How accurate are atomic clocks??

Tom,

You're obviously a little mixed up about all this stuff. Let me see if I can make it a bit simpler for you...

1. It is a <b>postulate</b> of special relativity that the speed of light is constant in all reference frames.
2. The word "postulate" means, essentially, "assumption". A postulate is not based on anything <i>a priori</i>. It is assumed. The consequences are then derived theoretically.
3. To see whether a postulate is ultimately justifiable or not, <b>the only true arbiter is experiment and observation</b>. We must look at the real world and compare the theory based on the postulates with what really happens. If the theory matches the reality, the postulate is correct (or at least useful). If not, the theory is wrong, and hence the postulate is wrong since the theory is derived from it.
4. In the case of special relativity, <b>all experiments and observations</b> support the constancy of the speed of light in all inertial reference frames, with no exceptions. Hence, we conclude that the postulate is true.
5. Time dilation and length contraction are <b>inevitable consequences</b> of the constancy of the speed of light in all inertial reference frames. Since they are directly observed, they provide further evidence for the constancy of the speed of light.
6. If, instead of the postulate of the constancy of the speed of light you choose to assume that time dilation and length contraction exist in the same mathematical form that Einstein et al. proposed, then the constancy of the speed of light follows inevitably from that alternative assumption.
7. It is incorrect that there is any circularity in relativity. The theory is justified by experiment, not by other theory or itself.

Now, to your comments...

<i>I attempted to prove that the speed of light was constant in all frames of reference by only using the formulas for time dilation and length contraction, since they are the only formulas that would "influence" the speed of light relative to the moving observer.</i>

See point 6, above.

<i>Unfortunately, I never came to the result that the speed of light is c for a moving observer.</i>

Then your maths is not very good. See point 6, above.

<i>Here is a simplified question with only one observer:

An observer is traveling at .90c. The observer turns on his/her flashlight, which is pointing in the same direction as his/her motion, while continuing to travel at that speed. After one second of a stationairy clock, how far is the light from the observer, and using only the formulas for time dilation and length contraction, explain why the light is travelling at c relative to the observer.

Do not make the assumption that light is travelling a c relative to the observer, instead, use only the formulas for time dilation and length contraction to prove that the light is traveling at c relative to the observer.</i>

1 second on a stationary clock corresponds to 2.29 seconds for the observer, using time dilation.

In the stationary reference frame, the speed of light is 300,000 km/s. The speed of the observer is 270,000 km/s, so after 1 second (measured in the stationary frame), the light is 30,000 km from the observer <b>as measured in the stationary frame</b>.

In the observer's frame, in the 2.29 seconds the point which the light started from moves backwards from the observer (who considers himself stationary) a distance of 270,000 &times; 2.29 = 619422 km. Due to the length difference the observer measures from the stationary frame, after 2.29 seconds, the light is a distance of 30,000 &times; 2.29 = 68,825 km from the observer at the end of the time interval, but in the opposite direction to the start point.

To calculate the speed of the light, the observer takes the total distance measured by him for the travel of the light divided by the total time measured by him.

The total distance is: 619422 + 68825 = 688247 km
The total time is: 2.29 seconds.

The speed of light, according to the observer is therefore 688247/2.29 = 300,000 km/s

Note that the only assumptions made in getting this solution were:

1. The speed of light in the stationary frame is 300,000 km/s.
2. The time dilation and length contraction formulae of special relativity apply.

From that, the constancy of the speed of light for the moving observer follows inevitably. See point 6, above.

Surely you must be satisfied now?

(I'm betting you'll now change your tune and claim that the speed of light is not constant in all frames of reference after all.)

 

Tom,

The way you're doing it you're not measuring the speed of light. To measure the speed your must mark off on your space axis the point where the light started and where it ended up. The distance between those two points is the distance the light has travelled.

What you're doing is marking off the position of the <b>observer</b> and the position the light ends up at and pretending that the distance between those points is the only distance the light has travelled.

Use your common sense.

 

Tom,

<i>Fact: The light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

Question 1: How far away is the light from the observer after
1 stationairy second in the observers frame of reference?</i>

68,700 km. The stationary frame sees a contracted distance of 30,000 km.

<i>Question 2: How much time passed by for the observer during that one second of stationairy time?</i>

0.229 seconds. The stationary frame sees the moving clocks as running slow.

<i>Question 3. By dividing the answer from question 1 by the answer from question 2, how fast is the light traveling in the observer's frame of reference?</i>

68,700 km/ 0.229 seconds = 300,000 km/second.

<i>To calculate the speed of light in the observer's frame of reference, you do not need to know the total path of the light, you only need to know how far the light is from the observer, and how much time it took to get there.</i>

That is correct. Here's how I worked it out, in detail:

Consider 2 events:
Event A: The observer turns on the light.
Event B: The position of the light pulse is measured after 1 stationary second.

In the <b>stationary</b> frame, the spacetime co-ordinates of the two events are:
Event A: (x,t) = (0 km, 0 seconds)
Event B: (x,t) = (300,000 km, 1 second)

The speed of light in this case is the change in the distance divided by the change in the time, which gives 300,000 km/s.

In the <b>observer's</b> frame, we must use the Lorentz transformations to get the observer's spacetime co-ordinates for events A and B. The Lorentz transformations from the (x,t) co-ordinate system to the (x',t') co-ordinate system are:

x' = <font face="symbol">g</font>(x - vt)
t' = <font face="symbol">g</font>(t - vx/c²)

(which you can confirm by looking them up on any relativity web site).

For event A, we find:
(x',t') = (0 km, 0 seconds).

For event B, we find:

x' = 2.29 (300,000 - 270,000 &times; 1) = 68,700 km
t' = 2.29 (1 - (0.9c)(300,000)/c²)
= 2.29 (1 - 270,000/300,000) = 2.29 &times; 0.1 = 0.229 seconds.

Again, the speed of light in this case is the change in the distance divided by the change in the time, which gives:

68,700 km/ 0.229 seconds = 300,000 km/s

 

The Chosen,

You never stop climbing the education ladder. There's always more to learn.

I haven't seen the MUS tapes, but thanks for the tip.

 

Tom,

I've thought this problem through properly now. My previous post (where I used the Lorentz transformations) is the correct solution to your problem. Please disregard my previous posts on that issue. There's mostly truth there, but it's mixed in with a couple of mistakes. Relativity is tricky. You need to be very careful about which reference frame you are using at all times. I played it a bit fast and loose initially, but now I've tightened up the reasoning. As I say, my post immediately before this one is the definitive and correct answer, so let's just concentrate on it and go from there.

I said the observer measures a distance of 68,700 km. The stationary observer measures a shorter distance of 30,000 km.

You said: <i>The stationairy observer does not see a contracted distance, because he/she is at rest. The moving observer sees the contracted distance. Therefore the distance is 13,080 km for the moving observer.</i>

Remember that relativity is symmetrical. The stationary observer sees the moving observer's clocks running slow and the moving observer's lengths contracted. The moving observer sees the stationary observer's clocks running slow and the stationary lengths contracted. This can cause a lot of confusion when you try to apply the length contraction and time dilation formulae. As I said, you need to be VERY careful that you get things the right way round. I stuffed it up in my first couple of posts on this problem, which is why I decided to use the Lorentz transformations instead. That way, I could be guaranteed I was keeping track of everything properly.

Anyway, what all this boils down to is that your statement above is incorrect, and my statement is correct.

<i>If the the time for the moving observer is running slower than the stationairy frame, then 1 second in the stationairy frame is 2.29 seconds for the moving observer.</i>

No, that statement would be wrong regardless of my calculation. Think about it. The stationary observer sees the moving observer's clocks as running slow. Therefore, in the time that the stationary clocks tick off 1 second, we'd expect that the moving clocks, according to the stationary observer, would tick off <b>less</b> than 1 second, since the moving clocks are ticking slower from this point of view. Therefore, it cannot possibly be the case that the stationary observer sees the moving clock tick off 2.29 seconds in the time that the stationary observer's clock ticks off 1 second.

<i>Funny, just a few posts ago, you said it was 2.29 seconds as well. You must have changed your calculations when they didn't give you the correct result, right??</i>

Yes. Right. I was wrong the first time, but now I am correct. My previous paragraph above makes clear where I made my mistake.

<i>And how exactly did you get 0.229 seconds from the formula:

t = t0/sqrt(1-(v^2/c^2))

Was it those magic fairies again???</i>

I didn't use that formula. I used the Lorentz transformations, which are clearly set out above. The formula you have quoted turns out to be not directly applicable in your example. That fact in itself had me confused at first, which is another reason why I made some mistakes initially. In fact, you need the full Lorentz formulae to handle your problem. I have used the correct formulae to come up with the correct answer. No magic fairies required. Just careful thought.

Thankyou for continuing to push me until I got my thinking correct on this. I am now 100% confident that my solution is correct and we can discuss it from there.

So, I imagine you will now take issue with the Lorentz transformation. Can you see any problem with what I have done there? (There isn't any, I assure you.)