How accurate are atomic clocks??

James R, Crisp, Thed, Q, and anyone else who cares,

The proof that I have provided in my previous post may have been a little too complicated for some people to understand. Therefore, here is the same problem, only more simplified:

Case 1) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it forward. After 1 second of the stationairy clock, how far away is the light from the moving observer?

Case 2) An observer, moving at .90c, flys over a stationairy clock. When he is directly above the stationairy clock, he turns on his flashlight and points it behind him(in the opposite direction of his motion). After 1 second of the stationairy clock, how far away is the light from the moving observer?

In the first case, since the light is moving with the observer, the distance between the light and the observer, in the stationairy frame of reference is:

v= c - .90c
v=300,000 km/s - 270,000 km/s
v=30,000 km/s
d=v/t
d=30,000 km/s/1 second
d=30,000 km

Therefore, the the light is 30,000 km away from the moving observer, in the stationairy frame of reference, after one second.

In the second case, since the light is moving away from the observer, the distance between the light and the observer, in the stationairy frame of reference is:

v= c + .90c
v= 300,000 km/s + 270,000 km/s
v= 570,000 km/s
d= v/t
d= 570,000 km/s/1 second
d= 570,000 km

Therefore, in the second case, the light is 570,000 km away from the moving observer, in the stationairy frame of reference, after one second.

Let's review, after 1 second, in the stationairy frame of reference, the distance between the light and the observer is:

Case 1) 30,000 km
Case 2) 570,000 km

Unfortunately, these values can't be used in the moving observer's frame of reference because of the length contraction and time dilation that the moving observer experiences. Therefore, we have to convert these figures so that they are valid in the moving observer's frame of reference.

Let's assume that the length contraction ratio for the moving observer is equal to x, and that the time dilation ratio for the moving observer is equal to y.

Therefore, to get the speed of light in the moving observer's frame of reference, you need to multiply the stationairy length by x, to get the length in the moving observer's frame of reference. Then you need to multiply the stationairy time by y to get the time in the moving observer's frame of reference. You would then divide the length by the time to get the speed of light in the moving observer's frame of reference.

Since relativity claims that the speed of light is always c in a moving observer's frame of reference, the mathematical formula for case 1 would be:

30,000 km/1 second * x/y = 300,000 km/s

In case 2 it would be:

570,000 km/1 second * x/y = 300,000 km/s

Because the speed of the observer is equal in both cases, the length contraction ratio(x) is equal in both cases, and the time dilation ratio(y) is also equal in both cases.

The conclusion: If you adjust the time dilation and length contraction ratios in one case so that the speed of light is equal to c, the same time dilation and the length contraction ratios would not give c as the result in the other case. There are no two numbers(x and y) that would give a result of 300,000 km/s in both cases.

In other words, the speed of light is not always equal to c in all frames of reference. The principle of invariance of light is incorrect.

Crisp: I noticed that you still didn't respond to my earlier post. Am I right or am I wrong??

Tom

 

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Hi Tom,

"Crisp: I noticed that you still didn't respond to my earlier post. Am I right or am I wrong??"

Neither, I am busy

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... Will get to it as soon as I have the time, at the moment I just drop in for brief comments and keeping up with what happens. Perhaps friday 28th of June I can do some of the calculations myself (yes, it requires scheduling in my agenda at the moment

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).

Bye!

Crisp

 

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Tom,

What you have is an incorrect way of measuring light speed. It's not enough for you to sit on top of the "stationary" clock and observe the moving wavefronts. The problem is that once the flashlight emits its photons, it's never going to see them again. So how is the "moving" flashlight supposed to calculate the speed of light when it can't make the measurement?

What it would have to do, is perhaps bounce the light off a mirror. But you will observe that any such "bouncing" would completely cancel out the contributions of the flashlight's "velocity", in the end always coming out to be c even from the flashlight's perspective.

So... As far as you are concerned, the speed of light is c. And... As far as the "moving" flashlight is concerned, the speed of light is also c! See?

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Overdoze,

You're right. If you want the flashlight to truely measure the speed of the light, then the light would have to bounce off a mirror and return to the flashlight. And you are correct, in that case the discrepencies would balance each other out.

However, we don't need to do that because we already know the formulas for time dilation and length contraction. All we have to do is use these formulas to obtain the resulting speed. And as you can see from my example, when the light does not bounce off any mirrors, it's relative speed to the observer is not c.

I hope that you see, from my example, why I believe that light does not travel at c in all frames of reference. This is what I was talking about on the other thread when I was referring to the unrelative components of atomic clocks.

Tom

 

Tom,

Let's rephrase. Forget for a moment about length contraction and time dilation, as these are not the problem in your case and you can ignore them. Let's say three bodies, A B and C are arranged along their velocity vector, like this:

<tt><pre>
C
/\
A | (velocity)

B
</pre></tt>

Assume AB = AC.

What you're saying is, looking from a "stationary" FOR, if A emits a spherical light pulse the light will reach B before C. That is absolutely true -- from a particular FOR registering that particular velocity for A, B and C. But you aren't looking from the FOR of A. If you were co-moving with A, you would see B and C light up at the same time (simultaneously). Remember that as far as A is concerned, it's standing still while you are flying away in the opposite direction.

That's one of the more mind-bending consequences of relativity -- relativity of simultaneity. IOW, there is no such thing as absolutely simultaneous events. Depending on your FOR, for two events X and Y you can either perceive X occur before Y, X and Y occur simultaneously, or Y occur before X. You will always observe events, but in what order depends entirely on how you're moving relative to them.

For example, if you were looking from an FOR that actually travels in the same direction as A but even faster (so that your measured velocity of A is negative), you would in fact perceive C light up before B. It's perfectly symmetrical, and depends only on the velocity that you are measuring given your FOR.

So now take your claim:

How does this follow? In your "stationary" frame of reference the light is travelling at c. In A's frame of reference the light is still travelling at c (as measured by A). You have yet to provide a frame of reference from which the speed of light would not be observed to be c.

So for example, even if in an imaginary "absolute" reality B truly lights up before C, A would never know it -- it would have no way of ever finding out. Of course, if you can come up with a way for A to find out, you'd become the most celebrated scientist of this century.

 

I really don't know what to say.

I took the stationairy distance and the stationairy time that the light travelled in two cases:

1) If the observer shines the flashlight forward.
2) if the observer shines the flashlight back.

In case 1, the light is 30,000 km away from the observer after 1 second in the stationairy frame of reference.

In case 2, the light is 570,000 km away from the observer after 1 second in the stationairy frame of reference.

Relativists will argue that the moving observer experiences time dilation and length contraction, so I included these in my calculations.

The fact is the when the results of one frame of reference are adjusted for time dilation and length contraction, they should give the results for the other frame of reference. In other words, the results of two frames of reference should be equal to each other when they are adjusting for time dilation and length contraction.

The fact is that if you give time dilation and length contraction a value of z, then there is no mathematical solution for both of the equations:

30,000 km * z = 300,000 km/s

and

570,000 km * z = 300,000 km/s

As I stated before, if you modify z to give 300,000 km/s for one formula, then using the same z will not give you 300,000 km/s for the second formula. In other words, only one formula, at most, can give the result of 300,000 km/s. The other formula will give a result larger or smaller than c.

If you still don't believe me, give me what you believe is the length contraction and time dilation ratios so that I may incorporate them in my formulas to demonstrate that light can't travel at c under both circumstances.

By the way, a few posts back I included the time dilation and length contraction figures provided by James R in my formulas, and proved that the speed of light is not c in both cases. Crisp and James then told me that the time dilation and length contraction figures that James R provided were incorrect.

I am still waiting for James R or Crisp to provide me with the correct time dilation and length contraction ratios, so that I can prove that Einstein was wrong once and for all.

If you do not wish to wait for James R or Crisp, please feel free to give me your length contraction and time dilation figures.

Tom

 

I am still waiting for James R or Crisp to provide me with the correct time dilation and length contraction ratios, so that I can prove that Einstein was wrong once and for all.

You will of course be publishing your results so that others will peer review your theory. There is a Nobel Prize waiting for you.

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Tom,

A couple more corrections are in order:

That is not true. In addition to time dilation/length contraction, you still have to take into account the relative motion of the FORs. The latter goes back to Newton, and Relativity is not exempt (remember, it is merely a generalization of Newtonian mechanics.)

That is incorrect. Please ponder this question: from which FOR is the measurement being made? What is the actual speed of light in that FOR? Take my diagram above, and imagine yourself sitting on A. What do you observe? (And remember, to observe anything information must be incoming as opposed to outgoing.)

I believe you! We all believe you. There is nothing wrong with your math. But you are drawing the wrong conclusion because you are misinterpreting the math. That's what I'm trying to point out.

 

Overdoze,

You're wrong. The only thing we are measuring is distance and time. Relative motion does not effect distance or time(excluding the effects of length contraction and time dilation).

If you are positive that relative motion effects this experiment in addition to length contraction and time dilation, please state your formulas so that I can include them in by calculations.

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If you throw a ball and after 1 second it goes 5 meters, how fast is the ball moving from you. Let me guess 5 m/s??? Opps, I'm wrong, I'm misinterpreting math again, right?

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The simplist way to calculate the speed of an object is to measure the distance it travelled in a certain amount of time. This is just not logical, it is common sense as well. After all, if you have a distance and the time that distance was travelled, how would you calculate speed??

I'm so sorry if using the basic formula for speed contradicts Einstein's theories. But don't blame me, it's not my fault, blame reality.

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Tom

 

Tom,

Here we go again:

Think, which observer are you talking about, and how exactly is this "observer" observing the lightfront?

The distances you calculated are in the frame of reference of someone who is sitting "still". That "stationary" observer could have placed little markers along the way that light up as they are reached by the lightfront, and that is how this observer would actually observe these unequal distances with respect to the emitter following an elapsed time of 1 sec (plus whatever additional time needed for reflected light from the markers to reach this observer).

Now, how does the emitter observe the distance the lightfront has travelled? If the emitter looks at those markers, it would see them light up at exactly the same time, progressively, front and back, and so would conclude they were equidistant from it at the time they reflected the propagating spherical lightfront it emitted. Remember, the effects of the emitter's relative velocity with respect to the markers are cancelled out as the light bounces back toward the emitter. Remember also that emitter assumes speed of light to be equal in all directions, and has no way of determining that it isn't.

It seems you are finding it impossible to picture yourself as the emitter rather than the "stationary" observer, and so you are perpetually stuck in the "stationary" FOR. Well great, in the stationary FOR, the distance light travels in either direction is not measured with respect to the travelling emitter, but with respect to the precise point in space where the light was emitted. Where the emitter has gone since then doesn't matter. So in the stationary FOR the light travels exactly the same distance forward and backward from the point of emission in 1 second, namely the roughly 300,000 km.

However, as far as the emitter is concerned the point of emission is located on its premises at all times, and does not move. Rather, it's the coordinate "markers" flying past the emitter that move. So the emitter is going to emit the light forward and backward, and record the lidar return from the markers at what it believes is 300,000 km front and back after exactly 2 seconds. The emitter can easily calculate that from the "stationary" FOR the marker in the back was a lot closer to it than the marker in the front (since the emitter knows that the "stationary" FOR is receding from it at the same exact velocity as the coordinate markers.) But the emitter has no reason to suspect that the "stationary" FOR really is stationary and somehow superior to its own.

[edit: corrected some mistakes (I was somewhat tired when I first wrote this) -- such as changing "farther" to "closer" and such in a couple of places.]

 

Overdoze,

Let's get down to basics:

If the light in case 1 is 30,000 km away after 1 second in the stationairy frame of reference:

1) What is the 30,000 km of stationairy distance equal to in the moving observer's frame of reference??

2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??

3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.

If the light in case 2 is 570,000 km away after 1 second in the stationairy frame of reference:

1) What is 570,000 km of stationairy distance equal to in the moving observer's frame of reference??

2) For 1 second of stationairy time, how much time passed in the moving observers frame of reference??

3) Divide the answer from question 1 with the answer from question 2 to get the speed of light for the moving observer.

Do the calculations and you will find that answer 3 from the first case can't equal answer 3 from the second.

You keep on implying that the only way to measure the speed of the light is to bounce it off something. I agree with you that if you take the reflected light into effect, the speed of light averages out to give c. However, Einstein did not claim that the speed of light averages out to c after reflection, he claimed that it always travels at c in every frame of reference.

Through your suggestion that the speed of light averages out to c as a result of reflection, you are actually admitting that the speed of light is not equal to c in the moving observer's frame of reference. If it were, it wouldn't have to "average out".

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Tom

 

Sure, let's get down to basics.

In your case, with relative velocity of 0.9c, we can use the length contraction formula in reverse to figure it out:

L' = L/sqrt(1-v^2/c^2) = 30000/sqrt(1-0.81)

which is approximately 68824.720 (km).

Using the time dilation formula:

t' = t*sqrt(1-v^2/c^2) = 1*sqrt(1-0.81)

which is approximately 0.43588989 (s).

approximately 157894.74 (km/s)

approximately 1307669.7 (km)

approximately 0.43588989 (s)

approximately 3000000.1 (km/s)

You're absolutely correct. However, what we've been calculating is not the speed of light as the moving observer would measure it.

There are plenty of other ways, but they are all equivalent to bouncing light off something (to which the distance is known). If you have a non-equivalent way, I'm waiting to hear of it.

If we used a proper method to measure lightspeed for the moving observer (i.e. using lidar), we would indeed come away with c as the answer. Therefore, we would claim that light travels at speed c in the moving frame of reference.

I've admitted from the get-go that relativity does not exclude absolute frames of reference. What you have to admit, is that it doesn't matter whether such an absolute frame of reference exists or not. What matters is your own FOR, and what you measure in it.

The point is, it does average out. Always. 100%. No matter what experimental setup you contrive. That's the real message of the theory -- not that there is a separate universe for each distinctly moving observer, but that there is precisely 1 universe that we all share, and no matter what (inertial) perch you look from you still see that same universe with the same constants and the same laws of physics.

 

Overdoze, Crisp, Thed, Q, James R, Xev, C'est Moi, and 137,

Overdoze:

Thank you for your posts. I finally understand what the principle of invariance of light really means. I just wish that someone explained this to me sooner.

Everyone Else:

My formulas were correct in both cases. It is assumed in relativity that the observer actually has to recieve the light he/she transmitted, from a reflection, in order for the observer to measure the speed of light. If the observer measures the speed of light by reflection(the speed of light to a mirror or object, and back), the observer will always measure the speed of light to be c. This is because, as in case 1, the transmitted light is slower than c for the observer, but the reflected light is faster than c, averaging out the total speed of light to c. This "averaging out" to c happens in all cases, regardless of the speed of the observer. However, this does not mean that the one-way speeds of light are equal to c. Logic dictates that the one-way speed of light is always more or less than c in a moving frame of reference.

In other words, the observer perceiving the round trip of the light to be c, does not mean that all one-way speeds of the light are equal to c in a moving frame of reference.

So finally, what are the one-way speeds of light in a moving frame of reference?? We can't measure them directly, so what could they possibly be??The answer is something that none of you (except Overdoze, C'est Moi, and 137) want to hear:

The one-way speeds of the light in a moving frame of reference are related to the speed that the relative frame of reference is traveling in relation to the absolute frame of reference. Only in an absolute frame of reference is the one-way speed of light always equal to c.

As I stated before, there isn't a direct way to measure the one-way speed of light in a moving frame of reference. However, there is an indirect way. It is called a light clock:

http://www.physics.wustl.edu/~visser/physics-216/notes-light-clock.html

I already posted this link on another thread.

This light clock slows down the faster the clock is traveling in relation to the absolute frame of reference. It can be used to measure it's own absolute motion. Once the absolute motion of the clock is known, it can be used to derive the one-way speed of light that is traveling with the clock.

One more important thing to consider is that sometimes the round trip speed of light is not as important as the one-way speed of light. This is especially true in atomic clocks(where the electromagnetic radiation stimulates and seperates the caesium atoms, this same radiation DOES NOT need to get reflected back in order for it to influence the caesium atoms). This effect may cause the traveling atomic clock to give the appearance that time is slowing down, when in reality, it isn't.

Tom

 

Hi Tom,

I think you jumped to your conclusion too fast, overdoze also said:

Also, you said:

"In other words, the observer perceiving the round trip of the light to be c, does not mean that all one-way speeds of the light are equal to c in a moving frame of reference."

I disagree and (for now) proclaim without further evidence that the "one way" speeds are also c. I'll get back to this as soon as I have more time (quantum field theory eats all time at the moment

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). I will also attempt to prove that the formula's you used to formulate your contradiction ("that no two numbers x and y can be found to ... ") are not valid. That should be fun to do

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, just a little patience...

Bye!

Crisp

[edited to fix bold/italic statements]

 

Crisp and Overdoze,

A lidar would not be able to measure the one-way speed of light. It would only be able to measure the average speed of the transmitted and the reflected light.

Crisp: I will be looking forward to your evidence.

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Tom

 

Can someone explain lidar briefly to me please?

 

Adam,

A lidar is a laser radar. Instead of bouncing low frequency electromagnetic radiation off an object to determine it's location like a regular radar, a lidar would bounce a pulse of light off the object.

Tom

 

Why use light part of spectrum? Why not just use radar like normal?

 

Adam,

I'm not sure.

Maybe it's more accurate because the wavelength of light is shorter.

Tom