# Time Dialation Equation

1. Jul 16, 2008

### PatPwnt

Can someone give me a simple derivation of the ux/c^2 in the numerator of the Lorentz time contraction equation? u = relative velocity

Why isn't the equation just (t-x/c)/gamma because of the time delay in light going from x to O. Im guessing because the source is moving away from O, we have to multiply x/c by u/c? I hope someone can understand what I am trying to ask.

2. Jul 16, 2008

### bernhard.rothenstein

Do you mean "in the time transformation equation"?

3. Jul 16, 2008

### JesseM

What kind of derivation are you looking for? A derivation of the Lorentz transformation from just the two postulates of SR, or a derivation of the time transformation equation which already assumes the length contraction and time dilation equations?

4. Jul 16, 2008

### PatPwnt

Yes, I mean if you look at the time transformation equation, in the numerator there is a
t - ux/c^2

How do we get the ux/c^2?

I think I may have got it on my own because it seems similar to the dopplar effect: When an object, emitting a certain frequency, is moving away from an observer, it appears to run at a slower frequency with respect to that observer. So, the same thing happens with the apparant passage of time of an object when it travels away from you. You get the time the light takes to get to you (x/c) multiplied by the apparant slow down factor (v/c). Then, after all that, you put in the simple Lorentz transformation/gamma factor.

5. Jul 16, 2008

### humanino

There are a few derivations on wiki

6. Jul 16, 2008

### JesseM

You didn't answer my question, do you want an answer that takes for granted the time dilation equation for moving clocks and the length contraction equation for moving rulers, or do you want a derivation of the time transformation equation which starts from only the two basic postulates that Einstein started with (the postulate that the laws of physics are the same in all inertial frames, and the postulate that the speed of light is the same in all inertial frames)? The latter will be more complicated and probably less easier to follow conceptually, but a derivation of the Lorentz transformation from the two postulates can be found in any SR textbook.

Also, are you familiar with the "relativity of simultaneity"? It may help to realize that if I see a rod moving at speed u with length x in its own rest frame (so in my frame its length is $$x*\sqrt{1 - u^2/c^2}$$, then if there are two clocks at either end which have been synchronized in the rod's own rest frame, in my frame one clock's time will be ahead of the other clock's time by exactly ux/c^2.

7. Jul 16, 2008

### PatPwnt

Sorry, I understand how we derive the time dialation equation in that we should get
t' = (gamma)t for x = 0 but, I was wondering where the ux/c^2 came from, and that was all that I was asking. I didn't need a derivation of how we get the gamma factor, just why we put that ux/c^2 in the time dialation equation. "u" being relative velocity.

Why isn't it just
t' = (gamma)(t-x/c)? then, I realized that what I had said in my second post may be the reason why we multiply x/c by u/c. But, I am still unsure.

8. Jul 16, 2008

### JesseM

OK, so does this mean it's all right to assume we already know the time dilation equation (and the length contraction equation) when I show you where the ux/c^2 comes from? And you didn't answer my question about whether you were familiar with the idea of the relativity of simultaneity, namely that clocks at different locations which are defined as "synchronized" in their own rest frame will be out-of-sync in another frame.

9. Jul 16, 2008

### PatPwnt

I am familiar with simulaneity being relative and the time/length contraction equations. I just want you to show me where the ux/c^2 comes from, in the time dialation equation.

10. Jul 16, 2008

### JesseM

Well, do you understand the reasons for the specific fact about relative simultaneity I mentioned earlier, namely that if two clocks are synchronized and a distance x apart in their own rest frame, then in a frame where they are moving at speed v along the axis between them, they'll be out of sync by ux/c^2? If not I can show how to derive this from time dilation, length contraction and the fact that each frame synchronizes their own clocks using the assumption that light moves at c in all directions in that frame. Once we already have ux/c^2 in the context of the relativity of simultaneity, it's not too hard to explain why the same factor appears in the time transformation equation (keep in mind that the coordinates of each frame are supposed to be based on local readings by a system of rulers and clocks which are at rest in that frame, with the clocks synchronized according to that frame's definition of simultaneity).

11. Jul 16, 2008

### PatPwnt

When you say the that two clocks are sychronized, do you mean that if I am in the middle of a train compartment and there is a clock at both ends of the compartment that they both read the same times to me standing mid way(x/2) between them? And that they will appear out of sync with each other by ux/c^2 if I am at rest watching the train compartment drive by?

Yes, I want you too show me why this is.

12. Jul 16, 2008

### JesseM

It's not about visual appearances, but about how different observers assign time-coordinates to events. In SR the concept of an inertial reference frame is usually introduced by assuming each observer has a network of rigid rulers and clocks at rest with respect to them, and that the time and position coordinates of distant events are based on local readings on this system. For example, if when my clock reads 2008 I notice a distant explosion in my telescope, and I see that it happened right next to the 4-light year mark on my ruler, and that a clock attached to that mark showed a reading of 2004 as the explosion was happening, then I would assign the event of the explosion a time-coordinate of t=2004, not t=2008. But in order to do things that way, each observer must have a way of defining what it means for clocks at different positions along their rulers to be "synchronized", so this is where Einstein suggested that each observer could define clock synchronization based on the assumption that light moves at the same speed in all directions in that observer's frame, so that an observer could synchronize two clocks by setting off a flash at the midpoint between them, and having them both show the same reading at the moment the light from the flash reaches each one. A consequence of this definition is that different observers will disagree about simultaneity. For instance, if I'm on a ship with clocks at each end, I can synchronize them in the ship's rest frame by setting off a flash at the midpoint, and setting them to read the same time when the light reaches them. But if the ship is moving forward in your frame, then since your frame's coordinates are constructed around the assumption that light moves at the same speed in both directions relative to you, then naturally since the back of the ship is moving towards the point where the flash was set off while the the front of the ship is moving away from that point, in your frame the light will catch up with the back clock before it catches up with the front clock, so the back clock's reading will be ahead.

We can see how much it'll be ahead by using the time dilation and length contraction equations. Say my ship is moving forward along your x-axis at speed u in your frame, and in my frame it has a length of x', so using the Lorentz contraction equation, in your frame the length is only $$x' * \sqrt{1 - u^2/c^2}$$. Say a flash is set off at the midpoint of the ship when the midpoint is next to the x=0 mark on your ruler, so at that moment in your frame, the back clock is at position $$x= \frac{-x' * \sqrt{1 - u^2/c^2}}{2}$$ while the front clock is at position $$x = \frac{x' * \sqrt{1 - u^2/c^2}}{2}$$. Since both clocks are moving foward at speed u, the back clock's position as a function of time is given by:

$$x(t) = \frac{-x' * \sqrt{1 - u^2/c^2}}{2} + ut$$

and the front clock's by:

$$x(t) = \frac{x' * \sqrt{1 - u^2/c^2}}{2} + ut$$

Meanwhile, the light beam going in the -x direction has position as a function of time x(t) = -ct, while the light beam going in the +x direction has x(t) = ct. So, to figure out when the light hits the back clock, we solve for t in this equation:

$$-ct = \frac{-x' * \sqrt{1 - u^2/c^2}}{2} + ut$$

...which gives $$t = \frac{x' * \sqrt{1 - u^2/c^2}}{2*(c + u)}$$

And to figure out when light hits the front clock, we solve for t in:

$$ct = \frac{x' * \sqrt{1 - u^2/c^2}}{2} + ut$$

which gives $$t = \frac{x' * \sqrt{1 - u^2/c^2}}{2*(c - u)}$$

If you subtract the first time from the second, you get:

$$\frac{x' (c + u) * \sqrt{1 - u^2/c^2}}{2*(c - u)*(c + u)} - \frac{x' * (c - u) \sqrt{1 - u^2/c^2}}{2*(c + u)*(c - u)}$$ = $$\frac{x' * u * \sqrt{1 - u^2/c^2}}{c^2 - u^2}$$

So, this shows that in your frame, the light doesn't hit the front clock until this amount of time has passed since the light hit the back clock. If both clocks were set to read some time T when the light hit them, then the back clock has been ticking forward past T in this time, but it is slowed down by factor of $$\sqrt{1 - u^2/c^2}$$ thanks to time dilation, so when the light hits the front clock and it reads T, the back clock will read $$T + \sqrt{1 - u^2/c^2} * \frac{x' * u * \sqrt{1 - u^2/c^2}}{c^2 - u^2}$$ = $$T + \frac{x' * u * (1 - u^2/c^2)}{c^2 - u^2}$$. And since (1 - u^2/c^2) = (1/c^2)*(c^2 - u^2), this works out to $$T + \frac{x' * u * (c^2 - u^2)}{c^2 * (c^2 - u^2)}$$ = $$T + \frac{x' * u}{c^2}$$. So, this shows that in your frame, the back clock is ahead of the front clock by x'*u/c^2, where x' is the distance between the clocks in their own rest frame and u is their speed in your frame.

Let me know if you follow this, and if so I can then explain how to derive the time transformation equation from it.

13. Jul 17, 2008

### PatPwnt

Wow, nice work. Yea, I followed everything easily. Please continue with the derivation for the whole transformation; I appreciate the help.

14. Jul 17, 2008

### JesseM

OK, keep in mind what I said about each observer's coordinate system being defined in terms of local measurements on their own ruler/clock system (with their own clocks 'synchronized' using the light signal method). So imagine we have a ruler at rest in observer A's frame which defines his x-axis, and another ruler is sliding alongside it at velocity u in the +x direction, this second ruler defining observer B's x'-axis. It's also assumed in the Lorentz transformation that at the moment the origins of each observer's system are right next to each other (x=0 on observer A's ruler lining up with x'=0 on observer B's ruler), clocks at each observer's origin read a time of 0 (the clock at x=0 reads t=0, and the clock at x'=0 reads t'=0).

Suppose some event E occurs right next to these two rulers, and according to the measurements of observer B, it happens right next to some mark x' on her ruler, and the clock at that mark reads a time of t'. What position will this event be next to on observer A's ruler, and what time will his own clock at that position be reading when the event happens? Well, we know that the mark x' on observer B's ruler must be at a distance of exactly x' from the origin in B's own frame, although the distance is smaller in A's frame. And we know that in A's frame, if there's a clock at B's origin (x' = 0) and another clock at the mark x', then if they are synchronized in B's frame, in A's frame they must be out-of-sync by exactly x'*u/c^2, with the time on the clock at the origin being ahead of the clock at x' by this much (since the clock at the origin is the trailing clock in terms of their motion in A's frame). So, in A's frame, at the "same moment" (according to A's definition of simultaneity) that the event E occurs and B's clock at E reads at time of t', B's clock at the origin must read a time of t' + x'*u/c^2. And furthermore, we know that in A's frame, the clock at B's origin read t'=0 at the time t=0 in A's frame, and ever since then it has been ticking forward at a slowed-down rate of $$\sqrt{1 - u^2/c^2}$$. This means that at any time T in A's frame, the clock at B's origin only reads a time of $$T' = T * \sqrt{1 - u^2/c^2}$$. Turning this around, when the clock at B's origin reads a time of T', the time in A's frame must be $$T = \frac{T'}{\sqrt{1 - u^2/c^2}}$$, or gamma * T'. So, for example, when the clock at B's origin reads a time of (t' + x'*u/c^2), the time in A's frame must be t = gamma*(t' + x'*u/c^2). And remember that in A's frame, the event of the clock at B's origin reading (t' + x'*u/c^2) was simultaneous with the event of the clock at position x' on B's ruler reading t', which happened right next to the event E as it was occurring, so it must be true that in A's frame the event E has time coordinate t = gamma*(t' + x'*u/c^2). And that's the time transformation equation! You can easily switch things around and look at things from the perspective of B's frame, where A's ruler is moving in the -x' direction at speed u, and if you imagine some other event E2 which happens next to the mark x on A's ruler and A's clock there reads t, then you can translate back into B's frame to get the time transformation equation t' = gamma*(t - x*u/c^2).

15. Jul 18, 2008

### PatPwnt

Okay, I think I finally get it. Although it is more complicated than I thought it should be, I am satisfied and I thank you for that. However, I don't understand why we synchronize the clocks in B's fram according to the midpoint and not the origin x' = 0 or does it matter? Will the clocks always be out-of-sync in A's frame by ux/c^2 no matter which point we synchronize it in B's frame?

16. Jul 18, 2008

### JesseM

The point is just to synchronize the clocks using the assumption that light moves at the same speed in both directions in the clocks' own rest frame. One way of doing this is by setting off a flash at the midpoint; if you assume light moves at the same speed in both directions, it should reach both clocks at the same time. But you don't need to use the midpoint, no. For example, if you have two clocks A and B, you could send a flash of light from the position of A when that clock reads T_A1, note the time T_B on clock B when the light is reflected from its position, and note the time T_A2 on clock A when the reflected light returns to it; then the clocks are defined as "synchronized" if T_A2 - T_B = T_B - T_A1 (i.e. the time measured for the light to travel between the two clocks is the same in both directions). This was actually the procedure Einstein suggested in his original 1905 paper. But this synchronization procedure is equivalent to the one I suggested before about setting off a light flash at the midpoint, so both procedures will result in the fact that for an observer who measures the clocks moving at speed u, the clocks will be measured to be out-of-sync by u*x'/c^2.

17. Jul 18, 2008

### PatPwnt

Okay, so if I am standing next to one of the clocks in the B frame, and the ship is 2 light-seconds long, I am still allowd to say that the clocks are synchronized even if I look down at the other end of the ship and see that the clock there is 2 seconds behind the clock I am standing next to? Because if I walked to the center, the two clocks would appear to be in synchronization.

18. Jul 18, 2008

### JesseM

Yes, provided the ship is 2 light-seconds long in its own rest frame and not in some other frame, this definition of synchronization will match the others.

19. Jul 18, 2008

### PatPwnt

Alright then. I can finally sleep, thank you! :D