Yes i know. If you want to say at low speeds, or using first order approximation using binormial/taylor expansion for the time dilation factor, fair enough. But mathematically, it's a fact that both will not agree on time even if the difference is very small.zong wrote:Time dilation doesn't bother you until your speed nears the speed of light.. so just take it that the time A observes that B travels from the 1st post to 2nd is the same time B observes himself to travel the same distance. So they WOULD agree on the time.
Just for your info if you want to know how much time dilation can affect you:
If a taxi driver drives at 60kph every shift and continues every day of the year for 60 years, he would be time dilated by roughly 1 second. That is how insignificant time dilation is, at "normal" speedsBeware tho this is only a rough guide, not exact figures!
Special Relativity
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ChaosKnight
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Special Relativity
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ChaosKnight
- Posts: 293
- Joined: Sun Oct 05, 2003 6:54 pm
Here's the math i came up with.
Acceleration and Gravity in relation to Time
Imagine there are a series of frames labeled S, S1, S2, S3……
There are two observers, A and B.
B is accelerating with constant magnitude as as seen by A.
S is the stationary frame, where Observer A stays in.
S1, S2, S3….. are a series of frames with infinitesimally increasing velocities dV. We are going to pass Observer B from S1 to S2 to S3……..in order to derive an equation for acceleration.
Imagine there are a series of posts 1,2, 3……in S, which are spaced at equal infinitesimally small distances dLp apart.
Now, B is going from V = 0m/s to V = dV m/s, which means B is transiting from S to S1. When B is in S1, it passes from post 1 to 2.
The time recorded by B to go from 1 to 2 is dtp1.
The time recorded by A for B to go from 1 to 2 is dt1.
The distance A-B recorded by B is dL1.
V by both observers is V1.
( The events defined here are: B at post 1, and B at post 2. Since these events occur at the same place for B, B records proper time, hence its notation tp. A records improper time t.)
Now, B increases its velocity again, going from S1 to S2. It passes posts 2 and 3. Again,
The time recorded by B to go from 2 to 3 is dtp2.
The time recorded by A for B to go from 2 to 3 is dt2.
The distance A-B recorded by B is dL2.
V by both observers is V2.
As can be seen, B records a series of proper times, while A records a series of improper times.
We want to know the time interval between B starting and B ending at some arbitrary post.
With B in any frame, we can relate dtn and dtpn by:
dtn = g dtpn ---------------- (1)
where g is the time dilation factor:
g = 1 / [ 1 - Vn2/c2]1/2 ----------------(2)
n is the frame B is in. Vn is the velocity of frame n.
Since all frames can be described with equations (1) and (2), we drop the subscript n.
Therefore, substituting (2) in (1), we get
dt = [1 / (1 - V2/c2)1/2] dtp-----------------(3)
We also know that:
V = ast -----------------(4)
Where t is the time elapsed in frame S.
Substituting (4) in (3) and rearranging, we get
(1 – { ast }2/c2)1/2 dt = dtp --------------(5)
Integrating both sides, we get
(t/2) [1 – ( ast )2/c2]1/2 + (c/2as)[sin-1 ast/c] = tp + K -----------(6)
where K is a constant from integration.
Using initial conditions, at t = 0, tp must be zero. Therefore K = 0. Equation (6) is therefore
(t/2) [1 – ( ast )2/c2]1/2 + (c/2as)[sin-1 ast/c] = tp -----------(7)
Checking
At low velocities, i.e. ( ast )/c << 1, we can let
[1 – ( ast )2/c2]1/2 = 1
sin-1 ast/c = ast/c
Therefore, equation (7) reduces to
t/2 + t/2 = t = tp -------------(8)
which is what we expect, i.e. at low velocities, both A and B agree on time.
Now we prove that if B is stationary relative to A, both observers see the same acceleration.
Since observers must agree on the velocity and the magnitude in the change in velocity,
dV = asdt = aBdtp --------------(9)
where aB is the “relative” acceleration of A as observed by B.
Substituting equation (3) without subscript the n, we get
as[1 / (1 - V2/c2)1/2] dtp = aBdtp ---------- (10)
Canceling common factors dtp on both sides, we get
as[1 / (1 - V2/c2)1/2] = aB ---------- (11)
Therefore acceleration is subject to time dilation as well. If V = 0, i.e. both A and B are stationary, both observers see the same magnitude of acceleration, i.e.
as = aB ---------- (12)
Gravity and Acceleration
Using the Principle of Equivalence, we can substitute as = G, where G is the acceleration due to gravity, in equation (7). We get
(t/2) [1 – ( Gt )2/c2]1/2 + (c/2G)[sin-1 Gt/c] = tp -----------(13)
Note that G is the acceleration due to gravitation felt by B. A does not feel any acceleration.
(To understand why, imagine A at a space station, where he would not feel his own weight. B is on the space station elevator. When the elevator starts up, B feels his weight on the floor, while A still feels nothing. B is therefore in acceleration but not A. Acceleration is therefore not “relative”.)
Quantitative Analysis of Equation (13)
Lets play around with some numbers:
For t = 1s,
G = 9.81m/s2 tp = 1s
G = 9e2m/s2 tp = 1s
G = 9e3m/s2 tp = 0.9999999999s
G = 9e4m/s2 tp = 0.999999985s
G = 9e5m/s2 tp = 0.9999985s
G = 9e6m/s2 tp = 0.9998499797s
G = 9e7m/s2 tp = 0.9847906907s
G = 9.1e7m/s2 tp = 0.9844458801s
G = 9.2e7m/s2 tp = 0.9840970316s
G = 9.3e7m/s2 tp = 0.9837441371s
-------------------------------------------------------------------------------------
G = 2.5e8m/s2 tp = 0.3264434512s
G = 2.6e8m/s2 tp = 0.2995066309s
G = 2.7e8m/s2 tp = 0.2680126944s
G = 2.8e8m/s2 tp = 0.2295783722s
G = 2.9e8m/s2 tp = 0.1780972954s
As can be seen, as the gravitational field experienced by B increases, time slows down compared to A, which is not subjected to any gravitational field.
(For an event, A records 1s, while B records less than 1s. Therefore B’s clocks have slowed.)
There are some problems with the theory, so don't use this info for anything.
Acceleration and Gravity in relation to Time
Imagine there are a series of frames labeled S, S1, S2, S3……
There are two observers, A and B.
B is accelerating with constant magnitude as as seen by A.
S is the stationary frame, where Observer A stays in.
S1, S2, S3….. are a series of frames with infinitesimally increasing velocities dV. We are going to pass Observer B from S1 to S2 to S3……..in order to derive an equation for acceleration.
Imagine there are a series of posts 1,2, 3……in S, which are spaced at equal infinitesimally small distances dLp apart.
Now, B is going from V = 0m/s to V = dV m/s, which means B is transiting from S to S1. When B is in S1, it passes from post 1 to 2.
The time recorded by B to go from 1 to 2 is dtp1.
The time recorded by A for B to go from 1 to 2 is dt1.
The distance A-B recorded by B is dL1.
V by both observers is V1.
( The events defined here are: B at post 1, and B at post 2. Since these events occur at the same place for B, B records proper time, hence its notation tp. A records improper time t.)
Now, B increases its velocity again, going from S1 to S2. It passes posts 2 and 3. Again,
The time recorded by B to go from 2 to 3 is dtp2.
The time recorded by A for B to go from 2 to 3 is dt2.
The distance A-B recorded by B is dL2.
V by both observers is V2.
As can be seen, B records a series of proper times, while A records a series of improper times.
We want to know the time interval between B starting and B ending at some arbitrary post.
With B in any frame, we can relate dtn and dtpn by:
dtn = g dtpn ---------------- (1)
where g is the time dilation factor:
g = 1 / [ 1 - Vn2/c2]1/2 ----------------(2)
n is the frame B is in. Vn is the velocity of frame n.
Since all frames can be described with equations (1) and (2), we drop the subscript n.
Therefore, substituting (2) in (1), we get
dt = [1 / (1 - V2/c2)1/2] dtp-----------------(3)
We also know that:
V = ast -----------------(4)
Where t is the time elapsed in frame S.
Substituting (4) in (3) and rearranging, we get
(1 – { ast }2/c2)1/2 dt = dtp --------------(5)
Integrating both sides, we get
(t/2) [1 – ( ast )2/c2]1/2 + (c/2as)[sin-1 ast/c] = tp + K -----------(6)
where K is a constant from integration.
Using initial conditions, at t = 0, tp must be zero. Therefore K = 0. Equation (6) is therefore
(t/2) [1 – ( ast )2/c2]1/2 + (c/2as)[sin-1 ast/c] = tp -----------(7)
Checking
At low velocities, i.e. ( ast )/c << 1, we can let
[1 – ( ast )2/c2]1/2 = 1
sin-1 ast/c = ast/c
Therefore, equation (7) reduces to
t/2 + t/2 = t = tp -------------(8)
which is what we expect, i.e. at low velocities, both A and B agree on time.
Now we prove that if B is stationary relative to A, both observers see the same acceleration.
Since observers must agree on the velocity and the magnitude in the change in velocity,
dV = asdt = aBdtp --------------(9)
where aB is the “relative” acceleration of A as observed by B.
Substituting equation (3) without subscript the n, we get
as[1 / (1 - V2/c2)1/2] dtp = aBdtp ---------- (10)
Canceling common factors dtp on both sides, we get
as[1 / (1 - V2/c2)1/2] = aB ---------- (11)
Therefore acceleration is subject to time dilation as well. If V = 0, i.e. both A and B are stationary, both observers see the same magnitude of acceleration, i.e.
as = aB ---------- (12)
Gravity and Acceleration
Using the Principle of Equivalence, we can substitute as = G, where G is the acceleration due to gravity, in equation (7). We get
(t/2) [1 – ( Gt )2/c2]1/2 + (c/2G)[sin-1 Gt/c] = tp -----------(13)
Note that G is the acceleration due to gravitation felt by B. A does not feel any acceleration.
(To understand why, imagine A at a space station, where he would not feel his own weight. B is on the space station elevator. When the elevator starts up, B feels his weight on the floor, while A still feels nothing. B is therefore in acceleration but not A. Acceleration is therefore not “relative”.)
Quantitative Analysis of Equation (13)
Lets play around with some numbers:
For t = 1s,
G = 9.81m/s2 tp = 1s
G = 9e2m/s2 tp = 1s
G = 9e3m/s2 tp = 0.9999999999s
G = 9e4m/s2 tp = 0.999999985s
G = 9e5m/s2 tp = 0.9999985s
G = 9e6m/s2 tp = 0.9998499797s
G = 9e7m/s2 tp = 0.9847906907s
G = 9.1e7m/s2 tp = 0.9844458801s
G = 9.2e7m/s2 tp = 0.9840970316s
G = 9.3e7m/s2 tp = 0.9837441371s
-------------------------------------------------------------------------------------
G = 2.5e8m/s2 tp = 0.3264434512s
G = 2.6e8m/s2 tp = 0.2995066309s
G = 2.7e8m/s2 tp = 0.2680126944s
G = 2.8e8m/s2 tp = 0.2295783722s
G = 2.9e8m/s2 tp = 0.1780972954s
As can be seen, as the gravitational field experienced by B increases, time slows down compared to A, which is not subjected to any gravitational field.
(For an event, A records 1s, while B records less than 1s. Therefore B’s clocks have slowed.)
There are some problems with the theory, so don't use this info for anything.
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ChaosKnight
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Destructos
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QuantumGravity
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Destructos
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carlogambino
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some wrong concept here. you could never reach the speed of light, because it is "the" speed limit for all motion possible. Even if you were very near the speed of light, you would need an infinite amount of energy to actually reach the speed of light, which is not possible. Think of an exponential graph. For the graph of e^x, the graph never touches the x-axis, even if you were to extend it forever, though it would approach the x-axis forever. This applies to objects approaching the speed of light too.
For tachyons, the same principle applies. Tachyons are a totally different concept altogether, but similary, they can never reach the speed of light too. Meaning, they can approach the speed of light but they can never actually reach the speed of light, c. This is why if you were to plot a velocity-time graph for both tachyons and relativistic objects increasing or decreasing in speed, both objects would approach v=c, but never reach it.
One of the laws of special relativity states that c is a constant. This constant can never be reached by moving objects. Thus, time cannot "stop" for you and me, which is why we cant travel back to the past either
For tachyons, the same principle applies. Tachyons are a totally different concept altogether, but similary, they can never reach the speed of light too. Meaning, they can approach the speed of light but they can never actually reach the speed of light, c. This is why if you were to plot a velocity-time graph for both tachyons and relativistic objects increasing or decreasing in speed, both objects would approach v=c, but never reach it.
One of the laws of special relativity states that c is a constant. This constant can never be reached by moving objects. Thus, time cannot "stop" for you and me, which is why we cant travel back to the past either
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Destructos
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I completely understand that it is physically possible to have a piece of matter achieve the speed of light however i just want to know what is it like... According to the bible, God is light so therefore time does not apply to him, so i'm wondering, if particles(or God for that matter) is/are travelling at c, will it observe the universe 3 dimensionally or 4 dimensionally?
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ChaosKnight
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I don't know enough to give a good answer, but maybe a suggestion.
If you look at the equations of special relativity, and if you consider a 1-D case excluding time, i.e. the particle can only move along a line, the particle is....well......everywhere at once.
As for time, perhaps it could be argued that the dimension of time has been reduced so much until it doesn't exist?
zong is better in relativity than me. Perhaps he could provide an answer.
If you look at the equations of special relativity, and if you consider a 1-D case excluding time, i.e. the particle can only move along a line, the particle is....well......everywhere at once.
As for time, perhaps it could be argued that the dimension of time has been reduced so much until it doesn't exist?
zong is better in relativity than me. Perhaps he could provide an answer.