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The two postulates of SR are:

1) the laws of physics are the same in every inertial frame

2) light is measured travelling isotropically at c in every inertial frame

I intend to derive SR by applying only the second postulate alone, that the speed of light is measured isotropically at c in every inertial frame. We will start with a reference frame where the light is always measured to travel isotropically at c. Whether this particular frame is considered absolute or just a preferred arbitrary frame is of no consequence since we will only be determining the results according to how the physics directly relates between observers, which is how SR would relate the physics since there is otherwise nothing else to relate it to within the philosophy of SR. All other frames are to be viewed from the perspective of this preferred frame. We will apply certain properties that might occur to observers as they move relative to that frame of reference, along with their clocks and rulers. The clocks of observers in motion to this frame may time dilate by a factor of z, lengths contract in the line of motion by a factor of Lx and perpendicularly to the line of motion by a factor of Ly. Each of these ratios are to be compared to that of the reference frame, so if any of these are found not to occur, then the value for that property will be 1.

Alice will always be considered the reference frame observer that remains stationary to that particular frame of reference for the purposes of this demonstration. We will have Bob and Carl travel away from Alice with a speed of v with a distance of d between them, Carl before Bob, as these measurements are taken by Alice. Light always travels isotropically to the reference frame at c, and we want to find what must be true for Bob and Carl to measure the same isotropic speed. Let's say a pulse is sent from Bob to Carl. According to Alice, the light travels a distance of d plus the extra distance Carl has travelled away from the pulse in the time it took for the pulse to reach Carl, so c t_BC = d + v t_BC, t_BC = d / (c - v). The time Alice will measure for a light pulse to travel from Carl to Bob while Bob moves toward the pulse in that time is c t_CB = d - v t_CB, so t_CB = d / (c + v).

If the clocks of Bob and Carl are synchronized to each other as Alice views them, then Bob and Carl will say that their own clocks are unsynchronized because the time it takes for the pulse to travel from Bob to Carl is different from the time it takes to travel from Carl to Bob. So Bob and Carl establish a new simultaneity convention between their own clocks. This is done by having Bob turn his own clock forward or having Carl turn his back by an amount which will produce a time lag between their readings of tl. Bob's clock now reads a greater time than Carl's according to Alice, but Bob and Carl say their clocks are synchronized. When the pulse travels from Bob to Carl, it travels from Bob when Bob's clock reads TB = tl and Carl's clock reads TC = 0, then reaches Carl when Carl's clock reads TC' = z t_BC, so with a difference in times as measured by Bob and Carl of t'_BC = TC' - TB = z t_BC - tl. When the pulse travels from Carl to Bob, Carl sends it when the clocks read TC = 0 and TB = tl, and Bob receives it when his clock reads TB' = tl + z t_CB, so in a time measured by them of t'_CB = TB' - TC = tl + z t_CB. Since they must measure these two times to be equal, then t'_BC = t'_CB, z t_BC - tl = tl + z t_CB, 2 tl = z t_BC - z t_CB = z d / (c - v) - z d / (c + v) = z d [(c + v) - (c - v)] / [(c - v) (c + v)] = 2 z d v / (c^2 - v^2), whereby tl = z d v / (c^2 - v^2). This is now the simultaneity difference between the clocks of Bob and Carl according to Alice. In the frame of Bob and Carl according to Alice, their rulers have been contracted in the line of motion by Lx, so if Alice measures a distance between them of d, then they will measure d' = d / Lx . So when the pulse is sent from Bob to Carl, it will be measured to have a speed of c' = d' / t'_BC = (d / Lx) / (z t_BC - tl) = (d / Lx) / [z d / (c - v) - z d v / (c^2 - v^2)] = [(c^2 - v^2) / (z Lx)] / [(c + v) - v] = c (1 - (v/c)^2) / (z Lx), whereby if c' = c, then z Lx = 1 - (v/c)^2 . Likewise, the speed measured from Carl to Bob gives the same result.

Now let's say that in the frame of Bob and Carl, an apparatus has been set up where light is allowed to travel across the lengths of two perpendicular arms of equal lengths d' and back. Since the pulses must have the isotropic speed of c along equal lengths of d', then the times to travel both arms are measured the same in the frame of the apparatus with c = d' / t' which is to be accepted as the usual definition of speed with distance measured over time measured, not a law that must be derived, but a given definition for speed, so since t' = d' / c where c and d' are measured the same along both arms, then so must be t' be the same along both arms. Also, because the pulses coincide in the same place upon separating and then coincide in the same place again when returning, then all observers in all frames must agree that the times to traverse both arms is the same for whatever time they measure between these two events. Let's say that one arm travels directly in the line of motion of the apparatus to the reference frame. From Alice's frame, the apparatus is contracted in the line of motion by Lx and perpendicularly by Ly, so the lengths of the arms are dx = Lx d' and dy = Ly d'. The time that Alice measures for the pulse to travel the arm in the line of motion and back is t_forward = dx / (c - v) and t_back = dx / (c + v), whereby tx = dx / (c - v) + dx / (c + v) = dx [(c + v) + (c - v)] / (c^2 - v^2) = 2 (Lx d') c / (c^2 - v^2). In the perpendicular direction, Alice measures a time of (c t_away)^2 = (v t_away)^2 + dy^2, so t_away = dy / sqrt(c^2 - v^2). The pulse travels in the same way along the same angle away and back, so t_perp = 2 dy / sqrt(1 - (v/c)^2) = 2 (Ly d') / sqrt(1 - (v/c)^2). In order for these times to be the same as Alice measures them, then 2 (Lx d') c / (c^2 - v^2) = 2 (Ly d') / sqrt(c^2 - v^2), and from this we gain Lx / Ly = sqrt(1 - (v/c)^2).

Okay, here's where things get interesting. One might think that two observers measuring the same relative speed of each other would follow from the first postulate, since if the laws of physics is the same in all inertial frames, then with nothing else to relate the physics to except between the observers, then each must measure the same relative speed between them as the other does, but it actually follows from the second principle alone. Let's say that Bob and Carl pass Alice. Alice says that the time for Carl to pass and Bob to reach her is t = d / v, all as measured in her own frame. Now, from what is measured in the frame of Bob and Carl, when Carl passes Alice, his clock reads TC = 0 and Bob's reads TB = tl. When Bob passes Alice, his clock then reads TB' = tl + z t. Bob and Carl will read the difference in times that has passed between their clocks as t' = TB' - TC = tl + z t = z d v / (c^2 - v^2) + z d / v = [z d / (v (c^2 - v^2))] [v^2 + (c^2 - v^2)] = z d / (v (1 - (v/c)^2)) and the distance Alice has travelled of d' = d / Lx, giving a relative speed for Alice as Bob and Carl measure it of v' = d' / t' = (d / Lx) / [z d / (v (1 - (v/c)^2)] = v (1 - (v/c)^2) / (z Lx), but since we have already established earlier that z Lx = 1 - (v/c)^2, then we gain v' = v (1 - (v/c)^2) / (z Lx) = v, so the observers will measure the same relative speed of each other in both frames.

Now let's look at the addition of speeds. Let's say that according to Alice, who is stationary with the reference frame, Bob and Carl are travelling in one direction at v and Danielle is travelling past them in the other direction at u. According to Alice, it takes a time of t = d / (u + v) for Danielle to travel from Carl to Bob, so Bob and Carl measure their difference in times to be TC = 0 and TB' = tl + z t, so t' = TB' - TC = z t + tl = z d / (u + v) + z d v / (c^2 - v^2) = z d [(c^2 - v^2) + (u + v) v] / [(u + v) (c^2 - v^2)] = z (Lx d') [c^2 - v^2 + u v + v^2] / [(u + v) (c^2 - v^2)] = d' [c^2 + u v] / [(u + v) c^2] = d' [1 + u v / c^2] / (u + v). Therefore, the speed that Bob and Carl measure of Danielle is w = d' / t' = (u + v) / (1 + u v / c^2). If Danielle were to travel in the same direction as Bob and Carl at u, then Alice would measure a time for Danielle to travel from Bob to Carl of t = d / (u - v), whereby Bob and Carl would measure their difference in times to be TB = tl and TC' = z t, for a difference in times of t' = TC' - TB = z t - tl = z d / (u - v) - z d v / (c^2 - v^2) = [z d / ((u - v) (c^2 - v^2))] [(c^2 - v^2) - v (u - v)] = [z d / ((u - v) (c^2 - v^2))] [c^2 - u v] = [z (Lx d') / ((u - v) (1 - (v/c)^2)] [1 - u v / c^2] = [d' / (u - v)] [1 - u v / c^2]. The relative speed Bob and Carl measure for Danielle when travelling in the same direction, then, is w = d' / t' = (u - v) / (1 - u v / c^2).

Now let's say Bob and Danielle are both travelling in ships that they measure of a length of d' in their own frames. Let's determine what the length contraction Bob measures of Danielle's ship will be. Bob cannot measure the length of Danielle's ship at a distance or even directly by using his ruler while Danielle's ship is in motion to his, so he has to find another way. What he does is to find the difference in time that it takes for the front of Alice's ship to pass an antenna on his ship and then the back of her ship to pass the same antenna. At T=0 on his clock, the front of Alice's ship passes the antenna. According to Alice, the time that it takes for Danielle's ship to pass Bob's antenna is t = (Lx(u) d') / (u + v), where Lx(u) is the length contraction Alice measures of Danielle's ship. If t passes in Alice's frame, then z(v) t, where z(v) = z from before but now we are adding more speeds so must become more specific, passes for Bob and all observers must agree that this is Bob's reading when Danielle passes since the events of the readings upon his clock coincide in the same place as the front and back of Danielle's ship with Bob's antenna when the clock is placed in the same place as the antenna also. The length of Danielle's ship as Bob measures it, then, is d" = w t' = [(u + v) / (1 + u v / c^2)] [z(v) (Lx(u) d') / (u + v)] = z(v) Lx(u) d' / (1 + u v / c^2). The observed length contraction, then, is Lx(w) = d" / d' = z(v) Lx(u) / (1 + u v / c^2).

So now let's find out what Bob and Carl measure for the time dilation of Danielle's clock. We will place Carl in the front of the ship of proper length d' and Bob at the back. Alice says Danielle travels from Carl to Bob in a time of t = (Lx(v) d') / (u + v). When Danielle passes Carl, the readings upon the clocks according to Alice are TC=0 and TB = tl. When Danielle passes Bob, Bob's reading is TB' = tl + z(v) t, and again, all observers must agree since the events of the clock readings and Danielle directly passing the clocks coincide in the same places. Bob and Carl say the difference in times that has passed between their clocks is TB' - TC = tl + z(v) t = z(v) (Lx(v) d') v / (c^2 - v^2) + z(v) (Lx(v) d') / (u + v) = [z(v) Lx(v) d' / ((c^2 - v^2) (u + v))] [v (u + v) + (c^2 - v^2)] = d' [c^2 + u v] / (c^2 (u + v)) = d' (1 + u v / c^2) / (u + v) = d' / w. The amount of time that has passed upon Danielle's clock while travelling from Carl to Bob is t" = z(u) t = z(u) (Lx(v) d') / (u + v), so the time dilation Bob and Carl measure of Danielle's clock is z(w) = t" / t' = [z(u) (Lx(v) d') / (u + v)] / [d' / w] = [z(u) Lx(v)] [w / (u + v)] = z(u) Lx(v) / (1 + u v / c^2).

Now let's compare what we have gained for the time dilation and length contraction Bob and Carl measure of Danielle and dive into a little math logic for this part of the demonstration. We have found that Lx(w) = z(v) Lx(u) / (1 + u v / c^2) and z(w) = z(u) Lx(v) / (1 + u v / c^2), whereby (1 + u v / c^2) = z(v) Lx(u) / Lx(w) = z(u) Lx(v) / z(w), so by rearranging we gain [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = 1. The values here that are represented by u and v are what Alice measures for the relative speeds of Danielle and of Bob and Carl, respectively. w represents the relative speed that is measured by Bob and Carl of Danielle. Now, from Alice's perspective, u and v can have any arbitrary values for the relative speeds to the reference frame and w will be determined by what those values are. u and v can have the same value and still be arbitrary, so let's say that u = v. In that case, [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 so [z(w) / Lx(w)] = 1 also. Since w can still have any arbitrary value with any arbitrary values of u and v where u=v, then [z(w) / Lx(w)] = 1 for any arbitrary value whatsoever, therefore z(w) / Lx(w) always equals 1 for any relative speed of w. If that is the case, then [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 for any arbitrary values of u and v even when the speeds are not equal, and the only way they can do that when changing the speed of u slightly while keeping the speed of v the same, for instance, is if z(v) / Lx(v) = 1 and z(u) / Lx(u) = 1 always also.

So since we had z(v) Lx(v) = 1 - (v/c)^2 and Lx(v) / Ly(v) = sqrt(1 - (v/c)^2) as found at the beginning of the demonstration, and now we have z(v) / Lx(v) = 1, whereby z(v) = Lx(v), then z(v) Lx(v) = z(v)^2 = 1 - (v/c)^2, giving z(v) = Lx(v) = sqrt(1 - (v/c)^2), as well as Lx(v) / Ly(v) = sqrt(1 - (v/c)^2), giving Ly(v) = 1, so no contraction takes place perpendicularly to the line of motion. And there we have it. We have determined that all of the basic principles of SR can be determined from the second postulate alone.

1) the laws of physics are the same in every inertial frame

2) light is measured travelling isotropically at c in every inertial frame

I intend to derive SR by applying only the second postulate alone, that the speed of light is measured isotropically at c in every inertial frame. We will start with a reference frame where the light is always measured to travel isotropically at c. Whether this particular frame is considered absolute or just a preferred arbitrary frame is of no consequence since we will only be determining the results according to how the physics directly relates between observers, which is how SR would relate the physics since there is otherwise nothing else to relate it to within the philosophy of SR. All other frames are to be viewed from the perspective of this preferred frame. We will apply certain properties that might occur to observers as they move relative to that frame of reference, along with their clocks and rulers. The clocks of observers in motion to this frame may time dilate by a factor of z, lengths contract in the line of motion by a factor of Lx and perpendicularly to the line of motion by a factor of Ly. Each of these ratios are to be compared to that of the reference frame, so if any of these are found not to occur, then the value for that property will be 1.

Alice will always be considered the reference frame observer that remains stationary to that particular frame of reference for the purposes of this demonstration. We will have Bob and Carl travel away from Alice with a speed of v with a distance of d between them, Carl before Bob, as these measurements are taken by Alice. Light always travels isotropically to the reference frame at c, and we want to find what must be true for Bob and Carl to measure the same isotropic speed. Let's say a pulse is sent from Bob to Carl. According to Alice, the light travels a distance of d plus the extra distance Carl has travelled away from the pulse in the time it took for the pulse to reach Carl, so c t_BC = d + v t_BC, t_BC = d / (c - v). The time Alice will measure for a light pulse to travel from Carl to Bob while Bob moves toward the pulse in that time is c t_CB = d - v t_CB, so t_CB = d / (c + v).

If the clocks of Bob and Carl are synchronized to each other as Alice views them, then Bob and Carl will say that their own clocks are unsynchronized because the time it takes for the pulse to travel from Bob to Carl is different from the time it takes to travel from Carl to Bob. So Bob and Carl establish a new simultaneity convention between their own clocks. This is done by having Bob turn his own clock forward or having Carl turn his back by an amount which will produce a time lag between their readings of tl. Bob's clock now reads a greater time than Carl's according to Alice, but Bob and Carl say their clocks are synchronized. When the pulse travels from Bob to Carl, it travels from Bob when Bob's clock reads TB = tl and Carl's clock reads TC = 0, then reaches Carl when Carl's clock reads TC' = z t_BC, so with a difference in times as measured by Bob and Carl of t'_BC = TC' - TB = z t_BC - tl. When the pulse travels from Carl to Bob, Carl sends it when the clocks read TC = 0 and TB = tl, and Bob receives it when his clock reads TB' = tl + z t_CB, so in a time measured by them of t'_CB = TB' - TC = tl + z t_CB. Since they must measure these two times to be equal, then t'_BC = t'_CB, z t_BC - tl = tl + z t_CB, 2 tl = z t_BC - z t_CB = z d / (c - v) - z d / (c + v) = z d [(c + v) - (c - v)] / [(c - v) (c + v)] = 2 z d v / (c^2 - v^2), whereby tl = z d v / (c^2 - v^2). This is now the simultaneity difference between the clocks of Bob and Carl according to Alice. In the frame of Bob and Carl according to Alice, their rulers have been contracted in the line of motion by Lx, so if Alice measures a distance between them of d, then they will measure d' = d / Lx . So when the pulse is sent from Bob to Carl, it will be measured to have a speed of c' = d' / t'_BC = (d / Lx) / (z t_BC - tl) = (d / Lx) / [z d / (c - v) - z d v / (c^2 - v^2)] = [(c^2 - v^2) / (z Lx)] / [(c + v) - v] = c (1 - (v/c)^2) / (z Lx), whereby if c' = c, then z Lx = 1 - (v/c)^2 . Likewise, the speed measured from Carl to Bob gives the same result.

Now let's say that in the frame of Bob and Carl, an apparatus has been set up where light is allowed to travel across the lengths of two perpendicular arms of equal lengths d' and back. Since the pulses must have the isotropic speed of c along equal lengths of d', then the times to travel both arms are measured the same in the frame of the apparatus with c = d' / t' which is to be accepted as the usual definition of speed with distance measured over time measured, not a law that must be derived, but a given definition for speed, so since t' = d' / c where c and d' are measured the same along both arms, then so must be t' be the same along both arms. Also, because the pulses coincide in the same place upon separating and then coincide in the same place again when returning, then all observers in all frames must agree that the times to traverse both arms is the same for whatever time they measure between these two events. Let's say that one arm travels directly in the line of motion of the apparatus to the reference frame. From Alice's frame, the apparatus is contracted in the line of motion by Lx and perpendicularly by Ly, so the lengths of the arms are dx = Lx d' and dy = Ly d'. The time that Alice measures for the pulse to travel the arm in the line of motion and back is t_forward = dx / (c - v) and t_back = dx / (c + v), whereby tx = dx / (c - v) + dx / (c + v) = dx [(c + v) + (c - v)] / (c^2 - v^2) = 2 (Lx d') c / (c^2 - v^2). In the perpendicular direction, Alice measures a time of (c t_away)^2 = (v t_away)^2 + dy^2, so t_away = dy / sqrt(c^2 - v^2). The pulse travels in the same way along the same angle away and back, so t_perp = 2 dy / sqrt(1 - (v/c)^2) = 2 (Ly d') / sqrt(1 - (v/c)^2). In order for these times to be the same as Alice measures them, then 2 (Lx d') c / (c^2 - v^2) = 2 (Ly d') / sqrt(c^2 - v^2), and from this we gain Lx / Ly = sqrt(1 - (v/c)^2).

Okay, here's where things get interesting. One might think that two observers measuring the same relative speed of each other would follow from the first postulate, since if the laws of physics is the same in all inertial frames, then with nothing else to relate the physics to except between the observers, then each must measure the same relative speed between them as the other does, but it actually follows from the second principle alone. Let's say that Bob and Carl pass Alice. Alice says that the time for Carl to pass and Bob to reach her is t = d / v, all as measured in her own frame. Now, from what is measured in the frame of Bob and Carl, when Carl passes Alice, his clock reads TC = 0 and Bob's reads TB = tl. When Bob passes Alice, his clock then reads TB' = tl + z t. Bob and Carl will read the difference in times that has passed between their clocks as t' = TB' - TC = tl + z t = z d v / (c^2 - v^2) + z d / v = [z d / (v (c^2 - v^2))] [v^2 + (c^2 - v^2)] = z d / (v (1 - (v/c)^2)) and the distance Alice has travelled of d' = d / Lx, giving a relative speed for Alice as Bob and Carl measure it of v' = d' / t' = (d / Lx) / [z d / (v (1 - (v/c)^2)] = v (1 - (v/c)^2) / (z Lx), but since we have already established earlier that z Lx = 1 - (v/c)^2, then we gain v' = v (1 - (v/c)^2) / (z Lx) = v, so the observers will measure the same relative speed of each other in both frames.

Now let's look at the addition of speeds. Let's say that according to Alice, who is stationary with the reference frame, Bob and Carl are travelling in one direction at v and Danielle is travelling past them in the other direction at u. According to Alice, it takes a time of t = d / (u + v) for Danielle to travel from Carl to Bob, so Bob and Carl measure their difference in times to be TC = 0 and TB' = tl + z t, so t' = TB' - TC = z t + tl = z d / (u + v) + z d v / (c^2 - v^2) = z d [(c^2 - v^2) + (u + v) v] / [(u + v) (c^2 - v^2)] = z (Lx d') [c^2 - v^2 + u v + v^2] / [(u + v) (c^2 - v^2)] = d' [c^2 + u v] / [(u + v) c^2] = d' [1 + u v / c^2] / (u + v). Therefore, the speed that Bob and Carl measure of Danielle is w = d' / t' = (u + v) / (1 + u v / c^2). If Danielle were to travel in the same direction as Bob and Carl at u, then Alice would measure a time for Danielle to travel from Bob to Carl of t = d / (u - v), whereby Bob and Carl would measure their difference in times to be TB = tl and TC' = z t, for a difference in times of t' = TC' - TB = z t - tl = z d / (u - v) - z d v / (c^2 - v^2) = [z d / ((u - v) (c^2 - v^2))] [(c^2 - v^2) - v (u - v)] = [z d / ((u - v) (c^2 - v^2))] [c^2 - u v] = [z (Lx d') / ((u - v) (1 - (v/c)^2)] [1 - u v / c^2] = [d' / (u - v)] [1 - u v / c^2]. The relative speed Bob and Carl measure for Danielle when travelling in the same direction, then, is w = d' / t' = (u - v) / (1 - u v / c^2).

Now let's say Bob and Danielle are both travelling in ships that they measure of a length of d' in their own frames. Let's determine what the length contraction Bob measures of Danielle's ship will be. Bob cannot measure the length of Danielle's ship at a distance or even directly by using his ruler while Danielle's ship is in motion to his, so he has to find another way. What he does is to find the difference in time that it takes for the front of Alice's ship to pass an antenna on his ship and then the back of her ship to pass the same antenna. At T=0 on his clock, the front of Alice's ship passes the antenna. According to Alice, the time that it takes for Danielle's ship to pass Bob's antenna is t = (Lx(u) d') / (u + v), where Lx(u) is the length contraction Alice measures of Danielle's ship. If t passes in Alice's frame, then z(v) t, where z(v) = z from before but now we are adding more speeds so must become more specific, passes for Bob and all observers must agree that this is Bob's reading when Danielle passes since the events of the readings upon his clock coincide in the same place as the front and back of Danielle's ship with Bob's antenna when the clock is placed in the same place as the antenna also. The length of Danielle's ship as Bob measures it, then, is d" = w t' = [(u + v) / (1 + u v / c^2)] [z(v) (Lx(u) d') / (u + v)] = z(v) Lx(u) d' / (1 + u v / c^2). The observed length contraction, then, is Lx(w) = d" / d' = z(v) Lx(u) / (1 + u v / c^2).

So now let's find out what Bob and Carl measure for the time dilation of Danielle's clock. We will place Carl in the front of the ship of proper length d' and Bob at the back. Alice says Danielle travels from Carl to Bob in a time of t = (Lx(v) d') / (u + v). When Danielle passes Carl, the readings upon the clocks according to Alice are TC=0 and TB = tl. When Danielle passes Bob, Bob's reading is TB' = tl + z(v) t, and again, all observers must agree since the events of the clock readings and Danielle directly passing the clocks coincide in the same places. Bob and Carl say the difference in times that has passed between their clocks is TB' - TC = tl + z(v) t = z(v) (Lx(v) d') v / (c^2 - v^2) + z(v) (Lx(v) d') / (u + v) = [z(v) Lx(v) d' / ((c^2 - v^2) (u + v))] [v (u + v) + (c^2 - v^2)] = d' [c^2 + u v] / (c^2 (u + v)) = d' (1 + u v / c^2) / (u + v) = d' / w. The amount of time that has passed upon Danielle's clock while travelling from Carl to Bob is t" = z(u) t = z(u) (Lx(v) d') / (u + v), so the time dilation Bob and Carl measure of Danielle's clock is z(w) = t" / t' = [z(u) (Lx(v) d') / (u + v)] / [d' / w] = [z(u) Lx(v)] [w / (u + v)] = z(u) Lx(v) / (1 + u v / c^2).

Now let's compare what we have gained for the time dilation and length contraction Bob and Carl measure of Danielle and dive into a little math logic for this part of the demonstration. We have found that Lx(w) = z(v) Lx(u) / (1 + u v / c^2) and z(w) = z(u) Lx(v) / (1 + u v / c^2), whereby (1 + u v / c^2) = z(v) Lx(u) / Lx(w) = z(u) Lx(v) / z(w), so by rearranging we gain [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = 1. The values here that are represented by u and v are what Alice measures for the relative speeds of Danielle and of Bob and Carl, respectively. w represents the relative speed that is measured by Bob and Carl of Danielle. Now, from Alice's perspective, u and v can have any arbitrary values for the relative speeds to the reference frame and w will be determined by what those values are. u and v can have the same value and still be arbitrary, so let's say that u = v. In that case, [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 so [z(w) / Lx(w)] = 1 also. Since w can still have any arbitrary value with any arbitrary values of u and v where u=v, then [z(w) / Lx(w)] = 1 for any arbitrary value whatsoever, therefore z(w) / Lx(w) always equals 1 for any relative speed of w. If that is the case, then [z(v) / Lx(v)] [Lx(u) / z(u)] [z(w) / Lx(w)] = [z(v) / Lx(v)] [Lx(u) / z(u)] = 1 for any arbitrary values of u and v even when the speeds are not equal, and the only way they can do that when changing the speed of u slightly while keeping the speed of v the same, for instance, is if z(v) / Lx(v) = 1 and z(u) / Lx(u) = 1 always also.

So since we had z(v) Lx(v) = 1 - (v/c)^2 and Lx(v) / Ly(v) = sqrt(1 - (v/c)^2) as found at the beginning of the demonstration, and now we have z(v) / Lx(v) = 1, whereby z(v) = Lx(v), then z(v) Lx(v) = z(v)^2 = 1 - (v/c)^2, giving z(v) = Lx(v) = sqrt(1 - (v/c)^2), as well as Lx(v) / Ly(v) = sqrt(1 - (v/c)^2), giving Ly(v) = 1, so no contraction takes place perpendicularly to the line of motion. And there we have it. We have determined that all of the basic principles of SR can be determined from the second postulate alone.

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