Derivation of Lorentz Transformation

1. Dec 1, 2007

ObsessiveMathsFreak

On this page, and in many, many other resources both in textbooks and online, derivations of time dilation and length contraction are given which lead to;

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
$$t^{'}=\gamma t$$
$$x^{'}=\frac{1}{\gamma}x$$

Now a great many derivations of the Lorentz Transformations, such as the one on this page, use the above formulae in some way to arrive at

$$t^{'}=\gamma \left(t - \frac{v}{c^2}x\right)$$
$$x^{'}=\gamma \left(x- vt\right)$$

It seems to me that the second formulae do not agree with the first. Am I missing something here. Do they in fact agree with substitution of some correct values?

I noticed that in his original 1905 paper, Einstein assumed the transformation between two referance frames was linear, and derived the Lorentz transformations directly. Is this way, i.e. all in one gulp, the correct way to go about their derivation?

Are the original formulae strictly speaking incorrect? Under what circumstances are they valid, if at all? As an additional question, does dropping the linear transformation assumption lead, in part, to general relativity?

2. Dec 1, 2007

amolv06

If the transformations weren't linear, things would be extremely awkward. You would see different lengths for objects based on where you were in space.

3. Dec 1, 2007

JesseM

The formulas are giving you different things. In the first formula (well, the first one after the definition of gamma) t and t' are time intervals--t is the time between two events in a frame where they happen at the same location in space (like two ticks of a clock as measured in its rest frame), t' is the time between the same two events in a frame where they happen at different locations (like the same two ticks of the same clock in a frame where it's moving). In the second formula x represents the length of an object as measured in its own rest frame and x' is its length in a frame where it's moving (usually this formula is written with the symbols L and L' or L and l rather than x and x'). But in the third and fourth formula, x and t represent the coordinates of a single event in one coordinate system, while x' and t' represent the coordinates of that same event in a different coordianate system in motion relative to the first.

You can derive the first pair of formulas from the Lorentz transformation (the second pair of formulas)--To get the time dilation equation, take two events which have the same x-coordinate but different t-coordinates, then use the Lorentz transformation to find the coordinates of the same two events in the x', t' system and see how the time-interval has changed. Deriving the length contraction equation is slightly trickier, because the "length" of an object in any given frame depends on finding an event on the worldline of the back of the object and an event on the worldline of the front of the object which are simultaneous in the frame you're using (because you want to know where the back was at the 'same time' that the front was in some other spot), and finding the difference in x-coordinates between these events. And different frames define simultaneity differently--a pair of events which have the same t-coordinate (and different x-coordinates) in one frame will have different t-coordinates in another. So, what you'd need to do to derive it would be to pick three different events--say, one event A on the worldline of the object's back, the event B on the worldline of the object's front that's simultaneous with this event in the object's rest frame (where both the front and back have unchanging position-coordinates), and then the event C on the worldline of the object's front that's simultaneous with A in the second frame where the object is moving. Then you can compare the difference in x-coordinate between A and B in the object's rest frame, and compare with the difference in x'-coordinate between A and C in the second frame.

And you can also do the reverse and derive the Lorentz transform from the equations for length contraction and time dilation--you just have to realize that each observer's coordinate system is defined in terms of local readings on an imaginary set of rulers and synchronized clocks which are at rest relative to themselves (so, for example, if I see an explosion happening next to the x=10 light-seconds mark on my ruler, and a clock sitting on that mark reads t=12 seconds at the moment it happens, I assign this explosion coordinates x=10, t=12), and that each observer "synchronizes" his own clocks using the Einstein clock synchronization convention which is based on assuming that the speed of light is the same in all directions in that observer's own frame (so that if I set off a flash at the midpoint of two clocks relative to me, I define them as synchronized if they both read the same time at the moment the light from the flash reaches them).

Last edited: Dec 1, 2007
4. Dec 4, 2007

ObsessiveMathsFreak

Thanks for the insightful reply. I see now that the original formulae we for simple time intervals and lengths in different frames, not for space and time coordinates.

So the time dilation formulae works by setting the x(t)=vt+d, i.e. considering a point that is moving in the first frame, but static in the other. Then
$$\begin{equation*} \begin{split} t'(t)&=\gamma\left( t-\frac{v}{c^2}(vt+d) \right) \\ &=\gamma\left( t-\frac{v^2}{c^2}t -\frac{vd}{c^2} \right)\\ &=\gamma t\left(1-\frac{v^2}{c^2}\right) -\gamma \frac{vd}{c^2}\\ &=\frac{t}{\gamma} - \gamma \frac{vd}{c^2} \end{split} \end{equation*}$$

So a time interval is given by
$$t'(t_2)-t'(t_1)=\frac{t_2}{\gamma} - -\gamma \frac{vd}{c^2} -\frac{t_1}{\gamma} - +\gamma \frac{vd}{c^2}=\frac{t_2-t_1}{\gamma}$$
Which matches the first formulae

The second transformation for length contraction is a lot trickier, though it seems that it could be fudged by considering static objects in the second frame, i.e., disregarding the time difference.

It's interesting that the time coordinates in the moving frame actually depend upon their distance from the origin of the static frame. It's clear to me now how events with the same time t in one frame will have different times t' in another frame if their spacial coordinates x are different.

5. Dec 4, 2007

6. Dec 4, 2007

Fredrik

Staff Emeritus

I'm trying to show (just as an exercise for myself) that Einstein's two postulates imply that coordinate transformations must form a group, and that every coordinate transformation must take the form

$$x\mapsto\Lambda(x)+a$$

where $\Lambda$ is a linear operator. What's the easiest way to see that $\Lambda$ must be linear? (The linearity is the only detail I have a problem with).

I tried searching the relativity forum, but didn't find a good explanation. That doesn't mean that none have been posted, only that I didn't find one.

Feel free to only consider the proper, orthochronous part of the homogeneous Lorentz transformation in 1+1 dimensions, if that makes it easier. I would like to see the simplest possible proof.

7. Dec 4, 2007

Ich

Lambda may not be linear. You need homogeneity of spacetime as an additional postulate.
You can find the proof that parallel boosts form a group in Einstein's paper. For the Poincaré group, maybe one of the mathematically skilled here can help.

8. Dec 4, 2007

Fredrik

Staff Emeritus
Hm...the first postulate states that the laws of physics must take the same (mathematical) form in every inertial frame. Doesn't that imply homogeneity? What should we take as the definition of an "inertial frame" here?

I would usually define an inertial frame as an isometry of the Minkowski metric, but we can't do that here, since we're trying see how Einstein's postulates lead to all that stuff about Minkowski space.

9. Dec 5, 2007

Ich

Hmm...the concept of an inertial frame seems to imply spatial infinity and homogeneity. What about time? Physical laws could change with time, but be the same for all IF.