frame of the garage, the car's clocks are running more slowly. When the car stops, that is, when the frame of the car coincides with the frame of the garage, the two lengths L and L0 agree. Likewise, it we compare the ticks of the clocks, we find that again they agree – they occur at the same rate. However, and this was a key test of relativity, we find that while the frames of the car and garage did not coincide, ie, while the car is in motion as viewed from the garage, the clocks in the car really were running more slowly...just as observed and predicted.

This was finally verified I think in the early sixties using two atomic clocks; one on the ground, and one in a jet. After flying one of the clocks around for a while, and after accounting for the effects of gravity, the clock on the plane had indeed lost time as predicted to within the accepted margins of error. This has since be replicated in many other ways. Also, we see the lifespan of subatomic particles increase according to Relativity and their relative speed – since their clocks run more slowly, we see them live longer. This is seen in particle accelerators as well as in nature.

To the extent that isotopes have a 1/2t, does that in relativity,
an accelerating particle runs more slowly than than at rest, factor in when calculating the 1/2t of an isotope?
 
  • #36
Yes, a particle will appear to have a longer half-life when it is travelliong at relativistic speeds in some rest frame, indeed this is a famous example of a relativtic effect.
 
  • #37
Originally posted by Ambitwistor
I wouldn't say that we know that the Earth twin is "at rest", if that's what you're implying by defining a rest frame, but it is true that we know that acceleration breaks the symmetry between the two twins in this variant of the twin paradox. (There are other variants in which this is not the case.)

This was a poor choice of words on my part.

There are other variants in which acceleration does not break the symmetry between the two twins? I may not know what you mean. Could you give an example or two?
 
  • #38
No, acceleration isn't the critical point. You can construct variants of the twin paradox in which nobody accelerates, but the twins come back with different ages.

How exactly does this apply to the twins paradox? I thought a key assumption is that they start and end in the same frame of reference. By this it would seem that the twin set into motion must experience a force in order to start, and to end his trip.
 
  • #39
Originally posted by Ambitwistor
Not really; they just need to start and end at the same place. If you stick in an arbitrarily large acceleration, it changes the elapsed proper time by an arbitrarily small amount. In my example, if the traveling twin goes out and comes back, it still takes him only 6 subjective years to do that, regardless of whether he stops at the end or keeps going.

This poses an interesting question. IF, as you had said, there is both a theoretical model in which one of the twins does not accelerate, but non-the-less 'comes back' at a different time (or age, as the case may be), and also if it is so that an acceleration need not be 'large' (you have not defined a range for 'large'), but
an arbitrary 'small' (whatever 'small' may mean), the would, if string theory were true, the 'vibration' of the string be (by each string an element in a large dynamic consisting of the sum of all of the strings), would all matter exist in two or more times?
 
  • #40
If neither twin accelerates or decelerates then how do you determine which twin is younger since they can both argue the other is moving? Sorry if you already covered this.

Kaw
 
  • #41
Originally posted by kawikdx225
If neither twin accelerates or decelerates then how do you determine which twin is younger since they can both argue the other is moving? Sorry if you already covered this.

Kaw
That's what much of this thread has been about. The short version: the younger twin is the one who travels along a shorter path in spacetime. For the long version, read the thread.
Actually, if neither twin accelerates, then they are both sitting next to each other not moving with respect to each other for their entire lives. They stay the same age relative to each other.

The whole point of them being "twins" is that at the starting point in their lives they are sitting next to each other in the same frame. To get an age difference, one MUST accelerate.

I think the problem got so complicated, the initial intent of a "twins paradox" may have been lost.
 
  • #42
Originally posted by Ambitwistor
Not really; they just need to start and end at the same place. If you stick in an arbitrarily large acceleration, it changes the elapsed proper time by an arbitrarily small amount. In my example, if the traveling twin goes out and comes back, it still takes him only 6 subjective years to do that, regardless of whether he stops at the end or keeps going.

If we never break symmetry, then how does address the paradox? For example, suppose someone happens by planet Earth while traveling near the SOL. I look through my telescope at this person and see someone who appears to be twenty years old. Assume the traveller follows the proper path - only proper acceleration - and then passes by Earth in another eighty years or so and I look again. Through my ancient eyes I see a barely aged, 26 year old in the ship. He has also looked at me on first pass, and again on the return path. From his point of view, I was the one in motion. I am the younger. It seems that we have the original paradox. In other words, I don't see how we resolve the twins paradox without breaking symmetry.
 
  • #43
Originally posted by Ambitwistor
If there's no symmetry breaking, then there's no age difference...the end result depends on the entire history of the worldline, not just the endpoints.)

OK, this is the crux of the paradox to me. Under SR, we have no absolute frame of reference; and the relative motion between our two observers dates back ultimately to the big bang. So, it seems that we can never find a common or preferred frame no matter how far back we look. So it would seem that the answer is not that there is no age difference, the answer is that unless we break symmetry, there is no unique answer to the question: Who ages less quickly? To me, this also implies that the age of the universe depends on the observer. I have never been clear on this point.
 
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  • #44
Originally posted by Ambitwistor
the ones who see the universe as maximally isotropic. The Earth is close to, but not quite, one of those preferred observers.
Just to add to what Ambi said: these preferred frames are just those frames in which the cosmic microwave background radiation is uniform in all directions. If you have some relative motion with respect to the CMBR, you see it as blueshifted along the direction you're moving, and redshifted in the opposite direction. A "comoving" observer is one who is just sort of going along with the flow of the expansion of the universe.

An analogy are two people floating in a river. One guy just floats and let's the water carry him -- he's a comoving observer, and sees the water moving the same speed all around him: zero.

Another guy swims hard in some arbitrary direction. He sees the water moving faster in some directions than in others.

There's nothing that violates relativity theory: this CMBR rest frame is not an absolute rest frame or anything like that. There's nothing special about it relativistically. There is, however, something special about it cosmologically.

Cosmologists measure the age of the universe as the proper time experienced by such comoving observers.

- Warren
 
  • #45
Originally posted by Ambitwistor
If the situation is truly symmetric, you can uniquely answer the question: they both age at the same rate.

Truly symmetric? I'm not sure what exceptions you refer too here.

So you are saying that in my example, when we use our telescopes during each of our two passes near each other, we see each other aging at the same rate? That is, on his second pass by earth, I don't see a 26 year old in my scope, I see a 100 year old person in the ship?
 
  • #46
Originally posted by Ambitwistor
I can't say for sure, since I asked and you didn't clarify the details of how the other twin is returning. But I would guess that the situation is not symmetric and nor will be their ages. If the traveling twin returns by accelerating/decelerating, or by gravitational slingshot, or by circumnavigating a closed universe, then his worldline is very different from the Earth twin's, so there is no symmetry (regardless of whether he starts or ends at rest with respect to the Earth twin).

I mean the situation is which our world lines have never crossed since the big Band, and where our traveler follows a course having only proper accelerations - allowing a second pass by Earth and a second comparison of our clocks, by telescope..
 
  • #47
Originally posted by Ambitwistor
What does "having only proper accelerations" mean? That the traveller has a nonzero proper acceleration at all times? Does he accelerate and decelerate? Or what?

I mean that we never break symmetry.
 
  • #48
Originally posted by Ambitwistor
Can you describe a physical scenario in which the traveling twin is able to pass the Earth twice, without breaking symmetry?

From some of your earlier comments, I thought that you knew of such a trick. If this is not possible, then all is well.
 
  • #49
Originally posted by Ambitwistor
I can invent such a scenario (but not in pure special relativity); I just didn't know what scenario you were considering. For instance, both twins could circumnavigate a closed universe, with opposite velocities relative to a cosmological observer. In that case, their aging would be symmetric.

So in this case, if on some near pass one twin accelerates and joins the other, we find that the twins are [almost exactly] the same age?
 
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  • #50
Originally posted by chroot
Cosmologists measure the age of the universe as the proper time experienced by such comoving observers.

- Warren

Is there a maximum age for the universe as viewed by some class of observers; where all other observers will measure a lesser age?
 

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