Basic Lorentz transformation derivation

[jeremyfiennes]

The Lorentz transformations are mathematically simple. I had always imagined they could be easily derived. I however just found out from another PF thread that this is not so. Their originators Lorentz and Poincaré simply stated them without derivation. And the "proofs" I have seen to date have been complex and unconvincing. Since only simple geometry and the standard SR conditions (length contraction, time dilation and the constancy of the speed of light) are apparently involved, I attempted a derivation.
Consider single time and space dimensions. Let observer A see an event at point X=(x_a,t_x), Fig. 0‑1a. The question is: how does an observer B, moving at steady speed v relative to A, perceive the same event, Fig. 0‑1b?

Each observer has a clock. There is also one at point X. Define the time origin t=0 as the instant when frames A and B coincide. Synchronize all three clocks via a signal from the mid-point (x_a/2).
In frame A, an event photon takes time x_a/c to reach the observer A, who then sees it occurring at:
(x_a, t_x+x_a/c) (eq.1)
For observer B, frame A moves at steady speed v, meaning that its lengths are foreshortened by the Lorentz factor γ. Observer B's space point x_b is then:
x_b=(x_a – vt_x)/γ (eq.2)
where:
γ=1/√(1–β²); β=v/c (eq.3)
Clock X also moves for observer B, and for him runs slow by the factor γ. Including again the event photon travel time, B's time point t_b:
t_b = t_x/γ + x_b/c (eq.4)
Substituting and rearranging, and using 'prime' notation, B's points in terms of A's
x' = [(1+β)x – vt]/g; t' = [(1–β)t + vx/c2]/γ (eq.5)
These expressions however differ radically from the standard:
x' = γ[x – vt]; t' = γ[(t – vx/c²] (eq.6)
So something has gone wrong somewhere. But what and where? I checked the derivations with a numerical example, but could find no inconsistencies.

 

[jeremyfiennes]

The figure didn't show. I'll try again.
https://www.facebook.com/photo.php?...kkcVkGqXBVPofdBcJMYPk8reAaQrxpPiYIRgSfEYc83zX

 

[Nugatory]

And the "proofs" I have seen to date have been complex and unconvincing.

You don't need to derive it to prove it; whether you've derived it or you found it on a piece of paper in the gutter, you can check it for correctness. A convincing (note: "convincing" and "mathematically rigorous" are not always quite the same thing) demonstration of the correctness of the Lorentz transformation is fairly straightforward: show that the lines $x\pm{ct}=0$ transform to $x'\pm{ct'}=0$ and you have your proof that the Lorentz transforms preserve the speed of light in both frames.

 

[Nugatory]

Since only simple geometry and the standard SR conditions (length contraction, time dilation and the constancy of the speed of light) are apparently involved, I attempted a derivation.

Going about it the derivation this way is vaguely suspect because both length contraction and time dilation are usually derived from the Lorentz transforms and not the other way around; it has to be possible to derive the transforms just from the postulate of the invariant speed of light.

However, you can use that postulate to come up with very convincing arguments for time dilation and length contraction without going through the Lorentz transforms so you can legitimately assume time dilation and length contraction in your derivation if you prefer. Apparently you do, because that's the approach that you've taken... But given that, what's wrong with the derivation at https://en.wikipedia.org/wiki/Deriv...rmations#Time_dilation_and_length_contraction?

 

[Ibix]

I can't see your diagram - it appears to require a Facebook account to view. Use the upload button below the post editor to embed images.

Hopefully the diagram completes the explanation, because with just your maths I am completely unable to understand what you think you are doing. Changing notation half way through without saying what is what (is $t'$ supposed to replace $t_a$, $t_b$ or $t_x$?) doesn't help at all.

 

[Dale]

For observer B, frame A moves at steady speed v, meaning that its lengths are foreshortened by the Lorentz factor γ.

Seems like you should derive the Lorentz factor, not just assume it. I find this approach, sort of like just listing the thing you want to prove as one of the assumptions.

I like this one best:
https://arxiv.org/abs/physics/0302045

And I find the geometrical concept of Bondi’s k-calculus approach appealing.

 

[haushofer]

show that the lines $x\pm{ct}=0$ transform to $x'\pm{ct'}=0$ and you have your proof that the Lorentz transforms preserve the speed of light in both frames.

Isn't that circle reasoning? You already assume c'=c in your equation. I'd say c=c' is a postulate, and only after assuming that postulate you show that your line is indeed preserved by Lorentz transformations.

 

[jeremyfiennes]

What's wrong with the derivation at https://en.wikipedia.org/wiki/Deriv...rmations#Time_dilation_and_length_contraction?

Thanks. The "From physical principles. Time dilation and length contraction" is evidently the bit I want.
Observer F sees the point X as (x,t), Fig,(a). F' sees it as (x',t'), Fig,(b). But due to foreshortening, a length x' in frame F' corresponds to γx' in frame F, Fig.(c). Giving x=vt+γx', and not the Wiki value x=vt+x'/γ: And I don't see the time a light photon takes to get from the event to an observer considered.

 

[PeroK]

Thanks. The "From physical principles. Time dilation and length contraction" is evidently the bit I want.
Observer F sees the point X as (x,t), Fig,(a). F' sees it as (x',t'), Fig,(b). But due to foreshortening, a length x' in frame F' corresponds to γx' in frame F, Fig.(c). Giving x=vt+γx', and not the Wiki value x=vt+x'/γ: And I don't see the time a light photon takes to get from the event to an observer considered.

View attachment 234849

Wiki is correct on that first point.

The time it takes a light signal (or a sound signal, or a telegram) to get to the origin of the coordinate system is irrelevant. The event could take place in a closed room from which no light escapes, but it would still have coodinates $(t', x')$.

The "eye" on the y-axis is misleading. If I did those digrams I would have the eye at the point of the event. That's where the local observation is made or recorded.

PS Technically, the Lorentz Transformation is not about observers but about reference frames.

If you want to take light signals into account you'd have to do that for classical, Gallilean relativity as well. (And don't say that Newton thought the speed of light was infinite! He had a good estimate of the speed of light.)

 

[Nugatory]

Observer F sees the point X as (x,t), Fig,(a). F' sees it as (x',t'), Fig,(b). But due to foreshortening, a length x' in frame F' corresponds to γx' in frame F, Fig.(c).

Except that length contraction, like time dilation, is symmetrical: A distance the measures $\Delta{L}$ in frame F will measure $\Delta{L}/\gamma$ in in frame F' but also a distance the measures $\Delta{L}$ in frame F' will measure $\Delta{L}/\gamma$ in in frame F - not $\gamma\Delta{L}$.

Relativity of simultaneity is at work here, in a rather unobvious way. Are the vertical lines representing the T axes the same lines in diagrams (a) and (b)? In frame F the distance between O and O' is $vt$, and in frame F' the distance between O and O' is $vt'$, but $t$ and $t'$ are not equal.

And I don't see the time a light photon takes to get from the event to an observer considered.

That's not needed in any derivation of the Lorentz transformations. Instead of having one observer receiving light signals from remote events and subtracting out the light travel time to assign a time coordinate to the remote event, imagine that you have multiple observers with synchronized clocks, all at rest relative relative to one another and strategically positioned so that one of them is present at each interesting event. They make timestamped notes of whatever happens where they are (no light travel time correction needed), and after the fact we gather up all their notes and work out what happened.

 

[jeremyfiennes]

You don't need to derive it to prove it."

Physics is based on 1) experimental results, 2) rational deduction. Length contraction and time dilation can be considered experimental results (muons, etc.). So it should be possible to derive the Lorentz transformations from these and logical deduction.

 

[PeroK]

Physics is based on 1) experimental results, 2) rational deduction. Length contraction and time dilation can be considered experimental results (muons, etc.). So it should be possible to derive the Lorentz transformations from these and logical deduction.

Here's an idea. If an event happens at location $x'$ in the primed frame, that means (by definition) that the event happened at a length of $x'$ from the origin. You could imagine a measuring stick of proper length $x'$ protruding from the origin. The events happens at the end of that stick. That's what gives it the $x'$ coordinate, by definition.

What does that look like in the unprimed frame? First, the origin is at $x = vt$. Second, the measuring sticking is contracted to $x'/ \gamma$, so we have:

$x = vt + x'/ \gamma$

And, if you rearrange that you get the LT for the x'-coordinate.

For time, you could imagine a clock at the end of the measuring stick. The time on that clock records the time of the event, $t'$.

Again, switch to what this look like in the unprimed frame. In particular, clocks in the primed frame:

A clock at origin of the primed frame differs from the clock at the event by $vx'/c^2$. The origin clock reads $t' + vx'/c^2$ at the time of the event.

But, this clock was synchronised with the clocks in the unprimed frame at $t = 0$ and has been dilated ever since. So:

$t = \gamma (t' + vx'/c^2)$

And, in fact, that gives you directly the inverse transformation.

The rest can be done either by a symmetry argument, or by manipulating these two equations.

 

[Nugatory]

So it should be possible to derive the Lorentz transformations from these and logical deduction.

Of course it is (although there is an intermediate inductive step: getting from the experimental results to the postulate that light speed is constant so that you have something to deduce from). The point that I'm making is that derivation and validation are not the same thing. "Is this the right answer?" and "Can I improve on the way I came up with this answer?" are different questions to be answered separately.

 

[PeroK]

Of course it is (although there is an intermediate inductive step: getting from the experimental results to the postulate that light speed is constant so that you have something to deduce from). The point that I'm making is that derivation and validation are not the same thing. "Is this the right answer?" and "Can I improve on the way I came up with this answer?" are different questions to be answered separately.

Yes, and personally I wouldn't be satisfied simply to derive the LT from arguments of isotropy and homogeneity etc and not double check that time dilation, length contraction, RoS produce the same thing. The LT is way too important not to double check it!

 

[stevendaryl]

The Lorentz transformations are mathematically simple. I had always imagined they could be easily derived. I however just found out from another PF thread that this is not so. Their originators Lorentz and Poincaré simply stated them without derivation. And the "proofs" I have seen to date have been complex and unconvincing. Since only simple geometry and the standard SR conditions (length contraction, time dilation and the constancy of the speed of light) are apparently involved, I attempted a derivation.

I'm not sure I understood your derivation, but the Lorentz transformation can be thought of (although I'm not sure if this is the best way to think about it) as the combination of three effects:

1. Length contraction
2. Time dilation
3. Relativity of simultaneity: clocks that are synchronized in one frame are not synchronized in another frame.

#3 is necessary if you want to have light have speed c in every inertial coordinate system. But starting with 1 & 2 is kind of unsatisfying, since you have to assume the mathematical form of length contraction and time dilation, when in other derivations, that is derivable.

Anyway, I can sketch how you can get the Lorentz transformations from these three.

Let's make a coordinate system discrete, to make it more concrete. In the moving frame, let's assume that we have a line of clocks at rest along the x-axis, with a distance $L$ between them. The location of clock number $j$ is just:

$x'_j = j L$

In the stationary frame, there are two differences: the length between clocks is shrunk to $\frac{L}{\gamma}$, and the clocks are moving at speed $v$. So in this frame, the location of clock number $j$ is:

$x_j = \frac{jL}{\gamma} + v t$

What about the time on clock $j$? Let $\Delta T_j$ be the offset of clock number $j$ at time $t=0$ (according to the stationary frame). By time dilation, each clock is running slower than a stationary clock by a factor of $\gamma$. So the time on clock $j$ is given by:

$t'_j = \frac{t}{\gamma} + \Delta T_j$

We're almost done, except for figuring out the offset, $\Delta T_j$. Let's assume that clock number 0 is started at $t=0$, so the offset is 0. For $j > 0$, let's assume that at $t=0$, a light signal is sent from clock 0 to clock $j$. We can compute the time that it arrives in the two different frames.

In the moving frame, the moving observers measure the distance between clock number 0 and clock number $j$ is $j L$. So if clocks are synchronized so that light has speed $c$, then the signal arrives when:

$t'_j = \frac{j L}{c}$

In the stationary frame, the distance between the clocks is contracted, to just $\frac{j L}{\gamma}$. Also, in this frame, clock number $j$ is moving away from the light signal, so it takes longer for the light signal to reach clock number $j$. That time is given by:

$t = \frac{jL}{\gamma (c-v)}$

Putting the facts about $t$ and $t'_j$ together:

• $t = \frac{jL}{\gamma (c-v)}$
• $t'_j = \frac{jL}{c}$
• $t'_j = \frac{t}{\gamma} + \Delta T_j$

So we conclude:

$\frac{jL}{c} = \frac{1}{\gamma} \cdot \frac{jL}{\gamma (c-v)} + \Delta T_j$

$\Delta T_j = -jL (\frac{1}{\gamma^2 (c-v)} - \frac{1}{c}) = -jL (\frac{1-\frac{v^2}{c^2}}{c-v} - \frac{1}{c}) = -\frac{jL v}{c^2}$
(using $\frac{1}{\gamma^2} = 1-\frac{v^2}{c^2}$)

So we have our transform:

$x'_j = j L$
$t'_j = \frac{t}{\gamma} - \frac{jL v}{c^2}$

To get rid of $L$, remember that $x_j = \frac{j L}{\gamma} + vt$, so $jL = \gamma(x_j - v t)$. So in terms of $x_j$:

$x'_j = \gamma (x_j - vt)$
$t'_j = \frac{t}{\gamma} - \gamma (\frac{x_j v}{c^2} - \frac{v^2}{c^2} t)$
$= \gamma (t (\frac{v^2}{c^2} + \frac{1}{\gamma^2}) - \frac{v x_j}{c^2})$
$= \gamma (t - \frac{v x_j}{c^2})$
(again using $\frac{1}{\gamma^2} = 1-\frac{v^2}{c^2}$)

 

[jeremyfiennes]

I can't see your diagram. With just your maths I am completely unable to understand what you think you are doing. Changing notation half way through without saying what it is supposed to replace doesn't help at all.

Sorry, I was trying to use the "image" icon. Ref the notation, I chose t_a and t_b to differentiate from t_x, in fact not necessary. Correcting for these, my original post becomes:
"Consider a universe comprising single time and space dimensions, and two observers A and B moving at steady relative speed v. Each observer has a clock, and there is also one at X. Define the time origin t=0 as the moment when frames A and B coincide. Synchronize all three clocks via a signal from the mid-point (x/2) at this instant.

Consider an event (x,t_x) occurring at point X in frame A, Fig. 0‑1a. t_x is the time on the clock at X. The question is: what is observer B's perception of this event in terms of A's, Fig. 0‑1b?
In frame A, an event photon takes time x/c to reach observer A, who sees it occurring at:
(x,t) where t = t_x+x/c) ) (eq.1)
For observer B, frame A moves at speed v, meaning that its lengths are foreshortened by the Lorentz factor γ. Observer B then perceives the event as ocurring at distance x':
x' = (x – vt_x)/γ (eq.2)
where:
γ=1/√(1–β²); β=v/c (eq.3)
Relative to observer A's clock, B's runs slow by the factor γ. Including the photon travel time, B perceives the event occurring at time t':
t' = t_x/γ + x'/c (eq.4)
Substituting and rearranging, B's points in terms of A's
x' = [(1+β)x – vt]/γ; t' = [(1–β)t + vx/c²]/γ (eq.5)
Because these expressions differ radically from the standard:
x' = γ[x – vt]; t' = γ[(t – vx/c²] (eq.6)
something has gone wrong somewhere. But what and where?

 

[jeremyfiennes]

Seems like you should derive the Lorentz factor, not just assume it.

The Lorentz factor follows from Einstein's station and traveling observers thought experiement.

 

[Nugatory]

something has gone wrong somewhere. But what and where?

At the point where your derivation diverges from the one in the wikipedia section linked to above.

 

[jeremyfiennes]

I'm not sure I understood your derivation,

It is based on 1) length contraction, time dilation and c constant as experimental facts; 2) geometrical reasoning. But this doesn't give the standard result, and I want to know why. I reposted my original derivation in standard notation.

 

[PeroK]

In frame A, an event photon takes time x/c to reach observer A, who sees it occurring at:

No, no, no, no, no. The event takes place in a darkened room. There are no photons.

What you should do is go back to classical physics and introduce the finite speed of light and see all your equations fall apart. An event takes place at time $t$ regardless of who observes it, how and when.

 

[PeroK]

No, no, no, no, no. The event takes place in a darkened room. There are no photons.

What you should do is go back to classical physics and introduce the finite speed of light and see all your equations fall apart. An event takes place at time $t$ regardless of who observes it, how and when.

For example, consider an object moving along the x-axis at speed $v$. Each local observer sees it pass at speed $v$. But, at time $t$ it is a distance $vt$ from the origin, so an observer at the origin sees that position event at $t + vt/c$.

At time $t + 1$, the object is at position $vt + v$, and the observer at the origin sees that event at $t + 1 + v(t+1)/c$.

So, for an observer at the origin we have $\Delta x = v$ and $\Delta t = 1 + v/c$

So, the speed of that object to an observer at the origin is $v/(1 + v/c) \ne v$.

In other words, if you factor in the time for different observers to receive light signals from events, then all of the equations of classical mechanics fall apart. You no longer have a well-defined speed of an object in an inertial reference frame.

 

[jeremyfiennes]

At the point where your derivation diverges from the one in the wikipedia section linked to above.

It diverges where Wiki gets x=vt+x'/γ and I get x=vt+γx'. Lengths are contracted in the moving F frame that F' sees (Fig.(b)). The F-frame correlate of x' (Fig.(c)) must therefore be greater than the F' value. Meaning that the Lorentz factor γ (always >1) must be in the numerator (where I put it) and not in the denominator (where Wiki puts it).

 

[PeroK]

It diverges where Wiki gets x=vt+x'/γ and I get x=vt+γx'. Lengths are contracted in the moving F frame that F' sees (Fig.(b)). The F-frame correlate of x' (Fig.(c)) must therefore be greater than the F' value. Meaning that the Lorentz factor γ (always >1) must be in the numerator (where I put it) and not in the denominator (where Wiki puts it).

It depends on which frame you are in. $x'$ is a length in primed frame: the distance from the origin of the primed frame to the event. This distance is contracted when measured in the unprimed frame.

Again, imagine a stick of length $x'$ from the origin to the event. You want to dilate that stick in the unprimed frame to a length of $\gamma x'$. That would be length dilation.

 

[Nugatory]

It diverges where Wiki gets x=vt+x'/γ and I get x=vt+γx'. Lengths are contracted in the moving F frame that F' sees (Fig.(b)). The F-frame correlate of x' (Fig.(c)) must therefore be greater than the F' value. Meaning that the Lorentz factor γ (always >1) must be in the numerator (where I put it) and not in the denominator (where Wiki puts it).

And you have it wrong. As we pointed out above, length contraction is symmetrical and you have fallen into a somewhat unobvious relativity of simultaneity trap.

You could take advantage of the fact that you know that the Lorentz transforms are correct even though you cannot (yet) derive them, and use them to transform all of the relevant events, and you'll likely see the error in your analysis.

 

[jeremyfiennes]

The time it takes a light signal (or a sound signal, or a telegram) to get to the origin of the coordinate system is irrelevant.

We are talking about observers perceiving events in their reference frames. Distant events are perceived later than corresponding close events.

 

[PeroK]

We are talking about observers perceiving events in their reference frames. Distant events are perceived later than corresponding close events.

No, no, no, no, no, no, no!

We are talking about events as measured in a reference frame.

The speed of signals has nothing to do with this. When the Titanic sank, no one in London knew until the next day. But, that doesn't mean that the ship sank a day later in a London reference frame.

And, certainly,no one in London actually saw it sink.

 

[jeremyfiennes]

It depends on which frame you are in. x′x′x' is a length in primed frame: the distance from the origin of the primed frame to the event. This distance is contracted when measured in the unprimed frame.

This is the crux of it. Unprimed distances move, and are contracted, in the primed frame (Fig.(b)). And not in the unprimed frame as you maintain.

 

[jeremyfiennes]

No, no, no, no, no, no, no!

Explain. Does light not have a finite velocity?

 

[PeroK]

Explain. Does light not have a finite velocity?

It also has had. Einstein was not the first to discover that. Why light? Bats navigate by echo location. So, does SR for bats need to take the speed of sound into account?

 

[jeremyfiennes]

The Lorentz Transformation is not about observers but about reference frames.

Is not a reference frame the viewpoint of an observer?

 

[Ibix]

Explain. Does light not have a finite velocity?

Because relativity assumes you correct for light travel time when determining when something happened. Otherwise there's no reference frame - two people at relative rest but not in the same place would assign different coordinates to events.

You are also treating the time coordinate differently from the spatial one. You add on the travel time of light to the time coordinate, but not the travel distance to the spatial coordinate. You shouldn't do either, I hasten to add, but you might like to reflect on why you're doing one and not the other.

 

[Ibix]

Is not a reference frame the viewpoint of an observer?

No. It's the combined viewpoint of an infinite array of observers filling spacetime. Alternatively, it's one observer who carefully corrects for light travel time, so the observer's location is not involved in the maths.

 

[Nugatory]

Is not a reference frame the viewpoint of an observer?

Absolutely not. It is a convention for assigning coordinate values to events.
Here we are talking about the Lorentz transformations, and they are about how we transform the (x,t) values assigned using one coordinate system into the (x',t') values assigned using another coordinate system given that the origins of the two coordinate systems are in relative motion.

You've heard me strongly recommend Taylor and Wheeler before; one of the reasons is that they make this clear in the very beginning when they model a reference frame as a collection of observers each recording only the events that happen where they are - that is, directly recording the coordinates instead of inferring them.

 

[PeroK]

Is not a reference frame the viewpoint of an observer?

No. Not in the sense that it is what an observer sees. There may be no observers. Just devices to measure where and when an event takes place: local measurements. These measurements could be collated and sent by email for analysis the next day.

A reference frame is a system of coordinates, where the origin is essentially arbitrary. It has no special designation as the place at which all events must be directly observed by EM radiation.

 

[jeremyfiennes]

Does SR for bats need to take the speed of sound into account?

I leave SR for bats to bats. An observer can't be everywhere at once. He is normally taken to be at the origin of his reference frame. He observes events at the origin instantaneously. And those further away with a time delay. That is why one needs to distinguish between event time (t_x) and the time perceived by different observers on their own clocks(t, t').