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Russell
2003-Dec-12, 12:52 PM
Hello:
I'm new to this and I'm not a pro scientist so please excuse me.
Can anybody please give me a good analogy for visualizing how the speed of light will always travel at a constant speed no matter how fast you are traveling? Though I understand the consequences of both Special and General Relativity this concept just has'nt sunk in. Thank You very much.

Glom
2003-Dec-12, 01:11 PM
Welcome.

The truth is that visualising this is tricky as it all depends on giving up a subconsious concept that we all hold dear: an ethereal frame of reference. After the M-M experiment, Lorentz derived his translations mathematically. But, to explain the results physically, he still made use of the quintessential ether. He postulated that as objects move through the ether, their length shrinks and their time dilates. Einstein did away with the ether and said that everything is relative.

It is not an attractive prospect. We like the idea of an ethereal frame of reference. It gives us a feeling of security, knowing we can define our exact position, time, etc. So, when we consider that things seem different in different frames, and that their is no way to determine which is the "true" way it should look, is a little disorienting.

Consider: You observe a couple of spaceships moving relative to you, exchanging messages via flashes of light in the direction of travel. You see the light travel at c and are happy. They are travelling at a certain speed, v and as such, you perceive them travelling c-v slower than the light.

However, from the perspective of the people on the ships, the light is travelling at c as well, when if we were Newtonian, we would expect it to be c-v. How can they measure the same speed of light? The answer is that while you think they should measure c-v for the speed of light, you observe the clock on their ship and see that it is ticking more slowly that in should. So one of their seconds is longer than yours, and hence the time they'll measure for the light to travel between ships is less than the time you would measure it, because their clocks are running slow. Since v=s/t, and the value of t onboard the ships is shorter than from your position, they calculate the speed relative to them to be faster than the speed you think they should measure. The maths surrounding these translations make it such that the faster speed is in fact c.

The concepts are pretty abstract because in everyday life, we are capable of establishing a reasonable "absolute" coordinate system.

Russell
2003-Dec-12, 01:33 PM
Glom,
I understand. Thanks. There is nothing bad about being a RAF pilot.

Glom
2003-Dec-12, 01:36 PM

I'm aware of that. That's the problem.

Russell
2003-Dec-14, 06:50 AM
Please don't take offense, I work RAF and Royal Marines down here in Naples and love it. Like I said before, I'm new to this site so you have to excuse me. I have more questions than answers and look at this site as a good forum for learning. I'm getting ready to tackle "The Meaning of Relativity" and I'm quite scared of the math. I've read many books on the subject but never anything on the mechanics. So I may need help. Gotta get to work, thanks.

Glom
2003-Dec-14, 01:47 PM
The topics can be very heavy. I often don't have the attention span to read them properly.

Sam5
2003-Dec-14, 02:13 PM
Consider: You observe a couple of spaceships moving relative to you, exchanging messages via flashes of light in the direction of travel. You see the light travel at c and are happy. They are travelling at a certain speed, v and as such, you perceive them travelling c-v slower than the light.

However, from the perspective of the people on the ships, the light is travelling at c as well, when if we were Newtonian, we would expect it to be c-v. How can they measure the same speed of light? The answer is that while you think they should measure c-v for the speed of light, you observe the clock on their ship and see that it is ticking more slowly that in should. So one of their seconds is longer than yours, and hence the time they'll measure for the light to travel between ships is less than the time you would measure it, because their clocks are running slow. Since v=s/t, and the value of t onboard the ships is shorter than from your position, they calculate the speed relative to them to be faster than the speed you think they should measure. The maths surrounding these translations make it such that the faster speed is in fact c.

Spacecraft clocks run slow because of their motion? The mainstream sites say that spacecraft clocks speed up when the craft get out of the strong gravitational field that’s near the surface of the earth.

In the Hafele Keating flying clocks experiment, all the clocks’ averaged time change showed an overall speed up, not an overall slowdown.

How can the relative motion of two spacecraft slow down the two clocks in the two different spacecraft? How can “looking” at a clock slow it down?

If you think they are traveling away from a light source and they encounter the light that overtakes them from the rear at c-v, then if their clocks slow down the right amount, that might cause them to think that the light speeded up to c, but what of the light they receive from the front? From a source they are moving toward? If you think they encounter that light at c + v, and if their clocks slow down, then they would see that light from the front traveling much faster than c + v.

Do their clocks slow down for the light coming from the rear and speed up for the light coming from the front?

Glom
2003-Dec-15, 12:57 AM
Spacecraft clocks run slow because of their motion? The mainstream sites say that spacecraft clocks speed up when the craft get out of the strong gravitational field that’s near the surface of the earth.

In the Hafele Keating flying clocks experiment, all the clocks’ averaged time change showed an overall speed up, not an overall slowdown.

How can the relative motion of two spacecraft slow down the two clocks in the two different spacecraft? How can “looking” at a clock slow it down?

If you think they are traveling away from a light source and they encounter the light that overtakes them from the rear at c-v, then if their clocks slow down the right amount, that might cause them to think that the light speeded up to c, but what of the light they receive from the front? From a source they are moving toward? If you think they encounter that light at c + v, and if their clocks slow down, then they would see that light from the front traveling much faster than c + v.

Do their clocks slow down for the light coming from the rear and speed up for the light coming from the front?

Did I say anything about Earth or gravitational fields? Your straw man tactics will not work with us. SR is very clearly explained (http://www.fourmilab.ch/cship/). We don't need yet another thread focusing on your discontent with it. Time dilation is not direction dependant. Only speed dependant. The dilation works the same for both separating and approaching frames.

Sam5
2003-Dec-15, 04:19 AM
Glom,

Well, since I’m so stupid and you are so smart, I hoped you could answer my question. You seem to know all about this stuff. So, if you slow down the clocks of the people on the moving spacecraft so they will see a light beam coming toward them from the rear at the speed of “c” (when it would be c-v if their clocks don’t slow down), then how will they see light coming at them from the front at “c”, since their clock slow-down would seem to cause them to see the oncoming light from the front at a speed of much faster than “c” (since without the clock slow-down the speed would be c + v)?

Eroica
2003-Dec-15, 09:12 AM
When dealing with relativistic problems like this one, there are three factors that must be borne in mind: Fitzgerald-Lorentz Contraction, time dilation, and the relative nature of simultaneity.

Let’s do a thought-experiment. Tom, Dick and Harry are sitting out in deep space in three spaceships, which are aligned along the x-axis, with Tom to our left, Dick in the middle and Harry to our right. To begin with, they are at rest relative to one another:

TOM--------------------DICK-------------------HARRY

Dick’s ship is sausage-shaped and 300,000 km long. At the stern are two clocks, one on the port side and one on the starboard side. At the bow are two identical clocks. Dick has a fifth clock in his hand. He stands in the middle of his ship - equidistant from the bow and stern clocks - and sets the fifth clock going. It has been programmed to send out a signal that starts the other clocks. The bow and stern clocks are now synchronized.

Tom sends a light pulse to Dick. As it passes the stern clock on the port side of Dick’s ship, it stops that clock. The pulse then races the length of Dick’s ship and as it passes the bow clock on the port side, it stops that clock too. Dick now examines the two port clocks at his leisure. He notices that the bow clock was stopped precisely one second after the stern clock. So, he calculates, the light must have been travelling at 300,000 kps (distance = speed x time) = c.

The experiment is repeated, but this time it is Harry who sends out a pulse and it’s the starboard clocks which are stopped. Naturally, the result is the same.

Okay. That was easy. Now comes the tricky part. Let’s say that at the start of the experiment, Tom and Harry are at rest relative to each other, but Dick is moving towards Harry at 150,000 kps. Dick’s clocks are all set at zero and are not yet going. As far as Dick is concerned, he’s the stationary frame of reference and it’s Tom and Harry who are moving to our left at 150,000 kps. Dick stands in the middle of his ship, equidistant from the four stern and bow clocks, and sets them all going at the same moment, just as before, by sending them a signal from his fifth clock. From his point of view, the four clocks are now synchronized.

But - and this is the crucial point - from Tom and Harry’s point of view, the clocks are not synchronized at all. Since Dick was moving towards Harry when his fifth clock sent out the signal, that signal had to chase after the bow clocks, but was met by the stern clocks. So the stern clocks would have received the signal first and would have started ticking before the bow clocks.

But how long before? Let’s work it out from Tom’s point of view. Remember, in Tom’s stationary frame, Dick’s ship is no longer 300,000 km long. Fitzgerald-Lorentz contraction means it is only 300,000/Gamma km long, where Gamma = 1/√(1 - v²/c²). Also remember that Dick’s clocks (once they’ve started) will be ticking more slowly than Tom’s clocks by that same Gamma factor. Now, Dick’s speed, v, is 150,000 kps, so Gamma = 1.1547005…, and Gamma² = 4/3.

From Tom’s point of view, the signal from the fifth clock should have been catching up with the bow clocks at 150,000 kps. The distance it had to traverse was 150,000/Gamma. So the undilated time it took was 1/Gamma. The dilated time is 1/Gamma² = ¾ seconds. In other words, the fifth clock in Dick’s hand will tick ¾ seconds before the bow clocks are started.

Now from Tom’s point of view, the signal from the fifth clock should have been approaching the stern clocks at 450,000 kps. The distance is again 150,000/Gamma, so it should take it 150,000/(Gamma²x450,000) dilated seconds, or ¼ seconds. In other words, by the time the bow clocks receive the signal and start ticking, the stern clocks will have already ticked ½ a second!

So Dick sees the stern and bow clocks as synchronized. But to Tom and Harry, the stern clocks are half a second ahead of the bow clocks.

Now let’s repeat the original experiment. Tom sends out a pulse of light. It passes the stern clock on the port side of Dick’s ship and stops it. Let’s say that the time on that clock at that moment is 0. From Tom’s point of view, the bow clock now reads minus 0.5 seconds. The light pulse races after the bow clock, which is moving towards Harry at 150,000 kps. From Tom’s point of view, the light should be catching the bow clock at 150,000 kps. The distance it must traverse is 300,000/Gamma km. The undilated time it takes to traverse this distance is 2/Gamma. The dilated time it takes is 2/Gamma² = 1.5 seconds. In other words, after the stern clock stops, bow clock ticks for a further 1.5 seconds before it too stops. And since the bow clock read -0.5 when the stern clock stopped, it will read +1 second when it stops.

So when Dick examines the port clocks, he will discover that the light from Tom’s ship took precisely one second to traverse his ship, which is still 300,000 km long as far as he’s concerned. (Since he’s at rest relative to himself, he can detect no Fitzgerald-Lorentz contraction or time dilation.) Therefore the speed at which Tom’s light pulse overtook him was 300,000 kps = c!

Phew! Now let’s see what happens to Harry’s pulse. It rushes towards the oncoming ship. As it passes the bow clock on the starboard side, it stops it. Let’s say that the time on this clock when it stops is 0 seconds. Then from Harry’s point of view, the starboard stern clock reads +0.5 seconds at the moment the bow clock stops. The light pulse races on towards the stern clock. From Harry’s perspective it should be approaching it at 450,000 kps. The distance it must travel is 300,000/Gamma km. The dilated time it takes to cover that distance is 300,000/Gamma²x450,000 = 0.5 seconds. Since the stern clock already read 0.5, it will read +1 when the light pulse stops it.

So when Dick examines the starboard clocks, he will discover that the light from Harry’s ship took precisely one second to traverse the length of his ship: 300,000 km as far as he is concerned. Therefore, the speed at which Harry’s pulse passed his ship was 300,000 kps = c!

QED!

Russell
2003-Dec-15, 11:47 AM
Eroica:
Thank you for the reply to my original question. I'm going to have to spend some time with this and study it. I'm at work right now and it is difficult to think about this with 1000 people bugging. When I get home I have a seven month old who needs me. So maybe after I get her to sleep I can have some time with this. Thanks again.

swansont
2003-Dec-15, 12:10 PM
In the Hafele Keating flying clocks experiment, all the clocks’ averaged time change showed an overall speed up, not an overall slowdown.

The Hafele-Keating experiment was a measurement primarily of the gravitational redshift effect, not time dilation. Those explanations typically ignore gravitational effects - the example given by Eroica specifically points out that it's in "deep space" which is synonymous with "no gravity to worry about."

Moving clocks run slow. So do clocks in gravitational fields. Both effects are present in clocks in orbit, such as in GPS satellites, though we see the sign as opposite for the gravitational effect, since the clock is in a weaker gravitational field compared to the surface of the earth.

daver
2003-Dec-15, 05:30 PM
I saw this as a kid in grade school on a Bell science film.

Imagine Tom is in a spaceship, with mirrors on opposing walls and a light pulse bouncing back and forth between them. Tom sees the light pulse moving (of course) at c, it takes w/c = T1 seconds to make one complete circuit of the room, where w is twice the width of the room.

Now, from Harry's point of view, Tom is speeding past at a considerable rate of speed (umm, Harry thinks that Tom's velocity vector is pointing "up"--Tom's head is pointing towards where Tom is going, and his feet are pointing where he's been), and the path of the light pulse is a zig-zag; during one complete cycle the light pulse moves along the hypotenuse of a triangle w on the base and v*T2 on the side, where it takes T2 seconds for Harry to see the pulse of light make a complete circuit. Throwing a bit of algebra at the problem, we see that (cT2)**2 = (vT2)**2+w**2, or w = T2*sqrt(c**2-v**2). Substituting w = cT1, we get T1 = T2 * sqrt(1-(v/c)**2), which might look familiar.

There's a bit of sleight of hand going on in this substitution--Harry could presumably have seen w change (some sort of width contraction); you could set up a different thought experiment to show that this is unlikely.

You can put mirrors on the ceiling and floor of Tom's spaceship and repeat the experiment; if you do so, you'll find that the time dilation from the first experiment isn't sufficient--Tom's spaceship will appear foreshortened with respect to Harry (length contraction).

Something else that falls out from the mirrored ceiling experiment is that simultaneity isn't conserved. Imagine two light pulses bouncing between the ceiling and floor of Tom's space ship--say a red one and a blue one. The pulses are arranged so that they are bouncing out of phase--Tom (on the floor for now) sees the same time between the red pulse and the blue pulse as between the blue pulse and the red pulse. So, Tom thinks that at the same time the red pulse is bouncing off the floor mirror the blue pulse is bouncing off the ceiling mirror, and vice versa. Harry, watching from the outside, disagrees--he thinks that when the red pulse is bouncing off the floor mirror the blue pulse is still heading towards the ceiling mirror.

Sam5
2003-Dec-15, 10:43 PM
Moving clocks run slow.

Certain kinds of moving clocks experiencing acceleration run slow. That was known as early as the 16th Century. That’s why early sailors couldn’t determine their longitude while riding with a clock aboard a rocking ship. A special clock had to be designed that counteracted the effect of the constantly changing acceleration aboard a ship.

Glom
2003-Dec-15, 10:49 PM
I think you're confusing the mechanical effects of old style mechanical clocks on a boat in choppy waters with relativistic effects. It would take the accuracy of an atomic clock to measure relativistic time dilation, whether special or general, due to the motion of a boat at sea.

Sam5
2003-Dec-16, 02:22 AM
I think you're confusing the mechanical effects of old style mechanical clocks on a boat in choppy waters with relativistic effects. It would take the accuracy of an atomic clock to measure relativistic time dilation, whether special or general, due to the motion of a boat at sea.

Glom, my hypothesis is that “relativistic time dilations” in atomic clocks is nothing more than a basic “electrodynamical mechanical effect” felt at the atom that experiences a change from less to more gravitational potential.

With the further hypothesis being that different kinds of clocks slow down at different rates when exposed to more gravitational potential.

And don’t forget that atomic clocks can “speed up” too, when they experience less gravitational potential.

The 1905 term “time dilation” became obsolete in 1911, when Einstein figured out that atomic clocks could both speed up or slow down, when placed under less or more gravitational potential. I don’t know of any clock that just “slows down” but never can “speed up”, due to gravitational potential changes.

And, you can’t successfully use an atomic clock on a ship at sea in choppy waters, for the very same reason the old sailors’ clocks changed rates in choppy waters, because of the constantly changing acceleration. The old sailors’ mechanical clocks ran slow, fast, slow, fast, etc., just as atomic clocks do when experiencing choppy seas and changing accelerations.

Musashi
2003-Dec-16, 02:35 AM
And, you can’t successfully use an atomic clock on a ship at sea in choppy waters, for the very same reason the old sailors’ clocks changed rates in choppy waters, because of the constantly changing acceleration. The old sailors’ mechanical clocks ran slow, fast, slow, fast, etc., just as atomic clocks do when experiencing choppy seas and changing accelerations.

Eh?

Could you refrence that with something please? Sailors clocks were unreliable instruments because of their acceleration? What if I put a clock in a race car? What happens to my watch on a rollercoaster? Does it speed up and slow down?

Musashi
2003-Dec-16, 02:43 AM
I believe the problem was that clock were unreliable at that time period. Here is something to look at.

http://www.sailnet.com/sailing/96/f&amp;bapr96.htm
http://www.fordfound.org/news/view_reflection_detail.cfm?reflection_index=19
http://www.najaco.com/books/aviationchorus/longitude.htm

Clocks of that era were totally unreliable and shipboard clocks were wildly inaccurate because of the rolling seas and weather changes that expanded or contracted their parts.

Harrison’s innovation was the creation of a series of clocks that were virtually friction-free, needing only the natural lubrication that a few wood parts provided (which he knew about from his woodworking experience). The parts did not rust and they stayed in perfect balance, no matter how violently a ship tossed or was battered in storms. He combined different materials inside the clockworks in ways that precisely complemented each other. If temperature changes made one expand, the other contracted to exactly counteract the change, keeping the clock’s rate constant.

[My bold]

So, it seems that acceleration wasn't the problem, crappy clocks were the problem. Clock on a ship were subjected to horrible conditions that prevented even the crappy clocks from working well. See, weather rusts metal parts, and choppy seas play havoc with the balance of pendulums (pendulii?).

Sam5
2003-Dec-16, 04:59 AM
Eh?

Could you refrence that with something please? Sailors clocks were unreliable instruments because of their acceleration? What if I put a clock in a race car? What happens to my watch on a rollercoaster? Does it speed up and slow down?

Yes, mechanical and atomic clocks will drift considerably when subjected to constantly changing accelerations, like on board a ship at sea. If you wear a mechanical watch, its rate will change if you go on a rollercoaster or a race car, but only while you are going up and down or sideways.

If you put an electronic watch in your refrigerator freezer, it will lose several seconds over several days. That’s due to the thermodynamic time slowdown factor.

This was a problem of steady speed control on all early portable tape recorders, caused by moving the recorder while it was running, especially when it was rotated, until Sony invented one in the 1960s that had counter-rotating flywheels that tended to cancel out the effects acceleration had on each flywheel.

You can look the think up about the early ship’s clocks on any “clock history” website.

Sam5
2003-Dec-16, 05:08 AM
I believe the problem was that clock were unreliable at that time period.

No, sorry.

Pendulum clocks were very accurate in the old days, but they couldn’t be used on board ships because of the constantly changing acceleration problem. Harrison's chronometers contain devices designed to counter-act the effects of the ship’s constant rocking. It was the rocking of the ship that caused the constant acceleration changes, that made the old type land-based clocks useless at sea.

Sam5
2003-Dec-16, 05:15 AM
So, it seems that acceleration wasn't the problem,

John Harrison was a carpenter from Yorkshire. Already at the age of 20, he built a clock with a pendulum made of two metals that could compensate each other's temperature expansion. When he heard of the prize offered by the Board of Longitude, he spent several years developing his first chronometer (later named H1). It had 2 symmetrically connected pendulums that could compensate for each other's acceleration caused by the movement of the ship.

www.kellnielsen.dk/bol.htm+Harrison+chronometer+acceleration&amp;hl=en&amp;ie =UTF-8]SOURCE (http://216.239.57.104/search?q=cache:lOQsdZgsA1oJ:[url) OF QUOTE[/url]

Musashi
2003-Dec-16, 06:10 AM
I think you are intentionally missing the point...

You:

Pendulum clocks were very accurate in the old days

The problem at the time was that there was no accurate way of determining the time at two different places at once. Pendulum clocks had already been invented and used, but on the deck of a rolling ship, such clocks would slow down, or speed up, or stop running altogether. Warm and cold temperatures thinned or thickened a clock's lubricating oil and made its metal parts expand or contract, thus slowing or speeding up tremendously. A rise or fall in barometric pressure, as well as variations in Earth's gravity from one latitude to another, caused a clock to gain or lose time. These watches typically gained or lost as many as fifteen minutes a day. Since one degree of longitude equals four minutes of time, fifteen minutes a day would equal to almost four degrees of longitude. Near the equator, that would translate to almost two hundred and seventy-two miles. In only one day of navigation, the ship would be two hundred and seventy-two miles off course! Imagine a trip across the Atlantic of many months.

Maybe this is hard for you to understand. Have you ever seen a pendulum clock? The changing of the inclination of the deck, away from horizontal, would cause a pendulum clock to lose or gain time. It has nothing to do with the aceleration of the ship. The acceleration of the two intenal mechanism inside Harrison's clock were there to keep the mechanism moving properly on a pitching deck. Sam, you have got to be pulling my leg...

Russell
2003-Dec-16, 07:01 AM
Guys:
This has gone way past the point that I hoped it would. With that I have a new question. According to relativity energy and matter are one in the same, right? As matter approaches the speed of light it becomes energy. This is true because we can now bake the earth because of it. If your moving through empty space with no reference point and you are accelerating towards the speed of light will you become energy? After all, there is nothing to refer your speed to. And, light is always moving away from you at a constant speed, so you can never get going that fast anyways, right? Or is it that you have to keep adding energy, hence more mass?

Sam5
2003-Dec-16, 07:34 AM
Maybe this is hard for you to understand. Have you ever seen a pendulum clock? The changing of the inclination of the deck, away from horizontal, would cause a pendulum clock to lose or gain time. It has nothing to do with the aceleration of the ship. The acceleration of the two intenal mechanism inside Harrison's clock were there to keep the mechanism moving properly on a pitching deck. Sam, you have got to be pulling my leg...

Read the quote from the website I gave you. It is acceleration caused by the pitching ship that causes the inaccuracies in the clocks.

A “pitching deck” causes “acceleration” to the people and clocks on board. That’s why passengers get slammed into the walls of a rocking ship. That’s why your coffee cup goes flying off the dashboard of your car when you make a sudden turn.

You need to look up the scientific definition of “acceleration”. It doesn’t only mean “blasting off in a rocket”. It means a change in velocity. And you need to look up the definition of velocity. The rocking ship causes the acceleration of the frame of the clock that is resting on the deck or on a desk in the ship. When the frame suddenly jolts, it is experiencing sudden acceleration, and that messes up the timing of the pendulum swing, and it can also mess up the steady motion of a balance clock flywheel.

Musashi
2003-Dec-16, 07:46 AM
So, if I tip my grandfather clock (which uses a pendulum) at a 45% angle, it will run ok? I mean, it should right? Since I am not accelerating it once it is moved.

Eroica
2003-Dec-16, 08:27 AM
According to relativity energy and matter are one in the same, right?
Wrong. Energy and mass are the same. Matter is "stuff," like atoms and molecules.

As matter approaches the speed of light it becomes energy.
Not quite. As matter (a spaceship, say) approaches the speed of light relative to a given frame of reference, its kinetic energy relative to that frame increases indefinitely. It's still matter, and its rest-mass (ie the mass or energy it has when it is not moving relative to the frame of reference) is unchanged.

If you're moving through empty space with no reference point and you are accelerating towards the speed of light will you become energy?
You must remember that mass/energy is relative. While you may be accelerating close to the speed of light relative to someone else's frame of reference, you are at rest relative to your own frame of reference. So you will observe no change in your own situation. To an outside observer, your kinetic energy is increasing to enormous levels, but you're still a man travelling in a spaceship, made of the same atoms and molecules as before.

And light is always moving away from you at a constant speed, so you can never get going that fast anyways, right?
Correct. Whatever is accelerating you is giving you the extra energy to go faster and faster. But because this energy is increasing your inertial mass (which is a measure of your reluctance to be accelerated), it's becoming increasingly difficult to accelerate you any more. You get closer and closer to the speed of light, but you will never reach it.

In all of this, it's important to remember that there is no such thing as absolute energy. The energy you have depends on the frame of reference relative to which you are doing the measuring. While you may be travelling close to the speed of light relative to one frame - and therefore have an enormous amount of kinetic energy relative to that frame - you might be at rest relative to another frame, and have no kinetic energy!

swansont
2003-Dec-16, 11:58 AM
Read the quote from the website I gave you. It is acceleration caused by the pitching ship that causes the inaccuracies in the clocks.

A “pitching deck” causes “acceleration” to the people and clocks on board. That’s why passengers get slammed into the walls of a rocking ship. That’s why your coffee cup goes flying off the dashboard of your car when you make a sudden turn.

You need to look up the scientific definition of “acceleration”. It doesn’t only mean “blasting off in a rocket”. It means a change in velocity. And you need to look up the definition of velocity. The rocking ship causes the acceleration of the frame of the clock that is resting on the deck or on a desk in the ship. When the frame suddenly jolts, it is experiencing sudden acceleration, and that messes up the timing of the pendulum swing, and it can also mess up the steady motion of a balance clock flywheel.

You need to differentiate from specific mechanical effects from general physics affects. All clocks will be uniformly affected by their position in a gravitational field. All clocks will be uniformly affected by their speed relative to an observer. These are both relativistic effects (i.e. physics). Some clocks will respond differently than others to being jostled - that is a mechanical/design effect. Mechanical effects can be eliminated with proper design. Physics effects cannot.

There are atomic clocks on ships and boats, and they work quite well, thank you.

Wally
2003-Dec-16, 02:28 PM
Eroica. An excellently explained example of SR time/length dilation in your tom/dick/harry post! =D>

And a bit of a warning for those of you who haven't been following Sam5's SR arguments in the other posts here. His rationale on many issues border on troll-like. Enter at your own risk!

Sam5
2003-Dec-16, 03:59 PM
You need to differentiate from specific mechanical effects from general physics affects.

The only difference between “specific mechanical effects” and “general physics effects” is that within atoms, inside the atoms, the “specific mechanical effects” are called “electrodynamical effects” because they involve not only the gravitational field, but the electric and magnetic fields as well.

Outside the atoms, the overall effects of the acceleration of a pendulum clock’s collection of atoms that we call the “weight”, which the old-timers called the “bob” or the “plumb bob”, we tend to think of that as a simple “mechanical effect”. But inside the atoms, when the same acceleration is felt by all the individual atoms, and their internal harmonic oscillation rate slows down, that is considered to be an “electrodynamical effect”. But it is also a “mechanical” effect on the small internal atomic level.

On that small atomic level there are other “fields” at work inside the atoms, fields other than the basic gravitational field that causes the bob of the pendulum clock to swing. These other fields are the electric and the magnetic fields. This is probably why we think of the internal “mechanical” workings of atoms in “electrodynamical”, rather than “mechanical”, terms.

All clocks will be uniformly affected by their speed relative to an observer.

No, this is incorrect and you need to get it out of your mind. “Relative motion” and "observers" never cause any clock to “slow down” or change rates in any way, since the various clocks “don’t know” the other clocks or "observers" are moving relative to them. What you said is a myth that grew out of the fundamental errors of the Kinematical part of the SR theory, which Einstein changed as he developed the General Relativity theory. What is often mistaken for a “relative motion” effect is actually an electrodynamical effect cause by atoms moving through fields.

All clock rate changes are “relativistic” effects, even the change noticed in the 16th Century when pendulum clocks were moved to higher altitudes. Einstein didn’t invent the basic concept of “relativity”. He applied previously known relativistic concepts to the inner workings of atoms. His best work was in taking 16th-19th Century large-scale mechanical effects and inquiring as to what kind of small-scale mechanical effects are at work inside atoms.

Most of his best earliest papers are about atoms and how they work. He had already published many papers about atoms before he ever published the SR paper in 1905, but this is generally not known by most people today.

There are atomic clocks on ships and boats, and they work quite well, thank you.

Show me some websites that say there are atomic clocks on ships and boats. I couldn’t find any. All I could find is that commercial ships and boats use the GPS time-code signals for their “atomic clocks”. But they don’t actually take atomic clocks on board the ships and boats.

There are some types of commercial clocks today that are called “atomic clocks” but they really aren’t atomic clocks. They are regular clocks that contain a radio receiver that receives a time coded transmission from Boulder Colorado that is linked to a stationary atomic clock in Boulder, these cheaper clocks are re-set at least once a day by the time code transmissions out of Boulder. You can buy these kinds of “atomic clocks” at Wal-Mart for \$14.95.

swansont
2003-Dec-17, 12:05 PM
You need to differentiate from specific mechanical effects from general physics affects.

The only difference between “specific mechanical effects” and “general physics effects” is that within atoms, inside the atoms, the “specific mechanical effects” are called “electrodynamical effects” because they involve not only the gravitational field, but the electric and magnetic fields as well.

This would imply that if one type of clock didn't work at sea, then no clocks would work at sea, which is false.

There are atomic clocks on ships and boats, and they work quite well, thank you.

Show me some websites that say there are atomic clocks on ships and boats. I couldn’t find any. All I could find is that commercial ships and boats use the GPS time-code signals for their “atomic clocks”. But they don’t actually take atomic clocks on board the ships and boats.

I wasn't referring to commercial or private ships and boats. Just because you can't find it on the internet doean't make it untrue.

Russell
2003-Dec-17, 02:00 PM
Eroica:
thank you very much, but I'm still having a little problem. Please keep in mind I play guitar for a living, I'm not a physicist. If my Stratocaster is traveling through space appraoching the speed of light, and BB King is holding it, to BB my strat is as it has always been. But, if i'm traveling much slower relative to them, then to me they appear to become energy. This is strange, am I correct in thinking that they are one or the other? or both?

swansont
2003-Dec-17, 03:47 PM
Eroica:
thank you very much, but I'm still having a little problem. Please keep in mind I play guitar for a living, I'm not a physicist. If my Stratocaster is traveling through space appraoching the speed of light, and BB King is holding it, to BB my strat is as it has always been. But, if i'm traveling much slower relative to them, then to me they appear to become energy. This is strange, am I correct in thinking that they are one or the other? or both?

No, not as such. How would you describe something as "becoming energy?"

BB King and your strat will be length contracted, as measured by you, and their clocks will run slow (again, measured by you) and they will have a certain amount of energy, which is a certain amount of rest mass energy (which is the same in all frames) and kinetic energy.

Sam5
2003-Dec-17, 03:58 PM
I wasn't referring to commercial or private ships and boats. Just because you can't find it on the internet doean't make it untrue.

Well, show me some source at all, books, personal experience, photos of atomic clocks on boats, anything at all to verify your claim.

It might be possible for an atomic clock on a submerged submarine to be accurate for a while, if the submarine is not rocking or experiencing acceleration.

swansont
2003-Dec-17, 07:30 PM
I wasn't referring to commercial or private ships and boats. Just because you can't find it on the internet doean't make it untrue.

Well, show me some source at all, books, personal experience, photos of atomic clocks on boats, anything at all to verify your claim.

It might be possible for an atomic clock on a submerged submarine to be accurate for a while, if the submarine is not rocking or experiencing acceleration.

Conversation with someone who had done a synchronization experiment with a ship, and others.

While it is true that GPS reduces the requirements for the quality of a shipboard chronometer, the military must have contingencies for a loss of GPS. Also, acquiring the military GPS signal (as for a surfacing sub) is faster if you already know what time it is. It's possible that you could get by with a good quartz oscillator, but basic cesium beam atomic clocks are rackmount items and only 3U or 4U high, or even smaller for rubidiums. The Office of Naval research has been overseeing the development of a "matchbox" atomic clock and smaller, for more portable applications. (google on matchbox atomic clock and I'm sure you'll find a press release)

Russell
2003-Dec-19, 07:51 AM
Swansont:
Please define lenght contraction for me. Thank You.

Glom
2003-Dec-19, 09:27 AM
Swansont:
Please define lenght contraction for me. Thank You.

l = l' × sqrt(1-v²/c²)

where l is the length as measured in a frame moving at v relative to the object and l' is the length as measured in a frame stationary relative to the object.

Russell
2003-Dec-19, 10:03 AM
Glom:
Thank You for your reply, but I must stress again, I play guitar for a living. I am fairly OK with math, but I still can not see the consequences in my head. I guess this is part of the reason why I have trouble with this. Please don't become frustrated with me here. If you could give me an analogy of this it would help me.

swansont
2003-Dec-19, 11:25 AM
Swansont:
Please define lenght contraction for me. Thank You.

Light from e.g. the front and back of an object take different amounts of time to reach an observer. But if I make a measurement of that object, I'm relying on photons that reach me at the same time. If the object isn't moving, there's no problem, but if it is, the photon from the front that reaches me at the same time as a photon from the back didn't originate at the same time, and thus didn't originate at the same position (relative to a stationary object). The object will be shorter in the direction of motion, as measured by any observer in a different frame of reference.

Russell
2003-Dec-22, 07:46 PM
Swansont:
Is this referring to red and blue shift?

Sam5
2003-Dec-23, 04:18 AM
Swansont:
Please define lenght contraction for me. Thank You.

Light from e.g. the front and back of an object take different amounts of time to reach an observer. But if I make a measurement of that object, I'm relying on photons that reach me at the same time. If the object isn't moving, there's no problem, but if it is, the photon from the front that reaches me at the same time as a photon from the back didn't originate at the same time, and thus didn't originate at the same position (relative to a stationary object). The object will be shorter in the direction of motion, as measured by any observer in a different frame of reference.

Actually, Einstein wrote in 1907:

“The shape of a body in the sense indicated we will call its ‘geometrical shape’. The latter obviously does not depend on the state of motion of a reference frame.

It is clear that observers who are at rest relative to a reference system S can ascertain only the kinematical shape with respect to S of a body that is in motion relative to S, but not its geometrical shape.

In the following, we will usually not distinguish between explicitly between the geometrical and kinematical shape; a statement of geometric nature refers to the kinematic or geometric shape, respectively, depending on whether the latter refers to a reference system S or not.”

So it is just an optical illusion, that is caused by the postulates and terms of his 1905 paper. It doesn't happen in real life, not based on "relative motion" alone.

freddo
2003-Dec-23, 04:53 AM
Einsteins words reconcile perfectly with what swansont said.

The object in motion observed from the reference frame S isn't actually shorter, but it will be observed as shorter. Optical illusion is sort of an accurate label for this (very loosely). It does, however, happen in real life.

Sam5
2003-Dec-23, 04:58 AM
Einsteins words reconcile perfectly with what swansont said.

The object in motion observed from the reference frame S isn't actually shorter, but it will be observed as shorter. Optical illusion is sort of an accurate label for this (very loosely). It does, however, happen in real life.

Optical illusions happen in real life. But geometrical length contraction due only to relative motion does not happen in real life. Young people should be taught the truth about this true fact of nature, and we should make absolutely sure they understand that there is no such thing as “real length contraction” due only to “relative motion”.

freddo
2003-Dec-23, 05:21 AM
Because, of course Sam5, the entire scientific establishment is perpetuating this fraud....

Sam5
2003-Dec-23, 05:39 AM
fredo,

Well, I think science teachers and physics professors sometimes get into a bad habit of teaching old myths, just as I was taught in the 1950s that electrons “orbit” the nucleus of an atom just as planets “orbit” the sun. I later learned that this was not only not true, but it was known in the mainstream physics community to not be true as early as 1925. But even now, some high-tech science company logos show electrons orbiting an atom’s nucleus just as planets orbit the sun.

freddo
2003-Dec-23, 05:46 AM
Sure, but that's not exactly saying anything now is it?

Eroica
2003-Dec-23, 09:46 AM
So it is just an optical illusion, that is caused by the postulates and terms of his 1905 paper. It doesn't happen in real life, not based on "relative motion" alone.What about the Rockets-and-Rope experiment (http://www.badastronomy.com/phpBB/viewtopic.php?t=15&amp;start=0)?

We tend to think that something like Fitzgerald-Lorentz Contraction is either real or an illusion. Surely it can't be both? But it is. The contraction that the stationary observer observes when he looks at a moving frame produces all the effects that a very real contraction would produce. In the rope experiment, the rope does break. There is nothing the observer can do to prove that it's just an illusion. Equally, there's nothing he or anyone else can do to prove that it's real.

I may be wrong, but I think the whole point is that in the weird world of relativity you cannot distinguish between the two cases. In one frame of reference it looks real, and therefore it is real.

Sam5
2003-Dec-23, 06:00 PM

That doesn’t make any sense at all. If the rope breaks, then the rockets would break apart too, all along their lengths from nose to tail.

If the 2-rocket and rope frame “contracts” to a “stationary” observer, as per SR theory, all that would happen is that the front of the first rocket would become closer to the rear of the second rocket. The rockets and the rope would “length contract” all together, and nothing would break. The rockets would not grow further apart since their entire 2-rocket/rope frame would contract. The rockets don’t contract in the theory while the space in between them doesn’t contract.

But here, this Einstein paradox is more fun to try to figure out. He published this in 1907:

”Thus, the temperature of a moving system is always lower with respect to a reference system that is in motion relative to it than with respect to a reference system that is at rest relative to it.”

So if both astronauts, each in two separate ships, is holding a glass of water, as the relative velocity between them increases, which glass of water will freeze first while the other doesn’t freeze?

Sam5
2003-Dec-23, 06:34 PM
We tend to think that something like Fitzgerald-Lorentz Contraction is either real or an illusion. Surely it can't be both? But it is.

No.

You apparently don’t understand what the Fitzgerald-Lorentz Contraction is or the reason for it in the Lorentz theory.

The reason for it is that Lorentz proposed the Michelson Morley apparatus, the big stone disc and one metal arm of it, were contracted very slightly in the direction of motion through the “ether” that many physicists thought existed. Lorentz hypothesized that the “ether” put up some kind of physical resistance to the motion of the apparatus in the direction of its motion, and this could have physically shortened the x-axis diameter of the disc and the x-length of the metal arm.

This would be like trying to push an inflated balloon through the air a high speeds, and observing a “contraction” in the diameter of the balloon in its direction of motion.

The SR theory has no such ether or resistance. There is no physical cause for the “length contraction” and that’s why Einstein said in 1907 it didn’t really exist, not due to “relative motion” alone.

Real contractions in objects do occur, for a variety of physical reasons, such as the contraction of a metal bar when it becomes cold, or the contraction of an object fired out of a cannon, during the sudden and brief acceleration period. The same thing happens to a car during sudden “acceleration” (“deceleration”) when it hits a brick wall at high speed. But these are real forces of nature at work, not hypothetical optical illusions in hypothetical thought experiments.

In the rope experiment, the rope does break.

No it doesn’t, not even hypothetically. The two rockets and the rope are in the same “frame”. The rope accelerates because of the pull of the first rocket on it. We’d have to give a slight extra thrust to the first rocket to counteract the added mass/inertia of the rope it is pulling, but in the SR theory, the two rockets and the rope automatically “length contract” all together because their entire "frame" is contracting (in the theory, but not in real life).

It would be the same as the two rockets being rigidly attached to each other with the rear rocket providing all the thrust to move both of them and the rope, just as each part of both rockets is rigidly attached to other parts of the same rocket.

This rope thought experiment and the incorrect results is a perfect example of how special relativity is not understood by the vast majority of improperly-educated masses of the world.

Glom
2003-Dec-23, 06:37 PM
But if there was an ether, then the amount of contraction would vary depending on motion relative to it and not any observer. However, the observation is that length contraction is the same for any two inertial frames moving relative to each other at a certain speed.

Sam5
2003-Dec-23, 06:42 PM
I may be wrong, but I think the whole point is that in the weird world of relativity you cannot distinguish between the two cases. In one frame of reference it looks real, and therefore it is real.

In the “weird world” of Einstein Special Relativity, parts of his first Special Relativity theory were just flat out wrong. Nature is not as weird as the wrong parts of the theory are.

From my frame, the moon looks like a small flat disk to me that revolves around the earth every day. But I’ve read in science books that it is actually a big spheroid that revolves around the earth about every 28 days, while the earth turns on its axis once every 24 hours.

But when studying Special Relativity theory, especially the Kinematical part, one has to be very careful about what one “believes” to be true, because nature doesn’t necessarily conform to all statements made in the SR theory.

Sam5
2003-Dec-23, 07:00 PM
But if there was an ether, then the amount of contraction would vary depending on motion relative to it and not any observer. However, the observation is that length contraction is the same for any two inertial frames moving relative to each other at a certain speed.

No, that’s not “the observation”, that’s “what the incorrect part of the 1905 theory says”. So many websites say the entire theory has been “proven over and over again”, most people think that what you just said is “the observation”, but it isn’t.

The SR theory contains two very different parts, the Kinematical part and the Electrodynamical part. The Kinematical part is filled with errors and paradoxes. But the Electrodynamical part is more correct, for a basic reason. That part has atoms and electrons moving through a local “ether”, and that local “ether” turns out to be the local electric, magnetic, and gravitational fields that are located at the surface of the earth. These are the very fields that influenced the results of some of the 19th Century electrodynamical experiments.

For example, when NASA did its flying tether experiment, the experiment worked. In the 19th Century it was predicted to work by Lorentz and his “Lorentz force” theory.

Now, consider these questions: Did it work because the tether was moving “relative to” the surface of the earth? Or did it work because the tether was moving at high speed through the earth’s magnetic field?

The “relative motion” had nothing to do with it. What caused the electrons to flow in the tether was becaue the tether was moving through the earth’s magnetic field.

In the second part of the SR theory, Einstein had the electrons of one “frame” moving through the fields of the other frame, and that is why that part of the theory is more correct than the first part of the theory.

So, a legitimate physicist can say, “SR theory is correct,” meaning the Electrodynamical part is correct, but, in my opinion, he should actually say, “The Electrodynamical part of SR theory is correct, while the Kinematical part contains a lot of errors.” It wouldn’t hurt anything at all for them to say that, and it would clear up a lot of misconceptions about both nature and the SR theory.

Eroica
2003-Dec-24, 03:21 PM
You apparently don’t understand what the Fitzgerald-Lorentz Contraction is or the reason for it in the Lorentz theory.Welcome to the world of SR. Appearances can be deceiving. :D

The SR theory has no such ether or resistance. There is no physical cause for the “length contraction” and that’s why Einstein said in 1907 it didn’t really exist, not due to “relative motion” alone.

Real contractions in objects do occur, for a variety of physical reasons, such as the contraction of a metal bar when it becomes cold, or the contraction of an object fired out of a cannon, during the sudden and brief acceleration period.
I think we are arguing semantics here. I haven't read Einstein's 1907 paper, so I don't know what he meant, but I suspect he meant that the Lorentz Contraction is not a physical reduction in the size of an object, such as occurs when you cool a metal bar. And I agree. It is a contraction of the actual dimension of space itself. Apologies if I seemed to imply anything else in my last post.

You are sitting at rest in space in your spaceship. You measure it and it is 100 metres from bow to stern. I zip past at 0.6c in my spaceship, and as I do so I measure the length of your ship. I find it is only 80 metres long.

Neither of these is an optical illusion. In your frame of reference your ship is 100m long. That is its real and actual length. In my frame of reference, the same ship is 80m long. That is its real and actual length. There is nothing either of us can do to prove that what we observe is not real but an illusion.

Sam5
2003-Dec-24, 05:23 PM
To everyone,

Can anyone answer the question as to why the people in the photo appear to be slanting in one direction while the car appears to be slanting in the other direction? This is a simple basic space-time relativity question involving only “relative motion”, no gravitational field and no acceleration, and it has a common-sense classical-relativity answer that involves Euclidean space-time but not Riemann space-time.

SeanF
2003-Dec-24, 07:14 PM
To everyone,

Can anyone answer the question as to why the people in the photo appear to be slanting in one direction while the car appears to be slanting in the other direction? This is a simple basic space-time relativity question involving only “relative motion”, no gravitational field and no acceleration, and it has a common-sense classical-relativity answer that involves Euclidean space-time but not Riemann space-time.

In case anyone's interested, I responded to this over in the Twin Paradox (http://www.badastronomy.com/phpBB/viewtopic.php?t=9731&amp;start=680) thread. Once again, Sam5 doesn't understand what he's reading.

Sam5
2003-Dec-24, 08:47 PM
Once again, Sam5 doesn't understand what he's reading.

Once again, you gave the wrong answer. I told you that "acceleration" was not involved. (See the relativity thread.)

I think I'll give you a while to correct your answer about the "acceleration". Think about it really hard. Measure the angle of the car slant as opposed to the angle of the people slant, and maybe you can figure out what caused the difference.

Musashi
2003-Dec-24, 08:48 PM
A black hole!!

Sam5
2003-Dec-24, 08:58 PM
A black hole!!

No, there are no curves due to the motion of the camera or the car. Just diagonal lines, so we are just dealing with Euclidean geometry and classical relativity. No acceleration and no gravity. Lol, SeanF thinks there’s “acceleration” involved.

Grashtel
2003-Dec-24, 09:19 PM
To everyone,

Can anyone answer the question as to why the people in the photo appear to be slanting in one direction while the car appears to be slanting in the other direction? This is a simple basic space-time relativity question involving only ?relative motion?, no gravitational field and no acceleration, and it has a common-sense classical-relativity answer that involves Euclidean space-time but not Riemann space-time.

Einstein's Relativity has nothing to do with it, its an artifact of how camera shutters work and the motion of the camera and the car. Camera shutters open from the bottom to the top and close in the same way so objects moving relative to the field of view of the camera appear distorted. In that photo the car is driving to the right and the camera is panning to follow it, though not quite fast enough, so as the open part of the shutter moves up the car is moving across the field of view to the right and the background is moving to the right (from the point of view of the camera) producing the apparent slants in that picture. Sorry if the description isn't overly clear, if anyone likes I can try and do an ASCII art illustration of the effect.

Musashi
2003-Dec-24, 09:24 PM
Don't worry about it Grashtel. Sam likes to use all kinds of things in his quest to prove his point (if he has one). You understand how the camera works, I understand how the camera works, SeanF understands how the camera works.... even Sam himself might actually understand how it works, he just chooses to misapply it to relativity.

Sam5
2003-Dec-25, 12:55 AM
Don't worry about it Grashtel. Sam likes to use all kinds of things in his quest to prove his point (if he has one). You understand how the camera works, I understand how the camera works, SeanF understands how the camera works.... even Sam himself might actually understand how it works, he just chooses to misapply it to relativity.

There is no need for you to continue to post such gratuitous insults. [-X

Sam5
2003-Dec-25, 01:05 AM
Einstein's Relativity has nothing to do with it,

Uhh, That’s why I said we are dealing with Ecludean space-time rather than Riemann space-time.

Camera shutters open from the bottom to the top and close in the same way so objects moving relative to the field of view of the camera appear distorted.

It depended on the cameras. Many old-time focal plane shutters (and 35 mm fps) went from right to left. That is, unless the camera is turned sideways, then they can go from bottom to top or top to bottom, depending on how the camera is rotated. This photo was taken with the camera oriented so the slit went from the bottom to the top.

In that photo the car is driving to the right and the camera is panning to follow it, though not quite fast enough, so as the open part of the shutter moves up the car is moving across the field of view to the right and the background is moving to the right (from the point of view of the camera) producing the apparent slants in that picture.

The background is moving to the left, relative to the camera rotation to the right. Remember, the camera is panning to the right, so this produces the visual effect of the background appearing to move to the left. And you are correct about the car moving to the right faster than the camera is panning to the right.

Russell
2003-Dec-25, 09:55 AM
Hello:
We should not interprete theories right, we should understand them. This is how I understand it. The more I read here the more confused I become. I've actully read quite a few books on relativity, and I'm beginning to wonder if I should trust them. I set out asking for some help in seeing a concept in my head, which seems to me has turned into an argument in interpretations. Marry Christmas all.

Russell
2003-Dec-26, 06:00 PM
Hello:
OK, I have started rereading a book I read along time ago, About Time-Paul Davies. The original problem was with the Newtonian view and the electromagnetic theory, right? In Newtonian physics, the speed of light would change in relation to your speed, right? But that would change the properties of that light beam, this is what Einstein had a problem with, am I correct? "Speed is distance traveled per unit time, so the speed of light can only be constant in all reference frames if distances and intervals of time are somehow different for different observers, depending on their state of motion". This I understand.

Sam5
2003-Dec-26, 10:28 PM
Hello:
OK, I have started rereading a book I read along time ago, About Time-Paul Davies. The original problem was with the Newtonian view and the electromagnetic theory, right? In Newtonian physics, the speed of light would change in relation to your speed, right? But that would change the properties of that light beam, this is what Einstein had a problem with, am I correct? "Speed is distance traveled per unit time, so the speed of light can only be constant in all reference frames if distances and intervals of time are somehow different for different observers, depending on their state of motion". This I understand.
Russell,

Unfortunately the “mass media” books like that won’t tell you exactly what’s true and what’s not. They will give you a lot of opinions. And here is mine:

In the original Newton theory days, and in the 19th Century, they thought one big giant “ether” controlled the speed of light in space. They thought the universe was pretty much “fixed” and not moving, and so they thought that ether was “fixed” with the universe.

In a special experiment conducted in 1886, Michelson and Morley found no evidence of a “universe stationary” ether, so the old ether theory was thrown out by Einstein in 1905, but he sort of brought it back, somewhat, in 1911, when he developed his gravitational redshift theory. His “ether” was the local gravitational field of an astronomical body, so it was a local ether.

But, there were complications with that theory, since it sort of depended on the rates of atomic clocks to measure the local “speed of light”.

A strange effect of the gravitation field is that a strong one slows down atomic clocks, while a weak one speeds them up. Also, a strong field slows down light while a weak one speeds it up. So, if you measure the “speed of light” at you, while the photons are arriving at you, and you measure their speed with an atomic clock located at you, resting on the surface of the planet where you are, then that particular clock is supposed to measure those particular photons traveling at “c”, right there locally.

But if you can measure or calculate the speed of the photons while they are passing near the sun, you will find that they slow down slightly. However, if you are “resting on the sun” with an atomic clock that is “resting on the sun”, then that atomic clock will measure those photons passing by the sun to be traveling at “c” there locally at the sun, but “faster than c” when they travel away from the such, such as at the earth.

It’s hard to keep all of this stuff straight and it’s difficult to explain, and, the worst part is that some people disagree with this 1911 Einstein theory. Another problem is that light-speed slows down and speed up such a very small amount, it’s difficult to measure the speed changes.

Regarding the "state of motion" of the observer, that complicates the issue even more.

Cougar
2003-Dec-27, 12:39 AM
Are you sure Russell didn't say essentially that, Sam?

I was about to say you're crazy for claiming light slows as it passes the sun, but after having it explained by Steve Carlip, (http://www.phy.ncku.edu.tw/mirrors/physicsfaq/Relativity/SpeedOfLight/speed_of_light.html) I'd have to agree that it depends on where you're sitting (I think).

Sam5
2003-Dec-27, 01:55 AM
Are you sure Russell didn't say essentially that, Sam?

I was about to say you're crazy for claiming light slows as it passes the sun, but after having it explained by Steve Carlip, (http://www.phy.ncku.edu.tw/mirrors/physicsfaq/Relativity/SpeedOfLight/speed_of_light.html) I'd have to agree that it depends on where you're sitting (I think).

Hi. Glad to know I’m not crazy.

"Einstein went on to discover a more general theory of relativity which explained gravity in terms of curved space-time and he talked about the speed of light changing in this new theory. In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity . . . cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position."

That’s part of what I’ve been trying to explain here. This stuff is in his 1911 paper. It is difficult to find in the paper, however, since he babbles a lot about different clocks, different kinds of clocks, atomic clocks, and “U” clocks. So, the 1911 paper says two significant things, which few people realize or understand today: 1) The internal harmonic oscillation rates of atoms resting in a gravitational field is influenced by the gravitational potential at the place where they are resting (ie atomic clocks slow down in strong gravitational fields), and 2) The speed of light waves/photons slow down when they pass through strong gravitational fields.

The two effects = this: An atomic clock resting in a gravitational field does not notice the slowdown of light waves/photons if the speed of the photons are measured at that clock, when they reach or pass by that clock, since the speed of the photons and the rate of the clock are slowed down the same amount by the gravitational field. Then it follows that the speed of waves/photons traveling in a weak gravitational field, when measured by an atomic clock resting in a strong gravitational field, will appear to that clock to be faster than “c”, while the speed of waves/photons traveling in a strong gravitational field, when measured by an atomic clock resting in a weak gravitational field, will appear to that clock to be slower than “c”.

If we use a distant rapidly spinning pulsar as our “clock”, then it’s rate should not be affected by our observation of it from within a gravitational field, and by use of that kind of clock, then we can determine that the speed of light does speed up and slow down as it travels through various parts of space, when it travels through areas of higher or lower gravitational potential.

The reason light “bends” as it passes massive objects like the sun is because the light “rays” (groups of photons traveling side by side) act like plane waves of light when they enter glass or water. They not only slow down but they change direction. Thus, light rays are “refracted” by the gravitational field of massive bodies, when they pass near to those massive bodies. The light is not “changing media”, like when it goes from air to glass, but it is gradually going from a less dense medium through a more dense medium, and that single medium is the gravitational field of the astronomical body through which the light passes, and the light's local speed regulator is that medium.

I got all of this right out of his 1911 paper, but it took me about 12 years to understand what the heck he was trying to say.

Uhh, you might want to print out this post for future reference. You won’t find it explained any better than this.

swansont
2003-Dec-27, 05:15 PM
A strange effect of the gravitation field is that a strong one slows down atomic clocks, while a weak one speeds them up.

I think perhaps you mean this relatively, as it were. A weak gravitational field slows clocks, but less than a strong one, so a clock in a field will speed up as the field weakens. But it's still slowed relative to no field at all.

Sam5
2003-Dec-27, 05:39 PM
A strange effect of the gravitation field is that a strong one slows down atomic clocks, while a weak one speeds them up.

I think perhaps you mean this relatively, as it were. A weak gravitational field slows clocks, but less than a strong one, so a clock in a field will speed up as the field weakens. But it's still slowed relative to no field at all.

That’s basically what I said. The weaker the field the atomic clock “feels”, the faster an atomic clock will “tick”.

But I and Einstein are talking only about “atomic clocks”, and NOT about other kinds of clocks. (See my posts on the SR thread.) For example, a pendulum clock will “speed up” in a strong gravitational field, not “slow down”. So we aren’t talking about “all of time”, we are talking specifically about the rates of atomic clocks.

swansont
2003-Dec-28, 01:58 PM
But I and Einstein are talking only about “atomic clocks”, and NOT about other kinds of clocks. (See my posts on the SR thread.) For example, a pendulum clock will “speed up” in a strong gravitational field, not “slow down”. So we aren’t talking about “all of time”, we are talking specifically about the rates of atomic clocks.

No, it applies to all clocks. Pendulum clocks have a specific dependence on gravity that dominates - the relativistic effect is about a part in 10^16 near the surface of the earth - i.e. it's small.

Einstein couldn't have been talking about atomic clocks, as the hadn't been invented yet.

Sam5
2003-Dec-28, 07:44 PM
No, it applies to all clocks.

Sorry, wrong.

It does NOT apply to “all clocks”.

Einstein couldn't have been talking about atomic clocks, as the hadn't been invented yet.

Sorry, wrong again.

He WAS talking about “atomic clocks” in the 1911 theory. This is why he said:

“Let fo be the vibration-number of an elementary light-generator...”

His “elementary light-generator” was an individual “atom” of any particular “element” that radiated light.

In order that you can understand EXACTLY what he was talking about, you need to read Maxwell’s two 1873 books: “A Treatise on Electricity and Magnetism”, in which Maxwell said:

”In astronomy a year is sometimes used as a unit of time. A more universal unit of time might be found by taking the periodic time of vibration of the particular kind of light whose wave length is the unit of length.”

See? Maxwell’s 1873 “particular kind of light” is emitted by his the atom of his particular “periodic time of vibration”, and this is Einstein’s 1911 “elementary light-generator”.

Of course, since Einstein was right in 1911, then when the more complex “manufactured atomic clocks” were invented and manufactured, they followed Einstein’s rule of “slowing down” their “tick rate” while resting in a strong gravitational field and “speeding up” their “tick rate” while resting in a weaker gravitational field. But this “gravitational redshift” effect DOES NOT influence ALL KINDS of clocks in the same way, as Einstein noted in his 1911 theory, when he said:

”But from what has just been said we must use clocks of unlike construction, for measuring time at place with differing gravitational potential.”

I.E., we must NOT use “atomic clocks”, because if we use atomic clocks, their rates will be affected by the differing gravitational potential.

As a matter of fact, I called the atomic clock physicists – the world’s top experts – in Boulder, Colorado, to confirm this, and they told me that our world standard of “atomic clock time” is “averaged” among several atomic clocks, to "average out" all differences in “tick rate” caused by the various clocks resting at different gravitational potentials at different elevations. And their “average atomic time tick rate” is constantly compared to the revolution and rotation rates of the earth. This is why they have to add “leap seconds” to the atomic clocks every now and then.

I think you need to read a little more about 19th Century electrodynamics and kinematics, so you can understand what Einstein was actually saying in 1911.

swansont
2003-Dec-29, 01:06 PM
No, it applies to all clocks.

Sorry, wrong.

It does NOT apply to “all clocks”.

No, really, it does. I'm not kidding. It'll work on quartz oscillators and spring-driven flywheels.

As a matter of fact, I called the atomic clock physicists – the world’s top experts – in Boulder, Colorado, to confirm this, and they told me that our world standard of “atomic clock time” is “averaged” among several atomic clocks, to "average out" all differences in “tick rate” caused by the various clocks resting at different gravitational potentials at different elevations.

With whom did you speak at NIST? Because he/she is wrong, or your interpretation of it is wrong. And why didn't you call an expert at the US Naval Observatory, where they have even more atomic clocks than at NIST? (S)He would have told (and in fact is now telling) you that the elevations are factored in. The averaging happens because no two clocks are identical, and never independently keep synchronous time (even clocks that are nominally at the same frequency). Both the stability and stationarity of a clock ensemble improves with more clocks. You need to have multiple clocks to do pair-wise comparisons and discern any biases in individual clocks. And you use different kinds of clocks (e.g. H masers, cesium beam, fountains, ion trap clocks) so you aren't prone to a common systematic error.

Sam5
2003-Dec-29, 07:47 PM
No, it applies to all clocks.

Sorry, wrong.

It does NOT apply to “all clocks”.

No, really, it does. I'm not kidding. It'll work on quartz oscillators and spring-driven flywheels.

No, not just due to “relative motion” alone, since no physical “force” is placed on the oscillating mechanism of the clocks due only to “relative motion”. Your atomic clock rate changes due to elevation are GR, not SR changes.

Anyway, a greater gravitational potential slows down atomic clocks, while it speeds up pendulum clocks. Synchronize one of your atomic and a pendulum clock in DC, make them run as synchronously as possible, then take both of them to Denver and set them up. The atomic clock will run fast, and the pendulum clock will run slow. If you happen to ever find a pendulum clock that “runs fast” in Denver and “slow” in DC, please let me know.

As a matter of fact, I called the atomic clock physicists – the world’s top experts – in Boulder, Colorado, to confirm this, and they told me that our world standard of “atomic clock time” is “averaged” among several atomic clocks, to "average out" all differences in “tick rate” caused by the various clocks resting at different gravitational potentials at different elevations.

With whom did you speak at NIST? Because he/she is wrong, or your interpretation of it is wrong. And why didn't you call an expert at the US Naval Observatory, where they have even more atomic clocks than at NIST? (S)He would have told (and in fact is now telling) you that the elevations are factored in.

Hey! You work at the US Naval observatory??

Do you know if they “average” the rate to “sea-level” elevation or not?

The averaging happens because no two clocks are identical, and never independently keep synchronous time (even clocks that are nominally at the same frequency). Both the stability and stationarity of a clock ensemble improves with more clocks. You need to have multiple clocks to do pair-wise comparisons and discern any biases in individual clocks. And you use different kinds of clocks (e.g. H masers, cesium beam, fountains, ion trap clocks) so you aren't prone to a common systematic error.

[see added note at bottom of post]

Perhaps I did not explain it thoroughly, since they told me there are other reasons for the averaging too. I’ve said in other posts that they told me that the “averaging” is also done to eliminate some of the glitches in individual clocks that often occur, “biases” and “drifts”, such as sudden “leaps” ahead or back, but the averaging also factors in the different elevations, since if they only used DC atomic clocks for the averaging, the “world standard” would be running too slow (or too fast for sea-level time), and if they used only Boulder clocks, the “world standard” would be running too fast.

So yes, I think we can say either that the elevations are “factored in” or we can say that the different clock-elevation rate “averaging” is done to “average” the different rates due to the different elevations. And as I mentioned, I suspect that the “world standard” time rate is probably “sea-level” atomic clock time, but you should know the answer to that.

They also told me that no two atomic clocks can be manufactured to run at exactly the same rate. In other words, when being manufactured side by side, every one of the clocks will have a slightly different rate.

Would you agree with this or not?

Ok, I’ve thought about it some more, and I think your term “factored in” is better than my term “averaged out”, because my term could be misconstrued to mean that if 5 clocks in Boulder are “averaged out” with 5 clocks in DC, then the “average” would represent the rate of an ideal atomic clock placed at the elevation of about 2,600 feet, and that is not what I intended to imply. I should have said “factored in”.

swansont
2003-Dec-30, 12:30 AM
No, not just due to “relative motion” alone, since no physical “force” is placed on the oscillating mechanism of the clocks due only to “relative motion”. Your atomic clock rate changes due to elevation are GR, not SR changes.

Anyway, a greater gravitational potential slows down atomic clocks, while it speeds up pendulum clocks. Synchronize one of your atomic and a pendulum clock in DC, make them run as synchronously as possible, then take both of them to Denver and set them up. The atomic clock will run fast, and the pendulum clock will run slow. If you happen to ever find a pendulum clock that “runs fast” in Denver and “slow” in DC, please let me know.

If you go back to what I wrote, I think you'll see that I never implied otherwise. Gravitation is a GR effect on all clocks, and there is another, far larger dependence that affects pendulum clocks.

Hey! You work at the US Naval observatory??

Do you know if they “average” the rate to “sea-level” elevation or not?

It's corrected to the geoid, which is essentially the idealized sea level.

[see added note at bottom of post]

Perhaps I did not explain it thoroughly, since they told me there are other reasons for the averaging too. I’ve said in other posts that they told me that the “averaging” is also done to eliminate some of the glitches in individual clocks that often occur, “biases” and “drifts”, such as sudden “leaps” ahead or back, but the averaging also factors in the different elevations, since if they only used DC atomic clocks for the averaging, the “world standard” would be running too slow (or too fast for sea-level time), and if they used only Boulder clocks, the “world standard” would be running too fast.

So yes, I think we can say either that the elevations are “factored in” or we can say that the different clock-elevation rate “averaging” is done to “average” the different rates due to the different elevations. And as I mentioned, I suspect that the “world standard” time rate is probably “sea-level” atomic clock time, but you should know the answer to that.

They also told me that no two atomic clocks can be manufactured to run at exactly the same rate. In other words, when being manufactured side by side, every one of the clocks will have a slightly different rate.

Would you agree with this or not?

Ok, I’ve thought about it some more, and I think your term “factored in” is better than my term “averaged out”, because my term could be misconstrued to mean that if 5 clocks in Boulder are “averaged out” with 5 clocks in DC, then the “average” would represent the rate of an ideal atomic clock placed at the elevation of about 2,600 feet, and that is not what I intended to imply. I should have said “factored in”.

Averaging corrects for biases, drifts (which are measured and removed) and the range of frequencies, but really not for "glitches" - if a clock's frequency changes in a non-statistical way, it will be removed from the mean until it can be re-characterized. The weird thing is that once you are convinced a clock has been working correctly, you can re-do all of your averaging as if the clock were in tyhe mean the whole time. You do calculations that are not real-time, as it were, so if you decide that your ensemble is off by e.g. a nanosecond (after you've added the clock back in), you can slowly steer the clock output to correct for that, and do it such that it can't be seen by any user.

Also, even if two clocks did have the same frequency at some point, the best you can realistically hope for is that there will be white noise in the system. Time is the integral of frequency, and the integral of white frequency noise is a random walk in time - so two clocks, synchronized and at the same frequency, will still exhibit a random walk and diverge from each other.

Sam5
2003-Dec-30, 12:54 AM
If you go back to what I wrote, I think you'll see that I never implied otherwise. Gravitation is a GR effect on all clocks, and there is another, far larger dependence that affects pendulum clocks.

I think we are both a little mixed up about what the other one means about this. I was originally talking to Russell and others about the SR and about “relative motion” NOT changing the rates of clocks at all. Gravitation will slow down some kinds of clocks, but not all at the same rates, and not all for the same reasons. In fact, many mechanical clocks will slow down only because of the “friction” on their bearings caused by the gravitational field potential pulling down on the gears and the balance wheel. And some kinds of electronic clocks can be speeded up more by slight heating than they are slowed down by gravitational effects.

As far as I’m concerned, “GR” refers specifically to atomic clocks or the “harmonic oscillation rates of atoms”, as per the 1911 theory. Gravitation will speed up pendulum clocks, and that was discovered 500 years ago and it’s not part of GR theory, it’s part of Newtonian relativity theory. In the 19th Century some guys used to measure elevation levels by how much a pendulum clock slows down at higher elevations.

Averaging corrects for biases, drifts (which are measured and removed) and the range of frequencies, but really not for "glitches" - if a clock's frequency changes in a non-statistical way, it will be removed from the mean until it can be re-characterized. The weird thing is that once you are convinced a clock has been working correctly, you can re-do all of your averaging as if the clock were in tyhe mean the whole time. You do calculations that are not real-time, as it were, so if you decide that your ensemble is off by e.g. a nanosecond (after you've added the clock back in), you can slowly steer the clock output to correct for that, and do it such that it can't be seen by any user.

Also, even if two clocks did have the same frequency at some point, the best you can realistically hope for is that there will be white noise in the system. Time is the integral of frequency, and the integral of white frequency noise is a random walk in time - so two clocks, synchronized and at the same frequency, will still exhibit a random walk and diverge from each other.

Very interesting stuff. I’d like to read an entire book by some person who works with this stuff. Most of what we members of the general public read in books is a mix of real stuff and SR and GR theory. Tell us more about atomic clocks and what they do and what you know about them. Tell us more about this "random walk" and "white noise".

Russell
2003-Dec-30, 09:49 PM
Hello:
When I think of white noise, I think acoustic white noise. White noise combines equal parts of all audio frequencies, which is analogous to white light, which combines equal parts of the visible light spectrum.

Sam5
2003-Dec-30, 10:01 PM
Hello:
When I think of white noise, I think acoustic white noise. White noise combines equal parts of all audio frequencies, which is analogous to white light, which combines equal parts of the visible light spectrum.

Me too. That's why I want to know what the "white noise" is in relation to atomic clocks.

swansont
2003-Dec-31, 12:33 AM
I was referring to a gaussian white noise, i.e. noise in a measurement. Please forgive the imprecision. ("White noise" isn't really noise from a measurement standpoint - it's a uniform frequency distribution)

swansont
2003-Dec-31, 12:46 AM
Tell us more about atomic clocks and what they do and what you know about them.

The biggest revelation for me was the timescale calculations. There is no perfect clock, so how you decide what the right time is has a small degree of arbitrariness. The fact that the calculations are done after the fact - a month goes by between some of the data reported to the international bureau of weights and mesures (BIPM), so you are trying to figure out what time it was and predict what time it will be, when all the numbers are crunched. Plus the ability to decide that a clock was good/bad, and add/drop it from the calculation dating back for months, and decide that your currrent time needs to be adjusted.

Here (http://tycho.usno.navy.mil/clockdev/cesium.html) is what I've been working on. Also this (http://tycho.usno.navy.mil/clockdev/Rubidium%20Fountain.html), but the web page is rather sparse since it's lower priority than the actual design and construction.

Sam5
2003-Dec-31, 01:06 AM
Here (http://tycho.usno.navy.mil/clockdev/cesium.html) is what I've been working on. Also this (http://tycho.usno.navy.mil/clockdev/Rubidium%20Fountain.html), but the web page is rather sparse since it's lower priority than the actual design and construction.

Very interesting, thanks!

Do you ever compare the clock rates with the earth’s rotation rate, to be sure all the clocks haven’t drifted in the same direction for some reason?

swansont
2003-Dec-31, 11:41 AM
Do you ever compare the clock rates with the earth’s rotation rate, to be sure all the clocks haven’t drifted in the same direction for some reason?

Well, that's what you do to determine whether leap seconds have to be added (or subtracted). You use different types of clocks to try to make sure there is no common mode systematic error like that, but then, I don't know if you'd notice. The earth rotation is variable, and some of the fluctuations can't be predicted, plus the fact that the rotation rate changes are much larger than clock drifts, so you'd have trouble using that as a measure. Earth rotation isn't a very good clock, compared to state-of-the-art, and that statement has been true for decades now.

Sam5
2003-Dec-31, 04:14 PM
Do you ever compare the clock rates with the earth’s rotation rate, to be sure all the clocks haven’t drifted in the same direction for some reason?

Well, that's what you do to determine whether leap seconds have to be added (or subtracted). You use different types of clocks to try to make sure there is no common mode systematic error like that, but then, I don't know if you'd notice. The earth rotation is variable, and some of the fluctuations can't be predicted, plus the fact that the rotation rate changes are much larger than clock drifts, so you'd have trouble using that as a measure. Earth rotation isn't a very good clock, compared to state-of-the-art, and that statement has been true for decades now.

Are the leap seconds put into the system so that we won’t eventually have darkness at atomic-clock “noon” and the sun high overhead at “midnight”?

In other words, while the atomic clock average is very steady on an hour to hour basis, doesn’t the whole world basically still go by astronomical time on a day to day and a year to year basis, with the atomic clocks ticking out nearly perfect “seconds” during each day and hour of the year, with the "leap seconds" being added to keep up with long-term astronomical time.

russ_watters
2003-Dec-31, 05:24 PM
Are the leap seconds put into the system so that we won’t eventually have darkness at atomic-clock “noon” and the sun high overhead at “midnight”?

In other words, while the atomic clock average is very steady on an hour to hour basis, doesn’t the whole world basically still go by astronomical time on a day to day and a year to year basis, with the atomic clocks ticking out nearly perfect “seconds” during each day and hour of the year, with the "leap seconds" being added to keep up with long-term astronomical time. Yes, but there is a little more to it than that. The earth doesn't rotate or revolve at a constant rate. So even if we tried, we couldn't re-define a second in such a way as to make every day exactly 24 hours long.

swansont
2003-Dec-31, 09:13 PM
Are the leap seconds put into the system so that we won’t eventually have darkness at atomic-clock “noon” and the sun high overhead at “midnight”?

Long-term, the answer is yes. But near-term, since not many people care if noon is off by a few seconds, the most vocal group for having leap seconds seems to be astronomers, so they don't have make time corrections for observing (which would be a royal pain)

As I understand it, there have been some heated discussions about whether to abolish leap seconds and go to something like leap minutes, amongst a select group of people who give a hoot.

Sam5
2003-Dec-31, 10:55 PM
Are the leap seconds put into the system so that we won’t eventually have darkness at atomic-clock “noon” and the sun high overhead at “midnight”?

Long-term, the answer is yes. But near-term, since not many people care if noon is off by a few seconds, the most vocal group for having leap seconds seems to be astronomers, so they don't have make time corrections for observing (which would be a royal pain)

As I understand it, there have been some heated discussions about whether to abolish leap seconds and go to something like leap minutes, amongst a select group of people who give a hoot.

Interesting, thanks!

How much of a bother is it to add the leap seconds?

Sam5
2004-Jan-01, 12:54 AM
Ok, this is a “light” thread, and I’ve got a question. Does anyone see anything wrong with the standard model or illustration of how light colors immediately diverge from a white beam that goes into a prism, with the blue part of the beam immediately becoming a thin blue line at the bottom of the beam inside the glass, and the red part immediately becoming a thin red line at the top of the beam inside the glass, with the blue and red lines diverging inside the glass and diverging more when they exit the glass?

LIKE IN THIS ILLUSTRATION (http://micro.magnet.fsu.edu/primer/java/scienceopticsu/newton/)

Sam5
2004-Jan-01, 03:57 AM
Ok, so here’s the question: How does a white light beam know to split up into blue and red beams immediately upon entering a prism, with the red beam diverging toward the top and the blue beam diverging toward the bottom of the expanding rainbow beam inside the glass, when the white beam doesn’t know if its entering a prism or a pane of window glass.

LIKE IN THIS ILLUSTRATION (http://micro.magnet.fsu.edu/primer/java/scienceopticsu/newton/)

This illustration also suggests that the red and blue beams somehow know that if the glass through which they are traveling is window glass, with parallel sides, then the diverging of the red and blue beams stops at a slanted angle in the middle of the glass, and the red and blue beams began to converge again, and they emerge on the other side all united as a single white beam again. And, if so, how do the red and blue beams know where the “middle” of the window glass is?

Maybe there’s something very simple that I’m just overlooking.

Kaptain K
2004-Jan-01, 12:20 PM
Let me get this straight. You don't even understand simple refraction, but have spent 30+ pages 'splaining relativity to us? #-o

It is really, really, really, very, very, very simple. When light hits the interface between media of different indices of refraction (at an angle other than perpendicular) it begins to diverge. When it hits the second interface (back into the original medium) the divergence will change again (assuming the ray is not perpendicular to the interface). In the case of a prism, it will diverge even more. In the case of parallel interfaces (a window pane, for example) the ray will coverge back into a parallel ray. The ray will have a slight chromaticity, but it is usually too slight to be noticed. You can prove this to yourself with a pinhole and a piece of glass held at an angle to the beam.

Sam5
2004-Jan-01, 01:53 PM
It is really, really, really, very, very, very simple. When light hits the interface between media of different indices of refraction (at an angle other than perpendicular) it begins to diverge.

I’m asking you specifically about the standard illustration like the one I linked to. It shows a thick white beam going into the glass, and it shows red emerging from the top of the white beam and blue emerging from the bottom of it. Are you saying that in a beam of white light, all the red photons travel at the top of the beam and all the blue ones travel at the bottom, as in this illustration? You don’t really believe that, do you?

I understand that a prism separates the red from the blue, but it doesn’t do it in the way shown in the standard illustrations. So, can you provide us with an accurate illustration showing what happens inside the glass, and exactly how the two colors diverge?

Sam5
2004-Jan-01, 02:03 PM
Let me get this straight.

If the blue and the red photons immediately begin to diverge when they enter the glass, then all the blue photons from the bottom to the top of the beam should be bent toward the bottom of the prism, and all the red ones from the top to the bottom of the beam should be bent toward the bottom of the prism too, but at less of an angle than the blue photons, so the colors should diverge, but not in the way they are shown in the standard illustrations.

The standard illustrations show all the red photons emerging from top of the white beam and bending a little, with all the blue photons emerging from the bottom of the white beam and bending more. I am reasonably sure that white beams of light do not have all the blue photons traveling at the bottom of the beam and all the red ones traveling at the top. If they worked that way, we wouldn’t need a prism to see the separation of the colors.

Sam5
2004-Jan-01, 02:39 PM
I think a better illustration would show a much thinner white beam going into the prism, with all the blue photons, top to bottom of the white beam, diverging slightly more toward the bottom of the prism than the red ones. The blue ones diverging in the glass from the top of the white beam would actually cross over the red ones diverging in the glass from the bottom of the white beam.

When a thicker white beam is used, what emerges from the other side of the prism is not a pure rainbow, but a blob of light that is white in the center and that has different colored fringes on the outside edges of the white area.

But in standard illustrations, when a thick white beam is shown going into the prism, the blue is often shown emerging only from the bottom of the thick white beam, and the red is shown as emerging from the top of the thick white beam, and this is not correct.

Sam5
2004-Jan-01, 04:24 PM
Ok, I’ve got it now.

Many of the standard “wide white beam” prism illustrations we’ve been seeing in books for years are wrong.

For example, this illustration is wrong:

LIKE IN THIS ILLUSTRATION (http://micro.magnet.fsu.edu/primer/java/scienceopticsu/newton/)

If a wide white beam is shined into a real prism, then what will show up on the white screen to the right of the prism will not be the entire wide white beam split up into separate colors. What would actually show up on the screen will be a large central area of white light, with a thin red fringe at the top and a thin blue one at the bottom.

This is because the divergence of the red and blue is not as great as is shown in these illustrations, and the white beam is shown as being too large, too thick, and that doesn’t allow all the red and blue photons to separate and diverge enough inside the prism.

In order to see the proper effect of the full spectrum at the white screen, we must use a very thin white beam going into the prism. Then, all the blue photons refract downward inside the prism more than the red ones do, and some of the red and blue photons cross over each other, because the blue ones are going more downward to the right than the red ones.

Then what emerges on the other side of the prism is the spectrum split into several thin colored lines, each of which represents the full thickness of the original white beam that originally entered the prism. The white card onto which the colored lines are projected, must be some distance from the right face of the prism, to allow more time and space for the colored lines to diverge more. Then the thin red line will be at the top on the screen, and the thin blue line will be at the bottom, with the other colors displayed as other thin lines in the middle. Each of the thin lines of color represents the full thickness (or “thinness”) of the original thin white beam that entered the prism.

Since a prism is much thicker than window glass, when a thin white beam of light travels at an angle through window glass, since the divergence of the red and blue photons is very slight inside the glass, and since the glass is very thin when compared to a prism, we see no “rainbow” effect on the right side of the window glass.

SeanF
2004-Jan-01, 04:59 PM
Like I said a long time ago (http://www.badastronomy.com/phpBB/viewtopic.php?t=9446&amp;postdays=0&amp;postorder=asc&amp;star t=358):

A diagram drawn on a website to demonstrate a concept does not necessarily represent a literal representation of the physical reality

The way light is shown inside a prism in most diagrams is not absolutely perfectly correct, but it is not fair to categorize it as "wrong" - especially given that you can't see the light diverging inside the prism anyway.

Sam5
2004-Jan-01, 05:12 PM
Like I said a long time ago (http://www.badastronomy.com/phpBB/viewtopic.php?t=9446&amp;postdays=0&amp;postorder=asc&amp;star t=358):

A diagram drawn on a website to demonstrate a concept does not necessarily represent a literal representation of the physical reality

The way light is shown inside a prism in most diagrams is not absolutely perfectly correct, but it is not fair to categorize it as "wrong" - especially given that you can't see the light diverging inside the prism anyway.

If an illustration is not “right”, then it’s “wrong”, and that confuses students. It is just as easy to draw a correct illustration than it is to draw a wrong one.

You are not defending “truth”, you are defending the commercial educational media industry that occasionally publishes incorrect illustrations and incorrect theory explanations, and those errors confuse students rather than educating them.

No prism illustration should ever show a thick white beam going into the prism.

If we used a wide white beam going into the prism, this is what we would see on the white screen to the right side of the prism:

Sam5
2004-Jan-01, 06:13 PM
Here are the correct drawings for a THIN white beam and a THICK white beam going into the prism. Note that most of the THICK white beam remains white during it’s travel, both inside the prism and to the right on the screen outside the prism.

Please excuse the crooked lines. I did this in about 30 seconds without a straight-edge.

There is no reason why text books and internet websites can’t illustrate these phenomena correctly.

Students need to know that the THIN beam will allow a small prism to split the white light up into a spectrum better, and they need to know that when the red and blue (and the other colored) waves/photons overlap, then that area on the screen will be white. The students also need to know that some of the spectrum colors overlap on the screen and produce mixed-color bands, and some of the color bands will be wider than others.

LINK TO CORRECT SCHEMATIC DRAWINGS – (not to scale) (http://im1.shutterfly.com/procserv/47b4df20b3127cce857b7b3551e80000001610)

This is NOT correct:

NOT CORRECT (http://micro.magnet.fsu.edu/primer/java/scienceopticsu/newton/)

swansont
2004-Jan-01, 06:17 PM
How much of a bother is it to add the leap seconds?

For me it's no bother at all. :D

Someone has to be around to monitor it, but it's pretty simple. And it happens at midnight UTC, or 7PM EST, so the person checking up on it doesn't even have to stay until midnight. The people that say we add it at the stroke of midnight imply that it's done locally, and it's not.

Sam5
2004-Jan-01, 06:29 PM
How much of a bother is it to add the leap seconds?

For me it's no bother at all. :D

Someone has to be around to monitor it, but it's pretty simple. And it happens at midnight UTC, or 7PM EST, so the person checking up on it doesn't even have to stay until midnight. The people that say we add it at the stroke of midnight imply that it's done locally, and it's not.

Interesting. :D

I bought one of those new “atomic clocks” from a department store. It is NOT a real “atomic clock”, but it receives a radio signal from Ft. Collins, Colorado, and the Ft. Collins transmitter receives a time code signal from Boulder. So, several times a day, the regular electronic clock inside my “clock”, receives an “updated” signal from Boulder, via Ft. Collins, and that corrects any accumulated errors inside my “atomic clock”. It’s the only clock I’ve ever had that is “accurate” (at least several times a day and night).

Do you know if the guys in Boulder use some of your atomic clocks for their “averaging”? If so, I’ll think of you every time I look at my clock, and I’ll tell all my friends that I know a guy who makes sure my wall clock keeps proper time! Lol!! :D

Russell
2004-Jan-01, 10:06 PM
Hello:
Going back to time, why can't we use pulsars as a standard time? They seem to be the most accurate objects in the universe, accurate enough to back relativity.

swansont
2004-Jan-02, 01:32 AM
I bought one of those new “atomic clocks” from a department store. It is NOT a real “atomic clock”, but it receives a radio signal from Ft. Collins, Colorado, and the Ft. Collins transmitter receives a time code signal from Boulder. So, several times a day, the regular electronic clock inside my “clock”, receives an “updated” signal from Boulder, via Ft. Collins, and that corrects any accumulated errors inside my “atomic clock”. It’s the only clock I’ve ever had that is “accurate” (at least several times a day and night).

Do you know if the guys in Boulder use some of your atomic clocks for their “averaging”? If so, I’ll think of you every time I look at my clock, and I’ll tell all my friends that I know a guy who makes sure my wall clock keeps proper time! Lol!! :D

NIST keeps UTC(NIST), which is their realization of universal time. USNO keeps UTC(USNO), and we both report to BIPM, who takes data from all of the timing labs and computes the worldwide standard. We all steer to that, so in that sense the clocks are averaged. There's also a direct data link, but we don't directly average each others' clocks.

For virtually all situations USNO time and NIST time are equivalent.

swansont
2004-Jan-02, 01:36 AM
Hello:
Going back to time, why can't we use pulsars as a standard time? They seem to be the most accurate objects in the universe, accurate enough to back relativity.

The problem as I understand it is star quakes - pulsars "settle" occasionally and change their rate. Then you have to recharacterize them all over again. I recall being been told pulsars were the best clocks ( as far as long-term stability is concerned) until the HP5071A cesium beam clock came out.

Sam5
2004-Jan-02, 05:12 AM
The problem as I understand it is star quakes - pulsars "settle" occasionally and change their rate. Then you have to recharacterize them all over again. I recall being been told pulsars were the best clocks ( as far as long-term stability is concerned) until the HP5071A cesium beam clock came out.

Do you ever use this equation in any of your clock calculations?

√1 – (v^2/c^2)

Do atomic clocks anywhere ever change rate due to fluctuations in the earth’s magnetic field, due to sunspot activity, because of the auroras, or due to any kind of outside EM waves any outside field fluctuations? Do they have to be shielded from these effects?

Chip
2004-Jan-02, 08:34 AM
Hello:
Going back to time, why can't we use pulsars as a standard time? They seem to be the most accurate objects in the universe, accurate enough to back relativity.

Pulsation rates of binary pulsars drift, but as you implied, it is that drift effect which allowed for a wonderful test of relativity. The pulsar PSR 1913+16 in Aquila, with its fluctuation of 50 microseconds, provided a natural laboratory for such a test. Hulse and Taylor received the 1993 Nobel Prize in Physics for their work with this binary pulsar. Two neutron stars orbiting around one another with one being a pulsar provides an excellent clock that is also subjected to both motion and gravitation.

Sam5
2004-Jan-02, 04:28 PM
Pulsation rates of binary pulsars drift, but as you implied, it is that drift effect which allowed for a wonderful test of relativity.

It was also a great proof of the Classical 1842 Doppler relativity theory. They said:

“The pulse repetition frequency, that is, the number of pulses received each second, can be used to infer the radial velocity of the pulsar as it moves through its orbit. When the pulsar is moving towards us and is close to its periastron, the pulses should come closer together; therefore, more will be received per second and the pulse repetition rate will be highest. When it is moving away from us near its apastron, the pulses should be more spread out and fewer should be detected per second.

The pulsar arrival times also vary as the pulsar moves through its orbit. When the pulsar is on the side of its orbit closest to the Earth, the pulses arrive more than 3 seconds earlier that they do when it is on the side furthest from the Earth. The difference is caused by the shorter distance from Earth to the pulsar when it is on the close side of its orbit. The difference of 3 light seconds implies that the orbit is about 1 million kilometers across.”

Doppler predicted this effect with starlight in 1842, and Doppler relativity theory is something just about everyone uses nearly every day.