# Derivation of Lorentz Transformation

1. Sep 28, 2007

### kaotak

Hi,

I don't fully understand the derivation of the Lorentz Transform.

For example, in the transformation:

$$t' = \gamma t - \gamma\frac{vx}{c^2}$$

Where do you get the x-term from? I understand the derivation of gamma, and I see that when x = 0 (when the two frames coincide in origin) the tranformation is just gamma * t, but I don't understand how you derive the x-term. I've checked out numerous sites and these sites don't give a derivation for the equation for t'.

The x-term looks suspiciously like the time it takes to time the light's trip but that's not the way of thinking of special relativity, so far as I know...

Basically, I'm looking for (A) the x-term in the t' transformation equation and (B) the derivation of the Lorentz transform for motion in the x-direction.

Thanks in advance.

2. Sep 28, 2007

### meopemuk

kaotak,

there exist dozens of different ways to derive Lorentz transfrormations. Some examples are in

A. R. Lee and T. M. Kalotas, "Lorentz transformations from the first postulate", Am. J. Phys. 43 (1975), 434

J.-M. Levy-Leblond, "One more derivation of the Lorentz transformation", Am. J. Phys. 44
(1976), 271

D. A. Sardelis, "Unified derivation of the Galileo and the Lorentz transformations", Eur. J. Phys. 3 (1982), 96

H. M. Schwartz, "Deduction of the general Lorentz transformations from a set of necessary assumptions", Am. J. Phys. 52 (1984), 346

J. H. Field, "A new kinematical derivation of the Lorentz transformation and the particle description of light", (1997), KEK preprint 97-04-145

R. Polishchuk, "Derivation of the Lorentz transformations", http://www.arxiv.org/abs/physics/0110076

It would be easier to help you if you indicate which particular book/article you are reading.

Eugene.

3. Sep 28, 2007

### kaotak

Thanks for the reply.

I was using this article that I found online: http://galileo.phys.virginia.edu/classes/252/lorentztrans.html

I understand the setup of the thought experiment, but there is this one line in particular that seems to assert what I want proven:

"When the bomb explodes and the clock at x' reads t', O will see O''s origin clock to read:

$$t' + \frac{vx'}{c^2}.$$"

This quote is in the second paragraph of the "deriving the lorentz transformations" section.

It asserts the Lorentz transformation for time without proving it :( Can anyone tell me the derivation of this equation?

4. Sep 28, 2007

### Johan de Vries

You should not look it up, derive it yourself. When learning things it is essential that you learn how to derive the theory from fundamentals without any help. If you hadn't "checked out numerous sites" you would already have obtained the answer by now

Anyway, since you've wasted a lot of time instead of deriving the equation, let me do it for you this time

Suppose I'm moving away from you at velocity v in the x-direction. at time t = 0, I was at x = 0 and my clock was synchronized at
$$t' = 0$$. Suppose at time $$t = T$$, you send a light signal to me. I receive it at space-time coordinates $$\left(x_{p},t_{p}\right)$$ (in your frame):

$$\begin{equation*} \begin{split} t_{p} &= \frac{1}{1-v/c}T\\ x_{p}&=\frac{v}{1-v/c}T \end{split}$$

Suppose that part of this lightsignal reflects off me and comes back to you. When do you receive the reflected signal back? Since the reflected light signal has to travel back a distance $$x_{p}$$ from me to you, this is at time

$$\begin{equation*} t_{\text{ref}}=t_{p} + \frac{x_{p}}{c} = \frac{1+v/c}{1-v/c}T$$

Now, when do I receive the lightsignal according to my clock? Let's denote this by $$t'_{p}$$ and define a constant $$K$$ such that:
$$\begin{equation*} t'_{p}= K T$$

So, we assume that if you send me a signal at time $$T$$ on your clock, I'll receive it at time $$t'_{p}= K T$$ on my clock. Now, we can interpret the reflected signal originating from me as a signal that I send you. Since I can say that I'm at rest and you are moving away, the above equation implies that you receive the reflected signal back at:

$$\begin{equation*} t_{\text{ref}}= K t'_{p} = K^{2}T$$

And it follows that:

$$\begin{equation*} K = \sqrt{\frac{1+v/c}{1-v/c}}$$

Now suppose that not all of the lightsignal was reflected back and part of it passed me and reflected off a mirror at x-coordinate $$x$$ (in your frame). Clearly, in your frame this reflection happens at time $$t = T + \frac{x}{c}$$. The reflected signal then moves back, passes me and arrives at you. The signal will arrive back at you at time
$$\begin{equation*} t_{\tex{back}}=T +2\frac{x}{c}$$.

Note that the signal takes at much time to travel to the mirror as it takes to travel back from the mirror. This must also be true for me, because in my frame I see the pulse pass me, hit the mirror at some position and move back to me traveling the same distance.

This means that in my frame the time of reflection, $$t'$$, is the average of the time the signal passed me on its way to the mirror and the time it passed me on its way to you:

$$\begin{equation*} t' = \frac{1}{2}\left[KT + K^{-1}t_{\text{back}}\right]$$

Substitute in here the equation for $$t_{\text{back}$$ and express $$T$$ in terms of $$x$$ and $$t$$ using the equation before that one. Then you're done if you can figure out how to simplify $$K\pm K^{-1}$$

5. Sep 29, 2007

### kaotak

Thanks for the reply.

I don't understand why $$t_{p} = \frac{1}{1 - \frac{v}{c}} T$$, as $$1 - \frac{v}{c}$$ is not $$\gamma$$ or anything close to it.

Okay, my friend led me through a derivation, and through it we got the x-term for t', but I cannot get the full equation for t' from the Lorentz transformation through this derivation. Can you help me complete this derivation? Or point out anything that's wrong?

Consider a spaceship with an observer, T, at the tail with a clock and an observer, A, outside the spaceship with a clock. The spaceship is moving at a velocity v in the x-direction relative to A.

We define t = 0 when the nose of the spaceship coincides with the observer with clock A. (We will later find a contradiction regarding this.)

Now consider the event when the tail of the spaceship (observer T) coincides with observer A (their x coordinates are the same). Let's list what the clocks read after the time elapsed while the spaceship when from its nose at the origin to its tail at the origin.

Define x as the length of the ship according to T (the clock in the tail of the spaceship).

In A's frame of reference, we have:

A's clock reads $$\frac{1}{\gamma} \frac{x}{v}$$
T's clock reads $$\frac{1}{\gamma^{2}} \frac{x}{v}$$

In T's frame of reference, we have:

T's clock reads x / v.
(What A's clock reads is irrelevant)

We know that what T reads must agree for A and T, so

$$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v}$$

We see that this is impossible, thus they cannot have had the same initial times. The right-hand side is really a time interval, $$\Delta t$$ Thus,

(1) $$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v} - t_{0}$$

We subtract $$t_0$$ because T's clock is slower.

Solving for $$t_0$$:

$$t_{0} = \frac{x}{v} - \frac{1}{\gamma^{2}} \frac{x}{v}$$

$$t_{0} = \frac{x}{v}(1 - \frac{1}{\gamma^{2}})$$

$$t_{0} = \frac{x}{v}(1 - (1 - \frac{v^2}{c^2}))$$

$$t_{0} = \frac{vx}{c^2}$$

And there would really be a gamma in there, since that's x as A observes it, which is contracted, but I'm not going to bother revising everything to make it correct.

So this gets me the x-term in the Lorentz transformation for t'. But I can't get the equation itself from it. For if we plug $$t_{0}$$ back into equation (1), we get:

$$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v} - \frac{vx}{c^2}$$

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

Where t' is the time that T's clock reads in T's frame of reference, substituting it for the LHS above.

$$t' = t - \frac{vx}{c^2}$$

Where t is the time that A's clock reads in A's frame of reference, allowing the substitution.

But, this is not the equation for t' in Lorentz transform! The equation is

$$t' = \gamma(t - \frac{vx}{c^2})$$

I'm missing a gamma :(

Can anyone point out why I got the right x-term, and why I can't get the equation for t'. Can anyone fix this derivation?

Thanks in advance.

Last edited by a moderator: Sep 29, 2007
6. Sep 29, 2007

### bernhard.rothenstein

LT transformation

Start with Galileo's transformation
x-0=V(t'-0)+(x'-0). (1)
If you analyze what he is doing, you see that he expresses the length
(x-0) measured in I as a sum of two lengthts measured in I'. He can do that because in his relativity length is absolute.
In Einstein's theory of relativity length is relative and so we can add only lengths measured in the same reference frame say in I'. Due to length contraction in I' we have (x-0)sqrt(1-VV/cc). (2)
Expressing (1) as a function of lengths measured all in I' we have
(x-0)sqrt(1-VV/cc)=V(t'-0)+(x'-0) (3)
and the way is oppen to all the other three transformation equations in an one space dimensions approach.
Is that to simple?

7. Sep 29, 2007

### kaotak

Thanks for the reply.

I understand that; that's for length contraction. But how do you derive the time dilation from that?

8. Sep 29, 2007

### bernhard.rothenstein

Very good question. Start again with Galileo
x'=x-Vt

and take into account that x' is a proper length. Measured from I it is
sqrt(1-$$\beta$$^2) with which (1) becomes
sqrt(1-$$\beta^2$$)x'=x-Vt (2)
Solving (2) with the transformation equations from my previous thread, for t and t' respectively you have the answer to your question
If you are not satisfied I have other solutions.

9. Sep 29, 2007

### OOO

I haven't looked at your source because I don't like these lengthy explanations with spaceships an so on. The basic idea behind Lorentz transforms is utterly simple:

Electrodynamics forces you to do only coordinate transforms that leave the metric $$t^2-x^2$$ invariant (for simplicity I have set c=1 which you can reverse in the end by substituting t). Since this is one condition for the two unknowns t and x you will expect to find a one parameter transform. It is easy to show that the transform

$$t^\prime = a t + b x$$
$$x^\prime = b t + a x$$

does indeed fulfill the above metric condition if the two parameters a and b are bound by $$a^2-b^2 = 1$$ (if you don't believe it, substitute the transform into the metric condition, it's really easy to show that you will get t^2-x^2 again). Thus you have a one parameter transform as expected.

Now if you let a "spaceship" rest at the origin of the primed system then, of course, it has to obey the equation

$$x^\prime =0 = b t + a x$$

So its velocity as seen from the unprimed system is

$$v=-\frac{b}{a}$$

Finally, from the condition $$a^2-b^2 = 1$$ in connection with b=-av you immediately get "a" as a function of v:

$$a = \frac{1}{\sqrt{1-v^2}} = \gamma$$

and consequently $$b=-\gamma v$$. Substitution of "a" and "b" into the above transform gives you everything you've asked for.

Last edited: Sep 29, 2007
10. Sep 29, 2007

### robphy

Draw a spacetime diagram.
Equate the distances traveled by the traveller and by the light ray to him sent after a delay of T.

There is an algebraic relation between v (tanh), gamma (cosh), and K (exp).

11. Sep 29, 2007

### kaotak

ooo: Thanks, I understand your derivation, and it is pretty succinct and elegant, but I have a couple criticisms. The assumption that t^2 - x^2 is invariant is OK by me, since I believe you can know that without knowing the answer, but the assumption that a^2 - b^2 is invariant... doesn't that rely on you knowing the coefficients in the first place; or, in other words, on knowing the answer in the first place? Furthermore, you showed that the following equations:

$$t^\prime = a t + b x$$
$$x^\prime = b t + a x$$

could be correct, with the assumption that a^2 - b^2 = 1, since these equations satisfy the invariant t^2 - x^2. But that does not prove that they have to be correct; wouldn't you have to prove there are no other equations that satisfy the invariance of t^2 - x^2? I feel like you are saying, "t' must be in this form" when it's really, "t' could be in this form".

bernhard: I tried working from what you gave me and I got something slightly off:

Starting with Galileo
$$x' = x - vt$$

We know this is wrong, and that it is really

$$\frac{1}{\gamma} x' = x - vt (*)$$

In your first post, you gave as Eq (3)

Multiplying both sides of your Eq (3) by gamma, and substituting the RHS for x in Eq (*), you get:

$$\frac{1}{\gamma} x' = \gamma vt' + \gamma x' - vt$$

$$\gamma vt' = \frac{1}{\gamma} x' - \gamma x' + vt$$

$$\gamma vt' = (\frac{1}{\gamma} - \gamma) x' + vt$$

$$vt' = (\frac{1}{\gamma^{2}} - 1) x' + \frac{1}{\gamma} vt$$

$$t' = \frac{1}{\gamma} t - \frac{vx'}{c^2}$$

Which is incorrect :( First of all, I believe it should be x and not x', and second of all, I believe it should be gamma in front of t, not 1 over gamma. Can you pinpoint some mistake I made in the work above?

I am sorry that I am drolling on about this problem. Thanks for all your help!

12. Sep 29, 2007

### bernhard.rothenstein

Lt

[We know this is wrong, and that it is really

$$\frac{1}{\gamma} x' = x - vt (*)$$

In your first post, you gave as Eq (3)

Multiplying both sides of your Eq (3) by gamma, and substituting the RHS for x in Eq (*), you get:

$$\frac{1}{\gamma} x' = \gamma vt' + \gamma x' - vt$$

$$\gamma vt' = \frac{1}{\gamma} x' - \gamma x' + vt$$

$$\gamma vt' = (\frac{1}{\gamma} - \gamma) x' + vt$$

$$vt' = (\frac{1}{\gamma^{2}} - 1) x' + \frac{1}{\gamma} vt$$

$$t' = \frac{1}{\gamma} t - \frac{vx'}{c^2}$$

Which is incorrect :( First of all, I believe it should be x and not x', and second of all, I believe it should be gamma in front of t, not 1 over gamma. Can you pinpoint some mistake I made in the work above?

I am sorry that I am drolling on about this problem. Thanks for all your help![/QUOTE]
The system of equations you should use are the corrected Galileo transformations
sqrt(1-VV/cc)x=x'+Vt' (1)
sqrt(1-VV/cc)x'=x-Vt (2)
Solving them for t and t' you obtain
t=(t'+Vx'/cc)/sqrt(1-VV/cc)
t'=(t-Vx/cc)/sqrt(1-VV/cc)
As an exercise for myself I have repeated the calculus.

13. Sep 29, 2007

### kaotak

Ahah! That works. Thanks so much =)

$$\frac{1}{\gamma} x = x^{\prime} + vt^{\prime} \ (1)$$

$$\frac{1}{\gamma} x^{\prime} = x - vt \ (2)$$

Working from (1), we have

$$vt^{\prime} = \frac{1}{\gamma} x - x^{\prime}$$

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \frac{x^{\prime}}{v}$$

Substituting (2) into the equation, we get

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \gamma \frac{x - vt}{v}$$

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \gamma \frac{x}{v} + \gamma t$$

$$t^{\prime} = \gamma \frac{x}{v} (\frac{1}{\gamma^{2}} - 1) + \gamma t$$

$$t^{\prime} = \gamma(t - \frac{vx}{c^{2}})$$

14. Sep 29, 2007

### OOO

It's not an assumption. You get $$a^2-b^2=1,$$ (in words: one, not simply invariant) when you substitute the transformation into the metric condition $$(t^\prime)^2-(x^\prime)^2=t^2-x^2$$.

What you say is of course correct, at least partly. First let me say that my intention was to put the derivation as simple as possible. The most general linear Ansatz would have been

$$t^\prime = a t + b x$$
$$x^\prime = c t + d x$$

Giving you just the symmetric Ansatz is kind of cheating, that leaves you with the question: where does it come from. But on the other hand, giving a prepared Ansatz and proving that it is working and unique is common in the literature. In this way you may have probably been presented the solution of the classic harmonic oscillator. I have rarely seen a solution where the author sets up a power series, compares coefficients and in the end obtains the complex exponential. Usually you're just told to try the complex exponential and if it works, then fine.

So what remains to be shown is completeness, you mentioned it already. I've tried to show this by arguing about the one parameter group of transformations the Lorentz group really is. To illustrate this more clearly, think of the euclidean metric $$x^2 + y^2=r^2$$. You immediately see that this condition describes a circle for fixed r. So if you have one vector that is on the circle then you can reach all others by a rotation about a single angle. You know this from the fact that the circle is one-dimensional. So if you find a one-parameter transform that allows you to reach neighbouring points then you can reach every point on your circle by "analytic continuation" and thus the set of transforms is complete.

The only difference with planar Lorentz transforms is that the metric is indefinite so the characteristic curve of the metric is not a circle but four hyperbola. Although these hyperbola are disconnected they form a one-dimensional set. So what I was hoping you'd believe me () was that if your Ansatz yields a one-parameter group of transformations and since you can reach every element from the identity (a=1, b=0), the group of transformations is determined completely (you just have to "go" there).

Strictly speaking this is not a completely satisfactory argument because you can only "move" continuously on one of the four hyperbola with the proper orthochronous (space orientation and time preserving) Lorentz transforms you were initially referring to. We say the Lorentz group is multiply connected (you need some reflections in order to reach all vectors with unit length / all hyperbola). But on the other hand it doesn't hurt too much that I have swept this under the carpet because I guess the typical "spaceship" explanations don't mention this either. And as you can see this increases the required number of words considerably.

Last edited: Sep 29, 2007
15. Sep 29, 2007

### bernhard.rothenstein

back to Galileo

Very nice. My problem is how to go back to Galileo if the LT are in our hands? To consider v "very small" or to consider c infinite?

16. Sep 29, 2007

### kaotak

Well, if you're talking about going back to Galilean time, Galilean time is invariant. For small velocities v, we can consider v/c^2 as being close to 0, in which case gamma = 1 and vx/c^2 = 0, leaving us with t' = t, which is what we wanted.

As for the v versus c question, you really only have to consider v, i.e. whether it's non-relativistic or not.

17. Sep 30, 2007

### bernhard.rothenstein

back to Galileo

Thanks for your help. What I think is that c should go to infinity in order to perform the "clock synchronization" in Galileo's relativity. The discussion has an academic character but my students ask that question frequently.

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