# Deriving the Lorentz Transformation

#### toothpaste666

Hello everyone. I am attempting to teach myself about special relativity, and have learned a derivation for the Lorentz transformation. Before I go any farther I want to clear up the parts I don't understand right now and check that I am correct about the parts I do understand. Here I will derive the Lorentz transformation and ask a couple of questions as I go. I would greatly appreciate it if someone can help increase my understanding of these concepts.

There will be two observers, S and S' that will start out with their x axis' lined up and the clocks they are carrying synched at t=0 and t'=0. S' will begin moving with speed v away from S who will remain stationary so that there is distance vt in between S and S'. S' will be moving toward a box that is moving at the speed of light c in the same direction as S'. I have made a picture of the situation. The distance between S and the box is x and the distance between S' and the box is x'. x=ct.

Here is my first question: From what I have learned it seems that not only does x=ct but x'=ct'. Is this only true because c is the same in every reference frame or would that also be true in Newtonian physics?

According to S': x = x' + vt'
According to S: x' = x - vt

plugging in x=ct into the second equation you have:

x' = ct-vt
x'=(c-v)t

Since distance is a velocity times a time and we know the time, that means the velocity that S' sees the box moving is less than c. But we know that c is the same in every reference frame. So our equations must be off by some factor.

Next question: From derivations I have seen on the internet, they add an error factor (lets call it 'f') into the two equations, so that they become this:

x = (x' + vt')f
x' = (x - vt)f

My question is: how do we know that we can use the same error factor for both equations? Wouldn't we use f for one and lets say g for the other? How can we prove that we can use the same for both? Forgive me if this is obvious, I am fairly new to these concepts.

Anyways we multiply the two equations together. (Is there a clear way to know whether to multiply, add or subtract two equations?) After multiplying them we get:

$xx' = (x' +vt')(x-vt)f^2$

$xx' = (xx' - x'vt + xvt' - v^2tt')f^2$

we know that x=ct and x'=ct' so t=x/c and t'=x'/c. plugging in we get:

$xx' = (xx' - x'v(\frac{x}{c}) + xv(\frac{x'}{c}) - v^2(\frac{x}{c})(\frac{x'}{c}))f^2$

$xx' = (xx' - \frac{xx'v}{c} + \frac{xx'v}{c} - \frac{v^2xx'}{c^2}) f^2$

$xx' = (xx' - \frac{v^2xx'}{c^2})f^2$

$1 = (1 - \frac{v^2}{c^2})f^2$

$f^2 = \frac{1}{1-\frac{v^2}{c^2}}$

$f = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}$

$f = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

which is the lorentz transformation. Any feedback would be greatly appreciated. :)

#### Attachments

• 7 KB Views: 463
Last edited:
Related Special and General Relativity News on Phys.org

#### pervect

Staff Emeritus
It's not clear to me where you learned your Lorentz transformations from. So lets check to see if your source gave them correctly:

Do you have

$$x' = \frac{1}{\sqrt{1-\left( \frac{v}{c} \right)^2}}\left( x - v t \right) \quad t' = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2 }} \left( t - \frac{v\,x}{c^2} \right)$$

I noticed you didn't discuss the transformation of t to t' at all, there is an important difference in the behavior of time between relativity and pre-relativity physics,

#### stevendaryl

Staff Emeritus
I noticed you didn't discuss the transformation of t to t' at all, there is an important difference in the behavior of time between relativity and pre-relativity physics,
Actually, the time translations are contained in the pair of equations

$x' = f(x - vt)$
$x = f(x' + v t')$

Plugging the first equation into the second gives:

$x = f^2(x - v t) + fvt'$
which can be rearranged to give:

$t' = f[\dfrac{(1-f^2)}{f^2 v} x + t]$

which is the same as the usual LT transform, if you substitute $\dfrac{1}{f^2} = 1-\frac{v^2}{c^2}$ and $1-f^2 = \dfrac{-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}$

#### DiracPool

I like the derivation of the Lorentz transformations from Einstein's 4-vector.

$ds^2=c^2(dt')^2-(dx')^2-(dy')^2-(dz')^2$

Then we set the spatial component to zero to put us in a proper time frame..

$ds^2=c^2(dt')^2$

Now we have $ds^2=c^2(dt')^2=c^2dt^2-dx^2-dy^2-dz^2$.

So, in this instance we can call dt' (dtau), because it is a clock in its own inertial reference frame, yielding..

$ds^2=c^2(dtau)^2=c^2dt^2-dx^2-dy^2-dz^2$

Here is where the fun comes in.. Factor out the dt^2 and c^2 from the rhs of the previous equation, which gives you..

$c^2(dtau)^2=c^2(1 - \frac{1}{c^2}\frac{dx^2}{dt^2}-\frac{1}{c^2}\frac{dy^2}{dt^2}-\frac{1}{c^2}\frac{dz^2}{dt^2})dt^2=c^2(1 - \frac{v^2}{c^2})dt^2$

Where the velocity vector is just the dx,dy,dz.

So now you just cancel the C^2 from each side and you have your time dilation formula pulled strait from the 4 vector.

$dtau = {\sqrt{1-\frac{v^2}{c^2}}}dt$

Or, as it is more popularly seen,

$dt = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(dtau)$

Where $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is known as "gamma"

Neet huh?

#### m4r35n357

I like the derivation of the Lorentz transformations from Einstein's 4-vector.

Neet huh?
Well of course life becomes much easier when you start by postulating the answer ;) In a more logical (pedagogically, I like that word!) sequence the line element would be derived (observed, really) from invariance of the Lorentz Transform, not the other way around.

I salute the OP's approach to learning this!

#### toothpaste666

It's not clear to me where you learned your Lorentz transformations from. So lets check to see if your source gave them correctly:

Do you have

$$x' = \frac{1}{\sqrt{1-\left( \frac{v}{c} \right)^2}}\left( x - v t \right) \quad t' = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2 }} \left( t - \frac{v\,x}{c^2} \right)$$

I noticed you didn't discuss the transformation of t to t' at all, there is an important difference in the behavior of time between relativity and pre-relativity physics,
well for the spacial equation, if i plug the value for f back into x'= (x - vt)f I get:

$x' = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$

so that part checks out. As for the time equation we know that x=ct x'=ct' and t=x/c. We can plug these back into x'=(x-vt)f

$x'=(x-vt)f$

$ct' = (ct-v(\frac{x}{c}))f$

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

Plugging f back in:

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

so I am pretty sure that checks out too. I'm still confused as to whether using the same f for both earlier equations requires proof that f would be the same though. Is it enough that both equations are basically the same equation rearranged?

#### DrStupid

Next question: From derivations I have seen on the internet, they add an error factor (lets call it 'f') into the two equations, so that they become this:

x = (x' + vt')f
x' = (x - vt)f

My question is: how do we know that we can use the same error factor for both equations? Wouldn't we use f for one and lets say g for the other?
As far as understand it, the transformation should be symmetric and independent from orientation. This leads to the same factor in both directions. In general you can start with a linear transformation like

$t' = f_{tx} \left( v \right) \cdot x + f_{tt} \left( v \right) \cdot t$
$x' = f_{xx} \left( v \right) \cdot x + f_{xt} \left( v \right) \cdot t$

Than you need four conditions to get the four parameters. One of them is the required symmetry.

I have to admit that I can't explain why the transformation need to be linear. In his paper "Zur Elektrodynamik bewegter Körper" Einstein wrote it is clear that the equations must be linear due to homogeneity of space. But I'm not as smart as Einstein and can't see how the linearity results from the homogeneity.

#### stevendaryl

Staff Emeritus
My question is: how do we know that we can use the same error factor for both equations? Wouldn't we use f for one and lets say g for the other?
The principle of relativity, that all inertial frames are equivalent, requires that the factors be the same for both frames. We have two frames, F and F'. The transformation

$x' = f(x-vt)$

implies this:
If an object aligned with the x-axis has length $L$ in its own rest frame, and the object is moving in the x-direction at speed v as measured in frame F, then the object has length $L/f$ as measured in frame F.

So if that's true in frame F, it should also be true in frame F'. So the factor $f$ must be the same for transforming from F to F' as in the other direction.

There's actually another issue that you glossed over when you asked whether f = g. You have two observers, A and B. If B has speed v relative to A, then A has speed v relative to B. (Remember, speed is the magnitude of the velocity, so it's always positive.) If this weren't the case, then there would be an asymmetry between the two observers.

A point that people are sometimes confused about is the fact that in the two equations:

$x' = f (x - vt)$
$x = f (x + vt)$

there seems to be an asymmetry between the two equations, because $v$ appears with a negative sign in the first equation, and with a positive sign in the second equation. That's an artifact of the particular choice of the direction of the x-axis in the two frames.

#### toothpaste666

There's actually another issue that you glossed over when you asked whether f = g. You have two observers, A and B. If B has speed v relative to A, then A has speed v relative to B. (Remember, speed is the magnitude of the velocity, so it's always positive.) If this weren't the case, then there would be an asymmetry between the two observers.
So if I am understanding you correctly it is because of the principle of relativity that the error factor is the same? In other words since A is an inertial reference frame and B is moving with uniform velocity and direction relative to it it is also an inertial frame. So it would make no difference whether A was moving with -v relative to B with B stationary, or if B was moving with v relative to A with A stationary. The magnitude of the velocity between them is still the same?

#### toothpaste666

I am reviving this topic because there is something related I am still confused about. The principle of relativity states that the laws of physics are the same in every reference frame. In pre-relativity physics this is easy to show by differentiating the galileaon-newtonian transformations

$x' = x-vt$

$\frac{dx'}{dt} = \frac{dx}{dt} - \frac{d(vt)}{dt}$

the time derivative of position is velocity. lets call the velocities of the reference frames u and u'

$u' = u - v$

we differentiate again to get acceleration

$\frac{du'}{dt} = \frac{du}{dt} - \frac{dv}{dt}$

v is constant so its derivative is 0 so

$\frac{du'}{dt} = \frac{du}{dt}$

or

$a' = a$

since a given mass will also be the same for both reference frames, then it follows that F=ma will hold for both reference frames and the principle of relativity holds.
however when I try to do this for the lorentz transformations:

$x' = γ(x-vt)$

$x' = γx-γvt$

since v and c are constant , γ is constant

$\frac{dx'}{dt} = γ\frac{dx}{dt} - γ\frac{d(vt)}{dt}$

$u' = γu - γv$

$\frac{du'}{dt} = γ\frac{du}{dt} - γ\frac{dv}{dt}$

$a' = γa$

Since the accelerations are not the same, doesn't that also mean that F=ma is not the same for each reference frame and that the principle of relativity does not hold? I am confused about this if anyone can clear this up it would be much appreciated :)

#### PeterDonis

Mentor
Since the accelerations are not the same, doesn't that also mean that F=ma is not the same for each reference frame
Yes.

and that the principle of relativity does not hold?
No, it means that $\vec{F} = m \vec{a}$ is not a correct relativistic law of physics; it is only a non-relativistic approximation. The correct relativistic law is $F = dp / d\tau$, where $F$ is the four-force, $p$ is 4-momentum, and $\tau$ is proper time. For an object with constant rest mass, we have $p = m u$, where $u$ is the 4-velocity, and $F = m a$, where $a$ is the 4-acceleration; but this is still not the same as the ordinary 3-vector form of $\vec{F} = m \vec{a}$. To derive the latter, you need to use the non-relativistic approximation in the instantaneous rest frame of the object; so, unlike the relativistic laws above, the 3-vector law is frame-dependent.

#### toothpaste666

Yes.

No, it means that F = ma is not a correct relativistic law of physics; it is only a non-relativistic approximation. The correct relativistic law is $F = dp / d\tau$, where $F$ is the four-force, $p$ is 4-momentum, and $\tau$ is proper time.
tau being t and not t' in this case right?

#### PeterDonis

Mentor
tau being t and not t' in this case right?
No. $\tau$ is the object's proper time; $t$ and $t'$ are coordinate times in two different frames. Since the object is accelerating, its proper time is not equivalent to coordinate time in any inertial frame (since it is not at rest in any inertial frame for more than a single instant).

#### Nugatory

Mentor
tau being t and not t' in this case right?
Neither, unless you choose your coordinate system such that one of them is equal to $\tau$ along the world line of the object being accelerated, and this can only be true for a single moment.

#### DrStupid

when I try to do this for the lorentz transformations:

$x' = γ(x-vt)$

[...]

$a' = γa$
You forgot the transformation of time. The correct result is

$a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3$

Since the accelerations are not the same, doesn't that also mean that F=ma is not the same for each reference frame and that the principle of relativity does not hold?
The original definition of force is F=dp/dt and this remains the same in every frame of reference both with Galilei and Lorentz transformation.

#### toothpaste666

You forgot the transformation of time. The correct result is

$a' = a \cdot \left( {\frac{{\sqrt {1 - u^2 } }}{{1 - u \cdot v}}} \right)^3$

The original definition of force is F=dp/dt and this remains the same in every frame of reference both with Galilei and Lorentz transformation.
If its possible can you show me how to get that or where I went wrong in my calculations? '

also if anyone can answer: how is tau defined mathematically?

#### DrStupid

If its possible can you show me how to get that or where I went wrong in my calculations?
You started with the Lorentz transformation of x:

$x' = \frac{{x - u \cdot t}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}$

but that is only half the truth. The time needs to be transformed too:

$t' = \frac{{t - \frac{{u \cdot x}}{{c^2 }}}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}$

The transformation of velocity is

$v' = \frac{{dx'}}{{dt'}} = \frac{{\frac{{dx'}}{{dt}}}}{{\frac{{dt'}}{{dt}}}}$

With

$\frac{{dx'}}{{dt}} = \frac{{v - u}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}$

and

$\frac{{dt'}}{{dt}} = \frac{{1 - \frac{{u \cdot v}}{{c^2 }}}}{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} }}$

This results in the well known equation

$v' = \frac{{v - u}}{{1 - \frac{{u \cdot v}}{{c^2 }}}}$

The transformation of acceleration works the same way:

$a' = \frac{{dv'}}{{dt'}} = \frac{{\frac{{dv'}}{{dt}}}}{{\frac{{dt'}}{{dt}}}}$

With

$\frac{{dv'}}{{dt}} = a \cdot \frac{{1 - \frac{{u^2 }}{{c^2 }}}}{{\left( {1 - \frac{{u \cdot v}}{{c^2 }}} \right)^2 }}$

this results in

$a' = a \cdot \frac{{\sqrt {1 - \frac{{u^2 }}{{c^2 }}} ^3 }}{{\left( {1 - \frac{{u \cdot v}}{{c^2 }}} \right)^3 }}$

and with c=1 in my simplified version above.

#### Dale

Mentor
also if anyone can answer: how is tau defined mathematically?
$d\tau^2=dt^2-dx^2-dy^2-dz^2$
In units where c=1.

#### PeroK

Homework Helper
Gold Member
2018 Award
also if anyone can answer: how is tau defined mathematically?
You have already got an answer but I wonder how you are learning SR. If you have a text book it must cover the definition of proper time. Even online there are complete introductory texts. Or are you trying to piece things together for yourself?

#### toothpaste666

Ok I see where I went wrong. Thank you all. I am learning SR on my own, but I do have my textbook for my intro physics classes that has a chapter on SR (giancoli) although it does not go into very much detail. It seems to just gloss over alot of stuff. I also have tried out some online lectures from yale and stanford but these assume more knowledge than I have in some cases, they are probably discussing things from a reading I haven't read or are for more advanced students.

#### PeroK

Homework Helper
Gold Member
2018 Award
Imho you'd be better off getting a book aimed at learning SR from scratch. There are plenty out there.

#### toothpaste666

any good not that expensive/free ones? I still gotta buy all my books for next semester =\

Staff Emeritus
Gold Member

#### Philip Wood

Gold Member
Forgive me if this has already been covered, but the expression for $\gamma$ emerges very simply from your proposed transforms,
$$x' = \gamma \ (x-vt) \ \ \ \ and \ \ \ \ x = \gamma \ (x'+vt')$$
Simply insert the special case of light emitted from the origin of S and S' at t = t' = 0, for which $x = ct$ and $x' = ct'$.
$$ct' = \gamma \ (ct-vt) \ \ \ \ and \ \ \ \ ct = \gamma \ (ct'+vt')$$
Clearly multiplying the equations will enable one to get rid of the common factor tt', leaving $c^2= \gamma^2 \ (c^2-v^2)$.
Rindler gives this symmetry method in his Introduction to SR. Resnick gives a painstaking justification of the linearity of the equations in his Introduction to SR.

#### Dale

Mentor
In a more logical (pedagogically, I like that word!) sequence the line element would be derived (observed, really) from invariance of the Lorentz Transform, not the other way around.
Certainly it is no more nor less logical to start with the transforms and derive the line element than it is to start from the line element and derive the transform. Personally, pedagogically I prefer the line element -> transform approach since it prepares the groundwork quite well for GR.

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving