Understanding Relativity of Simultaneity in Resnick, Part I

  • #1
zenterix
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TL;DR Summary
I'd like to understand a section of a book I am reading ("Introduction to Special Relativity" by Resnick).

I will create a series of posts that go through snippets of the section "relativistic kinematics" of the book and try to interpret the content in my own words and perform certain calculations that were left for the reader to do.
I'm going to go step-by-step through the reasoning. If there is any mistake, then it is a true mistake because I am trying to make each statement that follows as accurate as I can.

Unambiguous time scale in a single frame of reference.

Consider that we have two inertial frames of reference, 1 and 2.

We also have two events, A and B.

1) If the two events occur at the same location in space, then one clock in frame 1 at that location can tell us if the events are simultaneous in that frame.

2) If two events occur at different locations in space in frame 1, and if we have one clock in each location, we need to make sure the clocks are synchronized.

Some "obvious" methods of synchronizing clocks turn out to be erroneous. For example, we can set the two clocks so that they always read the same time as seen by observer A. This means that whenever observer A looks at the B clock it reads the same to him as his clock.

The book seems to be mixing the names of the events here (A and B) with the names of the observers (ie, frames).

Spelling everything out with maximum detail I think this means the following

1) Observer A is a reference frame with origin at the space location of event A and at rest relative to this location.

2) Observer B is a reference frame with origin at the space location of event B and at rest relative to this location.

Since the space locations of events A and B are at rest relative to one another, then the frames of reference of observers A and B are also at rest relative to each other.

This means that whenever A looks at the B clock it reads the same to him as his clock. The defect here is that if observer B uses the same criterion (that is, that the clocks are synchronized if they always read the same time to him), he will find that the clocks are not synchronized if A says that they are. For this method neglects the fact that it takes time for light to travel from B to A and vice versa. The student should be able to show that, if the distance between the clocks is ##L##, one observer will see the other clock lag his by ##2L/c## when the other observer claims that they are synchronous. We certainly cannot have observers in the same reference frame disagree on whether clocks are synchronized or not, so we reject this method.

Let me try to show this result.

Observer A is at the location of event A and suppose the clock says zero.

A light ray takes ##L/c## to arrive from the location of event B.

Observer A "seeing" the clock at event B means that he sees light arriving from clock B (presumably through a telescope).

Thus, for observer A to see the clock at event B showing zero when clock A is showing zero, the light ray had to leave event B (showing time zero) when clock A time was ##-L/c##.

Thus, the clocks are not really synchronized: observer A simply sees the clocks as showing the same time at any given time in his frame, but he is actually seeing the time at clock B at ##L/c## in the past: clock B is really ahead by ##L/c##.

Now, whenever light moves from clock A to clock B, it starts at a time that is ##L/c## behind clock B and then takes ##L/c## to arrive, so that when it arrives it is ##2L/c## behind clock B.

Thus observer B observes clock A to be ##2L/c## behind clock B.

Here is a pictorial depiction of this

1698957613034.png


One question I have is about the statement

We certainly cannot have observers in the same reference frame disagree on whether clocks are synchronized or not, so we reject this method.

Is it technically correct that observers A and B are in the same frame?

I mean, I can see that the only difference between their frames is a constant term. It just bothers me to say they are the same frame. Are they?

The conclusion here is that if the definition of synchronized clocks is that one of the observers sees the clocks as showing the same time, then other observers in frames at rest relative to the first observer will not see the clocks as showing the same time.
 
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  • #2
zenterix said:
TL;DR Summary: I'd like to understand a section of a book I am reading ("Introduction to Special Relativity" by Resnick).

I will create a series of posts that go through snippets of the section "relativistic kinematics" of the book and try to interpret the content in my own words and perform certain calculations that were left for the reader to do.

1) If the two events occur at the same location in space ...
Before continuing reading, I need a clarification here: when you say, "at the same location", which reference frame you refer to? The locations are same in one and not same in other.
 
  • #3
zenterix said:
Unambiguous time scale in a single frame of reference.

Consider that we have two inertial frames of reference, 1 and 2.
Why are you introducing two frames here, when, as the book itself says in this section, it is looking at how we would set up an unambiguous time scale in a single frame of reference?

zenterix said:
We also have two events, A and B.
Yes, but the book at this point is looking at them in just one reference frame, not two. You are calling this frame "frame 1" in the quote that follows.

zenterix said:
) If the two events occur at the same location in space, then one clock in frame 1 at that location can tell us if the events are simultaneous in that frame.
Yes.

zenterix said:
2) If two events occur at different locations in space in frame 1, and if we have one clock in each location, we need to make sure the clocks are synchronized.
More precisely, that they are synchronized using the simultaneity convention of frame 1.

zenterix said:
The book seems to be mixing the names of the events here (A and B) with the names of the observers (ie, frames).
No. It is talking about two observers and only one frame (see my comment on this above). The book has previously used A and B to refer to events--so by implication, we expect observer A to be co-located with event A and observer B to be co-located at event B.

zenterix said:
1) Observer A is a reference frame with origin at the space location of event A and at rest relative to this location.

2) Observer B is a reference frame with origin at the space location of event B and at rest relative to this location.
No, this is not correct. See my comments above.

zenterix said:
for observer A to see the clock at event B showing zero when clock A is showing zero, the light ray had to leave event B (showing time zero) when clock A time was ##- L / c##.
Here you are assuming a simultaneity convention, i.e., a way to define at what "clock A time" an event that is not co-located with observer A happened. But we don't yet know, at this point in the book's exposition, how to define such a convention. Nor is the book using one at this point.

The actual argument underlying the book's statement that you quote is that, if observers A and B are at rest relative to each other (which is implied by the fact that they remain at spatial locations A and B in the same frame), and a distance ##L## apart, then light will take a time ##L / c## to travel between them. This says nothing about "clock A time" for the light leaving B. It just says how much time the light takes to travel, which is given by the known distance ##L## and the fact that light travels at speed ##c## in all inertial frames.

zenterix said:
Is it technically correct that observers A and B are in the same frame?
Yes. More precisely, they are at rest in the same frame. See above for why that is implied by what the book has said up to this point.
 
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  • #4
Hill said:
when you say, "at the same location", which reference frame you refer to?
The book has only introduced one frame at this point. See the first comment in my post #3.
 
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  • #5
zenterix said:
Is it technically correct that observers A and B are in the same frame?
It's sloppy; it should say "...are at rest in the same frame" because everything is in every frame, since all inertial frames cover all of spacetime in SR. With the correct wording it's more obvious that you need the same simultaneity criterion. Otherwise you don't have "an" inertial frame associated with a set of points at mutual rest - you have many.
zenterix said:
1) Observer A is a reference frame with origin at the space location of event A and at rest relative to this location.
No. An observer is not a reference frame. You may use an observer's state of motion as the beginnings of a reference frame, but an observer is only at one place, whereas a frame extends throughout spacetime. Many sources are careless about this distinction, but it is an important one.
 
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  • #6
zenterix said:
Is it technically correct that observers A and B are in the same frame?
As @Ibix has just explained, everything is always in all frames, so it is both true and a tautology to say that “A and B are in the same frame”. Is this your wording or Resnick’s? The short snippet you’ve quoted doesn’t seem to be using this sloppy and confusing terminology.
 
  • #7
PeterDonis said:
Why are you introducing two frames here, when, as the book itself says in this section, it is looking at how we would set up an unambiguous time scale in a single frame of reference?

The reason I introduced these two frames is because I initially intended to add another part to this post, namely, one that addresses the question: suppose one inertial observer does find that two separated events are simultaneous; will these same events be measured as simultaneous by an observer on another inertial frame moving with speed ##v## with respect to the first?

I decided to leave that for another post.

So, no, it is not necessary to have these two frames for the current post.
PeterDonis said:
Yes, but the book at this point is looking at them in just one reference frame, not two. You are calling this frame "frame 1" in the quote that follows.
Ok.
PeterDonis said:
More precisely, that they are synchronized using the simultaneity convention of frame 1.
And the simulteneity convention, in this section of the book specifically, is that two clocks are synchronized when they "read the same time as seen by observer A", correct?
PeterDonis said:
No. It is talking about two observers and only one frame (see my comment on this above). The book has previously used A and B to refer to events--so by implication, we expect observer A to be co-located with event A and observer B to be co-located at event B.
So if I understand correctly, we have a single frame. The only information we have about this frame is that it is at rest relative to locations of events A and B (equivalently, to the locations of observers A and B).

We do not know (nor do we need to know) the exact positions of these observers in the frame. The important point about the frame is that it is at rest relative to the observers.
PeterDonis said:
No, this is not correct. See my comments above.
Ok.
PeterDonis said:
Here you are assuming a simultaneity convention, i.e., a way to define at what "clock A time" an event that is not co-located with observer A happened. But we don't yet know, at this point in the book's exposition, how to define such a convention. Nor is the book using one at this point.
As I said above, did the book not define the simultaneity convention when it said "For example, we can set the two clocks so that they always read the same time as seen by observer A"?

With this "convention", an event not co-located with observer A (ie, clock B) happens at the clock A time of the light from clock B arriving at observer A, no?
PeterDonis said:
The actual argument underlying the book's statement that you quote is that, if observers A and B are at rest relative to each other (which is implied by the fact that they remain at spatial locations A and B in the same frame), and a distance ##L## apart, then light will take a time ##L / c## to travel between them. This says nothing about "clock A time" for the light leaving B. It just says how much time the light takes to travel, which is given by the known distance ##L## and the fact that light travels at speed ##c## in all inertial frames.

Again, my impression is that the book did introduce the convention.

PeterDonis said:
Yes. More precisely, they are at rest in the same frame. See above for why that is implied by what the book has said up to this point.
Yes.
 
  • #8
Hill said:
Before continuing reading, I need a clarification here: when you say, "at the same location", which reference frame you refer to? The locations are same in one and not same in other.
I introduced two reference frames but one is not being used in my post.

From reading other answers, my understanding now is that there is one reference frame in which both observers A and B are co-located with events A and B, respectively.
 
  • #9
zenterix said:
the simulteneity convention, in this section of the book specifically, is that two clocks are synchronized when they "read the same time as seen by observer A", correct?
"As seen by observer A" is given in the book as an example of a wrong simultaneity convention, i.e., one that doesn't work. This is done in what you yourself explicitly quoted from the book; note that the quote you explicitly gave ends with "we reject this method".
 
  • #10
Ibix said:
It's sloppy; it should say "...are at rest in the same frame" because everything is in every frame, since all inertial frames cover all of spacetime in SR. With the correct wording it's more obvious that you need the same simultaneity criterion. Otherwise you don't have "an" inertial frame associated with a set of points at mutual rest - you have many.

No. An observer is not a reference frame. You may use an observer's state of motion as the beginnings of a reference frame, but an observer is only at one place, whereas a frame extends throughout spacetime. Many sources are careless about this distinction, but it is an important one.
From what I understand at this point, there is one reference frame and two observers A and B in the reference frame co-located with events A and B, respectively.

The reference frame is at rest relative to these locations.

The next part of this whole reasoning, which I will address in another post, is when we have a second reference frame, moving with respect to the first (and hence with respect to the locations of events A and B.

We could replicate the entire reasoning of my OP to this second reference frame with its own events D and E at rest relative to that frame and observers at these locations. These observers would then be moving relative to the first frame, but not relative to the second frame.

I think that is it, right?
 
  • #11
PeterDonis said:
"As seen by observer A" is given in the book as an example of a wrong simultaneity convention, i.e., one that doesn't work. This is done in what you yourself explicitly quoted from the book; note that the quote you explicitly gave ends with "we reject this method".
Yes, exactly, but my entire question is about the calculations involved in this wrong simultaneity convention. In particular, the final result is that if you adopt this convention, then observer B's clock must be ahead by ##L/c## and what observer B sees is clock A being ##2L/c## behind his clock.

I quoted two snippets of the same paragraph in the book, and they both refer to this wrong convention. Here is the paragraph in full.

1698965703381.png
 
  • #12
Nugatory said:
As @Ibix has just explained, everything is always in all frames, so it is both true and a tautology to say that “A and B are in the same frame”. Is this your wording or Resnick’s? The short snippet you’ve quoted doesn’t seem to be using this sloppy and confusing terminology.
My sloppiness, and I can see that it is sloppy. Of course everything and everybody is in every frame. What I meant to ask was: is it technically correct that observers A and B are the same frame?

But at this point I have realized that in the context of my OP, there is only one frame, and the essential point is that observers A and B are at rest relative to that frame.

The underlying doubt I seemed to have is that if we define two different frames to have origin at the locations of A and B, respectively, then we do in fact have two different frames (at rest relative to each other but with different coordinates for events). Yet, now I realize that this is not what is being done in the book.
 
  • #13
zenterix said:
my entire question is about the calculations involved in this wrong simultaneity convention
Ok. That was not at all clear to me from your previous posts.

zenterix said:
the final result is that if you adopt this convention, then observer B's clock must be ahead by ##L / c##
More precisely, ahead by ##L / c## as compared to the standard simultaneity convention for inertial frames in SR.

zenterix said:
what observer B sees is clock A being ##2L / c## behind his clock.
Yes.
 
  • #14
zenterix said:
is it technically correct that observers A and B are the same frame?
No, that is always incorrect (and we don't need the word "technically"). An observer is not a frame so they can no more be the same than a zebra can be an emotion.

This being an I-level thread, we can do some oversimplification and say that a frame is a rule for assigning coordinates to events: event ##E## happened at time ##t## at point ##x##, ##y##, and ##z##. Observers don't come into it at all, we don't need them to define a frame, we don't need to have an observer at the origin of the coordinate system or anywhere else. You will often hear people saying things like "this object/observer is moving/at rest in that frame" - that's a somewhat sloppy but convenient linguistic shortcut for the more precise but unwieldy "using the coordinates assigned by that frame, this object/observer's spatial coordinates are not changing over time".

Here we have two observers. If we choose to use a frame in which one of them is at rest, the other one will be as well. If we wanted to describe this situation without introducing any frame at all we would say that they are "at rest relative to one another", a statement that is true regardless of our choice of frame.
 
  • #15
Nugatory said:
No, that is always incorrect (and we don't need the word "technically"). An observer is not a frame so they can no more be the same than a zebra can be an emotion.

This being an I-level thread, we can do some oversimplification and say that a frame is a rule for assigning coordinates to events: event ##E## happened at time ##t## at point ##x##, ##y##, and ##z##. Observers don't come into it at all, we don't need them to define a frame, we don't need to have an observer at the origin of the coordinate system or anywhere else. You will often hear people saying things like "this object/observer is moving/at rest in that frame" - that's a somewhat sloppy but convenient linguistic shortcut for the more precise but unwieldy "using the coordinates assigned by that frame, this object/observer's spatial coordinates are not changing over time".

Here we have two observers. If we choose to use a frame in which one of them is at rest, the other one will be as well. If we wanted to describe this situation without introducing any frame at all we would say that they are "at rest relative to one another", a statement that is true regardless of our choice of frame.
Here is, I think, an example of this sloppy but convenient usage of "frame" and "observer" interchangeably:

1698973365072.png

1698973430058.png


When "the observers" is first mentioned, it implicitly means observers located in each of the two frames. Then, later, it is stated that, more precisely, the observers are located at the origins of the two frames.
 
  • #16
zenterix said:
it implicitly means observers located at rest in using coordinates assigned by each of the two frames.
FIFY :smile:
And seriously, kidding aside, yes I am quibbling about the wording - but it's only a quibble to those who already understand the underlying concepts.
 
  • #17
zenterix said:
From what I understand at this point, there is one reference frame and two observers A and B in the reference frame co-located with events A and B, respectively.

The reference frame is at rest relative to these locations.
Not really, because you don't have a sense of "location" without defining "at rest", and you are again using "in a reference frame".

I'd say that you have two observers A and B who are inertial and at rest with respect to each other (and we can formalise "at rest with respect to each other" by saying that they can bounce radar pulses off each other and always get the same echo time). That's the physical reality. The rest is stuff we're layering on the top to make calculations easier. We then define a frame by saying that anything at rest with respect to A and B (in that same radar sense) is at rest in our frame. That's how we define "at rest". Then we say that time at a location in this frame is measured by clocks that are at rest at that location. The only thing remaining is a process to zero the clocks.
 

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