# Lorentz Transformation Equations, time transformation

1. May 6, 2015

### Grimble

t' = γ(t - vx/c2)
where t is the time in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
and x is the distance travelled in the time t in the stationary frame.
∴ x = vt
substituting this into the Lorentz time equation gives us:
t' = γt(1 - v2/c2) = t/γ

So why is the formula not written this, t' = t/γ, much simpler way?

2. May 6, 2015

### Orodruin

Staff Emeritus
Because it is only true when x = vt, i.e., for the world line of an observer at rest in S'.

3. May 6, 2015

### Grimble

So can you define what those terms refer to in the equation?

4. May 6, 2015

### Orodruin

Staff Emeritus
The x, x', t, and t' are simply coordinates that describe any event in space-time, not necessarily an event along a given world line. The Lorentz transformations tell you how coordinates in one frame relate to coordinates in another frame, much like x' = x cosθ - y sinθ, y' = x sinθ + y cosθ would tell you how coordinates transform under a spatial rotation.

5. May 6, 2015

### Grimble

Thank you, but that still leaves a question about the transformation equations; are the terms coordinates that are transformed (in which case there would surely be some factor that represnts the differences in the frames origins) or are they measurements between events?
I.e. is x really x1 - x2 and x', x'1 - x'2?

What is it that the LTE is transforming?
(I always thought it was how the relative velocity affected measurements made in one frame when converting them to be relative to another frame)

6. May 6, 2015

### Staff: Mentor

The things being transformed are coordinates, the coordinates of a single point in spacetime. And there is indeed a term in them that represents the difference in the origins of the two coordinate systems: it's the $vx/c^2$ term that you were asking about when you started this thread (and the $vt$ in the corresponding transformation for the $x$ coordinate).

The length contraction and time dilation formulas tell you how to convert measurements between events, but these are both derived from the coordinate transforms by looking at the coordinates of two events. The length contraction formula comes from looking at the difference between the x coordinate of the event "one end of the object is here at time T" and the event "the other end of the object is there at the same time", for example.

It is a really good exercise to derive the time dilation and length contraction formulas directly from the Lorentz transforms. It's just simple algebra, but I doubt that it is possible to make sense of SR without going through this exercise at some point.

Last edited: May 6, 2015
7. May 6, 2015

### Orodruin

Staff Emeritus
Just to clarify what Nugatory means here. A "single point" (or "event") in space-time is a point in space and time. It is not enough to specify a location, you need to specify a time as well. Given a location and time for one event in one system, they tell you which time and location the event occurs at in the other system.

8. May 7, 2015

### Grimble

So, if vt represents the spatial difference between the origins in the xx' transformation and x represents the same in the tt' transformation, then as x & vt represent the same value it still seems a reasonable substitution to simplify the formula as in my OP.

All I am doing is trying to reason from what you are telling me...

9. May 7, 2015

### harrylin

The much simpler way is like this, for small velocities (so that γ≈1): t' = t - vx/c2
That is called relativity of simultaneity and it dates from before SR.

It's very easy to work out if you think how someone who is in motion according to you will determine distant time by means of light or radio signals, if that person assumes to be in rest.

10. May 7, 2015

### PeroK

One of the good things about a formula is that you can plug some numbers in and see what you get! So, try the following. Take $v=\frac{3}{5}c$ and $\gamma = \frac{5}{4}$:

Find the coordinates in S' of:

1) The origin in S at times t = 0, 1, 2. I.e. find x' and t' for these "events".

2) The point at x = 1 in S at times t = 0, 1, 2. Again find x' and t' for these.

11. May 7, 2015

### Grimble

When t = 0 , 1, 2
x' = 1.25(0 - 0.6t), 1.25(1 - 0.6t), 1.25(2 - 0.6t) where t is the time elapsed since the separation of the origins. Or are you specifying that O and O' are coincident?
(t in this case is not 0 as vt is the separation of the Origins as is x in the time transformations
t' = 1.25(0 - 0.6x), 1.25(1 - 0.6t), 1.25(2 - 0.6t)

Please note that I am not trying to be difficult but to pull together all the different things I am told, because what seems right in one context seems problematical elsewhere.

12. May 7, 2015

### Grimble

Why? Are we not dealing with Events - they are surely fixed, not moving?
Light or radio signals from an event will travel at c and the time taken will be added to the time of the event...

13. May 7, 2015

### Orodruin

Staff Emeritus
What do you mean by fixed? Events are just a given place and a given time. There is no way of assigning a velocity to an event.

14. May 7, 2015

### PeroK

You're not alone in struggling with a formula and what it means and not realising that you can gain an insight just by plugging in some numbers. However, I've no idea what you've done above. What I meant was:

2) x = 1, t = 0:

$t' = \frac{5}{4}(0 - \frac{3}{5c}) = -\frac{3}{4c}$

So, the time of this event in S' is at t' slightly less than 0. Your simplified formula would not show this at all.

Perhaps a better thing to calculate next would be:

3) $x= 3 \times 10^8$, $t = 0$:

$t' = \frac{5}{4}(0 - \frac{3}{5}) = -\frac{3}{4}$

So, the further from the origin we go, the earlier the event (at t = 0) takes place in S'.

Finally:

4) $x= -3 \times 10^8$, $t = 0$:

$t' = \frac{5}{4}(0 + \frac{3}{5}) = +\frac{3}{4}$

So, times to left of the S origin (at t = 0) are actually ahead in S' (relative to S).

The point is that plugging in these numbers should give you an insight on what the LT means and why it's not a simple $t' = \gamma t$.

15. May 7, 2015

### Grimble

What do I mean by fixed? - I mean at a specific location in space at a specific point in time! i.e. Not moving! I.E having no velocity!

What can be confusing about the term fixed?

16. May 7, 2015

### Grimble

What I did above was to substitute values into the Lorentz Equations.
BUT, in the x transformation the term vt refers to the separation of the Origins so the t in vt refers to the time it would take for the moving frame to move from O to where O' was at time t = 0, t' = 0!
t in this case is NOT the time coordinate!

x' is the point on the X axis when t = t' = 0, when the two origins are not collocated.
therefore one has to subtract vt - the distance that would be travelled by O' from O before the start of the transformation. So obviously x' is x - vt or the separation of the origins at the start.
If the two origins were collocated then the term vt would disappear.

Come on now, that is not difficult to understand...

17. May 7, 2015

### PeroK

$t$ and $x$ in the Lorentz transformation are independent variables/coordinates. You can plug in any $t$ and $x$ and get $t'$ and $x'$.

You are restricting yourself to the case where $x = vt$. That is, in fact, the spatial origin of the S' frame over time: $x' = 0$ and $t' = \gamma t$.

The LT applies to any $x$ and any $t$; not just those that satisfy $x=vt$.

18. May 7, 2015

### harrylin

The term (t - vx/c2) was introduced by Lorentz based on the propagation of EM radiation, and he called it "local time".
Here's Einstein's clarification with the use of events: http://www.bartleby.com/173/9.html

19. May 7, 2015

### Staff: Mentor

No. In the Lorentz transform t is the time coordinate of the unprimed frame. Always.

The Lorentz transform tells you how coordinates (t,x,y,z) transform to coordinates (t',x',y',z') in a frame moving at v relative to the first frame.

Last edited: May 7, 2015
20. May 7, 2015

### Staff: Mentor

You're confusing yourself, by using the terms "location in space" and "point in time" as if they were absolute. They're not. They're frame-dependent.

Take a simple example: you and I are both way out in empty space somewhere in our spaceships. You look at your clock and it reads exactly 12 noon. At that instant, I fly past you at relativistic speed, and you get an image of my clock, also reading exactly 12 noon. So our worldlines share an event: the event of me flying past you, at an instant where both our clocks read 12 noon. Call this event O.

In your frame, I am moving, and you are fixed: so the "location in space" at which event O takes place is your spaceship. But in my frame, you are moving, and I am fixed; so the "location in space" at which event O takes place is my spaceship. These two "locations in space" are not the same; they are moving relative to each other. For event O itself, this does not matter; the spatial coordinates of event O in both of our frames are (0, 0, 0), because O is the event at which our two "locations in space" coincide. But to transform coordinates of any event other than O from your frame to my frame, it's not enough just to adjust the time; you have to also adjust the spatial position. And because time and space are interlinked, you have to make the adjustment in both time and space. That is why the equation for $t'$ includes the $vx / c^2$ term, and why the equation for $x'$ includes the $vt$ term.