2
$\begingroup$

I've been working a bit at understanding relativity a bit more, and understanding Lorentz transformations. I want to make sure I understand the meaning of a Lorentz transofmration, and when to use it in a workflow of figuring out a scenario.

I understand that Lorentz transformations are about how to take "Event Coordinates", which take place at a particular time and xyz coordinate within a reference frame, and convert them to another reference frame.

So, I have two closely related questions about this Lorentz transofmration:

The first is, the INPUT to the transformation should be the "real time" of the event in the base reference frame S. By "real time", I mean that if I "perceive" event A as having taken place at time 3s, but it happened 1 light-second away, and I perceive event B as having taken place at time 6s, but it happened 2 light-seconds away, then my time-coordinates for event A and event B should be 2 and 4, not 3 and 6 - I should be subtracting the time it took the light to get to me, and using the real time (if there's a better term for that, please let me know) of the events, and not my perceived time of having been hit by the light from the events.

So, my question 1 is about the input to the Lorentz transform, and my question 2 is the mirror question about the output of the Lorentz transform. When I transform events A and B from frame S to S', is the output the "real time" of each event? Or is the output the time that the event would be perceived, after the travelling of light, to the observer? I assume it's the "real time" (and boy would I like a better word for that), and thus if I wanted to also find out when the obsever in S' perceives the event, I'd then also have to do extra calculations to figure out how long it takes light to get from that event to the observer.

I'd like to be corrected if I'm wrong on anything above, thank you.

$\endgroup$
4
  • $\begingroup$ Good question! The answer is "no" to both, but this is a good time to learn about it. For the purposes of the Lorentz Transform, the time is the (synchronised & common) time in a "field" filled with (potential) observers at every point of interest. You are asking about a single observer, which is a layer (well two, as you have noted) of processing on top of the LT. If you are interested in what a real observer actually sees, you need to learn about the relativistic doppler effect, and aberration of light (both of which incorporate the effects of the LT). $\endgroup$
    – m4r35n357
    Commented Feb 5 at 12:21
  • $\begingroup$ @m4r35n357 I assume synchronized time is what I was referring to by 'real time', correct? So then, "no" as in, I'm not supposed to use the synchonized time as the input, and the output isn't the synchronized time? $\endgroup$
    – TKoL
    Commented Feb 5 at 12:30
  • 1
    $\begingroup$ Your assumptions are correct. $\endgroup$
    – WillO
    Commented Feb 5 at 12:31
  • $\begingroup$ Ironically, the best analogy for what you call "real time" is "Newtonian" or "universal" time in the frame you are working in! $\endgroup$
    – m4r35n357
    Commented Feb 5 at 12:34

3 Answers 3

2
$\begingroup$

You can go either way. Suppose you have two events, A and B, in different places some distance from you. You can perform a Lorenz transformation on the events themselves to find their coordinates in another reference frame. However, you seeing the light from event A is also an event, which we can call event C. Likewise we can define event D as you seeing the light from event B. If you want to, you can apply the Lorenz transformation to events C and D to find their coordinates in another frame.

$\endgroup$
7
  • $\begingroup$ but when I do that for C and D, I'm finding out when C and D happen to me still, just in another reference frame right? When I transform C and D, I'm not finding out what an observer in that other reference frame experienced, I'm finding out what they think I experienced? (or rather, when they think I experienced it) $\endgroup$
    – TKoL
    Commented Feb 5 at 12:32
  • $\begingroup$ I believe the OP was asking about trying to transform the coordinates of $A$ by inputting the coordinates of $C$, which is a third alternative that the OP correctly expected to be wrong. $\endgroup$
    – WillO
    Commented Feb 5 at 12:33
  • 2
    $\begingroup$ @TKoL ; Yes, what you are saying is exactly right. (Responding to your first comment.) $\endgroup$
    – WillO
    Commented Feb 5 at 12:35
  • 2
    $\begingroup$ @TKoL : If you are just now, at 11lpm, seeing an event that happened eight minutes ago on the sun, then the time to input into the Lorentz transformation for that event is 10:52. $\endgroup$
    – WillO
    Commented Feb 5 at 12:41
  • 2
    $\begingroup$ @TKoL : Yes, definitely. $\endgroup$
    – WillO
    Commented Feb 5 at 12:58
1
$\begingroup$

I call the “real time” as the time we "measure" and the time perceived as the time we "observe".

It's true that we should use the measured time in the Lorentz Transformation instead of the time we observed. In fact, mostly we just think of a situation in a totally objective frame using the time we measured as the time coordinate to calculate, and we believe that's exactly what happening at the place though we may haven't "observed" it yet.

If we are dealing with a situation in which our observation is what matters, then we just add the time the light take to arrive to us. And this leads to various interesting phenomena like Terrell Effect and apparent superluminal motion.

$\endgroup$
5
  • 1
    $\begingroup$ Terrell "rotation" is a misnomer, it was coined because the original analysis was done WRT a sphere, which preserves it shape under aberration. When you look at cubes, and bear in mind the effects of aberration, it becomes obvious that you are apparently seeing the "back" of the cube (which looks like it is in front of you) because you have already gone past it! In any case, describing the resultant visual distortion of the cube as a "rotation" is pretty sloppy terminology. $\endgroup$
    – m4r35n357
    Commented Feb 5 at 12:54
  • $\begingroup$ @m4r35n357: It's true that in our "objective" reference frame the cube doesn't rotate, but we observed a rotated cube in fact. So I think "rotation" is a fine description of the observed phenomenon as we can even calculate the angle(arcsin v/c). I just take this terminology because my textbook said so and I am not sure if it's widely used in English. Anyway, I will edit it to be "effect". $\endgroup$
    – you3
    Commented Feb 5 at 13:26
  • $\begingroup$ I was not picking your terminology in particular out for criticism, it was a general comment ;) $\endgroup$
    – m4r35n357
    Commented Feb 5 at 13:33
  • $\begingroup$ @m4r35n357: Oh... I am not so familiar with these. After all, thank you for your comment and kindness!(ノ>ω<)ノ $\endgroup$
    – you3
    Commented Feb 5 at 13:43
  • 1
    $\begingroup$ @m4r35n357 Thumbs up for your correct description of Terrell "rotation". $\endgroup$
    – KDP
    Commented Feb 5 at 21:17
1
$\begingroup$

The first is, the INPUT to the transformation should be the "real time" of the event in the base reference frame S. By "real time", I mean that if I "perceive" event A as having taken place at time 3s, but it happened 1 light-second away,

What you are calling the "real time" is the coordinate time and this is what is input into the Lorentz transformation equation. An inertial reference frame is hypothetically set up by creating an infinite network of rulers and synchronised clocks that are all at rest with the origin and all synchronised. Lets say we set up reference frame S with Earth at the origin. We could set up a clock on a space station that is 1 light second away and another on a light station that is 2 light seconds away, all in a straight line. We establish the distances and synchronise the clocks by sending light signals to and fro. Once we done that we can forget about light travel times and only use the times indicated by a cock that is at rest in the reference frame and adjacent to the event being considered. We can have as many clocks as we like, to make sure there is always one next to a given event. he times indicated by these clocks are the coordinate times. Now if we send a light signal from Earth to space station 1, at time zero, then we "measure" that it arrives at the station at coordinate time 2. By "measure" we sometimes mean "imagine"/"perceive"/"calculate"/"predict". In reality the earliest we can have confirmation the signal arrived is at 2 seconds when a confirmation signal returns. A better definition of measure, is what we what we have when we look at the records of the space station and the records of any other confirmation signals from the network of stationary observers and put it all together at some future time. So coordinate or measured time is what you call "real" time.

I mean that if I "perceive" event A as having taken place at time 3s, but it happened 1 light-second away, and I perceive event B as having taken place at time 6s, but it happened 2 light-seconds away

The observer at rest on the Earth measures the coordinates of event A as being (x,t) = (3,1) and event B as being (6,2). The time for an object to travel between those two events is 6-3 = 3 seconds according to the measured time the Earth observer. The Earth observer also measures the coordinate distance to be 2-1 = 1 light second. Another observer moving at v relative to the Earth has her own set of synchronised cloaks and she will measure a different coordinate time between those two events and a different coordinate distance between the same two events. All observers with different velocities relative to the Earth we disagree on the coordinate times and coordinate distances between the two events. The only thing they can all agree on is the proper time shown by a clock that is actually transported in a straight line and constant velocity between those two vents so it is the only clock that is actually present at both events. This is called the proper time between the two events or invariant space time interval and is probably what most people would call the "real time". Coordinate times between spatially separated events are made with two different clocks that are spatially separated and differences in the ticking rates of clocks, length contraction of rulers and the simultaneity of relativity cause coordinate times to be observer dependent. All observers also agree on the relative velocity v between frame S and S' and if you enter v, and coordinate distance x and coordinate time t of observer S in the Lorentz transformation, it spits out the coordinate distance x' and the coordinate time t' that will be measured by observer S' and the equation automatically allows for any issues about light travel times.

Usually in SR if it not specifically made clear, we are talking about times and distances where light travel travel times are already accounted for. If light travel times are what we trying to measure and analyse, then it should be made clear that is what we are doing and this is the exception in general. What we "observe" is not necessarily what we measure. Unfortunately what we actually see or observe is light that arrived from an event that may have happed much earlier, but unfortunately when we say observer S observed event A at (x,t) we usually mean "calculated"/"perceived"/"measured" which can be confusing. What we see is the "received time" or the coordinate time plus light travel time and the "observed time" is usually meant to mean the coordinate time or calculated time, unless specifically stated otherwise.

Terrell rotation is an optical illusion due to light travel times and is confusing because nothing actually physically rotates as correctly noted by @m4r35n357 in the comments.

Edit: Added the following diagrams to try and demonstrate how the Lorentz transforms can be related to moving clocks that are adjacent to events rather than the usual spacetime diagrams. The interactive app used for these diagrams is here.

enter image description here

In the diagram the blue points represent clocks at rest in frame S and the red points are clocks at rest in S' which is moving at 0.8c relative to S. The observer at the origin of S' sends out a signal as she passes the observer at the origin of S. The clocks are 8 light seconds apart in both frames.

enter image description here

At t' = 8 seconds (and x' = 8 ls) in S' the signal arrives at the second clock (hollow red circle) and this coincides with the blue clock at that location showing t = 24 seconds.

enter image description here

At t = 24 seconds (and x = 24 ls) in S the signal arrives at the second clock (hollow red circle) and this coincides with the red clock at that location showing t' = 8 seconds, agreeing with the reading seen by S' for that event.

enter image description here

At t = 50 in S and x = 40 in S the observer at the origin of S' (the red diamond) arrives at the blue clock indicated by a plus symbol and the time on her clock indicates 30 seconds at that location for that event. Both observers agree that t=50 and t'=30. However blue measures red to have travelled 40 light seconds and red measure the distance travelled by the blue plus sign clock as 24 light seconds. They differ about the distance due to length contraction.

All measurement are those made by clocks local to any given event and these are the times obtained when performing a Lorentz transformation.

No additions or subtractions due to light travel times are involved.

and thus if I wanted to also find out when the obsever in S' perceives the event, I'd then also have to do extra calculations to figure out how long it takes light to get from that event to the observer.

Short answer: Yes.

$\endgroup$
1
  • $\begingroup$ Usually in SR if it not specifically made clear, we are talking about times and distances where light travel travel times are already accounted for. - I thought so, thank you for confirming. $\endgroup$
    – TKoL
    Commented Feb 6 at 9:31

Not the answer you're looking for? Browse other questions tagged or ask your own question.