This is a common misconception. SR can handle accelerating frames of references quite well. Its gravitational fields that it cant deal with.
Time dilation and physical time differences due to traversing a different path are not the same. There are *two* distinct effects occurring in SR referred to under the same general terms, causing much confusion. Conventional SR "time dilation" is completely equivalent to an optical *illusion*. For example, two people looking through each side of a magnifying glass will both see the other as larger. Differences in aging of different observers are due to path differences travelled in space-time. This is a real physical effect.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
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The experiences *can* be symmetrical. How about large relative motion between the two but they are going different directions in a universe that is closed. So they pass each other going opposite directions and each sees the other's clock as slow. When they come back to the same meeting point after going a great circle trip through the universe each one would have seen the others time as slow so what do their clocks read (and what do they see the other's clock as?) as they pass each other on their way to another great circle trip?
You can get the same effect by both orbiting a large mass or black hole. With each having a large orbital velocity relative to the other.
I'm aware that they both *are* being accelerated by the bending of a "straight" line path into a great circle but it seems they would both see each others clock as slow when they pass at the limit of a very *large* great circle which would look like linear motion.
I read in sci.electronics.design that Kevin Aylward wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
But the velocity profile would be flat if there were no acceleration. This is getting into something like the 'voltage-control or current control' thing. I can just as well argue that the cause is the twice-integrated acceleration profile, and if that profile is the t axis (a = 0 for ever)......
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I read in sci.electronics.design that Robert wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
You are missing the point of the 'second twin paradox'. See below.
At this point you jump from assertion to question, and the question is off the mark.
The point is NOT that the twins PASS each other, but that they return to the same room on Earth, one from far away and one from elsewhere on Earth. If they both had identical dynamical experiences, they would be both younger AND older than each other, which is clearly not possible. The argument that they are still the same age denies that time-dilation occurs at all, whereas it is proven to occur.
The difference in the dynamical experiences is obvious; the traveller accelerated to high speed, decelerated to reverse course, accelerated again to high speed and decelerated again at Earth, **all in his own frame of reference**. He is not an 'inertial observer' in the SR sense. The stay-home had none of those experiences.
The apparatus for determining whether you are an inertial observer is rather simple - a bucket of water. If it sloshes about, you are not inertial.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I read in sci.electronics.design that Robert Monsen wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
It's good to know that Feynman agreed with me. (;-)
I don't see a difficulty in principle. The clocks can't become unobservable or change discontinuously, so their readings at any instant of either observer's observation must be computable.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I read in sci.electronics.design that Kevin Aylward wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
Yes, he does in the case I put forward, which is that the two observers meet in the same room on Earth after the journey taken by one of them. This requires four episodes of acceleration (if you count the course reversal as taking two), not experienced by the observer who remains on Earth. And, to avoid distractions, I restrict the motion of the traveller to 'straight line', whatever that means, so that during the whole journey there is no 'sideways' (orthogonal to the initial direction of motion) component of speed.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I agree that this is subtle. There is a lot of debate on this in the relativity NG.
One cant argue that it is the twice-integrated acceleration profile, because then that makes time a pure function of distance. That is, independent of how one got from a to b, the time contraction would be the same. This isnt true. The point is still that it is time contraction is only a function of velocity.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
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I read in sci.electronics.design that Robert Baer wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
In apples, yes, but the holes in mackintoshes are due to the Rubber Clothes Moth, Tinea elastophagia resultans (R. O'Shea).
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
That is obvious. I was just talking about time dilation.
It's an optical illusion that can slow the decay of particles.
Of course.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
I read in sci.electronics.design that Kevin Aylward wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
If you are going to argue against my example, you can't simply say 'we don't need it'. The thread developed to the point of putting forward the 'second twin paradox', which can be stated 'Since velocities are relative, and each twin sees the other's clock going slow, how does it come about that the travelling twin arrives back younger?'
My answer is that the twins have different dynamical experiences; the Earth twin remains an inertial observer but the travelling twin doesn't, and that provides the essential asymmetry.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
You were arguing that the twin paradox actually *depended* on one observer experiencing different speeds, i.e. accelerations with respect to speed, not direction. I simple gave an explain showing that this was not the case. Changes in *speed* of the observer are not necessary to effect aging differences.
As I indicated in another post, this is *two* separate issues. Time dilation, i.e the two observers *observing* each other's clocks running slow is equivalent to an optical illusion. The explanation as to why one observer experiences a net loss of real physical time, requires a
*different*, explanation.
As I said, its much more subtle then this. I did reference a web page on this,
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Simply defining an inertial frame, is itself a major problem.
Kevin Aylward snipped-for-privacy@anasoft.co.uk
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SuperSpice, a very affordable Mixed-Mode Windows Simulator with Schematic Capture, Waveform Display, FFT's and Filter Design.
The math behind the Lorentz transformation is dead easy. Simple algebra.
First, try very hard to mentally assume away any idea of an absolute (or 'preferred') reference frame and ask yourself how you might start out to describe the motion of a photon particle traveling at the speed of light from two different non-preferred perspective frames:
x = c*t coordinate system K
and,
x' = c*t' coordinate system K'
In other words, we make no presumptions about the two reference frames to each other, __except__ for the fact that they will each observe the same velocity for the photon. It is as though they are completely independent to each other.
To make this point come to home deeply and seriously, imagine that the photon is simultaneously observable in two completely different universes that have nothing in common to each other except that they both interact with shared photons.
The above two equations do express the fact that x and x' have nothing to do with each other except for the speed of light, and that t and t' similarly have nothing to do with each other except again for the speed of light. The only thing in common in those two equations is c, itself. Both independent frames will see that as the same value, whatever it turns out to be. Other than that, the two frames are completely without any other shared referents.
Now we write these same two, but with the photon traveling in the opposite direction, just to get all the possibilities:
x = -c*t
and,
x' = -c*t'
Then we can write:
x - c*t = 0, and also from the last case, x + c*t = 0
and,
x' - c*t' = 0, and also from the last case, x' + c*t' = 0
Of course. Nothing special about that.
But since these are zero, we can now make some statement that combines both frames:
x - c*t = A*(x' - c*t'), and, x + c*t = B*(x' + c*t')
We don't yet know what A and B are likely to be, but that's the problem to solve.
We can now add and subtract both of these from each other:
All this is pretty basic algebra, accessible to most anyone I think.
At this point, to get rid of the factor 2 (the constants A and B are just arbitrary ones, so we can create any new ones we want based on them for convenience and use those instead), we can define the following new derived constants:
R = (A+B)/2, and S = (A-B)/2
From that, we can re-write the above as:
x = R*x' - S*c*t' and c*t = S*x'-R*c*t'
Remember these two, we'll refer to them below.
Well, that's a start. Of course, we still haven't figured out what R and S are. So let's do so.
At the origin of the one of the frames, we have x = 0, so
x = R*x' - S*c*t' = 0
or,
x' = t' * [S*c/R]
This appears to be in a familiar x'=v'*t' form, if we defined the velocity v':
v' = [S*c/R]
If so, it would be the velocity for which the origin (or really, any point) in K is moving with respect to K'.
But in perhaps a little better form, we can write that the instantaneous change in x' is:
Same thing. And v' is the relative velocity of the two systems.
Now, we also know that length of a rest-object in system K observed from system K' must be the same as the length of a rest-object in system K' observed from system K (due to the assumption of relativity
-- no preferred point of view.) So let's take a picture, so to speak, at time t'=0, then
x = R*x'
Let's now look at two points in system K (x-axis) separated by a distance of exactly 1:
S = v'*R/c S = v'*[1/SQRT(1 - v'^2/c^2)]/c S = (v'/c)*[1/SQRT(1 - v'^2/c^2)] S = [1/(c/v')]*[1/SQRT(1 - v'^2/c^2)]/c S = 1/[(c/v')*SQRT(1 - v'^2/c^2)] S = 1/SQRT(c^2/v'^2 - 1)
So, we've solved for both R and S. Remembering,
x = R*x' - S*c*t' and c*t = S*x'-R*c*t'
And with R and S in hand, we are set to find:
x = (x' - v'*t') / SQRT(1 - v'^2/c^2) t = (t' - x'*v'/c^2) / SQRT(1 - v'^2/c^2)
And there you have it all.
All this is based on translation and not upon systems K that rotate with respect to K' or are under some acceleration, gravitational or otherwise. That is left for the general theory of relativity to deal with, where it makes the assertion that there is no difference between acceleration and gravity, from the point of view of the observer, and thus that inertial mass (which relates to acceleration) must be the same as gravitational mass (which, of course, relates to gravity) in a
1:1 relationship. With that assumption in hand, all the fun begins.
I read in sci.electronics.design that Kevin Aylward wrote (in ) about 'Wormhole theory', on Mon, 28 Mar 2005:
I didn't notice that; I'll read it up.
--
Regards, John Woodgate, OOO - Own Opinions Only.
There are two sides to every question, except
'What is a Moebius strip?'
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk
I'd get that if you explained the apostrophe. I know it has something to do with derivatives, but I don't know how to take derivatives quite yet. I'm getting up to the level at which I can learn to though.
Something to keep in mind about calculus is that it is really VERY simple, in basic concept. The older books I used to learn from made it lots more complex that it needed to be. I think there may be some newer ones based on something that I discovered entirely for myself, but then discovered that others had already discovered it -- namely "non-standard analysis." (Abraham Robinson)
But the very simple way to view things in calculus is that it is just like algebra, except that you get a new kind of special variable type that adds to what's already in algebra. This special kind of variable, unlike the usual ones, is ONLY ALLOWED to hold infinitely small values (which aren't exactly 0, but are smaller than any finite value.) In other words, in algebra, variables can only hold finite values. In calculus, you get to keep that type of variable and add a new one that can only hold infinitesimal values.
Take a look at a falling body under gravitation, assuming that at the start of time (t=0) the velocity is zero (v=0.) Now you release the object and it begins to fall. Under normal algebra, you often find the simple formula that looks like:
D = v * t
So that the distance some object travels is the speed it is going times the time it went at that speed. But this, of course, assumes that the speed is the same for the entire time or that the speed given is a useful average, at least.
In the case of a falling object, though, the velocity is constantly changing. So what do we use for V? It's never the same twice.
Now the cheating answer is to simply say that under constant acceleration, that the average velocity must be 1/2 of the final velocity and we can compute that final velocity as g*t. So by that, we can say:
D = [(1/2)*g*t]*t = g*t^2/2
Which is correct.
But if you want to think about things from a calculus perspective (a better one, in general), then what you do instead is ask yourself:
"Well, since the velocity is constantly changing all the time, exactly when is the velocity a precise value and not some average thing?"
In other words, looking at D=v*t, ask yourself for how long an exact velocity is still exact. If you think about that, you have to answer, "It is a particular finite value only for an instant of time."
Once you say this to yourself, now ask.. "Well, if the velocity is exact only for an instant, how far will it travel in that instant?"
And to this, you answer yourself, "For a very small distance, namely only an infinitely small distance." Kind of, well, an instant of distance. So you write:
dD = v * dt
Which is to say that the object travels an infinitely small fraction of the total distance, computed by multiplying the instantaneous and finite velocity at some moment by the duration of that moment, and that is itself infinitely small.
It makes sense that when you multiply a finite value by an infinitely small one, that you should also get another infinitely small result. Right?
Now remember, that dD isn't two things, but one thing. It's just a variable. We could have called it Q, if you prefer. But we need some way to keep track of the fact that it can ONLY HOLD infinitely small values. So if we just remember the little-d in front as kind of modifying the variable name so that it helps us remember this, then we are fine. So dD is a variable and dt is a variable. They are different variables, too.
In any case, keep in mind that these tiny, infinitesimally small values are NOT necessarily the same as each other. Just as two different variables in algebra are not necessarily the same as each other, despite both representing some finite value. In like fashion, the variables dD and dt, although each are infinitely small, they can be different from each other and yet both infinitely small.
It's perfectly possible to relate one infinitely small variable to another, like this:
dD/dt
In this case, all we are doing is dividing one infinitely small value by another. And this often produces a finite value. For example, it could be the case that:
dD/dt = 1
In this case, the two variables track each other in a 1:1 relationship. They are equally infinitely small. But we could have written:
dD/dt = 2
In this case, dD is twice as big as dt. They are both infinitely small, but it is still true that dD is twice as big as dt.
Now, as you already know about velocity, it is:
v = D/t
Well, an even better definition that works for objects that don't always travel at uniform speed, is this:
V(t) = dD/dt
In this case, we are just saying that the speed at some moment (t) is best computed as the instantaneous distance it traveled divided by the instant of time it traveled. This is always true, even for things that aren't always moving around at the same speed. So in that sense, it is a more powerful and better way to see speed defined. It is a more universal definition. The old one uses averages. The new one uses instantaneous precision.
Of course, that's all just useless unless at some point you can get back to finite values for real measurements.
How many infinitely small distances do you need to add up to make a finite distance? Well, an infinite number of them, of course. So calculus invents a new symbol, the integral sign, to indicate this particular kind of infinite sum.
If you had an infinite number of dD pieces, what does it add up to? Well, D of course! What else?
And like in algebra, what you do to one side of an equation you do to another side. So if you infinite sum one side, you infinite sum the other side, too. This is called "integration."
Going back to:
dD = v * dt
To clear it up and get rid of the infinitesimal variables, we just sum both sides:
SUM(dD) = SUM(v * dt)
Since we also know that v=g*t, we can substitute to get:
SUM(dD) = SUM(g * t * dt)
Now comes the fun part. We already know what the SUM(dD) is, so:
D = SUM(g * t * dt)
That was easy. But what about the rest?
Well, since 'g' is a constant (as given), then we already know that any common factor found in terms of a long sum can be extracted, just as in (a*b+a*c+a*d) = a*(b+c+d). So our sum can be adjusted so that:
D = g*SUM(t * dt)
Okay, now we are stuck. 't' itself is NOT a constant, so we cannot just pull it out. Similarly, dt, although a constant is infinitely small and needs to be treated by the SUM(). So both stay in there.
Any time you see a product, think of areas. In this case, imagine that you have an area that is 't' high and 'dt' wide. That inner part could be seen that way, just fine. So we are summing up an infinite number of tiny areas. What would that look like?
Well, imagine laying each tiny rectangle side by side, so that the width is 'dt' and they are sorted according to their heights, 't'. As in:
| | | | | | .. | | | |
And so on. Each is very slim, dt wide. But when you stack them side by side like that, they all add up to something that is 't' wide. Similarly, how high are they? Well, as we go from 0 to t, they grow from 0 high to 't' high. So in the end, we have a triangle that is 't' wide and 't' high. What is the total area of all that? Well, 1/2 of t^2, of course. So:
D = g*t^2/2
Just like before.
You can also cancel out various variables you see on different sides of the equation.
Getting this back to electronics, which I suppose I should do, you are told that the charge on a capacitor is:
Q = C * V
But if the V is changing, then the Q is changing. We could rewrite it more precisely as:
dQ = C * dV
Just like we did earlier for speed, time, and distance relationships.
And if we want to, we could compare both sides at each moment of time, so that:
dQ/dt = C * dV/dt
Dividing both sides by the same variable is just fine. It looks funny, in a way, but it is just another variable and we can divide both sides by this infinitesimal variable if we want to. Or we cancel them back out by multiplying both sides by dt and get back what we started with.
Just like in algebra.
By the way, dQ/dt is just I (current) so:
I = C * dV/dt
A useful thing to remember.
Anyway, calculus is just algebra with a new type of variable that can only hold infinitely small values. That forces you, at some point later on, to use an infinite sum if you want to see finite values. But you can delay that while you think and cancel things just as in regular algebra and, if you are lucky, there won't be that much left on which to have to use an infinite sum.
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