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Post by olmy on Oct 6, 2010 17:04:09 GMT 1
ALL real-world counters consist of detecting physical objects jiggling up and down in a gravitational field, as you put it, and counting those jiggles. Physical objects have mass, and there is no shield against gravity. Nope, spring based oscillators work horizontally too. Anyway, that's not what I meant by 'jiggling up and down'. You started talking about one clock being subject to changes in g, while the other was not. That would (say, on Earth) require enormous 'jiggles'. How did you imagine that one would get to experience a variable g and the other a constant one? Now taking STA's definition of a gravity field as a rate-of-change-of-g-field... No, that's totally backwards. The acceleration due to gravity (g) comes from the gradient of the gravitational potential. ...so a constant velocity trip in such a field will experience a continual rate-of-change of 'g'. Is this supposed to be an answer to my question about constant velocity time dilation? That happens with no gravity at all. It comes from Special Relativity which ignores gravity. It is also symmetrical. Each traveller sees the other's clock as slow... "When two observers are in relative uniform motion, and far away from any gravitational mass, the point of view of each will be that the other's (moving) clock is ticking at a slower rate than the local clock." en.wikipedia.org/wiki/Time_dilation#Relative_velocity_time_dilationWant to try again....?
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Post by Progenitor A on Oct 6, 2010 17:05:46 GMT 1
Hi Carnyx Your mate do go don't he? Here he is agin: [You can use it to measure g but g only affects the equilibrium position, not the period of oscillation. T = 2 pi sqrt(m/k) See? No g. An' 'e bin dun tole that that thur gravity do affects de period of oskillation in this way: With a spring oscillating under the influence of gravity if the gravity is weak then the decay period is small; if gravity is strong, the decay period is long. This means that the sinusoidal oscillations have an 'envelope' of exponential decay with the period of th envelope dependent upon (amongst other things) the force of gravity Fourier analysis shows that an 'envelope modulated' sine wave of that fashion contains frquencies higher than the fundamental period frequency; the faster the decay, the higher the amplitudes of these higher frequencies. Move the spring to weaker gravitaotional field and the spectrum of the oscillations willl change Additionally, as we have seen, if a spring is oscillating in a gravity dilated time then the fundamental frequency of oscillation will be higher than the same spring in a non-gravity dilated time So gravity do affect frequencies of oscilllation - no problem!
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Post by olmy on Oct 6, 2010 17:16:00 GMT 1
Progenitor AYou really should try to keep up. If not with me, then at least with the guy you are trying to agree with. carnyx thinks that time dilation is explained by purely Newtonian considerations such as the appearance of g in the formula for the period of a pendulum (although he did forget the 2 pi) - see the OP. My ignoring of things like decay of the oscillations is also entirely consistent with carnyx's use of the simple pendulum formula in the OP. Not that any of that has any relevance to trying to use it to measure time, anyway....
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Post by Progenitor A on Oct 6, 2010 17:59:43 GMT 1
What she is trying to say is that the gravitation of the earth varies at different heights so that when defining a potential energy you consider the mass of an object, its height, and the force of the gravitational field (which is represented by the constant value 9.8 as the differences in the force of the gravitational field is negligible at various heights), i.e. potential energy = mass of object in kilograms x 9.8 x height of object in meters. I do not think that that is quite the case that STA means Abacus (I am still awaiting clarification) Gravitational Potential Phi=-Gm/r where G=universal gravitational constant=6.67x10 -11 Nm 2kg -2m =earths mass= 6.0x10 24kg r = earths radius = 6.4x10 6m therefore gravitational potential at earth’s surface =-63x 10 6 j/kg Gravitational potential at 10 km above earths surface = -(6.67x10-11Nm2kg-2 x 6.0x1024kg)/6.41x106m=-62.9 x 10 6 j/kg Difference = 0.1 x 10 6j/kg
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Post by carnyx on Oct 6, 2010 18:05:04 GMT 1
eamonnshuteEr.. the caesium measurement IS the other counter! And as this counter is also affected by 'g' ... (... ask STA ...) then we have the situation as described
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Post by Progenitor A on Oct 6, 2010 18:16:52 GMT 1
First, I meant by gravitational potential what every basic physics text means by gravitational potential -- a function whose gradient is the gravitational field at a point. Hence zero of it undefined. Is this, then, STA, what you mean by gravitational potential? Gravitational Potential Phi=-Gm/r ?
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Post by eamonnshute on Oct 6, 2010 18:22:19 GMT 1
Er.. the caesium measurement IS the other counter! And as this counter is also affected by 'g' ... (... ask STA ...) then we have the situation as described You count only one thing, ie the number of waves. This gives you the time that has elapsed for the caesium atom. There is no 'other' counter. The counter is not affected by gravity - if there are a million waves then the counter will count to a million, regardless of gravity. Nor is the caesium atom affected by gravity; wherever you take your caesium atom you will still see the same number of waves per second.
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Post by Progenitor A on Oct 6, 2010 18:35:24 GMT 1
Nor is the caesium atom affected by gravity; wherever you take your caesium atom you will still see the same number of waves per second. But we have already seen that gravity causes a time dilation. Therefore a ceasium atom in a weak gravity field will have its seconds at a faster rate thana caesium atom in a strong gravity field. Threfore a caesium atom used as a timer in a weak gravity will surely count less items of the same frequency than a caesium atom in a weak gravity?
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Post by carnyx on Oct 6, 2010 18:50:55 GMT 1
olmyBut, in this orientation they will measure only the horizontal component of the gravitational field. To measure the G vector properly, you will need three orthogonal springbobs. As I said in the post; one counter stays at base and experiences g#1.. and the other counter goes with the traveller and experiences variable g#2 And this; So, when there is no gradient, there is no mass-attractive force? What happened to it? I think it is you who have it backwards. It is the physical position in the field that dictates what g-force is felt....the field being no more than a map, based on the 1/D-sqared relationship, of what an accelerometer would read at any position relative to the massive object. A 'g' map, in fact. But as I said earlier, STA gave the definition of a second-order map, of the derivative of accleration, or in simpler terms words the rate of change of g at any point. A g-rate map, in fact. (BTW in some circles, the jargon for rate-of-change-of-acceleration or g, is known as 'nudge') As for I'd thought that was obvious. Given that the fastest possible means of communication between the two is via em waves, then it is a matter of the ratio between the velocity of light and the velocity of separation. And finally, with reference to your later post, I merely gave the pendulum counter as an instance that could be readily seen. However, my substantive point in Post 1 is that as ALL sensible event counters that can be usd to measure 't' ...are affected by gravity.. then the gravitational force varies as 1/d squared is sufficient to show that measurements of 't' are affected by g..
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Post by carnyx on Oct 6, 2010 18:55:28 GMT 1
@eamonshute
And that is the nub of it ... in this measurement how do you know that the time-interval you call 'a second' is the same in another place?
I suspect you will find that the caesium atom is affected by gravity, as it has mass, and certainly the radiation will be. Anyway, as we know that differences in g equivalent to only one foot can be resolved by comparison between two of these kinds of counter, this shows that the effect is real.
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Post by eamonnshute on Oct 6, 2010 19:18:06 GMT 1
But we have already seen that gravity causes a time dilation. Therefore a ceasium atom in a weak gravity field will have its seconds at a faster rate thana caesium atom in a strong gravity field. Threfore a caesium atom used as a timer in a weak gravity will surely count less items of the same frequency than a caesium atom in a weak gravity? Alice on Earth sees her caesium atom produce 9,192,631,770 waves per second. Astronaut Bob, a million miles away and not moving relative to Alice, sees his own caesium atom also produce 9,192,631,770 waves per second. In that sense the caesium clock is not affected by gravity. But Bob will see Alice's atom produce slightly fewer waves per second because of gravitational time dilation.
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Post by speakertoanimals on Oct 6, 2010 19:21:13 GMT 1
First, I meant by gravitational potential what every basic physics text means by gravitational potential -- a function whose gradient is the gravitational field at a point. Hence zero of it undefined. Is this, then, STA, what you mean by gravitational potential? Gravitational Potential Phi=-Gm/r ? In the Newtonian case, and for a single centre-of-mass, yes. Although as I said, since the PHYSICAL effect (i.e., the force) depends on the gradient, I could define the zero to be wherever I wanted, and say that: Phi = 637.4 - Gm/r makes no difference to the DIFFERENCE in potential between two points. In the infinite slab/constant gravitational force case, the formula is different.
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Post by eamonnshute on Oct 6, 2010 19:28:24 GMT 1
And that is the nub of it ... in this measurement how do you produce that time-interval you call 'a second'? I have already given you the definition of a second. So you just count 9,192,631,770 waves, and that means that exactly one second has elapsed by definition!
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Post by carnyx on Oct 6, 2010 19:38:42 GMT 1
A most beautifully circular argument!
In fact, Alice and Bob are two counters .. each counting to 9.120.631,770. But, one finishes the count slightly earlier. Absent the effects of time-lag in communication, then the ONLY difference between the two is the difference in local forces of gravity!
And as a proof, we could check that the results with Bob at different distances conform to Newton's 1/D-sqared function.
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Post by eamonnshute on Oct 6, 2010 20:12:52 GMT 1
In fact, Alice and Bob are two counters .. each counting to 9.120.631,770. But, one finishes the count slightly earlier. Absent the effects of time-lag in communication, then the ONLY difference between the two is the difference in local forces of gravity! And as a proof, we could check that the results with Bob at different distances conform to Newton's 1/D-sqared function. There are two counters because there are two clocks. Each clock only requires one counter to give the time. And they both give the correct time for their own location. The fact that they disagree about a time interval is caused by the different frames of reference, not because one of them is not working properly due to gravity affecting the mechanism.
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