Variable speed of light

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[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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  • #2
chroot
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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
 
  • #3
jcsd
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Nothing to do with VSL theories (which are a far from conventional) or gravitational fields even, it's special relativty 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 travelling 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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  • #4
Hey, Thanx people! Now that's what I call service.

Remind me to visit this board more often.
 
  • #5
zoobyshoe
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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
 
  • #6
jcsd
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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.
 
  • #7
zoobyshoe
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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 sence 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 sence to me)?
 
  • #8
Ivan Seeking
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Originally posted by zoobyshoe
JCSD,

This being the case (which makes perfect sence 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 sence 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.
 
  • #9
zoobyshoe
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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
Ivan Seeking
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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
zoobyshoe
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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 occuring. When the relative motion stops the illusion stops.
 
  • #12
Ivan Seeking
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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 occuring. 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
zoobyshoe
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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
Ivan Seeking
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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
Ambitwistor
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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. (e.g., going around a closed universe, or a gravitational slingshot back home). 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.
 
  • #16
zoobyshoe
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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.
 
  • #17
Ivan Seeking
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Originally posted by Ambitwistor
or a gravitational slingshot back home

How do we do this without any acceleration?
 
  • #18
Ivan Seeking
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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].
 
  • #19
chroot
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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
 
  • #20
Ambitwistor
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Originally posted by zoobyshoe
In which case the one who takes the longer path ages less?

The longer path ages more, if you measure "length" using the spacetime interval.


This is in General Relativity, right?

Special relativity too.

Consider the case where the Earth twin stays at home for 10 years according to his own clock. The other twin travels at 80% of light speed, travelling 4 lightyears in 5 years according to the Earth twin, then returns the same way (after an instantaneous deceleration and acceleration back home).

The Earth twin's worldline is a line from (t,x) = (0,0) to (10,0). The travelling twin's worldline consists of two line segments, one from (0,0) to (5,4), the other from (5,4) to (10,0).

The proper time measured by the Earth twin is the spacetime length of his worldline,

[tex]
\tau = \sqrt{{\Delta t}^2-{\Delta x}^2} = \sqrt{(10-0)^2-(0-0)^2} = 10
[/tex]

The proper time measured by the travelling twin is the spacetime length of his worldline,

[tex]
\begin{equation*}
\begin{split}
\tau &= \tau_1+\tau_2 = \sqrt{{\Delta t_1}^2-{\Delta x_1}^2}+
\sqrt{{\Delta t_2}^2-{\Delta x_2}^2} \\
&= \sqrt{(5-0)^2-(4-0)^2}+\sqrt{(10-5)^2-(0-4)^2}\\
&= 3+3 = 6
\end{split}
\end{equation*}
[/tex]

The travelling twin ages 6 years to the Earth twin's 10 years.
 
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  • #21
Ivan Seeking
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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.
 
  • #22
chroot
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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
 
  • #23
Ivan Seeking
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Zooby, do you feel spoiled?
 
  • #24
Ambitwistor
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Originally posted by Ivan Seeking
How do we do this without any acceleration?

I was speaking of proper acceleration.
 
  • #25
Ivan Seeking
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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.
 
  • #26
Ambitwistor
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Originally posted by Ivan Seeking
We still have no preferred observers in the inertial frame.

I'm not sure what point you're trying to make.


If one accelerates, that frame is no longer valid under SR, and we can then make a distinction between the two systems.

However, acceleration is not necessary to distinguish between the two systems. As I said, the only thing that determines whether two twins age asymmetrically is whether the spacetime lengths of their worldlines are different. The symmetry can be broken by acceleration, or by other means.

We can resolve this using pure SR, by the way: my calculation was performed entirely in an SR inertial frame (that of the Earth twin).
 
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  • #27
chroot
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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
 
  • #28
Ivan Seeking
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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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  • #29
Ambitwistor
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Originally posted by Ivan Seeking
There is no absolute reference frame. There is no absolute state of rest.

So?


How do we determine which twin is younger?

The younger twin is the one whose worldline is shorter. Which twin that is depends on the physical situation. I gave an example.
 
  • #30
Ivan Seeking
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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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  • #31
zoobyshoe
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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 sence 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.
 
  • #33
Ambitwistor
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Originally posted by Ivan Seeking 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.

Yes. So?
 
  • #34
chroot
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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 [tex]\Gamma[/tex] is always the same. The interval is independent of observers and is a fixed quantity for any particular path through spacetime.

- Warren
 
  • #35
Ambitwistor
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Originally posted by zoobyshoe
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.

It's good to think in terms of geometry. Special relativity is just Euclidean geometry in disguise (with a slightly modified Pythagorean theorem).

Lorentz transformations are analogous to rotations. (Because of the minus sign in the Pythagorean theorem for distance, Lorentz transformations trace out hyperbolas in spacetime instead of circles in space.) So the statement that "there are no preferred inertial frames in SR" is really a disguised version of the statement, "there are no preferred directions in Euclidean geometry".

You can see that in this version of the twin paradox, the two worldlines make a triangle: the worldline of the Earth twin is one side, and the worldline of the travelling twin is the other two. People say, "why isn't the travelling twin's frame the same as the Earth twin's?" In Euclidean geometry, the analog of switching inertial frames is rotating. But you can see that just by a rotation, you can't turn the two bent line segments of the travelling twin into one line segment like the Earth twin's worldline: they are not geometrically equivalent to each other. Rotations don't affect lengths, so the two sides of the triangle will never be equal in length to the other side of the triangle, no matter what you do to it --- unless you start deforming it, but then that doesn't describe the same situation anymore.
 

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