Grav,

You shouldn't be surprised. This stuff works and is mathematically and physically consistent. What you did was find a Newtonian limit (low v and gx/c^2), and we know that relativity reduces to Newton-Galileo in that limit.

Grav, this is established theory that has been around for 100 years. It works. It is mathematically consistent and its predictions agree with empirical observation.

A recent paper by Nordtvedt that purports that lunar laser ranging data has verified GR's frame dragging predictions to the 0.1% level (we don't need no stinkin' Gravity Probe B says Nordtvedt), has given me something that I like to say, now:

The trajectory of the moon agrees with GR's predicted course to within *2cm* According to Nortdvedt, the frame dragging effects (and this is not "spin-spin" stuff of precessing spinning bodies, but the gravitomagnetic effects of the earth and moon's mass current in orbit around the sun) cause a perturbation of the moon's orbit of about 6m amplitude from the other terms. That's a tiny effect, but the precision of measurement is apparently so great as to easily to see it. So 2cm/6m is where the ~0.1% comes from.

Now, grav, wouldn't you say that is pretty darn good? All that comes from the same well as we're dealing with here, the way space and time work, and is a much rigorous test of how relativity theory says it works. And it comes through with flying colors.

So rather than fighting it every step of the way, accept it. If something doesn't seem right, that means you've made a mistake, got the wrong end of some stick as Grant said.

-Richard

I think what surprises me most is that GR cannot be worked out by integrating over the SR effects, as I've heard so much, at least not without incorporating Rindler. That is, it must always give a positive time dilation when acceleration is taken into account, so it cannot be worked out just by integrating for the time dilation over instantaneous velocities, which would only give a lesser value for the time of the other, but instead must include Rindler, although Rindler itself can be worked out through SR, so maybe it's just two halves of the same coin that play against each other to cancel out the effect. Whether it is or isn't, though, and however one might think about it, it does seem the way this must be performed for GR to work out correctly.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

As far as that last question goes, if two observers travel symmetrically away from a stationary observer at the end of a long line of stationary observers which we consider to be the x axis, so that the two moving observers are travelling at the same angle to it from the perspective of the stationary observer, then the relative speed between the moving ones will be different than what the stationary observer sees. Let's say all three see each of the others travelling away from them at the same rate, and each sees the others as travelling away at a relative speed of .1c. Normally, that would make a 60, 60, 60 degree triangle with sides of equal length, regardless of speed. In this case, however, the speed each sees between the other two can be found by perfoming the formula for adding speeds. We know the resulting relative speed should be .1c (w) between each two, so the observed speed in opposite directions from the symmetrical axis to the third observer in each case (u, v, where u=v) is

w = (u + v) / (1 + uv/c^2)

w = 2v / (1 + v^2/c^2)

(1 + v^2/c^2)*w = 2v

w + w*v^2/c^2 = 2v

(w/c^2)*v^2 - 2v + w = 0

v = [2 +/- sqrt(4 - 4w^2/c^2)]/(2w/c^2)

v/c = (c/w) * [1 - sqrt(1 - (w/c)^2)]

So for w=.1c, we find v=.05012563c . So to find the angle each observes between the other two, we simply take 2 * sin-1 (.05012563/.1) = 60.1663006 degrees, so the angle changes. They will each also see the distance between the other two as (2 * .0501563)/.1 greater than the distance between themselves and either of the others. now, if we take one of the moving observers and place them at the coordinates the stationary observers see them at, which should be the same as they see the stationary observers, each of the moving observers will see the other at this greater angle as well, so the other moving observer will appear to have travelled a greater distance, then, but at a different angle, and they will not be observed at the same x coordinates.

This, however, raises another question, if I have done this correctly. The time dilation should mean that each observes the other at a lesser time, and therefore, presumedly, at a lesser distance travelled over that lesser time. But each sees the others having travelled a greater distance from each other than they see themselves from each of the others, even though they see a lesser time passing for them. How is this possible?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

That's another one of those statements of yours I can't really construe, but it sounds like you've got the wrong of the stick again. An inertial observer (in flat space time) never sees any clock speed up. It is indeed a matter of integrating the instantaneous velocities. I repeat, an inertial observer never sees any clock rates higher than his own.

Now, a non-inertial observer, however, can very well see clock rates exceeding his own.

If we are inertial, and see two guys accelerate toward each other as you've described, we can calculate what their clocks read when they meet in the middle by doing nothing but integrating dtau = dt/gamma. And since it's the same event, we know they both see each other at that same clock reading.

However, if we want to do it from the perspective of the accelerating observer himself, we must use his non-Minkowski metric, Rindler in the case of constant proper acceleration.

Rindler is simply the name of the frame and metric of that observer, grav. Nothing special about it. It's just a particular coordinate system.

-Richard

Grav,

Now, for the second post there, I'm not sure exactly what you're puzzled over. But remember that formula for speed applies only to the 1 spatial dimension case, or coordinates chosen so that one axis is along the direction of relative velocity.

When you have three non-colinear velocities, you have to use the full
1T-3D version of the Lorentz transform. Velocity transforms get a bit more complex there.

Any text or reference on SR, such as that Wiki link I gave you will give you that full transform. It's rather tedious looking (except in the 4-vector form, where it is quite simple looking, save for when you split it out in components. ).

So, you need to do you equilateral triangle velocity transforms using that full version of the transform.

-Richard

Yes, I know. I was referring only to accelerating observers, as with the two observers that accelerate toward each other, where the other's clock may tick faster than their own. But even then, they will never actually see the other's clock read a greater time than their own, as we discussed earlier, because once the time of flight of light effects are figured in, the other's clock will always lag behind that of the observer's, right?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I'll post this here also since it is related to our discussion. In the "throwing a ball" thread, we considered two observers moving inertially toward each other with some relative speed between them and one throws a ball to the other when they are at some distance from each other. The catcher sees the ball travelling over the entire distance from the point of release while the thrower sees the ball only travel part of the distance since the catcher intercepts it while also travelling toward the ball. So each see a different distance travelled by the ball but reads the same for the time of travel.

Here I consider that same scenario for a pulse of light travelling from one to the other. Does this sound about right to everyone?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

I was hoping one of the others would tackle this last post, since it makes my head hurt to try to figure out what you're doing and thinking some times.

In there, you're talking about distances a light pulse travels. That is different between inertial frames. You can not think about this in a Galilean manner.

Grav, please read a derivation of the Lorentz transform starting out with the postulate that all observers see the speed of light the same.

Consider this. We mark off a length of 2L using our rod and sit in the middle. At each end we place a light "dinger" that registers when a light pulse hits it. Now, from the middle of that 2L distance, we fire a light beam in each direction. The light has to travel a distance of L, and the time of flight is then L/c. So, what does that mean?

It means the light beams hit the ends at the same time. Those events are simultaneous. And that allows us to synchronize clocks, start remote stop watches at the same time so to speak by sending light pulses and adjusting our zero points for the L/c time delay.

Are we hunky dory with that? Each end receives the light pulse at the same time.

Now, while we're doing that Rocket Man comes flying by at a velocity of 0.9c. He just happens to be right next to us in the middle of our laid off length when we fire that pulse.

Now, what does he see? He sees the speed of light as c in his frame, remember. So he sees the fore end of our rod moving *toward* the light pulse. He sees the other end of our rod moving *away* from that light pulse.

So what must happen according to him? The light must get to the fore end *before* it gets to the aft end. Those events cannot be simultaneous if both observers measure the speed of light to be the same thing, c. According to him, since we were moving, we screwed up our synchronization procedure.

That is the root of it all. Note that says nothing about relative clock *rates*, nor relative notions of length. This is about simultaneity. Two events that happen at the same time in one frame cannot happen at the same time in the other frame because both see c as constant.

Now, all else, time dilation and length contraction, follow logically from the above. The Lorentz transform is how space and time must "work" between those observers in order to be in sympathy with the above, to make it so.

-Richard

How about "Two *distant* events that *appear to* happen at the same time in one frame cannot happen at the same time in the other frame because both see c as constant."

In this case the distance is small, but it holds.

Maybe it's better to say two events with some real distance between them - which can be made infinitisemally small. Then the perceived times separating them approaches a limit at zero for all observers.

John,

Yes. To state that precisely, two events spatially separated along the line of relative motion that are simultaneous in one frame can NOT be simultaneous in another frame.

And by different frames, I mean two frames in relative motion in flat space-time. This is quite "real" for moving frames, because c is invariant.

This is not any light delay trick, as one could construe from the light travel time. Consider a family of stationary observers spatially separated. Light from spatially events will reach those at different times and so one might say they "appear" to occur at different times. But, if they have synchronized their clocks with each other, they can correct for the time of flight and find they all say the events occurred at the same times.

If the two events are simultaneous, they are simultaneous to all those stationary observers.

But this cannot be the case for observers in relative motion. Once they correct for time of flight, they still disagree on when those events happened according to their own sense of "now". And that is the thing one needs to appreciate. The invariance of 'c' across moving frames has this consequence. It's not a "trick" of light travel delay.

-Richard

You know, this is quite important. The relativity of simultaneity is *NOT* a trick of light travel time, and I think there is some confusion on this point.

Consider an observer on the mars and the other on earth. Relative velocities and gravitational effects are so weak as to be negligible for all but the most precise purposes. We can pretty much consider we're Galilean there without much error.

Now, those observers synchronize clocks, and both agree to blow up bright flash bomb at the same time. They set the timers and move a safe distance away. Now, the observer on Mars is going to see the light from his little flash bang long before he sees the light from the one on Earth. And vice versa.

That is not anything unexpected. And the relativity of simultaneity has nothing to do with that. I think some get in their heads it does. Galileo predicts that with a finite speed of light. Each observer synchronized his clock and so they occured at the same time, and each can verify that knowing the distance the light had to travel and the reception time. He can then say, "yep, the other one went off the same time mine did".

Now, Rocket man again is whizzing through at 0.9c on heading from Mars to Earth. When the flash bang on Mars goes off, its light travelling towards the earth takes a shorter time and travels less distance than the corresponding light pulse from earth.

Why? Because Rocket Man sees the speed of light as the same c relative to him as we do! The light from Mars leaves and heads towards earth at c. But earth is whizzing toward that light pulse at 0.9c. In the other direction, the light from earth leaves, travelling at c, but Mars is running away from it at 0.9c. The earth to mars light takes a longer time and travels a longer distance than mars to earth light.

That is the root of it all. And note that is not a light travel time effect. That is what *must happen* for Rocket Man if he is to see the speed of light the same as we do. There is no time dilation or length contraction, no doppler shift and all that coming in to play (although those follow logically) in that situation. That just the way it must be if Rocket Man's concept of distance and time are to be as good as ours, and his yardstick and his clock are as good as ours. Rocket Man is going to say that the flash on earth occured before the flash on Mars in order to sastify what that light must do in his frame.

Who is really moving, there? There is no way to answer that. There is only relative motion, and no way to say whose clock is "correct" and whose is screwed up somehow by motion.

-Richard

Please read it again and you'll see that is exactly what I have done. I have only considered it from both frames seeing the pulse travelling at c. I have not included time of flight effects in this even though the light is actually in flight, which sounds contradictory, but you can see the difference in the resulting formula for v=0, which makes to = t' regardless of the distance between observers. If time of flight were included in that, it would be to = t' - d/c. In other words, simultaneity effects can be seen through a different rate of time while flight of light effects can be seen through a lag in time. And according to that formula, different frames will disagree about what is simultaneous.

Oh, and it turns out, then, that Relativistic Doppler is purely simultaneity effects as well, with no time of flight involved, so I can use it for this. The reason is that it represents the difference in the rate of times observed, as the frequency observed would match the rate of time one would observe passing for the other. If flight of light effects were also included in this, one would also observe a lag in time over the distance.

The interesting thing, though, is that flight of light effects are not considered in Relativistic Doppler since we only want the ratio of the rates of times and frequencies, not the lag. But the lag would only be constant to stationary observers, not ones with a relative speed between them, because the distance between them would continually change, and then so would the amount of lag time, which itself would contribute to the rate of the times because of that continual change in lag.

So let's try something. Let's add in the flight of light effects to the simultaneity formula I came up with for each observer according to the other, switch frames, and subtract the flight of flight for the initial observer, if that makes sense, to see it in a round-about way from the other's perspective. This changes the rate of times due to the change in distance per time. The change in distance per time is just the relative speed, so the change in lag times is proportional to this. For the receiver, then, the rate of times for the flight of light is just tr = (v/c)(d/c). For the emitter, where the receiver contributes to part of the distance travelled, the rate of time for the flight of light is te = (v/c)(d/(c+v)). So adding that of the receiver to the formula, whereas 'to' was initially according to that of the emitter, but can be applied either way for either observer otherwise, regardless of who actually emits the light, we get to = (t' - (v/(c+v))d/c)/y + (v/c)d/c. Then switching frames, we find t' = y(to - (v/c)d/c) + (v/(c+v))d/c, and subtracting that observer's own flight of light observation, we arrive at the simultaneity formula "according to the other" of t' = y(to - (v/c)d/c). The formulas now match, even though we had to work through time of flight effects for what each actually sees to get there.

Now, something interesting. If Relativistic Doppler does not account for the changing flight of light effects, which would affect the rate of times overall, then it does not account for what one would actually observe, except over small distances, where a changing flight of light is neglible. Over very large distances, however, time of flight effects would become more noticable, then. Can anyone say "Pioneer anomaly"?

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

In the last post, I found that Relativistic Doppler does not include flight of light effects as far as the lag of time is concerned. It turns out that Relativistic Doppler, however, is only concerned with the ratio of the rate of frequencies in the first place, whose inverse matches that of the ratio of rates of the times observed. So Relativistic Doppler only depends upon a change in the lag per time for observers that always see light travelling at c, which is already incorporated into the formula, as well as the time dilation involved, not the actual lag itself, even though such a lag also exists for the time of the flight of light effect, but makes no difference for the ratio of the rate of times observed.

In other words, the actual lag is distance dependent. But the change in lag per time is not. This is because the change in lag depends upon the change in distance only, so it doesn't matter what the specific distance from the observer is as it would with lag, only the amount of change, and this is what we were adding and subtracting when switching frames before. The change in distance per time is proportional to the relative speed, so this and time dilation are all we are concerned with when it comes to Relativistic Doppler.

So that means that there is no correction over a large distance when it comes to Relativistic Doppler. For a constant relative speed, the change in lag and time dilation are also constant, and those are all we need to find the ratio of the frequencies. So if there is a correction to be made for Relativistic Doppler as far as the Pioneer anomaly is involved, it would not be due to lag or a change in lag, but to a change in the change of lag. Since the change in lag is proportional to the relative speed, then a change in the change of lag would depend upon a changing relative speed, an acceleration or a deceleration. I'll work through some integrations for this and see if I can come up with anything.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Well, I have been attempting to incorporate the equations for the relative speed and distance obtained with a constant acceleration to the simultaneity formula, and I have been getting strange results that include some lag, but I'm thinking now that a changing lag rate should only lead to a ratio of frequencies for Relativistic Doppler that varies directly with the instantaneous relative speed, since the rate of frequencies observed is precisely analogous to the ratio of the rate of times to the observer as the pulses are received. So I guess I can forget about the Pioneer anomaly for now, at least until I see something specific, and just concentrate upon trying to rework Relativity from the ground up. Sounds like fun. I want to thank everyone that has contributed to this thread for laying down the groundwork for this.

I have also found so far, though, that Relativity can only be applied as gamma, whether one thinks about it in terms of Lorentz contraction or time dilation. For example, if one compares the time of flight of light along a moving rod with clocks at each end that are synchronized with those of a stationary system at any instant at the places where they happen to be, as Einstein does in his 1905 paper, then we will find that the length of the rod has been shortened by a factor of gamma to the stationary system, due to the relative speed between them. However, we also know that clocks on a moving rod cannot be continually synchronous to the stationary system, even at the same instant in the places where they happen to be, because any relative speed will produce a time dilation between them, so each successive stationary clock that is passed, although synchronous to each other in the stationary system, will read a lesser and lesser time to the moving observer's clock. So we can think about it in terms of a contraction of length when time is continually measured according to the clocks of the stationary system, or we can think about it in terms of time dilation, where one would observe a lesser distance travelled in the other frame, contracted by gamma, but in accordance to the time that has passed in the other frame, so that one would see the full distance travelled for the full time on the other's clock, even though our own clock would read a greater time and so we might expect a greater distance travelled. So we can use one or the other, time dilation or contraction, but not both.

I also know that I am speaking in terms of distance travelled, not the length of an object, when it comes to contraction, but one can also see this with the initial M-M experiment. The path of the light beam travelling at a right angle to the central mirror does not experience this contraction in any way associated to the apparatus, as viewed by an observer with some relative speed, yet its path as viewed by an observer with some relative speed to the apparatus will appear at some angle whose distance in the forward direction must also be contracted. I do not think that we can consider the path of that light beam can be rigid in any way, in the way we were discussing in this thread, so the contraction must take place over the entire distance travelled in the forward direction to an observer with a relative speed.

Either way, then, if we consider contraction to take place over the entire distance travelled, or only that a time dilation has taken place, so that that lesser distances are travelled over lesser times in the other frame, IMHO, the rope does not break, regardless of tensile forces, Rindler, where the thrust is located upon the rockets, and so forth ( ducks and runs for cover ) . And since time dilation is observed to continue even after an object is accelerated and then brought to rest, resulting in an overall lag of times, while the observed distance travelled varies greatly between frames anyway, making it relative to the frame in which it is observed, I would say that time dilation is the best way to go overall with Relativity, so we can basically forget about the contraction altogether, except, of course, as a direct result of the time dilation observed in another frame, nothing else.

I will continue to attempt to work through the rest of what I have now begun, working my way up from simultaneity to relative speeds and distances involving acceleration, considering only time dilation issues, and see where it leads. I will base my formulas only upon two postulates, that the speed of light is measured the same regardless of the frame of reference, and that the physics of any observer remains the same in every frame. By this last postulate, I mean that it will be assumed that every observer sees the same thing using the same relative speeds and so forth, and the formulas will never include any formulas that deduce the "reverse psychology" of what one might think another measures by switching frames by simply rearranging the formulas, which I have never agreed with because the other would observe the same thing as the initial observer, and therefore becomes the observer, and should therefore go by the same original formulas, not the rearranged version, except in very specific asymmetric cases, as where one might physically accelerate while the other does not.

Again, I thank you all for your help here. When I get a little further with this, I will begin another thread.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I thought about this some more last night and realized that the central mirror would also Lorentz contract in the way we are considering in this thread, shortening the length of the mirror in the forward direction while keeping the same length cross-wise, and so changing the angle of the central mirror altogether, and therefore also changing the angle the light is reflected. The shortening of the path of the light beam travelling at a right angle to the direction of motion of the apparatus might be directly associated with the apparatus after all, then, caused by being reflected at a smaller angle from the central mirror, and it is possible that the physics of the angle of incidence to the angle of reflection might change somewhat anyway for an angled mirror with a relative speed. So it looks like I may have been just a bit too quick in saying that the rope does not break again, although I am still leaning that way, but I guess I will just have to keep working through the formulas the best I can from the ground up to find out.

__________________
Let's put together the pieces of The Grand Puzzle . (website)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."