- #1

- 457

- 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.

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

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.

**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.