# Relativity of time

1. Aug 29, 2006

### myoho.renge.kyo

it is now in my watch 8:02 pm (8/29/2006) in burbank, california.
this time is between 8:00 pm thru 9:00 pm.
i observe that this period of time is 3600 s according to my wrist watch.
but according to a watch (which is synchronized with my watch) moving at 0.9c relative to me, i would have observed that the period of time is 8258.97 s (about 2 h 17 m) .

let t0 = the period of time according to my watch.

let t = t0 / sqrt(1 - v^2 / c^2), where v = the velocity of the watch moving at 0.9c relative to me.

so, with my wrist watch, the present period of time (as i see it) is 8:00 pm thru 9:00 pm, but with the watch moving at 0.9c, the present period of time would have been (as i would have seen it) 8:00 pm thru 10:17 pm.

for an observer at rest relative to the watch moving at 0.9c relative to me, the present period of time would have been 8:00 pm thru 9:00 pm, but according to my wrist watch, the present period of time for him (or her) would have been 8:00 pm thru 10:17 pm.

please let me know what you think. if this idea is not consistent with Einstein’s theory of relativity, let me know what is wrong. thanks!

2. Aug 30, 2006

### Zeit

Hello,

t and t0 are the durations of a phenomenon for two differents inertial reference frames.

Let say that, as you said, t0 is the period of time according to your wrist watch (in your reference frame) and t, the period of time according to the watch (in its reference frame) moving at 0.9c for you (in your reference frame).

So, t = t0/sqrt(1 - v^2 / c^2) . It means that for you, the duration needed by the small hand of your wrist watch to go from the number 8 to the number 9, and would be in the other reference frame (and measured by this clock) the duration needed by the small hand of the clock to go from the number 8 to the number 10 and a bit more.

t0 is the period of time according to the clock for a phenomenon in the same reference frame. t is the period of time according to your wrist watch for a phenomenon in the clock's reference frame.

Am I clear? (Excuse me for my english...)

3. Aug 30, 2006

### myoho.renge.kyo

i think i understand.

now, the period of time for me is from 9:00 am thru 10:00 am (8/30/2006).

suppose that the phenomenon in the same reference frame is clock A (which is synchronized with my wrist watch). so the phenomenon would be the present period of time from 9:00 am thru 10:00 am. according to my wrist watch, the period of time of the phenomenon is 9:00 am thru 10:00 am.

suppose also that the phenomenon in the clock's reference frame is clock B (which is also synchronized with my wrist watch). so the phenomenon would also be the present period of time from 9:00 am thru 10:00 am according to the clock in that reference frame (as seen by someone in that reference frame).

but according to my wrist watch, the period of time of the phenomenon in the clock's reference frame would be from 9:00 am thru 11:17 am (as i see it in my reference frame).

is that right? thanks!

4. Aug 30, 2006

### Zeit

There is probably no other way to understand the meaning of time dilatation than to learn to find the equation of time dilatation.

You're A, an observer in the reference frame named $R_A$ and there's also B, another observer which has the reference frame $R_B$. You both have your wrist watch named respectively $w_A$ and $w_B$.

There is also a spaceship composed of two parallel mirrors. A light beam ($v = c = 299 792 458 ms^{-1}$) travels from one mirror to the other. The mirrors are distant from a distance $D$. Your experience is to measure the duration needed by light to come back to the first mirror.

In $R_B$, the spaceship is motionless and we note the duration $t_B$. In $R_A$, the spaceship moves at a speed $v$ and we note the duration $t_A$. We have $D_A$, half the distance traversed by light during $t_A$ and $D_B$, half the distance traversed by light during $t_B$.

$$t_A = \frac{2{D_A}}{c}$$

$$t_B = \frac{2{D_B}}{c}$$

$${D_A}^2 = {D_B}^2 + (v\frac{t_A}{2})^2$$

Using those those equations, you find :

$${t_A}^2 = \frac{4{D_A}^2}{c^2} = \frac{2{D_A}}{c} = \frac{4{D_B}^2}{c^2} + {v^2 \frac{{t_A}^2}{c^2}$$

Then,

$${t_A}^2 = \frac{4{\frac{c{t_B}}{2}}^2}{c^2} + {v^2 \frac{{t_A}^2}{c^2} = {t_B}^2 + {v^2 \frac{{t_A}^2}{c^2}$$

You put the $t_A$ together :

$${t_A}^2 (1-\frac{v^2}{c^2}) = {t_B}^2$$

And you have :

$${t_A}\sqrt{1-\frac{v^2}{c^2}} = {t_B}$$

So, if you know $t_A$ and $v$, you can calculate the duration $t_B$ measured by B in $R_B$.

It's call time dilatation because duration of a phenomenon is longer in a reference frame where the phenomenon is moving than in a reference frame where it is motionless. The equation above isn't completed, there's a larger version which I don't know correctly.

I hope that I have been clear

Last edited: Aug 30, 2006
5. Aug 30, 2006

### bernhard.rothenstein

time dilation

I think the way in which you teach time dilation is the best one. You do not mention how the mirror is orientated and that D_B=D is a relativistic invariant.
sine ira et studio

6. Aug 30, 2006

### cesiumfrog

There may be some event when both watches are at the same place and read the same hour, but you can't really keep two clocks synchronised unless they are stationary with respect to each other.

To unambiguously compare the time elapsed on the watches, they both must return to a common place at some second event..

7. Aug 31, 2006

### myoho.renge.kyo

$${D_A}^2 = {D_B}^2 + (v\frac{t_A}{2})^2$$

why? thanks! (11:00 pm thru 12:00 am, 8/30/2006)

8. Aug 31, 2006

### bernhard.rothenstein

....................................................................

I would rather say that a stationary clock and an uniformly moving one can be brought to display the same time say t=t'=0 (initialisation) when they are located at the origin O of the rest frame of the stationary clock. The moving clock moves in the positive direction of the OX axis along which we find the synchronized (a la Einstein) clocks of that frame. Let t' be the reading of the moving clock when he arrives in front of a stationary clock that reads t. Then t and t' are related by the time dilation formula. So your statement:
"To unambiguously compare the time elapsed on the watches, they both must return to a common place at some second event" is not compulsory.
sine ira et studio

9. Aug 31, 2006

### bernhard.rothenstein

because D=D_B is an invariant, because the mirrors of the light clock are parallel to the direction in which the light clock moves and because Pythagoras' theorem. You can find a transparent derivation in many places see for instance
Hans C. Ohanian
Special Relativity: A modern introduction
Physics Curriculum and Instruction Inc. pp.82-84

sine ira et studio

10. Aug 31, 2006

### myoho.renge.kyo

i was thinking along the following lines:

the current period for A is 1:00 am thru 2:00 am (8/31/2006) in RA.

wA = the wrist watch of A

the current period for B is also 1:00 am thru 2:00 am (8/26/2006) in RB.

wB = the wrist watch of B.

wA and wB synchrinize (The Principle of Relativity, p. 40, by A. Einstein).

v = the velocity of RB relative to RA.

D = the distance between two parallel mirrors at rest relative to RB.

c = the velocity of the beam of light (299,792,458 m / s).

tA = the time required by the light beam to come back to the first mirror (as measured by A with wA).

tB = the time required by the light beam to come back to the first mirror (as measured by B with wB).

allow me to define DA and DB as follows:

DA = the distance traversed by the beam of light during tA.

DB = the distance traversed by the beam of light during tB.

let tB = the period from 1:00 am to 2:00 am (8/31/2006).

tB = DB / c

tA = (DA + v*tB) / c, since c is a constant (assume that the direction of motion of the light beam from the second to the first mirror is the same as the direction of motion of RB relative to RA).

since DA = DB, tA = (DB + v*tB) / c

since DB = c*tB (tB = DB / c), tA = (c*tB + v*tB) / c

= tB*(c + v) / c

i can't get tA = tB / sqrt(1- v^2 / c^2)

i don't know what to do. thanks! (1:00 am thru 2:00 am, 8/31/2006)

Last edited: Aug 31, 2006
11. Aug 31, 2006

### bernhard.rothenstein

since DB = c*tB (tB = DB / c), correct (1)
t_A should be calculated using Pythagoras' theorem
(ct_A)^2=(DB)^2+(Vt_A)^2 (2)
eliminating DB between (1) and (2) you obtain the result ou are looking for.

12. Aug 31, 2006

### myoho.renge.kyo

i was instead thinking along the following line of reasoning:

the current period of time for A is from 11:00 am thru 12:00 pm (8/31/2006) in RA.

let's call RA the "stationary" system

wA = the wrist watch of A

the current period of time for B is also from 11:00 am thru 12:00 pm (8/31/2006) in RB.

let's call RB a moving body

wB = the wrist watch of B.

wA and wB synchronize

we have defined a common "time" for A and B when A and B were at rest relative to each other (we have established by definition that the "time" required by light to travel from A to B equals the "time" it requires to travel from B to A, The Principle of Relativity, p. 40, by A. Einstein)

v = the velocity of RB relative to RA.

D = the distance between the two parallel mirrors at rest relative to RB (the distance traversed by the beam of light from the second to the first mirror during tA and tB)

tB = the current period of time for A and B

d = the distance traversed by RB in RA during t

t = the current period of time for A and B

c = the velocity of the beam of light (299,792,458 m / s).

"any ray of light moves in the "stationary" system of coordinates with the determined velocity c, whether the ray be emitted by a stationary or by a moving body" (The Principle of Relativity, p. 41, by A. Einstein)

tB = D / c

t = d / v

tA = D / c, since c is the same for A and B (assume that the direction of motion of the light beam from the second to the first mirror is the same as the direction of motion of RB relative to RA).

if c was not the same for A and B, then tB = D / c, and tA = D / (c + v).

t = d / v (in either case)

the result of the line of reasoning above does not agree with the following:

tA = tB / sqrt(1 - v^2 / c^2)

why? thanks! (11:00 am thru 12:00 pm, 8/31/2006)

Last edited: Aug 31, 2006
13. Aug 31, 2006

### bernhard.rothenstein

because you do not take into account that the mirrors are parallel to the direction of relative motion the light signals propagating along the O'Y' axis of the frame where the mirrors are at rest.

14. Aug 31, 2006

### Zeit

Yes, but I have taken so much time to write my message and because English is not my best discipline I forgot to specify.

15. Sep 1, 2006

### myoho.renge.kyo

even if i do take into account that the mirrors are parallel and the propagation of the light beam perpendicular
to the direction of relative motion, is D the same in either case and is c still the same for A and B? if they are, tA = D /c and tB = D / c. if not, why not. i don't understand. thanks! (6:00 pm thru 7:00 pm, 9/1/2006)

16. Sep 1, 2006

### bernhard.rothenstein

D and c are the same for both observers as a consequence of the principle of relativity. There is an experiment which will convince you that D is the same, decribed in many introductory textbooks and which does not involve light signals at all. It involves two meter sticks perpendicular to the direction of relative motion and a piece of chalk. Other authors convince us simply stating that in the standard arrangement of the involved inertial reference frames, the relative velocity has no component in the direction perpendicular to the relative motion. Concerning the invariance of c it is the single assumption which leads to results in accordance with the experiments performed so far.