Question from Griffiths (EM): Synchronized clocks

Question

There's this vague conceptual question from the textbook Introduction to Electrodynamics by David Griffiths

Problem 12.5 Synchronized clocks are stationed at regular intervals, a million km apart, along a straight line. When the clock next to you reads 12 noon:

a.) What time do you see on the 90th clock down the line?

b.) What time do you observe on that clock?

The first question is easy. You just calculate the total time interval for the light to travel to you from the 90th clock. That time interval is how much earlier it is than 12 noon. $$ \Delta t = \frac{\Delta x}{c} = \frac{90 \times 1 km}{3 \times 10^8 m/s} = 300 s = 5 minutes $$

Thus the answer is 11:55 am.

The second question is now confusing. I don't really understand how the term "observe" differs from "see" as they are both just acts of measuring the time on the clock in question. But others tell me that the answer to (b.) is actually 12 noon. Anybody care to explain? Thanks in advance!

Answer 1

I don't really understand how the term "observe" differs from "see" as they are both just acts of measuring the time on the clock in question.

You're not the only one and understanding this difference is often crucial to truly understanding the results from Special Relativity (SR).

The essential difference is that an (inertial) observer in SR is non-local. Rather than think of an observer located at a particular point, think of an observer as the entire inertial reference frame that assigns to each event spacetime coordinates as determined by rods and clocks synchronized in that frame.

From the above linked Wikipedia article:

Physicists use the term "observer" as shorthand for a specific reference frame from which a set of objects or events is being measured.

Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from. The effects of special relativity occur whether or not there is a sentient being within the inertial reference frame to witness them.

In contrast, what a person (or camera) sees (or photographs) is an entirely different question. Here is a quote from a nice article on the difference.

Observation vs Photography

In Special Relativity there is a very specific meaning to the word observer. An observer is one of an infinite collection, in space and time, of robots whose sole task in life is to record the time and location of each detected event. When we say that an observation was made we simply mean that the space and time coordinates of an event have been recorded.

A photograph is a record of all of the photons that were received at the focal point of the camera at the time the photograph was taken.

These are fundamentally different processes and they give us very different information about the universe.

So, to specifically address your example, part (b) asks for different information than part (a).

Part (a) asks about the information contained in the light arriving at your location from the distant clock when your local clock reads 12 noon.

To answer this, one must calculate the time of flight from there to here and subtract that from the local time.

A photograph of the event would show the local clock reading 12 noon and the distant clock reading 11:55 am.

However, part (b) asks what the distant clock reads there at the same instant as the local clock here reads at 12 noon.

Since, it is stipulated that the distant clock there is synchronized with the local clock here, the answer requires no calculation: 12 noon.

Answer 2

In the text immediately above the question Griffiths says:

I belabour this point in order to emphasize that the relativity of simultaneity is a genuine discrepency between measurements made by competant observers in relative motion, not a simple mistake arising from a failure to account for the travel time of light signals.

His point is that when we do Lorentz transformations to switch between different frames the changes in the $t$ and $x$ coordinates of spacetime points are real, and not just because it takes longer to see something that's farther away.

Suppose you're standing by your clock when it shows 12:00 and I ask the question "what is the time on the clock 90 million km away right now?". The answer is 12:00, just as it is for a clock 900 or 9000000000 million km away. That's what we mean by the clocks being synchronised. You will see a different time if you look at the distant clocks, but as you've calculated above this is just because of the time taken for the light to reach you. So if you see a different time on distant clocks you know it's not a real difference. Griffiths is taking observe to mean measuring the time and allowing for the time the light takes to reach you.

The reason Griffiths is making big deal about this is because it's going to turn out that an observer moving relative to you will see different times on the different clocks, but this is not just due to the light travel time. Even allowing for light travel time the clocks really do show different times for a moving observer even when they show the same time for you.