# Derivation of lorentz transformation?

1. Apr 26, 2012

### Aziza

I would like to know a simple way of understanding how to derive the Lorentz transformation equations. My book states them without proof and on websites I only see complicated proofs that Im not mathematically ready for yet. Ok so I think i understand x'=γ(x-vt), but let me say it in my own words and have it confirmed as right just in case:

Suppose a reference frame with origin O' is moving at velocity v with respect to frame origin O and lets say an event happens at some point in space. O' has left O a time t ago (according to O), and O sees the event a distance x away, and O' sees the event a distance x' away. However, x' is actually shorter due to length contraction than it would have been had O' been at rest with respect to O. Therefore, to relate x to x', x must be the distance from O to O' (which is vt) plus the distance that x' would have seen had it been at rest with O, so we lengthen x' by multiplying x' by γ, so x=vt+x'γ and rearrangement leads to the conventional form.
Am i interpreting this right?

Now what i really dont understand is t'=γ(t-vx/c^2)...So if x' is the distance for O' to the event, and x is the distance for O to the event, and since in the previous equation, t was the time according to O that it took for O' to get a distance vt away, then in this equation, t' should be the time it took for O' to get that same distance away from O. So according to O', it took him the time t' to get the contracted distance vt/γ away from O. So t' is just t/γ....which is right but i dont see how it leads to above Lorentz equation...i feel im maybe misinterpreting what t' is supposed to mean???

note: i am trying to follow the diagram my book has drawn...i am assuming the labels it uses are supposed to correspond to the variables of the equations..but it doesnt specify what t' is, which is causing me confusion! :

http://af10.mail.ru/cgi-bin/readmsg?id=13354810280000000090;0;1&mode=attachment

Last edited: Apr 26, 2012
2. Apr 26, 2012

### elfmotat

3. Apr 26, 2012

### Aziza

4. Apr 26, 2012

### Aziza

wait no i am confused again now about the first equation...i wrote in my first post that x=vt+x'γ but obviously thats wrong since the last term should be x'/γ...but if x' is already the shortened distance, why are we shortening it even more?? I mean if L=Lp/γ, where L is the contracted length and Lp is the proper length, then in this case, the proper distance to the event from O' is x-vt and the contracted distance is x'. so x'=(x-vt)/γ.....but it should be x'=γ(x-vt) ?

Last edited: Apr 26, 2012
5. Apr 26, 2012

### elfmotat

x and t are measurements made in the unprimed frame, and x' and t' are measurements made in the primed frame. They both think they are at rest.

So in $x=vt+\frac{x'}{\gamma}$, the distance x (in the unprimed frame) is equal to the distance vt (in the unprimed frame) plus the contracted version of whatever the primed observer measures.

In $\frac{x}{\gamma}=vt'+x'$, the distance vt'+x' is the distance that they frames move apart according to the primed observer, plus the distance x'. According the the primed observer, this should be equal to the contracted version of whatever the unprimed observer measures as x (because the unprimed observer is moving according to him).

6. Apr 26, 2012

### robphy

Here's a nice physically-motivated derivation due to Bondi:
http://archive.org/details/RelativityCommonSense

start at p. 76 to understand the "k-factor" (the doppler factor)
see the diagram on p. 95
p. 102 has the formula for k in terms of v
p. 117 derives the Lorentz Transformation, followed by applications

7. Dec 9, 2012

### John Huang

The "contracted length" is still a confusing term for me. Right now, I use "contracted ruler" to understand it. I think, what actually contracted are all objects in the moving system which includes rulers. That means the moving observer uses the contracted ruler to measure distance between two points so that the result is actually a longer length. Do you think so?

If you think so, in your first post, to the stationary observer, the x will change to x=vt+(x'/γ) so that x'=γ(x-vt) ---(1).

There is another way to derive the time equation after the γ is proved. If you replace the x' in the inverse spatial equation x=γ(x'+vt') ---(2) by γ(x-vt) in (1), then you will get the time equation of Lorentz transformation.

That means, if he assumed (1) was correct, then, based on the principle of relativity, (2) must be correct and Mr. Lorentz could easily derive his transformation.

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