How to make two frames purely Galilean

  • #1

Main Question or Discussion Point

It is possible to synchronize two inertial frames with a relative speed between them such that what they measure of each other is purely Galilean. That is, according to each of the frames, they will measure no time dilation of each other, so their clocks will tick at the same rate, no length contraction, so they will measure the same lengths within each frame, and they will measure no simultaneity difference between frames. This will be done in a natural way, without tampering with the rate of time passing upon clocks or the lengths of rulers, but only depending upon how the clocks within each frame are synchronized. The first section below is mostly taken from another thread. From there, we will proceed to make the two frames Galilean.

Part l

Let's say we have two ships in frame A that are stationary to each other. One of the ships then accelerates to frame B along the x axis and cuts it engines, so that it is now travelling inertially. The speed of ship B is v as measured by A and v' is the speed of ship A as measured by B. Each frame, using their own rulers, measures the proper length of their own ship to be d. Frame A, however, using their own rulers and clocks, will measure a length contraction of L for ship B and frame B will measure a length contraction of L' for ship A. Likewise, each frame will observe a time dilation of the other ship's clocks of z for what A measures of B and z' for what B measures of A. We will place a clock at the front and back of each ship. Each frame observes that the clock at the back of the other ship reads some greater time than the clock at the front, tl for what A observes of B and tl' for what B observes of A.

Okay, so now we want to know how each ship will measure the other. Using A as our frame of reference, where A measures v, z, L, and tl of B, let's find the speed that B measures of A. B measures the speed of A as the ratio of the distance an observer in frame A travels in some time measured between B's clocks. If observer A travels the length of B's ship, then according to frame A, that is equivalent to the length of ship B travelling past observer A, which occurs in a time of (L d) / v according to A's clocks. According to A, however, B's clocks are time dilating by a factor of z, so a time of z L d / v passes upon B's clocks. The clock at the back of B's ship is also set to a greater reading of tl as observed by A, so the time that B measures passing between the clocks at the front and back of B's ship is z L d / v + tl over a distance of d, so B measures the speed of observer A to be

1) v' = d / (z L d / v + tl)

Now let's find the time dilation that B measures of A. Observer A carries a clock and travels the length of B's ship as before. Again, the time that passes upon A's clock is L d / v, while the time that passes according to B's clocks is z L d / v + tl, so the time dilation that B measures of A's clock is

2) z' = (L d / v) / (z L d / v + tl)

For measuring the length of A's ship, B observes ship A to pass a single clock on ship B. Multiplying the time that passes upon that single clock by the speed B measures of ship A gives the length of ship A as measured by B. An alternate method would be to mark off the front and back of ship A upon a ruler simultaneously in frame B according to clocks at both places, but this method is much more complex than the former method as found from the reference frame of A, so we will use the former method, although both are compatible. Frame A measures a time of d / v for a single clock on ship B to travel the length of A's ship. A also measures the single clock in frame B to be time dilating by a factor of z, so a time of z d / v passes upon the single clock in frame B. The length of ship A as frame B measures it, then, is (L' d) = v' (z d / v), or L' = z v' / v, where v' is found in equation 1, giving

3) L' = (z d / v) / (z L d / v + tl)

Finally, we will find the difference in readings that B measures between the clock at the front and back of ship A. Ship A passes a single clock in frame B. From A's perspective, the single clock travels the length of ship A. Clock B and the clock at the front of A's ship are synchronized to T=0 when they pass. When clock B reaches the clock at the back of A's ship, a time of d / v will have passed within frame A, so this what the clock at the back will read while B's clock reads z d / v when they coincide. According to B, A's clocks are time dilating by a factor of z', so with this fraction of the time that passes for clock B, and therefore only a time of z' z d / v has passed for A's clocks according to B, although the clock at the back of ship A reads d / v. Frame B, then, says that the clock at the back of A's ship is set to a greater reading than the clock at the front of tl' = d / v - z' z d / v, whereby taking z' from equation 2, we get

4) tl' = d / v - z' z d / v

= (d / v) (1 - z' z)

= (d / v) [1 - (z L d / v) / (z L d / v + tl)]

= (d / v) [(z L d / v + tl) - z L d / v] / (z L d / v + tl)

= (d / v) tl / (z L d / v + tl)

Examining the mathematical relationships between equations 1 through 4, we find that they all have the same denominator. If we further set K = (d / v) / (z L d / v + tl), we find from equation 1 that v' = K v, that z' = K L from equation 2, L' = K z from equation 3, and tl' = K tl from equation 4. Rearranging gives us

5) K = v' / v = z' / L = L' / z = tl' / tl

Part ll

So now we want to synchronize frame B such that no length contraction is measured of ship A. That is, frame B will measure d for the proper length of ship B and will measure d for the length of ship A as well. In that case, we would have L' = 1. So from equation 5, we have

K = (d / v) / (z L d / v + tl) = L' / z

K = (d / v) / (z L d / v + tl) = 1 / z

z d / v = z L d / v + tl

tl = z (1 - L) d / v

So if we synchronize the clocks of ship B so that frame A views the clock at the back to read a greater time of tl = z (1 - L) d / v, then frame B will measure no length contraction of ship A. Since we also have K = 1 / z, then we also gain

K = 1 / z = v' / v, v' = v / z

K = 1 / z = z' / L, z' = L / z, which for SR and Galilean kinematics alike, z = L, so both become z' = 1

K = 1 / z = tl' / tl, tl' = tl / z = (1 - L) d / v

Frame B now measures no length contraction or time dilation of frame A, but still so far measures a simultaneity difference, so now let's also synchronize frame A in the same way to measure no time dilation or length contraction of frame B. Let's say there is another ship Aa that is identical to ship A and that they lie side by side. We will change the synchronization of ship Aa by adding tla to the back, where A and Aa's ships face the opposite direction from that of ship B. Let's find the speed that ship Aa now measures of ship B. The front of ship B passes the front of ship Aa, a time of d / v later, it passes the back of ship Aa according to ship A. But the clock at the back of ship Aa is set to read tla later, so the time according to ship Aa is d / v + tla, and the speed of ship B according to ship Aa is

va = d / (d / v + tla)

Similarly, the time dilation measured of ship B by ship Aa is found by taking the ratio of the time that passes upon a single clock of ship B to the time that passes between the clocks of ship Aa, which again is tla greater than the time that passes for ship A, so

za = (z d / v) / (d / v + tla)

The length that ship Aa measures of ship B is found as the speed that Aa measures multiplied by the time ship B takes to pass. The time ship B takes to pass a single clock on ship Aa is the same as with ship A, which is L d / v, so

La d = va (L d / v)

La / L = va / v

and we want to synchronize such that La = 1, whereby

va = v / L = d / (d / v + tla)

d + tla v = L d

tla = (L - 1) d / v

from which we also gain

za = (z d / v) / (d / v + tla)

= (z d / v) / (d / v + (L - 1) d / v)

= z / (1 + (L - 1))

= z / L = 1

So if ship Aa adds a time of tla = (L - 1) d / v to the back of the ship, then ship Aa will measure no length contraction or time dilation of ship B and so if then the whole of frame A synchronizes in the same way, then frame A and frame B will measure no length contraction or time dilation of each other. Coming back to the simultaneity shift observed by each, before frame A re-synchronized, frame B observed a simultaneity difference of tl' = (1 - L) d / v between the clocks of ship A. But ship A then added (L - 1) d / v to the rear clock, so frame B now observes no simultaneity difference for ship A whatsoever, and for the whole of frame A, being synchronized in the same way, nor does frame A observe a simultaneity difference for frame B. What each frame observes of the other is now purely Galilean.
 

Answers and Replies

  • #2
Vanadium 50
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You cannot make two frames Galilean. At best you can make a subset of events in those frames Galilean. I also don't completely understand what you are doing, but "adding time to clocks" is not allowed if you want time to be the thing that clocks measure.
 
  • #3
You cannot make two frames Galilean. At best you can make a subset of events in those frames Galilean.
Each frame measures the same time for events as the other, the same lengths for objects as measured from either frame, and no simultaneity difference between frames. When synchronized in this way, the Lorentz transformations between the two frames become just t' = t and x' = x - v t as they would for Galilean transformations.

I also don't completely understand what you are doing, but "adding time to clocks" is not allowed if you want time to be the thing that clocks measure.
Rather than synchronizing using the Einstein simultaneity convention to measure light isotropically, the clocks are synchronized between the frames such that L = L' = 1, so there is no length contraction observed between frames. We can start with the frames being synchronized using the Einstein simultaneity convention and then change them accordingly. If z = L, then za = z' = 1 also, so there is no time dilation observed, and after re-synchronizing, neither frame observes a simultaneity difference for the other as well.
 
  • #4
JesseM
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It is possible to synchronize two inertial frames with a relative speed between them such that what they measure of each other is purely Galilean. That is, according to each of the frames, they will measure no time dilation of each other, so their clocks will tick at the same rate, no length contraction, so they will measure the same lengths within each frame, and they will measure no simultaneity difference between frames. This will be done in a natural way, without tampering with the rate of time passing upon clocks or the lengths of rulers, but only depending upon how the clocks within each frame are synchronized. The first section below is mostly taken from another thread. From there, we will proceed to make the two frames Galilean.

Part l

Let's say we have two ships in frame A that are stationary to each other. One of the ships then accelerates to frame B along the x axis and cuts it engines, so that it is now travelling inertially. The speed of ship B is v as measured by A and v' is the speed of ship A as measured by B. Each frame, using their own rulers, measures the proper length of their own ship to be d. Frame A, however, using their own rulers and clocks, will measure a length contraction of L for ship B and frame B will measure a length contraction of L' for ship A. Likewise, each frame will observe a time dilation of the other ship's clocks of z for what A measures of B and z' for what B measures of A. We will place a clock at the front and back of each ship. Each frame observes that the clock at the back of the other ship reads some greater time than the clock at the front, tl for what A observes of B and tl' for what B observes of A.

Okay, so now we want to know how each ship will measure the other. Using A as our frame of reference, where A measures v, z, L, and tl of B, let's find the speed that B measures of A. B measures the speed of A as the ratio of the distance an observer in frame A travels in some time measured between B's clocks. If observer A travels the length of B's ship, then according to frame A, that is equivalent to the length of ship B travelling past observer A, which occurs in a time of (L d) / v according to A's clocks. According to A, however, B's clocks are time dilating by a factor of z, so a time of z L d / v passes upon B's clocks. The clock at the back of B's ship is also set to a greater reading of tl as observed by A, so the time that B measures passing between the clocks at the front and back of B's ship is z L d / v + tl over a distance of d, so B measures the speed of observer A to be

1) v' = d / (z L d / v + tl)
This might be a little bit easier to follow with some numbers (for me, anyway!) Suppose each ship has a length of d = 30 light-seconds in their own frame, and in A's inertial frame B is moving at a speed of 0.6c, so the length contraction factor which you identify as L (somewhat confusing notation since L is usually a length rather than a factor, but never mind) would be 0.8, and the time dilation factor which you call z would also be 0.8. In this case, it's true that according to A, it takes a time of (L d) / v = 0.8*30/0.6 = 40 seconds between passing the two ends of B's ship. And it's true that in A's frame, only z times this amount of time passes on B's clocks, or 0.8*40 = 32 seconds. So if the clock at the back is ahead of the clock at the front by tl seconds in A's frame, then at the moment A passes the front the clock at the front reads 0 while the clock at the back reads tl, so then at the moment A passes the back the clock at the front reads 32 while the clock at the back reads 32 + tl, or (z L d / v) + tl. So in B's non-inertial frame (non-inertial because B is using a nonstandard simultaneity convention) the speed of A must indeed be d divided by this time, i.e. 30 / (32 + tl).

grav-universe said:
Now let's find the time dilation that B measures of A. Observer A carries a clock and travels the length of B's ship as before. Again, the time that passes upon A's clock is L d / v, while the time that passes according to B's clocks is z L d / v + tl, so the time dilation that B measures of A's clock is

2) z' = (L d / v) / (z L d / v + tl)
Yes, here (L d / v) = 40 seconds, the time that passes on A's clock traveling front to back, and (z L d / v) + tl = 32 + tl, the time on B's clock at the back when A reaches the back (I assumed the time on B's clock at the front was 0 when A passed the front, and A's clock read 0 there too).

Later you are going to decide to synchronize the clocks so tl = z (1 - L) d / v, which in this case would be 0.8*(1 - 0.8)*30/0.6 = 8 seconds, so B measures a time of 40 seconds for A to pass from the front end to the back, which is the same time that A measures to go from front to back (so no time dilation on A's clock in B's non-inertial frame, meaning z'=1). And since B measures A to take 40 seconds to travel the distance of the ship which is 30 light-seconds in B's non-inertial frame, B must measure A's speed to be v' = 0.75c according to your simultaneity convention.
grav-universe said:
For measuring the length of A's ship, B observes ship A to pass a single clock on ship B. Multiplying the time that passes upon that single clock by the speed B measures of ship A gives the length of ship A as measured by B. An alternate method would be to mark off the front and back of ship A upon a ruler simultaneously in frame B according to clocks at both places, but this method is much more complex than the former method as found from the reference frame of A, so we will use the former method, although both are compatible. Frame A measures a time of d / v for a single clock on ship B to travel the length of A's ship. A also measures the single clock in frame B to be time dilating by a factor of z, so a time of z d / v passes upon the single clock in frame B. The length of ship A as frame B measures it, then, is (L' d) = v' (z d / v), or L' = z v' / v, where v' is found in equation 1, giving

3) L' = (z d / v) / (z L d / v + tl)
Yes, in A's inertial frame, since B's clock at the front of the ship is moving at 0.6c and A's ship is 30 light-seconds long, it must take d/v = 30/0.6=50 seconds for the clock at the front of B's ship to go from the front to the back of A's ship, during which time B's clock only ticks forward by (z d / v) = 0.8*50 = 40 seconds. Then we multiply by the speed of A in B's non-inertial frame to get the length of A's ship in B's non-inertial frame, with the speed being v' = d / (z L d / v + tl)...multiplying by (z d / v) gives a length of:

(z d^2 / v) / (z L d / v + tl)

And L' is the "length contraction factor" in B's non-inertial frame, i.e. this length should be equal to d L', so dividing both sides by d gives your equation above. If B's clock ticks 40 seconds between passing the front and back of A's ship, and in B's non-inertial frame A is moving at 0.75c, then the length of A's ship in B's non-inertial frame should be 40*0.75c=30 light-seconds, so L' = 1. This fits with your equation, since (z d^2 / v) = (0.8 * 30 / 0.6) = 40 and (z L d / v + tl) = (0.8*0.8*30/0.6 + 8) = 40.

I want to make sure that B would get the same length by the method of "simultaneous" measurements of the back and front of A. If B measures the front of A's ship to pass its back clock when the clock there reads 40 seconds, then since B's ship is 30 seconds long in its own frame, in order for it to measure A's ship as 30 seconds long it should be true that the back of A's ship passes B's front clock when B's clock there also reads 40 seconds. In A's inertial frame, at t=0 the back of A's ship is 30 light-seconds away from the front of B's ship and B's clock reads 0 seconds, then traveling at 0.6c the back of A's ship reaches the front of B's after a time of 30/0.6 = 50 seconds, and B's clock at the front is dilated by a factor of 0.8 so it reads 0.8*50=40 at this moment, so it does work out.
grav-universe said:
Finally, we will find the difference in readings that B measures between the clock at the front and back of ship A. Ship A passes a single clock in frame B. From A's perspective, the single clock travels the length of ship A. Clock B and the clock at the front of A's ship are synchronized to T=0 when they pass. When clock B reaches the clock at the back of A's ship, a time of d / v will have passed within frame A, so this what the clock at the back will read while B's clock reads z d / v when they coincide.
Yes, in A's inertial frame the clock at the front of B's ship will take a time of d / v = 30/0.6=50 seconds to cross from the front of B's ship to the back, during which time the clock at the front of B's ship has elapsed z d / v = 0.8*50=40 seconds.
grav-universe said:
According to B, A's clocks are time dilating by a factor of z'
Which in the above example would be 1
grav-universe said:
so with this fraction of the time that passes for clock B, and therefore only a time of z' z d / v has passed for A's clocks according to B, although the clock at the back of ship A reads d / v.
So in the example, 40 seconds have passed for A's clocks according to B, but the clock at the back of ship A reads 50 seconds assuming it's synchronized with the front one in the standard inertial manner.
grav-universe said:
Frame B, then, says that the clock at the back of A's ship is set to a greater reading than the clock at the front of tl' = d / v - z' z d / v,
So here tl' = 50 - 40 = 10 seconds out-of-sync in B's non-inertial frame, again assuming A's clocks are synchronized in the usual manner of inertial frames.
grav-universe said:
whereby taking z' from equation 2, we get

4) tl' = d / v - z' z d / v

= (d / v) (1 - z' z)

= (d / v) [1 - (z L d / v) / (z L d / v + tl)]

= (d / v) [(z L d / v + tl) - z L d / v] / (z L d / v + tl)

= (d / v) tl / (z L d / v + tl)
Yes, (d / v) tl = (30/0.6)*8 = 400, and (z L d / v + tl) = 40 as before, so this gives 400/40 = 10 for tl' again.
grav-universe said:
Examining the mathematical relationships between equations 1 through 4, we find that they all have the same denominator. If we further set K = (d / v) / (z L d / v + tl), we find from equation 1 that v' = K v, that z' = K L from equation 2, L' = K z from equation 3, and tl' = K tl from equation 4. Rearranging gives us

5) K = v' / v = z' / L = L' / z = tl' / tl
Yes, this all works. In my example we would have K = 1.25, v'/v=0.75/0.6=1.25, z'/L=1/0.8=1.25, L'/z=1/0.8=1.25, tl'/tl=10/8=1.25.
grav-universe said:
Part ll

So now we want to synchronize frame B such that no length contraction is measured of ship A. That is, frame B will measure d for the proper length of ship B and will measure d for the length of ship A as well. In that case, we would have L' = 1. So from equation 5, we have

K = (d / v) / (z L d / v + tl) = L' / z

K = (d / v) / (z L d / v + tl) = 1 / z

z d / v = z L d / v + tl

tl = z (1 - L) d / v

So if we synchronize the clocks of ship B so that frame A views the clock at the back to read a greater time of tl = z (1 - L) d / v, then frame B will measure no length contraction of ship A. Since we also have K = 1 / z, then we also gain

K = 1 / z = v' / v, v' = v / z

K = 1 / z = z' / L, z' = L / z, which for SR and Galilean kinematics alike, z = L, so both become z' = 1

K = 1 / z = tl' / tl, tl' = tl / z = (1 - L) d / v

Frame B now measures no length contraction or time dilation of frame A, but still so far measures a simultaneity difference, so now let's also synchronize frame A in the same way to measure no time dilation or length contraction of frame B. Let's say there is another ship Aa that is identical to ship A and that they lie side by side. We will change the synchronization of ship Aa by adding tla to the back, where A and Aa's ships face the opposite direction from that of ship B. Let's find the speed that ship Aa now measures of ship B. The front of ship B passes the front of ship Aa, a time of d / v later, it passes the back of ship Aa according to ship A.
Yes, in ship A's inertial frame it takes 30/0.6=50 seconds for the front of B to go from the front to back of Aa.
grav-universe said:
But the clock at the back of ship Aa is set to read tla later, so the time according to ship Aa is d / v + tla, and the speed of ship B according to ship Aa is

va = d / (d / v + tla)
OK, and later you derive tla = (L - 1) d / v, in this example (0.8 - 1)*30/0.6 = -10. So the time that B's front reaches the back of Aa is d / v + tla = 50 - 10 = 40, meaning Aa measures B to have a speed of 30/40=0.75c just as B measures A and Aa to have a speed of 0.75c.

One thing worth noting here is that in A's inertial frame the back clock of Aa is set 10 seconds behind its front clock, whereas the back clock of B is set is set 8 seconds ahead of its front clock.
grav-universe said:
Similarly, the time dilation measured of ship B by ship Aa is found by taking the ratio of the time that passes upon a single clock of ship B to the time that passes between the clocks of ship Aa, which again is tla greater than the time that passes for ship A, so

za = (z d / v) / (d / v + tla)
Yes, and the time B's clock ticks forward between passing the two ends of Aa will be 0.8*30/0.6 = 40, and with the above tla we also have (d / v + tla) = 40, so according to Aa's measurements B's clock shows no time dilation.
grav-universe said:
The length that ship Aa measures of ship B is found as the speed that Aa measures multiplied by the time ship B takes to pass. The time ship B takes to pass a single clock on ship Aa is the same as with ship A, which is L d / v, so

La d = va (L d / v)
So, in this example the length is 0.75c * (0.8*30/0.6) = 30, so no length contraction. Let's again double-check that this fits with the method of simultaneous measurements. Aa measures the front end of B to line up with its back end at a time of 40 according to the clock at its back, so the clock at the front of Aa (which matches with A's front clock) should also show a time of 40 when the back of B passes it. And in A's inertial frame, when the front clock reads 0 the back of B is 30*0.8 = 24 light-seconds away, and it's traveling at 0.6c so it takes a time of 24/0.6 = 40 seconds to reach the front of A and Aa, so it does work out.
grav-universe said:
La / L = va / v

and we want to synchronize such that La = 1, whereby

va = v / L = d / (d / v + tla)

d + tla v = L d

tla = (L - 1) d / v
Right
grav-universe said:
from which we also gain

za = (z d / v) / (d / v + tla)

= (z d / v) / (d / v + (L - 1) d / v)

= z / (1 + (L - 1))

= z / L = 1

So if ship Aa adds a time of tla = (L - 1) d / v to the back of the ship, then ship Aa will measure no length contraction or time dilation of ship B and so if then the whole of frame A synchronizes in the same way, then frame A and frame B will measure no length contraction or time dilation of each other. Coming back to the simultaneity shift observed by each, before frame A re-synchronized, frame B observed a simultaneity difference of tl' = (1 - L) d / v between the clocks of ship A. But ship A then added (L - 1) d / v to the rear clock, so frame B now observes no simultaneity difference for ship A whatsoever, and for the whole of frame A, being synchronized in the same way, nor does frame A observe a simultaneity difference for frame B. What each frame observes of the other is now purely Galilean.
OK, for simultaneity let's think about regularly-spaced clocks along Aa, synchronized according to your convention, and how they would look in A's frame. If the front of Aa is at x=0 and the back is at x=-30 in A's inertial frame, then with the back clock 10 seconds behind the front clock in A's frame, this means if you had a row of clocks along Aa synchronized the same way, each time you move forward in the +x direction by 1, the clock there is 1/3 second ahead of the previous clock. So for example if you extended these clocks along a ruler attached the front of Aa, then when the clock at x=0 reads ta=0 the clock at x=3 reads ta=1, the clock at x=6 reads ta=2, and so on until you get to the clock at x=24 which reads ta=8. All of these are the readings at t=0 in A's inertial frame, and we also know that at t=0 the front of B's ship is at x=0 and the clock there reads tb=0, while the back of B's ship is at x=0.8*30=24 and the clock there reads tb=8 (since the back clock on B's ship is always 8 seconds ahead of the front clock in A's inertial frame). So at t=0 it should be true that all B's clocks are synchronized with the clocks of Aa they are next to at that moment, and since Aa's clocks aren't dilated in B's frame and B's clocks aren't dilated in Aa's frame, they should remain synchronized.

All in all, a very neat trick! It's also interesting to note there is even a symmetry in how they measure the speed of light rays. From A's perspective, a light beam traveling from the front to back of B would take 24/1.6 = 15 seconds to get there, and since B's clocks are running slow by a factor of 0.8 and the back one is ahead of the front by 8, B would measure a time of 15*0.8 + 8 = 20 seconds. A light beam traveling back to front would take 24/0.4 = 60 seconds in A's frame, so 60*0.8 - 8 = 40 seconds as measured by B. Meanwhile a light beam traveling front to back of Aa takes 30 seconds in A's frame to get from front to back, and since the clock at the back is 10 seconds behind Aa measures a time of 20 seconds, and similarly Aa measures a time of 40 seconds to go from back to front. But although this is a symmetry, this is actually a slight difference in how the laws of physics work in each frame, if we align which direction each labels as +x and which labels as -x (as is normally done in both the Lorentz transformation and the Galilei transformation); if +x lies in the back direction of B's frame and the front direction of Aa's frame, then in B's frame light travels at 30/20 light-seconds per second in the +x direction, while in Aa's frame light travels at 30/40 light-seconds per second in the +x direction. Of course you could reorient one frame's x-axis, but light might still behave differently in the two frames when it was traveling in a direction that had some component on the x-axis but also some component on one of the other axes.

Also, I would guess that this trick only works for a pair of frames, if you had 3 or more frames there wouldn't be any way to synchronize clocks in each frame so that any pair would be related by a Galilei transformation.
 
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  • #5
All in all, a very neat trick!
Thanks, and thank you very much also for working through all of that. I am so glad you did, as I was worried I would have problems convincing anyone otherwise unless they worked it out for themselves. That was excellent. :smile:

Also, I would guess that this trick only works for a pair of frames, if you had 3 or more frames there wouldn't be any way to synchronize clocks in each frame so that any pair would be related by a Galilei transformation.
Right, it only works between two frames at a time. I'm wondering how I might use it to determine more about kinematic theory, by finding relations between frames that are synchronized in this manner for different situations, like in yuiop's spinning arm thread, for instance, or a wheel rolling along a surface, although the re-synchronization from SR will most likely cause slight complications as far as rotation speeds are concerned, but we could also synchronize only one frame to measure no time dilation or length contraction of the frame of the wheel or spinning arm while leaving the other frame alone and see where that goes.
 
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  • #6
Isn't this essentially what ZapperZ said however?... that only subsets of events can be made Galilean? Don't get me wrong, it seems very cool, but I'm also prepared to be completely wrong here. So... help?
 
  • #7
Isn't this essentially what ZapperZ said however?... that only subsets of events can be made Galilean? Don't get me wrong, it seems very cool, but I'm also prepared to be completely wrong here. So... help?
Well, you can only make clocks that are stationary within each frame tick at the same rate and read the same times as observed from either frame and objects that are stationary within each frame have the same spatial dimensions as measured from either frame, and of course the transformation of events between frames is also Galilean.
 
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  • #8
One other interesting thing to note is that while we cannot synchronize more than two frames to be Galilean to each other, we can synchronize any number of other frames or all of them to measure the same things that a particular frame measures within its own frame as far as the rate of ticking of clocks and the spatial dimensions of objects, while that particular frame can still measure the same things as measured within one of the others. We can also synchronize such that frame A measures the same things as measured within frame B, while frame B measures the same things as within frame C, while frame C measures the same things as within frame D, while frame D measures the same things as within the original frame A in a circular fashion.
 
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  • #9
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So if ship Aa adds a time of tla = (L - 1) d / v to the back of the ship, then ship Aa will measure no length contraction or time dilation of ship B and so if then the whole of frame A synchronizes in the same way, then frame A and frame B will measure no length contraction or time dilation of each other. Coming back to the simultaneity shift observed by each, before frame A re-synchronized, frame B observed a simultaneity difference of tl' = (1 - L) d / v between the clocks of ship A. But ship A then added (L - 1) d / v to the rear clock, so frame B now observes no simultaneity difference for ship A whatsoever, and for the whole of frame A, being synchronized in the same way, nor does frame A observe a simultaneity difference for frame B. What each frame observes of the other is now purely Galilean.
So, what you are saying if you artificially change a frame clock such that it is not synchronized with the frame and is synched by time dilation with the other frame, then you think you have shown frame to frame clock synchronization which to this point in the mainstream has never been proven.

Here is your error.

Whatever point you are using and applying time dilation to that "clock" to "synch" with, the corresponding clock in the other frame will not sync it because the two combined will not be LT invertible.

In other words, apply LT back and forth with all the points of your problem and you will find it inconsistent with LT.

So, I know you will become defensive and claim you are correct.

Therefore, set up your entire experiment and apply LT back and forth to prove your case. It will fail.

I can say this because I know how to do frame to frame clock synchronization and I assure you, this is not it. My method is LT invertible.

Also, the solution is basically useless.
 
  • #10
Therefore, set up your entire experiment and apply LT back and forth to prove your case. It will fail.
Let's say frame A is synchronized using the Einstein simultaneity convention. Frame Aa is stationary to frame A but is synchronized with a lesser time of tla = (1 - L) x / v per distance x in the positive x direction. Frame B is also synchronized with a lesser time of tl = z (1 - L) x / v per distance L x in the positive x direction according to frame A. According to frame A, ship B travels in the positive x direction at a relative speed of v, the origin of frame B is at the front of ship B and with observer A in frame A, and both frames synchronize to T=0 upon passing.

According to observer A, an event occurs at coordinates t, x. According to frame Aa, then, with the same origin as A, the event occurs at

xa = x and ta = t - (1 - L) x / v, where

va = x / (x / v + tla)

va = x / (x / v - (1 - L) x / v)

(x / v) [1 - (1 - L)] = x / va

x / v = (x / va) / L

and solving for v also gives

v = va L

whereas ta then becomes

ta = t - (1 - L) (x / va) / L

Rearranging xa and ta gives

x = xa, t = ta + (1 - L) (xa / va) / L

According to frame B, the event occurs at

xB = (x - v t) / L

tB = z t - [(z (1 - L) x / v) / (L x)] (x - v t)

= z t - z (1 - L) (x - v t) / (v L)

= z (t - x / (v L) + t / L + x / v - t)

= z (t / L + x / v - x / (v L))

and substituting for x and t gives

xB = [xa - (va L) (ta + (1 - L) (xa / va) / L)] / L

= xa / L - va (ta + (1 - L) (xa / va) / L)

= xa / L - va ta - (1 - L) xa / L

= xa / L - va ta - xa / L + xa

= xa - va ta

tB = z [(ta + (1 - L) (xa / va) / L) / L + xa / (va L) - xa / (va L^2)]

where if z = L, then

tB = (ta + (1 - L) (xa / va) / L) + xa / va - x / (va L)

= ta + xa / (va L) - xa / va + xa / va - xa / (va L)

= ta

which are Galilean transformations between frames Aa and B. As for applying the inverse transformations, B measures the same speed of Aa as A, where v' = x / (z L x / v + tl) and tl = tl = z (1 - L) x / v, giving

v' = va'

= x / (z L x / v + z (1 - L) x / v)

= 1 / [z L / v + z / v - z L / v)

= v / z

I have been keeping all speeds positive, but with the transformations B observes Aa travelling in the negative x direction making v' negative as well, so

va' = - v / z

and rearranging and substituting gives us

v = - z va'

va L = - z va'

va = - z va' / L

where if z = L, then

va = - va'

and the transformations in terms of the speed B measures of Aa become

xB = xa - va ta

= xa + va' ta

xa = xB - va' ta

where of course the event is still measured at the times ta and tB according to each frame, so we still have

ta = tB

and so from the perspective of frame B, we have

xa = xB - va' tB
 
  • #11
Whatever point you are using and applying time dilation to that "clock" to "synch" with, the corresponding clock in the other frame will not sync it because the two combined will not be LT invertible.

In other words, apply LT back and forth with all the points of your problem and you will find it inconsistent with LT.
Now that I read your post again, I suppose a simpler way to reply is that the Lorentz transformations would not be applied to this, if that is literally what you mean by LT. The Lorentz transformations only apply when both frames are synchronized using the Einstein simultaneity convention, which they are not in this case, but a completely different convention for the synchronization is being applied. All transformations are invertible, however, if you are referring to the way events would be observed according to one frame or the other, since the transformations directly relate the coordinates according to either frame. Perhaps you mean time dilation, which often uses the same variables t and t' as with the transformations but is not the same thing.
 
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