How does reciprocal time dilation work?

[Matterwave]

Time doesn't depend on frames. Only time interval depends on the frames. (as in example above)

This statement suggest you still do not grasp the relativity of simultaneity. Consider for space like separated events $A,B$, I can always find 2 inertial reference frames, one in which $A$ has a coordinate time before $B$ and another one in which $B$ has a coordinate time before $A$.

 

[Dale]

[clock A = 80s] and [clock B = 100s]

In which frame? (And please use complete sentences)

[clock A = 100s] and [clock B = 80s].

In what frame? (And please use complete sentences)

clock a=80s -> clock a=100s - it's correct.

What is correct? Please use complete sentences.

But how is it possible that clock b = 100s -> clock=80s ?

So close to a complete sentence. What is the second clock here and what does the arrow mean?

The clock is running backwards.

Yay for a complete sentence. The sentence is false, but at least it is complete.

What makes you believe that any clock is running backwards? If you draw a spacetime diagram you will be able to see that neither clock ever runs backwards.

Even if you choose not to do the diagram, I am going to have to insist that you start writing in complete sentences.

 

[Sagittarius A-Star]

A simple explanation of the relativity of simultaneity is given in Morin's book "Special Relativity: For the Enthusiastic Beginner", chapter 1.3.1

Event 1 is “light hitting the left receiver”
Event 2 is “light hitting the right receiver”

In A's reference frame, E1 and E2 happen simultaneously.
In B's reference frame, E1 happened before E2 happens.

source:
https://davidmorin.physics.fas.harv...nuum12331/files/2025-10/relativity_chap_1.pdf
via:
https://davidmorin.physics.fas.harvard.edu/books/special-relativity

 

[robphy]

Case 1) Reference frame: A
E1: clock A shows 100s
E2: clock B shows 80s

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

I think you need to write out...

• I'm going to use "reads" to avoid possible misinterpretations
  when using anything suggesting what it "visually looks like".

According to observer A, E1 (clock A reads 100) and E2 (clock B reads 80) are simultaneous.

According to observer B, E3 (clock B reads 100) and E4 (clock A reads 80) are simultaneous.

 

[A.T.]

Time doesn't depend on frames. Only time interval depends on the frames.

You have it exactly backwards:
- Time coordinates assigned to events depend on reference frames, because they assign them.
- Time intervals measured by clocks between events local to that clock are frame invariant.

 

[Mike_bb]

A simple explanation of the relativity of simultaneity is given in Morin's book "Special Relativity: For the Enthusiastic Beginner", chapter 1.3.1

Event 1 is “light hitting the left receiver”
Event 2 is “light hitting the right receiver”

In A's reference frame, E1 and E2 happen simultaneously.
In B's reference frame, E1 happened before E2 happens.

View attachment 373528​

source:
https://davidmorin.physics.fas.harv...nuum12331/files/2025-10/relativity_chap_1.pdf
via:
https://davidmorin.physics.fas.harvard.edu/books/special-relativity

This is nice example. Thanks! But for our example you wrote that:

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 80s.

From frame B: clock B shows 100s(E3), clock A shows 80s. Then clock A shows 100s, clock B shows 125s. But E1 (clock A shows 100s), E2(clock B shows 80s). 125s isn't equal to 80s.

Where did I go wrong? THx.

 

[Matterwave]

From frame B: clock B shows 100s(E3), clock A shows 80s. Then clock A shows 100s, clock B shows 125s. But E1 (clock A shows 100s), E2(clock B shows 80s). 125s isn't equal to 80s.

Where did I go wrong? THx.

The same place you've been going wrong for the whole thread. You swap frames mid sentence and don't understand the difference between events and coordinate time. You don't grasp what the relativity of simultaneity means.

At some point, you will just have to sit down and work it out yourself. Everything to answer your misconceptions already exist within this thread, posted from multiple different angles by multiple posters.

Asking the same question over and over again will just hit a wall.

 

[Mike_bb]

The same place you've been going wrong for the whole thread. You swap frames mid sentence and don't understand the difference between events and coordinate time. You don't grasp what the relativity of simultaneity means.

At some point, you will just have to sit down and work it out yourself. Everything to answer your misconceptions already exist within this thread, posted from multiple different angles by multiple posters.

Asking the same question over and over again will just hit a wall.

If from frame B clock A shows 100s then in frame A clock A shows 100s. What's wrong?

 

[Mike_bb]

The same place you've been going wrong for the whole thread. You swap frames mid sentence and don't understand the difference between events and coordinate time. You don't grasp what the relativity of simultaneity means.

At some point, you will just have to sit down and work it out yourself. Everything to answer your misconceptions already exist within this thread, posted from multiple different angles by multiple posters.

Asking the same question over and over again will just hit a wall.

Well, see post #50:

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 $\gamma100s=125s$.

 

[Ibix]

If from frame B clock A shows 100s then in frame A clock A shows 100s. What's wrong?

Two straight roads cross each other, making an angle $\theta$. Two cars driving on the two roads at the same speed pass through the intersection at the same time. A time $t$ later, the driver of one car notes that he has proceeded a distance $vt$ along his road and the other car is falling behind because it is only doing $v\cos\theta$ in his forward direction. The other driver also notes that she has driven a distance $vt$ and the first car has fallen behind because it is only doing $v\cos\theta$ in her forward direction.

So both cars are behind and both are ahead. Is this a paradox?

 

[Mike_bb]

Two straight roads cross each other, making an angle $\theta$. Two cars driving on the two roads at the same speed pass through the intersection at the same time. A time $t$ later, the driver of one car notes that he has proceeded a distance $vt$ along his road and the other car is falling behind because it is only doing $v\cos\theta$ in his forward direction. The other driver also notes that she has driven a distance $vt$ and the first car has fallen behind because it is only doing $v\cos\theta$ in her forward direction.

So both cars are behind and both are ahead. Is this a paradox?

I provided example in post#69.

 

[Sagittarius A-Star]

Time doesn't depend on frames.

The time-coordinate of an inertial frame is defined by a grid of synchronized clocks, which are all at rest with reference to this frame.

The Einstein-synchronization is based on the definition, that the one-way-speed of light is isotropic.
But regardless which definition you use for synchronizing the clock-grid, simultaneity between events will always be frame-dependent.

This is nice example. Thanks! But for our example you wrote that:
From frame B: clock B shows 100s(E3), clock A shows 80s. Then clock A shows 100s, clock B shows 125s. But E1 (clock A shows 100s), E2(clock B shows 80s). 125s isn't equal to 80s.

Where did I go wrong? THx.

You missed the relativity (=frame-dependency) of simultaneity. See posting #50.

With reference to frame B the events [clock B = 100s] and [clock A = 80s] happen simultaneous to each other.
With reference to frame A the events [clock B = 100s] and [clock A = 125s] happen simultaneous to each other.

 

[Mike_bb]

With reference to frame B the events [clock B = 100s] and [clock A = 80s] happen simultaneous to each other.
With reference to frame A the events [clock B = 100s] and [clock A = 125s] happen simultaneous to each other.

Yes, I agree with this, there is no contradiction. But why did you write about these events? I asked about that from frame B clock A shows 80s (as you mentioned in post#50) then clock A shows 100s. Which time does clock B show?

 

[Sagittarius A-Star]

Yes, I agree with this, there is no contradiction. But why did you write about these events? I asked about that from frame B clock A shows 80s (as you mentioned in post#50) then clock A shows 100s. Which time does clock B show?

With reference to frame A the events [clock A = 100s] and [clock B = 80s] happen simultaneous to each other.
With reference to frame B the events [clock B = 100s] and [clock A = 80s] happen simultaneous to each other.

 

[Ibix]

I provided example in post#69.

We are 70+ posts into this thread and you are still banging your head on the same confusion.

Can you resolve the paradox of the cars being both ahead and behind? Because it is closely analogous to the issue you can't resolve in relativity.

 

[Mike_bb]

With reference to frame A the events [clock A = 100s] and [clock B = 80s] happen simultaneous to each other.
With reference to frame B the events [clock B = 100s] and [clock A = 80s] happen simultaneous to each other.

That's the point, [clock A = 100s] and [clock B = 80s]. I suggest following simultaneously events:
With reference to frame B the events [clock B = 125s] and [clock A = 100s]. Such way we have a contradiction because [clock A=100s] and [clock B=80s] are simultaneously events but [clock B = 125s] and [clock A = 100s] are also simultaneously events. How to be in this case?

 

[Sagittarius A-Star]

I suggest following simultaneously events:
With reference to frame B the events [clock B = 125s] and [clock A = 100s]. Such way we have a contradiction because [clock A=100s] and [clock B=80s] are simultaneously events but [clock B = 125s] and [clock A = 100s] are also simultaneously events. How to be in this case?

With reference to which frame is the highlighted statement valid?

 

[Mike_bb]

With reference to which frame is the highlighted statement valid?

frame A

 

[Sagittarius A-Star]

frame A

That means there is no contradiction, analog to what you wrote in posting #73:

Yes, I agree with this, there is no contradiction.

 

[robphy]

Here's my graphical attempt to help you @Mike_bb .

After a summary-and-diagram,
there is a second diagram with an exercise for you to do.


First, here's a simplified summary of radar measurements by Alice (in red) and by Bob (in blue).

• Let P' be the event "[on Bob's worldline] Bob's wristwatch reads 8".
  So, $t_{P',{\rm says\ Bob}}=8$.
  
  Alice assigns coordinates to distant-event P'
  by sending the appropriate light-signal (when her watch reads $t_s=4$) to reach P'
  and waiting to receive its echo (when $t_r=16$).
  She assigns the time $t_{P',{\rm says\ Alice}}=(16+4)/2=10$.
  Alice says the midpoint-event P ("[on Alice's worldline] Alice's wristwatch reads 10")
  is simultaneous with distant-event P' ("[on Bob's worldline] Bob's wristwatch reads 8").
• Similarly (swapping Alice and Bob, and Ps and Qs),
  Let Q' be the event "[on Alice's worldline] Alice's wristwatch reads 8".
  So, $t_{Q',{\rm says\ Alice}}=8$.
  
  Bob assigns coordinates to distant-event Q'
  by sending the appropriate light-signal (when his watch reads $t_s=4$) to reach Q'
  and waiting to receive its echo (when $t_r=16$).
  He assigns the time $t_{Q',{\rm says\ Bob}}=(16+4)/2=10$.
  Bob says the midpoint-event Q ("[on Bob's worldline] Bob's wristwatch reads 10")
  is simultaneous with distant-event Q' ("[on Alice's worldline] Alice's wristwatch reads 8").

---

Important exercise

Consider the event E "[on Alice's worldline] Alice's watch reads 10".
Use the radar construction on the diagram below to answer

1. What event on Bob's worldline does Alice regard as simultaneous with E?
2. What event on Bob's worldline does Bob regard as simultaneous with E?
   
   
   

View the next line after finishing the construction.



For the event E "[on Alice's worldline] Alice's watch reads 10",
Alice regards the event "[on Bob's worldline] Bob's watch reads 8" as simultaneous with E,
whereas
Bob regards the event "[on Bob's worldline] Bob's watch reads 12.5" as simultaneous with E.


Alice would choose the event on Bob's worldline such that
"its associated send-and-receive-pair-of-events has midpoint-event as E (which is on her worldline)".
- One could search: does the reflection-event associated with (say) 9-and-11 meet Bob's worldline? If not, try 8-and-12, etc...
- Alternatively, one could pick an arbitrary event on Bob's worldline, then note the associated send-and-receive-pair-of-events on Alice's worldline. If its midpoint-event isn't E, try another event on Bob's worldline. (Think "similar triangles".)

Bob would choose the event on Bob's worldline such that
it is the midpoint-event of the "send-and-receive-pair-of-events associated with E (which is not on his worldline)".
- Draw the lightcone-of-E (just the light-signals from E) and note the intersection-events with Bob's worldline.

 

[Dale]

That's the point, [clock A = 100s] and [clock B = 80s].

This is an incomplete sentence. You have named two events, so you have the subject of the sentence, but not the verb. Please use complete sentences.

I suggest following simultaneously events:
With reference to frame B the events [clock B = 125s] and [clock A = 100s].

Good. This is a complete sentence, you identified the reference frame, and made a correct statement.

Such way we have a contradiction because [clock A=100s] and [clock B=80s] are simultaneously events

This statement is incomplete, which makes it appear false. You have (again) made a statement about simultaneity without specifying the reference frame. In your previous statement you explicitly established that you are talking about simultaneity in frame B. So, without an explicit statement about the reference frame, the implicit frame is still frame B. In frame B the events [clock A = 100 s] and [clock B = 80 s] are not simultaneous. So, as written, this statement is false.

but [clock B = 125s] and [clock A = 100s] are also simultaneously events

Again, this is incomplete. In which frame are they simultaneous? You have not stated explicitly. Without an explicit statement the implicit frame would still be B. So this statement would be true, but unclear because of the lack of an explicit statement about which frame to use.

How to be in this case?

It cannot be. Your second statement is false due to your continued use of incomplete sentences.

You are not going to learn this material without drawing (preferably by hand) the diagram, either the one I suggested or the one @robphy suggested. But whether you choose to do that or not, you must start using complete sentences. Every single statement about simultaneity must include the reference frame to which it refers.

 

[Sagittarius A-Star]

The time-coordinate of an inertial frame is defined by a grid of synchronized clocks, which are all at rest with reference to this frame.

Let me visualize this.

Reference frame A:

Clock B:             [0s] -->
Frame A clock grid:  [0s]   [0s]   [0s]   [0s]   [0s]   [0s]   [0s]
                    --------> x

Clock B:                             --> [ 40s]
Frame A clock grid: [ 50s] [ 50s] [ 50s] [ 50s] [ 50s] [ 50s] [ 50s]
                    --------> x

Clock B:                                                  --> [ 80s]
Frame A clock grid: [100s] [100s] [100s] [100s] [100s] [100s] [100s]
                    --------> x

Reference frame B:

Clock A:                                                   <-- [0s]
Frame B clock grid:  [0s]   [0s]   [0s]   [0s]   [0s]   [0s]   [0s]
                                                                  --------> x'

Clock A:                             <-- [ 40s]
Frame B clock grid: [ 50s] [ 50s] [ 50s] [ 50s] [ 50s] [ 50s] [ 50s]
                                                                  --------> x'

Clock A:            [ 80s] <-- 
Frame B clock grid: [100s] [100s] [100s] [100s] [100s] [100s] [100s]
                                                                  --------> x'

When you define one frame as the "stationary frame", then the moving clock will always be directly compared to that clock of the stationary clock grid, which is next to it.

 

[A.T.]

How to be in this case?

More precise.

 

[Mike_bb]

Thanks to all! I solved my problems! Please don't ban me.

 

[robphy]

Thanks to all! I solved my problems! Please don't ban me.

Why would you be banned?

 

[berkeman]

Why would you be banned?

Let's don't go there.

Thanks to all! I solved my problems! Please don't ban me.

Great. Thread closed temporarily for moderation.

 

[robphy]

Just a postscript...
(thanks to a mentor for re-opening this to allow me to make this hopefully-helpful comment)

Since the OP recently posted a simiilar question at https://physics.stackexchange.com/q...t-time-7sec-and-15sec-reciprocal-time-dilatio )
and drew some clock faces, I thought the following spacetime diagrams (using a variant of the values in the SE post)
with those clock faces might help @Mike_bb.

Here are those diagrams.
Here the clock-readings are 15 and 9 (instead of 15 and 7 from SE and 100 and 80 here at PF).

and then here is the answer to the early exercise using 15 and 9:

The point of these diagrams is to say there is no situation (no spacetime event)
where a clock has two different readings.

The events E1 and E4 are distinct. The events E2 and E3 are distinct.

It is important to note that "events on a spacetime diagram" are like "points on a blackboard".

Events do not belong to frames-of-reference, to inertial-observers.
Events, however, may be local or distant with respect to a given observer.

It could be, however, that some event is of more interest to an inertial-observer, as opposed to another inertial observer.

In particular, in the exercise, consider the event E on Alice's worldline.

1. What event on Bob's worldline does Alice regard as simultaneous with E?
2. What event on Bob's worldline does Bob regard as simultaneous with E?

These two events on Bob's worldline are different events, corresponding to different clock-face readings of Bob's clock.
So, these two events and thus clock-face readings are distinct.
They do not coincide.
They never happen together.

I hope this helps.

 

[Sagittarius A-Star]

Just another postscript...

The OP recently posted a question about the reciprocal spacetime interval ($a=1$) at

M

Undergrad Thread 'Why does coordinate dependence violate homogeneity in spacetime intervals?'

Aug 1, 2026

Hello!

I have a problem with understanding homogeneity of the space.

In the book Landau&Lifshitz it was written that two spacetime intervals differ by constant $a$: $ds'^2=ads^2$

The coefficient $a$ can't depend on coordinates or time of interval because this violated homogeneity of the space.

My questions are:
1. Could anyone explain on the examples what does "violated homogeneity of the space" mean in this case?
2. How does dependence of coordinates and time of the interval violate homogeneity of the space?

As hard as I tried I couldn't understand it many times.

Thanks.

• Mike_bb
• Replies: 23
• Forum: Special and General Relativity

I thought a spacetime interval argument may help @Mike_bb.

I define some events as above:

• E0: clocks A and B are reset to $0s$
• E2: clock B shows $80s$
• E4: clock A shows $80s$

For convenience, I define, that all events have their y- and z-coordinates $=0$ and I omit these coordinates.

In observer B's frame (x', t') the time interval between E0 and E2 is
$\Delta t'_{20} = t'_2 - t'_0 = 80s \ \ \ \ \ (1)$
In observer B' frame (x', t') the distance between E0 and E2 is (the clock is at rest in this frame)
$\Delta x'_{20} = x'_2 - x'_0 = 0m \ \ \ \ \ (2)$

The invariant spacetime interval is
$s^2=c^2\Delta t^2 - \Delta x^2= c^2\Delta t'^2 - \Delta x'^2$

Inserting values from equations (1) and (2):
$c^2\Delta t_{20}^2 - \Delta x_{20}^2= c^2(80s)^2 - (0m)^2$

$c^2\Delta t_{20}^2 - v^2\Delta t_{20}^2= c^2(80s)^2$

$c^2\Delta t_{20}^2(1 - v^2/c^2)= c^2(80s)^2$

In observer A's frame (x, t), clock B shows $80s$ at the coordinate-time:
$\Delta t_{20}= \gamma (80s) = 100s$

----------------------------
Reciprocal calculation:

In observer A's frame (x, t) the time interval between E0 and E4 is
$\Delta t_{40} = t_4 - t_0 = 80s \ \ \ \ \ (3)$
In observer A's frame (x, t) the time interval between E0 and E4 is (the clock is at rest in this frame)
$\Delta x_{40} = x_4 - x_0 = 0m \ \ \ \ \ (4)$

Again, the invariant spacetime interval is
$s^2=c^2\Delta t^2 - \Delta x^2= c^2\Delta t'^2 - \Delta x'^2$

Inserting values from equations (3) and (4):
$c^2\Delta {t'_{40}}^2 - \Delta {x'_{40}}^2= c^2(80s)^2 - (0m)^2$

$c^2\Delta {t'_{40}}^2 - v^2\Delta {t'_{40}}^2= c^2(80s)^2$

$c^2\Delta {t'_{40}}^2(1 - v^2/c^2)= c^2(80s)^2$

In observer B's frame (x', t'), clock A shows $80s$ at the coordinate-time:
$\Delta t'_{40}= \gamma (80s) = 100s$