High School Basic Lorentz transformation derivation

[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').

 

[Ibix]

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 (tx) and the time perceived by different observers on their own clocks(t, t').

No. If you keep believing this you will be unable to derive the Lorentz transforms because you are not trying to solve the same problem the Lorentz transforms solve.

 

[PeroK]

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').

It's your prerogative to believe that. Just don't complain that the Lorentz Transformation never makes sense!

 

[jeremyfiennes]

A reference frame is a system of coordinates.

Right. Observer (or set of instruments) A perceives an event at distance x from him on his rod, and time t on his clock. Observer B, with respect to whom A is moving, perceives it at distance x' on his rod and time t' on his clock. Due to the finite speed of light, the perceived times depend on the distances of the event from the respective observers. The Lorentz transformation defines the relation between two such perceptions. And must therefore involve the event-to-observer travel times..

 

[Ibix]

Due to the finite speed of light, the perceived times depend on the distances of the event from the respective observers.

...which the observers then subtract out to get the coordinates.

The Lorentz transformation defines the relation between two such perceptions.

Not as you describe, no. With my correction, yes.

 

[jeremyfiennes]

You are not trying to solve the same problem the Lorentz transforms solve.

Do the Lorentz transformations not relate the different perceptions, (x,t) and (x',t'), of a single event by two relatively moving observers? Because if they don't, I have got it all wrong.

 

[Ibix]

Because if they don't, I have got it all wrong.

That is what PeroK, Nugatory and I have been telling you for the past few hours, yes.

The Lorentz transforms do not relate the times observers see events. They relate the coordinates assigned by frames (which have no particular observation point) that are in relative motion.

 

[jeremyfiennes]

...which the observers then subtract out to get the coordinates.

Where does it say they are subtracted out? I can't see it anywhere in the Wiki article Nugatory gave. And I still have no answer to why Wiki gives x=vt+x'/γ. When due to unprimed length contraction in the primed frame, the unprimed equivalent x should be greater than the primed x', giving x=vt+γx'.

 

[Ibix]

Where does it say they are subtracted out?

I haven't read the wiki article. I expect it just talks about assigning times to events and never mentions light travel time at all. If it did, it would only have to decide where the observer is standing, write down the coordinates of the event, add on the light travel time to the observer's location, then take it away again.

And I still have no answer to why Wiki gives x=vt+x'/γ. When due to unprimed length contraction in the primed frame, the unprimed equivalent x should be greater than the primed x', giving x=vt+γx'.

See post #12 by @PeroK.

 

[jeremyfiennes]

The Lorentz transforms do not relate the times observers see events.

<https://en.wikipedia.org/wiki/Frame_of_reference>: "In Einsteinian relativity reference frames are used to specify the relationship between a moving observer and or phenomena under observation, which implies an observer at rest in the frame, although not necessarily at its origin". According to this a reference frame is an observer's point of view, with arbitrary origin and axes.

 

[Ibix]

According to this a reference frame is an observer's point of view, with arbitrary origin and axes.

Then where does the position of the observer appear in the Lorentz transforms?

 

[jeremyfiennes]

See post #12 by @PeroK.

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. Lengths in the unprimed frame A are therefore greater than their correlates in the primed frame B, giving x=vt+γx', and not Wiki's x=vt+x'/γ.

 

[jeremyfiennes]

Then where does the position of the observer appear in the Lorentz transforms?

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

 

[Ibix]

It doesn't. That's the problem.

That's not a problem. That's the expected behaviour of coordinate transforms.

The co-ordinates in a reference frame depend on its origin, which is not specified.

The origin is an arbitrary agreed event where clocks at x=0 pass clocks at x'=0, and this time is taken to be t=t'=0. It can be any event in spacetime, but choosing an event that's significant to your physical setup would make sense.

 

[stevendaryl]

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.

Yes, and I'm saying that it doesn't make any sense to me. I used the same facts in my post, and I got the standard result.

But let's go through your argument. You say that there is some event at point $X$ with coordinates $(x,t_x)$. What does that mean? Coordinates in what coordinate system? Then you say

In frame $A$, an event photon takes time ##x/c## to reach observer $A$, who sees it occurring at:
##(x,t)$where$t = tx+x/c)##.

What do you mean, A sees it occurring at...? Are you talking about the coordinates of the event, again? The coordinates don't change because light takes time to propagate.

I think you're mixing up two different things: (1) The coordinates where an event takes place, and (2) the time when an observer sees the light from an event.

To figure out the coordinates of an event, you can do several different things:

Approach 1:

1. Set up an array of clocks that are at rest relative to your frame.
2. Pick one clock to be at the origin.
   
3. Measure the distances between the clocks.
4. Synchronize the clocks somehow.
5. Then an event has coordinates $(x,t)$ where $x$ is the location of the clock nearest that event, and $t$ is the time on that clock.

That's what I sketched out in my post.

Approach 2:

1. Again, set up an array, of distance-markers, rather than clocks.
2. Pick one distance-marker to be the origin.
3. Measure the distances between the markers.
4. Place a single clock at the origin.
5. Then an event has coordinates $(x,t)$ where $x$ is the location of the distance marker nearest that event, and $t$ is computed using the formula $t = t_{seen} - \frac{x}{c}$, where $t_{seen}$ is the time it takes for light to get back to the origin.

You can't just set the time for the even to be the time you see the event, because it takes time for light to travel.

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

 

[stevendaryl]

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).

I think it would make more sense for you to understand the standard derivation, rather than for everybody else to try to understand yours. Since yours is clearly wrong.

Pick a frame: The rest frame of $A$. We'll stick to this frame for describing what happens. As measured from the A frame, the measuring sticks at rest in frame B are length-contracted. They are shorter. That means that B will measure LONGER distances than A. If something is 3 meters long, and you measure it with a measuring stick that is only half a meter long, then you will measure the length to be 6, since it takes 6 of your measuring sticks to stretch from one end to the other.

As measured from the A frame, the clocks at rest in frame B are time-dilated. B's clocks run slower. That means that B will measure SHORTER times between events. His clock runs slower, so it shows less elapsed time.

 

[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:

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