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Grimble
2009-Aug-28, 12:45 PM
*** to the moderators ***
I am having troble enetring a title here! It will allow me three letters but as soon as I add a fourth - e it goes walk-about and leaves me with a blank screen. The title I was trying to pu was 'Time dilation formula?' but if you could help me to start this thread I would be happy with whatever title you chose to give it. Thank you.
********************

I know that this is a very basic question but what is the correct formula for time dilation?

In Wikipedia etc. I read t' = (gamma)t or at least (delta)t' = (gamma) x (delta)t; yet in this mathematical description (http://en.wikipedia.org/wiki/Twin_paradox#Difference_in_elapsed_time_as_a_resul t_of_differences_in_twins.27_spacetime_paths) 'phase 2' and 'phase 5' imply that the formula is t' = t/(gamma).

(sorry, but is there a guide anywhere concerning how to write special characters on these pages?)

Also, if a moving clock is seen to 'go slow' by a stationary observer, then one would expect that less time would be seen to pass in the transformed time, and t' = t/(gamma) seems to me to fit that scenario.

I have been looking at this for some time on the internet but taking heed of the warnings I have been given about believing all I read on there I have followed the arguments and read the 'derivations' and suchlike, but have a problem:

Whichever way I approach it the formula appears to be the latter viz. t' = t/(gamma) in the same way that x' = x/(gamma) the formula for length contraction.


where:
t is the time on the stationary observer's local clock and
t' is the travelling clock's time, transformed by the lorentz transformation formulae.

Or are there different formulae applied in different circumstances.

We talk of time dilation - expansion(?) yet also about the moving cock slowing (less time passing)?

:confused::confused::confused:

NEOWatcher
2009-Aug-28, 12:53 PM
*** to the moderators ***
You may want to contact the mods directly, a statement like this can get overlooked.

I am having troble enetring a title here! It will allow me three letters but as soon as I add a fourth - e it goes walk-about and leaves me with a blank screen. The title I was trying to pu was 'Time dilation formula?'
The word "time" in a title is a known problem on the board. It's been discussed in the about threads if you're interested in searching.

Swift
2009-Aug-28, 01:04 PM
As NEOWatcher said, there is a bug in the software that doesn't like the word "time" in thread titles. Not like we talk about time in a physics and astronomy forum :D. Anyway, I changed the title to what you were after, and throw a couple of spaces in to fool the software.

tusenfem
2009-Aug-28, 01:13 PM
Dear Grimble

the V-bulleting system has something funny that it does not allow the word "time" in the title of a thread. Don't ask me how or why, it just is.

About special characters. You can either write your message in e.g. word and the copy the text into the window, or if you use Firefox as a browser I would advise to install the abcTajpu add-on, which will give you all the special characters you will ever need with a click of the right button of your mouse.

Then for the time dilation and Lorentz contraction
Stationary observer: x and t
Moving observer : x' and t' @ velocity v

Lorentz factor: (1 - v2/v2)-1/2, note that this is smaller than 1 (I get the impressoin that you forget that gamma it is 1 divided by the squar root

Now the stationary observer sees the clock of the moving observer as:

t' = γ t (time dilation)

When the stationary observer sees a distance of L to e.g. a planet, then the moving observer sees a distance:

L' = γ L (Lorentz contraction)

And then to do the calculation on the wiki page you have to be very careful to get the right time and velocity etc. It is just regular math, but you have to keep your mind on it.

Grimble
2009-Aug-28, 01:49 PM
Then for the time dilation and Lorentz contraction
Stationary observer: x and t
Moving observer : x' and t' @ velocity v

Lorentz factor: (1 - v2/v2)-1/2, note that this is smaller than 1 (I get the impressoin that you forget that gamma it is 1 divided by the squar root



I'm sorry but I'm becoming confused again; surely the square root of a number that is <1, (and 1 - anything except zero has to be <1), and the reciprocal has to be >1. And, as you indicate, gamma, the Lorentz factor, is a reciprocal?
:confused::confused::confused:

tusenfem
2009-Aug-28, 05:21 PM
Methinks I made a typo there

Perikles
2009-Aug-28, 05:42 PM
the V-bulleting system has something funny that it does not allow the word "time" in the title of a thread. Don't ask me how or why, it just is.
Um, off-topic and unimportant, but I post on another forum run by vBulletin and I have just found 87 threads with 'time' in the thread title. Weird.

Grimble
2009-Aug-31, 08:39 AM
Methinks I made a typo there

OK, we all do that from time to time:lol:
But does that mean that the formulae are therefore the other way round?
i.e. t' = t/γ and L' = L/γ?
as the obvious one, length contraction would then read, if I am not getting muddled again,:whistle: L' (i.e.transformed/co-ordinate time) is less than L (Proper time in the stationary observer's frame of reference), i.e. contracted?:doh:

And thanks for the assistance I use Firefox and that is a BIG help!:lol::lol::lol:

Grimble
2009-Oct-05, 10:27 AM
I have been working on producing a diagram that pulls the multiplicity of elements constituting time dilation and length contraction together to see if I can find a visual representation that enables me to get a feel for just what is happening and this is what I have arrived at so far:

http://img41.imageshack.us/img41/5448/specialrelativitydiagra.jpg

A Special Relativity diagram to demonstrate the relationship between two Inertial Frames of Reference moving with a constant relative velocity.
A and B represent a single axis (time or length) of Minkowski spacetime for each of the two IFoRs. They are drawn against a common background representing Proper Units
An observer at rest within each IFoR will be experiencing proper units (length and time) within that system; as shewn by the horizontal lines, labelled A and B.
But from each IFoR, the other frame's axis -- rotated according to their relative velocity -- will be reckoned in co-ordinate units, as shewn by the perpendicular projections from the coloured diagonals onto the observer's own axis.
The diagram is drawn to scale to represent two IFoRs with a constant relative velocity = 0.6c, giving γ = 1.25 and 1/γ = 0.8

From this we can see exactly what Einstein was saying in chapter XII (http://www.bartleby.com/173/12.html), when he writes:

But the metre-rod is moving with the velocity v relative to K. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is
http://www.bartleby.com/173/M2.GIF

of a metre.

For the metre rod moving with the velocity 0.6c relative to B would be represented in the diagram by the number 1 on the red diagonal and we can see the projection onto B's x axis (as it would be in this scenario) where it would be the green 1 co-ordinate unit.
And this agrees with x' = x/γ
i.e. x' = 0.8x

Similarly, in the second part of that same chapter, Einstein writes:

**Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and
http://www.bartleby.com/173/E5.GIF

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but
http://www.bartleby.com/173/M5.GIF

seconds, i.e. a somewhat larger time.

And again we can see just how this works, for this time he is converting the time from the observer's frame, t (proper time units) into the observed time t' (co-ordinate time units) which would be to take the blue, 1 proper unit and project it upwards onto the red diagonal line or, indeed, one could read it off the green co-ordinate scale on B's axis.

Not surprisingly this agrees with Einstein's own equation:
http://www.bartleby.com/173/E5.GIF

or t = γt' = 1.25 co-ordinate units

I am fairly certain that this diagram meets all the conditions, references and relationships between the elements that compose SR as far as I have understood it.

For instance, the co-ordinate units are greater in number but reduced in size.

The principal problem that is brought to light here is that there are far more elements than at first appear.
Consider if you will:
1) We start with a solitary IFoR where the measurements are all, by definition, in proper units.
2) We add a 2nd IFoR moving at a constant velocity with respect to the first: both are measured in proper units and their times are identical and synchronous.
3) They then observe one another and upon doing so we find that the observed frames are rotated with respect to their observers, as shewn, but their units are still proper units.
4) When observing the rotated frames, their proper units are projected vertically onto the observer's frame of reference, being there-by transformed into co-ordinate units.
5) So we have the axis of each IFoR and its rotation, both measured in Proper units and its projection onto the other's axis measured in co-ordinate units.
6) And as all this is matched reciprocally by the other IFoR we have this duplicated giving 4 measures in proper units and two in co-ordinate units.

Is it any wonder that we become confused when trying to deal with this using only primed and unprimed symbols?
And this is with it all reduced to two dimensions...

This exercise has certainly helped me to understand how it all fits together I only hope that I have understood it correctly?