How does reciprocal time dilation work?

[Mike_bb]

These are not the same moment, there is no universal same moment. That is exactly the point of the relativity of simultaneity.

Do you want to say that if I see in the frame A to frame B(to clock B) and if observer from the frame B see in the frame A(to clock A) then this is not the same moment? But what then?

 

[Dale]

I see on the clock B from the frame A and clock B shows 80s. My clock in frame A shows 100s.

Clock A reads 100 and clock B reads 80 at the same moment in frame A.

Observer from the frame B see on my clock in the frame A that shows 80s. Observer's clock in frame B shows 100s.

Clock B reads 100 and clock A reads 80 at the same moment in frame B.

It is possible because “at the same moment in frame A” means a different set of events than “at the same moment in frame B”. This is the relativity of simultaneity. You should draw a spacetime diagram showing the different lines of simultaneity to understand this. Others have already posted them, but you should go through the exercise of drawing it yourself to understand

 

[A.T.]

... that if I see in the frame A to frame B (to clock B) ...

A reference frame is not a location that you look to from another frame. Frames of reference extend infinitely.

And relativity is not about what you actually see, because that includes signal delay. Relativity is about what remains after you have accounted for signal delay, based on a specific clock synchronisation convention.

You seem hung up on these common misunderstandings, plus some language issues.

 

[Ibix]

Do you want to say that if I see in the frame A to frame B(to clock B) and if observer from the frame B see in the frame A(to clock A) then this is not the same moment? But what then?

Draw a pair of axes, and label the horizontal one $x$ and the vertical one $t$. Mark tick marks 1s apart on the $t$ axis and 1ls apart on the $x$ axis. This is a map of spacetime (or at least two dimensions of it). You have drawn axes on it.

Now add more horizontal lines through $t=1,2,3,\ldots$. These represent "all of space at $t=1$, $t=2$, etc.". You have foliated spacetime into a stack of slices that are space at one time.

Now let's add what another frame calls space. $t'=\gamma(t-vx/c^2)$ so $t=vx/c^2+\gamma^{-1}t'$. Plot those lines for $x'=1,2,3,\ldots$. (Note that $c=1$ in these units and pick either $v=0.6$ or $0.8$ for a rational $\gamma$.) Those are lines of constant $t'$, so they are the slices of spacetimethat the primed frame calls "space at one instant".

Notice that the lines the frames call "at the same time" are slanted with respect to each other. Pick a point where one pair of lines cross. Imagine that something happens at that event - perhaps a firecracker goes off. Track along a horizontal line. This is the set of events the unprimed frame calls "at the same moment as the firecracker". Track along the slanted line. This is the set of events the primer frame calls "at the same moment as the firecracker". They are different events.

This is the same thing we keep telling you: spacetime is the real thing, and what part of it you call space is up to you. And if you make a different decision then the meaning of "at the same moment" changes.

 

[Dale]

@Mike_bb are you familiar with the Lorentz transform equations?

 

[Mike_bb]

@Mike_bb are you familiar with the Lorentz transform equations?

Yes. But I used them several times.

 

[Mike_bb]

Dale, I've read about the relativity of simultaneity but I can't understand how does it apply to example with observers.

 

[Matterwave]

Dale, I've read about the relativity of simultaneity but I can't understand how does it apply to example with observers.

Maybe we can try another way. Can you say in your own words, without quoting, what "relativity of simultaneity" means?

It seems to me that your core confusion is not understanding what the relativity of simultaneity means.

 

[Mike_bb]

Can you say in your own words, without quoting, what "relativity of simultaneity" means?

Let we have 2 events. In the 1st IRF these events happen simultaneity, but in the 2nd moving IRF these events happen at the different time.

 

[Dale]

Yes. But I used them several times.

So, I want you to find some graph paper (or make your own). The normal grid is your $x,t$ coordinates. Use the Lorentz transform equations with $v=0.6\ c$ and draw the $x’,t’$ grid lines. Label them $x’=0$, $x’=10$, $t’=0$, $t’=10$, etc. Traditionally $x$ is horizontal and $t$ is vertical, but as long as you are precise you will be able to see the relativity of simultaneity clearly, as well as time dilation and length contraction.

Post the picture when you have done it.

 

[Matterwave]

Let we have 2 events. In the 1st IRF these events happen simultaneity, but in the 2nd moving IRF these events happen at the different time.

Pretty good starting point, but let's make things a bit sharper.

Consider

1. Is it possible to do what you say for any two events? (Hint: it's not, but why not?)
2. What does it mean for two events to happen simultaneously in one frame of reference?
3. In your description of you and your friend, what are the events? How many events are you considering?

You may refer back to my post #16.

 

[A.T.]

Dale, I've read about the relativity of simultaneity but I can't understand how does it apply to example with observers.

Note that 'observers' is often used as a confusing synonym for 'reference frames', which is exactly what relativity of simultaneity applies to.

If that's not what you mean, then you should use the term 'detector' instead.

 

[Herman Trivilino]

I can't understand how is it possible that one clock shows two measurements of time

That can never happen. You have misunderstood something.

For example, my friend's clock shows 8sec [...]. But my friend says that him clock shows 10sec

That can never happen, either. A reading on a clock, like any measurement, is a relativistic invariant. Meaning all observers agree on its value, regardless of their speed relative to each other.

I think you are talking about the passage of time, which is different than a single clock reading. These clocks are moving relative to each other and to compare their readings they have to be in the same place. Since they are each moving in straight lines this will happen at most once.

So how is one to compare the passage of time on the clocks?

That's the question you need to ask yourself. The answer is in any decent textbook on the topic.

 

[robphy]

Here are operational definitions using a radar method (which was my "a-ha moment"[1][2] in relativity).

---

To assign coordinates to a distant event Z,
an inertial observer (wearing a wristwatch) can use a radar method:

• note the wristwatch-reading $t_s$ when sending the light-signal that can reach event Z
• note the wristwatch-reading $t_r$ when receiving the echo of that light-signal from event Z
• assign displacements from the observer's local "wristwatch reads t=0"-event
  • $\Delta t_Z=\left(\displaystyle\frac{t_r+t_s}{2}\right)$ (the half-way time of the wristwatch-readings)
  • $\Delta x_Z/c=\left(\displaystyle\frac{t_r-t_s}{2}\right)$ (half of the elapsed roundtrip-time)
  • Note that $(t,x)=(\Delta t_Z,0)$ is the midpoint-event
    between the send and receive events on the observer's worldline.
• Since $t_r=\left(\displaystyle\frac{t_Z+x_Z}{2}\right)$ and $t_s=\left(\displaystyle\frac{t_Z-x_Z}{2}\right)$, we have
  • $\frac{V_{rel}}{c}=\frac{\Delta x_Z/c}{\Delta t_Z}=\displaystyle\frac{\left(\frac{t_r}{t_s}\right)-1}{\left(\frac{t_r}{t_s}\right)+1}$ is the relative-velocity of the worldline-along-$OZ$,
  • so, $\displaystyle\left(\frac{t_r}{t_s}\right)=\frac{1+V_{rel}/c}{1-V_{rel}/c}=k^2$,
    which is the square of the relative-Doppler-factor between the observer and the worldline-along-$OZ$
    
    
  • $(\Delta t_Z)^2-(\Delta x_Z/c)^2=t_r t_s$ is the square-interval of $OZ$
• (more technical details at https://www.physicsforums.com/threa...eity-and-effects-on-time.1083894/post-7297133 )

---

"Simultaneous"
An inertial observer says that
"two events $Z_1$ and $Z_2$ are simultaneous according to the observer"
when they have the same time-coordinate according to the observer: $\Delta t_{Z_1}=\Delta t_{Z_2}$.


Let's decorate my earlier diagrams with radar experiments.
Inertial-observer Alice assigns coordinates to event $P'$
using light-signals sent at $t_s=4$ and received at $t_r=16$ according to her wristwatch.
So,
$\Delta t_{P'}=(16+4)/2=10$ and $\Delta x_{P'}=(16-4)/2=6$:
that is $(t,x)=(10,6)$, as you can count-off using Alice's diamonds.
Note that for all events E on $PP'$, we have $t_E=10$ according to Alice.
(Consider the event that is radar-measured by $t_r=13$ and $t_s=7$.)
So, $PP'$ is a line segment of simultaneous events according to Alice,
and is in fact simultaneous with local event $P=(10,0)$, the midpoint of $(16,0)$ and $(4,0)$.
(Note that $PP'$ is parallel to "Alice's stick", the spacelike diagonal of Alice's diamonds.)

---

What about Bob, another inertial observer who met Alice
at event O, when both had their wristwatches read 0?
(I'm not going to worry about issues concerning the the positive-x axis, the forward and backward spatial directions of the radar signals, and the appropriate sign to assign for the spatial-displacement.)

[I am swapping Alice and Bob , and swapping Ps and Qs.]

Inertial-observer Bob assigns coordinates to event $Q'$
using light-signals sent at $t_s=4$ and received at $t_r=16$ according to his wristwatch.
So,
$\Delta t_{Q'}=(16+4)/2=10$ and $\Delta x_{Q'}=(16-4)/2=6$:
that is $(t,x)=(10,6)$, as you can count-off using Bob's diamonds.
Note that for all events E on $QQ'$, we have $t_E=10$ according to Bob.
(Consider the event that is radar-measured by $t_r=13$ and $t_s=7$.)
So, $QQ'$ is a line segment of simultaneous events according to Bob,
and is in fact simultaneous with local event $Q=(10,0)$, the midpoint of $(16,0)$ and $(4,0)$.
(Note that $QQ'$ is parallel to "Bob's stick", the spacelike diagonal of Bob's diamonds.)

So, with these constructions, we have:
$PP'$ is simultaneous according to Alice, but not to Bob, and
$QQ'$ is simultaneous according to Bob, but not to Alice.

It can be shown (see the link above for more details) that
these "lines of simultaneity" are related to
the tangent-lines to the "circle of Minkowski spacetime, centered at an event on the observer-worldline."

This radar-construction with light-signals
operationally defines how an inertial observer
assign coordinates to events
(that is, how that observer slices-up spacetime into "his sense of space at different instants of his time").

 

[Matterwave]

This radar-construction with light-signals
operationally defines how an inertial observer
assign coordinates to events
(that is, how that observer slices-up spacetime into "his sense of space at different instants of his time").

I have no problem teaching this radar-construction method for inertial observer coordinates. I think it gives a very nice illustration of how one could go about trying to construct coordinate systems.

However, this method, by definition, can only construct coordinates for events which are causally connected (on or in the forward light cone) to the observer since you are sending radar signals around. Yet, the coordinates typically employed in SR (e.g. those which are "Lorentz transformed") are assumed for the entire spacetime.

It's why I tried to emphasize how idealized (non realistic) the coordinate system (the one I constructed in post #16) was.

 

[Dale]

It's why I tried to emphasize how idealized (non realistic) the coordinate system was.

I would say the radar coordinates are the realistic ones for that reason and the standard coordinates are the idealized ones for that specific reason.

 

[Matterwave]

I would say the radar coordinates are the realistic ones for that reason and the standard coordinates are the idealized ones for that specific reason.

Yes I would agree. To be clear, what I meant was that the construction I made in post #16 is the idealized / non realistic one.

Nevertheless, when we work in the standard SR formalism and give coordinates to events, we don't typically demand that we only patch things together from realistic coordinate patches created from radar measurements.

 

[robphy]

I have no problem teaching this radar-construction method for inertial observer coordinates. I think it gives a very nice illustration of how one could go about trying to construct coordinate systems.

However, this method, by definition, can only construct coordinates for events which are causally connected (on or in the forward light cone) to the observer since you are sending radar signals around. Yet, the coordinates typically employed in SR (e.g. those which are "Lorentz transformed") are assumed for the entire spacetime.

It's why I tried to emphasize how idealized (non realistic) the coordinate system was.

Once I have defined a light-clock diamond (in my local region),
and make reasonable assumptions of homogeneity and isotropy,
I can idealize tiling my spacetime with my diamonds.

This is analogous to starting with a unit square on my sheet of graph paper
and idealize extending it out to the whole plane.

In Galilean physics, we just extrapolate to all values of t and all values of x.

In all of these cases, when my linear approximation isn't good enough to describe the real situation,
I have to develop something fancier to work with calculus and differential geometry.


It should be pointed out that
radar-measurements by inertial observers are essentially using light-cone coordinates,
associated with the eigenbasis of the Lorentz boost.
These radar-measurements are arguably more-physical
than imagining a rigid-ruler extended into space,
and clocks slowly transported those locations in space
in order to construct the lattice of clocks and rulers.


The point of the construction is to give an operational definition of simultaneity,
as opposed to (say) a functional definition
(e.g. the set of events that transform to a constant-value of t for an inertial observer).


By the way,
the a-ha moment in grad school came when the radar-experiment construction
was connected to the decomposition of vectors and tensors by that observer,
using only tensor algebra. (No coordinates and no explicit lorentz transformations.)
See Ch 5 through Ch 7 of Geroch's "General Relativity" notes
from https://home.uchicago.edu/~geroch/Course Notes .
(Those notes were written up long before I was in his class... I only discovered them much later.)

Then I found the books by Synge and by Bondi that likely influenced these introductory chapters.
The construction to define the line of simultaneity with radar
was influenced by Ellis & Williams "Flat and Curved Space-times".
(Before grad school, I already had my maroon Taylor & Wheeler "Spacetime Physics".
So, I saw spacetime diagrams and rapidity....
but I couldn't appreciate what the coordinates meant until I saw the radar-construction
and then really began to appreciate the light-cone and causal-relations.)

 

[Matterwave]

(e.g. the set of events that transform to a constant-value of t for an inertial observer).

If I had an "a-ha" moment for special relativity, it was after learning General Relativity. Specifically chapter 8 in Wald on causal structure. It helps me immensely to think of things in coordinate invariant ways. Thinking in this way, the relativity of simultaneity of SR becomes a consequence of the light cone structure. Time-like or null separated events can not be be made simultaneous via any Lorentz transformation while space-like separated events can (in a globally Minkowski spacetime).

 

[Sagittarius A-Star]

Yes. But how is it possible that clock B shows 80s (case 1) and at the same moment clock B shows 100s(case 2)?

That's not happening at the same moment.

With reference to Observer B's rest-frame:
Clock B first shows 80s (E2) and $20s$ later clock B shows 100s (E3). Clock A shows then $100s/\gamma=80s$.

With reference to Observer A's rest-frame:
Clock B first shows 80s (E2) and $\gamma20s$ later clock B shows 100s (E3). Clock A shows then $\gamma100s=125s$.


I call the event, at which both clocks met and were zeroed: E0.

Case 2) Reference frame: B
E3: clock B shows 100s
E4: clock A shows 80s

The time-interval between E0 and E4 with respect to frame B (x', y', z', t') is given as $\Delta t'_{40} = t'_4 - t'_0 = 100s.$
The distance between E0 and E4 with respect to frame A (x, y, z, t) is $\Delta x_{40} = 0$ (clock A is at rest in frame A).

LT:
$\Delta t'_{40} = \gamma (\Delta t_{40} - v \Delta x_{40} /c^2) = \gamma (\Delta t_{40} - 0)$
$100s = \gamma \Delta t_{40}$
Clock A shows: $\Delta t_{40} = 100s / \gamma = 80s$.

 

[Mike_bb]

That's not happening at the same moment.

With reference to Observer B's rest-frame:
Clock B first shows 80s (E2) and $20s$ later clock B shows 100s (E3). Clock A shows then $100s/\gamma=80s$.

With reference to Observer A's rest-frame:
Clock B first shows 80s (E2) and $\gamma20s$ later clock B shows 100s (E3). Clock A shows then $\gamma100s=125s$.


I call the event, at which both clocks met and were zeroed: E0.

Case 2) Reference frame: B
E3: clock B shows 100s
E4: clock A shows 80s

The time-interval between E0 and E4 with respect to frame B (x', y', z', t') is given as $\Delta t'_{40} = t'_4 - t'_0 = 100s.$
The distance between E0 and E4 with respect to frame A (x, y, z, t) is $\Delta x_{40} = 0$ (clock A is at rest in frame A).

LT:
$\Delta t'_{40} = \gamma (\Delta t_{40} - v \Delta x_{40} /c^2) = \gamma (\Delta t_{40} - 0)$
$100s = \gamma \Delta t_{40}$
Clock A shows: $\Delta t_{40} = 100s / \gamma = 80s$.

Big thanks!! This solved my problem! But is your last edited version correct? I mean "Clock A shows then $100s/\gamma=80s$." and "Clock A shows then $\gamma100s=125s$."? Clocks can't show two different time.

 

[Sagittarius A-Star]

Clocks can't show two different time.

They can. A clock can (and shall) for example show at 2 P.M. another time than it showed at 1 P.M.

 

[DaveE]

https://www.khanacademy.org/science...fed5276339e80:relativity-and-particle-physics