Is the Lorentz transformation about observers?

[jeremyfiennes]

From a previous, now closed thread (Perok): "Technically, the Lorentz Transformation is not about observers but about reference frames."
Sorry, I still don't get this. In frame A with observer A at the origin, x is the distance of the event X he sees measured on his rod, i.e. as observed/perceived by him. t is similarly the time of the event measured on his clock, i.e. as perceived by him. So (x,t) is observer A's perception of event X. Similarly, (x',t') is observer B's perception. The Lorentz transformations relate (x,t) and (x',t'). And therefore relate A's and B's perceptions of the event X.

 

[Ibix]

A frame is used by all inertial observers who are at rest with respect to one another and have agreed a particular time and place (not necessarily where any of them are standing) as the origin. Their location does not matter at all because the coordinates assigned are the times and places things happen, not when they are seen by any particular person.

Different frames have different conventions for simultaneity and different notions of rest.

As noted in that thread, they cannot relate observations because there is no $|x/c|$ anywhere in the maths.

 

[DrGreg]

t is similarly the time of the event measured on his clock, i.e. as perceived by him.

No it isn't. This is the mistake you have kept making time and time again in the other thread. t is the time measured by a clock located at X which is at rest in frame A and which has been appropriately synchronised to the observer's clock, or, equivalently, it is the time measured by the observer's clock and then deducting the time taken for light to travel from the event to the observer.

When we look at the moon, we see what the moon looked like about 1½ seconds ago. The time we assign to an event we see on the moon is 1½ seconds before the time that we see it.

If your version were true, we would have to say that we are seeing the moon as it is now, and that the time taken for light to travel from the moon to Earth is zero, and the time taken for light to travel from the Earth to the moon is 3 seconds (because we can measure the round trip, there-and-back, takes 3 seconds). That would imply the speed of light moon-to-Earth is infinite and from Earth-to-moon is c/2, 149,896,229 m/s. That contradicts the postulate of relativity that the speed of light is always the same value.

 

[jeremyfiennes]

The coordinates assigned are the times and places things happen, not when they are seen by any particular person.

An event X happens in frame A. It is seen by observer A₁ as happening at position and time (x₁,t₁), by an observer A₂ in a different position as happening at position and time (x₂,t₂), and so on. What in this case are the event's 'absolute' coordinates, the ones that you say "are the times and places things happen"?

 

[Ibix]

An event X happens in frame A. It is seen by observer A1 as happening at position and time (x1,t1), by an observer A2 in a different position as happening at position and time (x2,t2), and so on.

No. If they are at rest with respect to each other they agree on the x and t coordinates. Observers are not required to be at the origin.

As has been pointed out to you time and time again you are constructing a coordinate system that does not work the way the coordinates related by the Lorentz transforms work. You are free to do this, but you will not be able to derive the Lorentz transforms because you are not trying to solve the problem that the Lorentz transforms solve.

 

[Nugatory]

An event X happens in frame A. It is seen by observer A₁ as happening at position and time (x₁,t₁), by an observer A₂ in a different position as happening at position and time (x₂,t₂), and so on. What in this case are the event's 'absolute' coordinates, the ones that you say "are the times and places things happen"?

There are no "absolute coordinates". A frame is a rule for assigning coordinates to events; different frames will assign different coordinates but as long as you pick one frame and stick with it, each event will have a single set of coordinates, for the same reason that a specific point on the surface of the Earth has only one latitude and longitude no matter where the navigator is.

Let's start by considering the case in which $A_1$ and $A_2$ are rest relative to one another. In that case, it is convenient to use the frame F in which they are at rest. Assume for the sake of argument that $A_1$ is to the left of where the event X happens and $A_2$ is to the right (this just saves us some absolute-value and plus/minus signs in the expressions below) and at distance $D_1$ and $D_2$ respectively (using frame F).

There are three relevant events:
1) Event X happens at coordinates $(x_1,t_1)$ using this frame. It's one event, it has one set of coordinates for the same reason that a specific point on the surface of the Earth has only one latitude and longitude no matter where the navigator is.
2) Light from event X reaches $A_1$; we'll call this event $E_1$ and it will have coordinates $(x_1-D_1, t_1+D_1/c)$
3) Light from event X reaches $A_2$; we'll call this event $E_2$ and it will have coordinate $(x_1+D_2, t_1+D_2/c)$

Suppose we want to use some other frame to describe this situation: For example we might want to know how $A_3$, moving relative to the first two, describes the situation. In this case, it will be convenient to use the frame F' in which $A_3$ is at rest. We have three events, each with their $(x,t)$ coordinates assigned using frame F, and we use the Lorentz transformations to calculate three sets of $(x',t')$ values for the coordinates of the three events using frame F'.

 

[PeterDonis]

When we look at the moon, we see what the moon looked like about 1½ seconds ago. The time we assign to an event we see on the moon is 1½ seconds before the time that we see it.

And implicit in this is "in a frame in which the Earth is at rest". Which, given the issue that has been ongoing in this discussion, seems worth making explicit.

 

[PeterDonis]

An event X happens in frame A.

No. Events happen; they do not happen "in" a particular frame. Different frames are different ways of assigning coordinates to events; different frames are not different sets of events, each of which only happens "in" a particular frame.

What in this case are the event's 'absolute' coordinates

There aren't any. Coordinates are frame-dependent. What @Ibix was saying was that the coordinates assigned by a particular frame are the times and places things happen according to that frame. There are no absolute times and places things happen.

 

[Nugatory]

What in this case are the event's 'absolute' coordinates, the ones that you say "are the times and places things happen"?

- We have a bunch of hypothetical observers, all of whom are at rest relative to one another.
- Because they are at rest relative to one another, the distances between them remain constant.
- Pick one of them and declare position $x=0$ to be the point in space where that one is. It doesn't matter which one you pick as long as once you've chosen you stick with that choice;.
- Whichever one you pick, the observer $D_1$ meters to the left of it will be at position $x=-D_1$ and the one $D_2$ meters to the right will be at position $x=D_2$, and similarly for all the other hypothetical observers.
- Let them all set their clocks to zero at the same time; this is possible because they are all at rest relative to one another so there is no problem with relativity of simultaneity.
- Because they are at rest relative to one another, their clocks will stay synchronized.

And now we have the reference frame - that is, a rule for assigning coordinates to events - in which this ensemble of observers is at rest:
- The origin of the coordinate system is the event "clock of the $x=0$ observer reads zero".
- When the observer at position $x=X$ is present at some event $E$ (that's "present" - they're right there on the spot, not seeing the light from somewhere distant), they look at their clock and see that it reads $T$. Using this frame in which our ensemble of observers are at rest, the coordinates of event $E$ are $(x=X, t=T)$.

And that's where the coordinates come from.

In practice, we don't need all these observers. If the observer at $x=X_1$ finds themselves present at the event "light from event E reaches me at time $T$" (this is, of course, a different event than $E$) and event $E$ happened at at a distance $D$ away to the left, then we can calculate that the coordinates of $E$ are $(X_1-D,T-D/c)$.
Exercise: Satisfy yourself that this is the same result as we'd get if we had an observer at $x=X_1-D$.
Exercise: Satisfy yourself that all observers at rest anywhere in this frame will come up with the same result.

We can construct other reference frames as well: just start with another bunch of observers, also all at rest relative to one another and therefore all moving with constant speed relative to the first. Go through the same steps and we'll find that using this new frame event $E$ will be assigned completely different coordinates $(x',t')$ - but the Lorentz transformation will give the relationship between $(x,t)$ and $(x' ,t')$.

 

[Nugatory]

At this point there's not much more to say, except to repeat the advice from the earlier thread about reading the first few chapters of "Spacetime Physics" by Taylor and Wheeler, where the ideas above are explained in more detail. This book is a fine reference even in a B-level thread (I pulled it off the shelf in my high school library when I was in the eleventh grade).

The thread will stay closed because we're running out of things to say. As with all thread closures, anyone can PM any mentor to have the thread reopened if there is need for further discussion - but please please please make a serious good-faith effort to understand what's been said here and in the earlier thread first.