B Basic Lorentz transformation derivation

[stevendaryl]

It doesn't. That's the problem. The co-ordinates in a reference frame depend on its origin, which is not specified.

Just a suggestion: Wouldn't it be a better use of time for you to try understand the correct derivation, instead of for the rest of humanity to try to understand your incorrect derivation?

 

[stevendaryl]

Sorry. It got hidden away so I hadn't seen it. Perok says: "the measuring stick is contracted to x′/γ. Here we part ways, so let's at least get this one sorted. In the primed frame B the unprimed frame A is moving, and its lengths are contracted.

Right. So consider an object that is at rest relative to A, but is moving at velocity $-v$ relative to B. Let $x$ be its distance from the origin, according to $A$. Then its distance from the origin, according to B is $x/\gamma$. But it's moving, according to B. So B would describe its position this way:

$x' = x/\gamma - v t'$

Initially, when $t'=0$, the object is at $x' = x/\gamma$. Later, when $t' > 0$, B will find that it has moved a distance of $v t'$ in the negative-x direction. So it's new location is:

$x' = x/\gamma - v t'$

Which can be rewritten as:

$x = \gamma (x' + v t')$
$= \gamma v t' + \gamma x'$

What you wrote is:

$x = v t + \gamma x'$

That's almost the same, except that you have replaced $\gamma t'$ by $t$. That isn't correct.

 

[jeremyfiennes]

I really can't understand what you are doing in your derivation. You haven't really set up a coordinate system.

I set up a very specific coordinate system. Repeating: "Consider two observers A and B moving at steady relative speed v. Each observer has a clock, and there is also one at the event point 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."
Can anything be more defined than this?

 

[stevendaryl]

I set up a very specific coordinate system. Repeating: "Consider two observers A and B moving at steady relative speed v. Each observer has a clock, and there is also one at the event point 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."
Can anything be more defined than this?

First of all, why is the clock at X relevant to the relationship between A's coordinates and B's coordinates?

Second, by saying "the mid-point (x/2) at this instant", you are assuming that there is a meaning to "at this instant" that is frame-independent. There isn't. "at this instant" means "with the same time coordinate". If different frames are using different time coordinates, then they have different notions of "at this instant".

Third, you say "A sees it occurring at $(x_a, t_x + x_a/c)$." What does that mean? Do you mean that those are the coordinates that A assigns to the event? If so, that's not true. A knows that light doesn't travel instantaneously, so if he sees a photon at time $t$, then the event where the photon was created must have been earlier.

So no, I don't consider what you're saying to be very well-defined, at all.

 

[jeremyfiennes]

Right. So consider an object that is at rest relative to A, but is moving at velocity −v−v-v relative to B. Let xxx be its distance from the origin, according to AAA. Then its distance from the origin, according to B is x/γx/γx/\gamma. But it's moving, according to B. So B would describe its position this way:
x′=x/γ−vt′

Right. But I think not quite. The 1/γ is correct. Stationary-in-A lengths are contracted in B's frame. But only as B sees them, i.e measured from his origin. Giving:
x′=(x−vt′)/γ
whence
x=vt′+γx′
as I hold. And not the
x = vt + x'/γ
given in Wiki.

 

[jeremyfiennes]

First of all, why is the clock at X relevant to the relationship between A's coordinates and B's coordinates?

It can be synchronized together with A's and B's clocks, and so forms a link between them. What is the problem with having a clock at X?

 

[stevendaryl]

It can be synchronized together with A's and B's clocks, and so forms a link between them. What is the problem with having a clock at X?

I'm just saying that it's not relevant, so it complicates things with no benefit.

 

[jeremyfiennes]

Second, by saying "the mid-point (x/2) at this instant", you are assuming that there is a meaning to "at this instant" that is frame-independent. There isn't.

The "instant" is the one I defined as t=0, when A's and B's space origins coincide. It is perfectly definable in terms of these two frames.

 

[jeremyfiennes]

I'm just saying that it's not relevant, so it complicates things with no benefit.

On the contrary, it facilitates things very much by providing a link between the perception of the point in the two frames, which is basically what we (well, at least I) are after.

 

[stevendaryl]

Right. But I think not quite. The 1/γ is correct. Stationary-in-A lengths are contracted in B's frame. But only as B sees them, i.e measured from his origin. Giving:
x′=(x−vt′)/γ

No, that's not correct. The $vt'$ should not be divided by $\gamma$.

Look, let's suppose that $A$ places a marker at the point $x$. In A's frame, the marker is stationary, so it's location is always at $x$.
In B's frame, however, the marker starts off (when $t' = 0$) at the point $x' = x/\gamma$. But then it moves at speed $v$ in the negative x-direction. So for B, its location at a later time will be given by:

$x' = x/\gamma - v t'$

 

[stevendaryl]

On the contrary, it facilitates things very much by providing a link between the perception of the point in the two frames, which is basically what we (well, at least I) are after.

The evidence shows that it doesn't help, because: (1) Nobody besides you understands what the point is, and (2) You, the person who does understand it, gets the wrong answer for the transformations.

 

[jeremyfiennes]

Third, you say "A sees it occurring at (xa,tx+xa/c)(xa,tx+xa/c)(x_a, t_x + x_a/c)." What does that mean?

There's a revised version of my original post in prime notation, posted a day or two later in response to a comment. What I say in its terms is "A sees the event occurring at (x,t_x+x/c)", the time t being the time on clock X plus the delay for the photon to reach observer A, here at the origin of his frame.

 

[jeremyfiennes]

A knows that light doesn't travel instantaneously, so if he sees a photon at time ttt, then the event where the photon was created must have been earlier.

We're talking about A's perception of the event, and not his reasoning as to what it would have been if he had not been where he was but somewhere else.

 

[stevendaryl]

The "instant" is the one I defined as t=0, when A's and B's space origins coincide. It is perfectly definable in terms of these two frames.

I'm telling you that "at this instant" doesn't make any sense unless you specify a coordinate system. You have two different events:

• $e_1$: A and B are together, at the same location.
• $e_2$: Synchronizing photons are sent from the location at coordinate $x/2$

Those are two different events, but you're saying that they are at the same instant---the same time. You can't say that without specifying WHOSE coordinate system is used to determine that the times are the same. $e_1$ has a time in A's coordinate system, $t_1$. It has a time in B's coordinate system, $t'_1$. Event $e_2$ has a time in A's coordinate system, $t_2$. It has a time in B's coordinate system, $t'_2$. If you say that events $e_1$ and $e_2$ are at the same instant, what does that mean?

1. Does it mean $t_1 = t_2$?
2. Does it mean $t'_1 = t'_2$?

 

[jeremyfiennes]

Nobody besides you understands what the point is,

How do you know? have you asked everybody? Science is based on experimentally verifiable fact. Is this fact verified?.

 

[stevendaryl]

There's a revised version of my original post in prime notation, posted a day or two later in response to a comment. What I say in its terms is "A sees the event occurring at (x,t_x+x/c)", the time t being the time on clock X plus the delay for the photon to reach observer A, here at the origin of his frame.

That doesn't make any sense. There are two different events:

• $e_{send}$: a photon is sent from the point at coordinate $x$
• $e_{receive}$, that photon is received by A

When you talk about "the event occurring at $(x, t_x + x/c)$", which event are you talking about? Because those coordinates don't make sense for either one.

$e_{send}$ has coordinates $(x, t_x)$
$e_{receive}$ has coordinates $(0, t_x + x/c)$

 

[jeremyfiennes]

If you say that events e1e1e_1 and e2e2e_2 are at the same instant, what does that mean?

I do NOT say that events occur at the same instant. Please quote where I said that or anything like it. I talk of clocks synchronized at the instant when the A and B axes coincide, defining this as time t=0. The synchronizing signal is sent from the mid-point between clocks A,B and X at that instant, all in A's frame.

 

[stevendaryl]

We're talking about A's perception of the event, and not his reasoning as to what it would have been if he had not been where he was but somewhere else.

That doesn't make any sense. A only sees the photon when it enters his eye. That event has coordinates:

$(0, t_x + x/c)$

The x-coordinate is ZERO, because A is located at x=0.

 

[stevendaryl]

How do you know? have you asked everybody? Science is based on experimentally verifiable fact. Is this fact verified?.

Okay, I'll make it a little softer: There is no evidence that anybody understands what you are saying. I don't think that what you're saying is very coherent, and I think that the fact that you get a result that contradicts what everyone else has gotten for the last 113 years is good evidence that you are the one who is making a mistake.

 

[m4r35n357]

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

Bearing in mind the fate of this thread perhaps you might want to revisit some of these. Basically, the $\gamma$ factor emerges in a natural way from the Lorentz transform, together with relativity of (non-)simultaneity. It is not an easy matter to (un-)wrangle these things back together into a derivation of the LT, and I would consider it a fatal distraction to a newcomer, who already has plenty to unlearn and then learn again.

Also, it is very common to see misunderstandings between what you see (light beams) and what you measure (events). This is IMO 100% the fault of sloppy and irresponsible teaching and evangelising of SR from about 1905 onwards (https://archive.org/details/RelativityCommonSense is a notable exception).

In short, I would recommend abandoning your particular approach, and learning one of the "unconvincing" ones. Here is my favourite, most others are more complicated and/or abstract, to me at least.

 

[Ibix]

We're talking about A's perception of the event, and not his reasoning as to what it would have been if he had not been where he was but somewhere else.

This, IMO, is the basic problem. You are talking about this but also saying you want to derive the Lorentz transforms, which have nothing to do with the perceptions of an observer. So you are talking about two different things as if they were the same thing.

You can, of course, derive transforms between any non-degenerate coordinate systems. But note that your basic transforms are non-linear (if you do them correctly, there's a $|x/c|$ in there), so it's probably rather messy.

 

[jeremyfiennes]

That doesn't make any sense. A only sees the photon when it enters his eye.

An observer doesn't see the photons. They are part of his 'seeing'. Photons from the event arrive at his retinas, send neural impulses to the visual cortex of his brain, and he subjectively experiences seeing the event. There is one event and two inertial observers views of it, (x,t) and (x',t').

 

[stevendaryl]

An observer doesn't see the photons. They are part of his 'seeing'. Photons from the event arrive at his retinas, send neural impulses to the visual cortex of his brain, and he subjectively experiences seeing the event. There is one event and two inertial observers views of it, (x,t) and (x',t').

Whatever. But what does it mean to say that A sees the event occurring at $(x, t_x + x/c)$? That is nonsensical. Or at least, I don't know what it means. It seems that it is mixing up two different things: The first coordinate, $x$ is where the event happened, and the second coordinate, $t_x + x/c$ is when A received the image of event. Why are you pairing those two things, and what does pairing them have to do with the derivation of the Lorentz transformations?

 

[Ibix]

Whatever. But what does it mean to say that A sees the event occurring at $(x, t_x + x/c)$? That is nonsensical.

And it should be $(x,t_x+|x/c|)$, just for extra fun.

 

[stevendaryl]

An observer doesn't see the photons. They are part of his 'seeing'. Photons from the event arrive at his retinas, send neural impulses to the visual cortex of his brain, and he subjectively experiences seeing the event. There is one event and two inertial observers views of it, (x,t) and (x',t').

I'm thinking that you don't understand that when people talk about the coordinates of event being $(x,t)$, they don't mean that $t$ is the time they saw the light from the event. $t$ is the time that the event took place.

The time that somebody sees an event is different for different observers, even if they are at rest relative to each other.

 

[stevendaryl]

I'm thinking that you don't understand that when people talk about the coordinates of event being $(x,t)$, they don't mean that $t$ is the time they saw the light from the event. $t$ is the time that the event took place.

The time that somebody sees an event is different for different observers, even when those observers are at rest relative to each other.

If I say that an event happened at longitude 21 degrees East, latitude 11 degrees North, at 5:21 pm Greenwich time, I don't mean that that's when I saw the light from the event. I can't see the light of something on the other side of the world.

 

[weirdoguy]

You are talking about this but also saying you want to derive the Lorentz transforms, which have nothing to do with the perceptions of an observer.

And that has been stated in this thread quite a few times @jeremyfiennes and it seems that you saw it, so I don't know why you are unwilling to accept this fact.

 

[jeremyfiennes]

So we at least agree that the Wikipedia x=vt+x'/γ is wrong, and that the γ should be
somewhere in the numerator?

 

[jeremyfiennes]

"t is the time that the event took place."

This implies an absolute time, which you have rightly observed is a meaningless idea. There is the time t on observer A's clock, and time t' on observer B's clock, and the Lorentz transform that gives the relationship between them: t' = ??(x,t).

 

[stevendaryl]

So we at least agree that the Wikipedia x=vt+x'/γ is wrong, and that the γ should be
somewhere in the numerator?

No. You have two equations that are both true:

1. $x = x'/\gamma + vt$, which implies $x' = \gamma (x - v t)$
2. $x' = x/\gamma - vt'$, which implies $x = \gamma (x' + v t')$

The equations with $\gamma$ in the denominator are just rearrangements of the equations with $\gamma$ in the numerator.

 

[Nugatory]

So we at least agree that the Wikipedia x=vt+x'/γ is wrong, and that the γ should be
somewhere in the numerator?

Are you talking about the $x=vt+x'/\gamma$ in the section "Time dilation and length contraction"? Wikipedia is correct and the $\gamma$ belongs where it is, in the denominator.

See posts #10, #12, #23, #24 above... And I strongly recommend that you try the exercise I suggested in #24. Of course using the Lorentz transform to derive the Lorentz transform is unacceptably circular, but that's not what we're asking you to do. The point is that once you look at the transformed coordinates it will become clear how they relate to one another and you'll see why the $\gamma$ belongs where it does.

[Edit: You might also find it helpful to draw a Minkowski diagram showing the event M and the worldline of the origins of the two coordinate systems. Pay particular attention to the distance between the origin of F' and M in the two frame - this is where the relativity of simultaneity is misleading you].

 

[stevendaryl]

This implies an absolute time,

$t$ is the time in a particular COORDINATE system. It's not the time on any particular clock. A coordinate system is not a single clock.

 

[stevendaryl]

There is the time t on observer A's clock, and time t' on observer B's clock, and the Lorentz transform that gives the relationship between them: t' = ??(x,t).

That is absolutely NOT what the Lorentz transformation tells you. The Lorentz transformations don't tell you how A's clock relates to B's clock. They tell you how A's COORDINATE SYSTEM relates to B's coordinate system. A coordinate system is a way of assigning (in the case of one spatial dimension) two numbers, $x$ and $t$ to every event. So here's one event:

• A's clock shows $t = 12$

The coordinates for this event in A's coordinate system are: $(x=0, t=12)$. The coordinates for this event in B's coordinate system is:
$x' = - \gamma v \cdot 12$ (that is, $x' = \gamma (x - v t)$ with $x=0$ and $t=12$) and $t' = \gamma \cdot 12$ ($t' = \gamma (t - \frac{vx}{c^2})$).

Suppose that $v = 0.866 c$ so that $\gamma = 2$. Then that means that B's coordinates for that event are $(x' = -20.784, t' = 24)$

This is NOT relating what's on B's clock to what's on A's clock. It is NOT saying that "When A's clock shows time 12, B's clock shows time 24". That would be contradictory, because we can also look at a different event:

• B's clock shows $t' = 12$

B's coordinates for this event are $(x'=0, t'=12)$. A's coordinates for this event are $(x=+20.784, t=24)$.

The Lorentz transformations transform between coordinates for an event in one frame and coordinates for an event in another frame. It is NOT a transformation between the time on one clock and the time on another clock.

 

[PeterDonis]

At this point this thread seems to be going nowhere, so it is now closed.

 

[Nugatory]

This thread is closed because it is clearly going nowhere.
If the answers already supplied are not enough for the original poster to work out for themself why $x=vt+x'/\gamma$ and not $x=vt+\gamma{x'}$ is the correct expression all the way back at post #8, we can have another thread devoted to only that question.