How does this equation imply that time slows down?

  • #1

Main Question or Discussion Point

I've studied SR previously and, after a lot of work, grew to understand it. Now I have to re-work through it using a different textbook and can't figure out the deal with time dillation.

I derived the equation

T = T0γ

With relative ease. In this equation, T0 is measured by a person at rest (in his own frame) while T is measured by a person moving (in the rest frame of the first observer. Due to the nature of γ, T will always be larger than T0.

At this point, my textbook states that this means that the observer at rest observes time slowing down for the moving person. I don't understand this.

Say a spaceship was passing by Earth. I'm laying in the grass, looking up. In my frame, I'm at rest. The spaceship passes over, quickly. I measure the time I can see it. Since I'm at rest in my own frame, this time is T0.

The same chain of events could be measured from the spaceship. Staying in my frame, my equation tells me that the spaceship measures a higher time. Perhaps I measured 4 seconds, while the spaceship measured 8!

In other words, more time passed for the ship. Doesn't that mean that I observe a spaceship where time is moving much faster! After all, I "saw" twice as much as time passing on their clocks than on mine!
 

Answers and Replies

  • #2
PeterDonis
Mentor
Insights Author
2019 Award
29,684
8,953
The most important thing to remember about time dilation is that you can't just look at time dilation. You have to also take into account relativity of simultaneity--or, to put it another way, you have to look at both space and time together, not just time. The equation ##T = T_0 \gamma## does not do that, so if you just use that equation by itself, you will be misled.

My advice would be to first assign coordinates in your rest frame to the two events of interest (the ones separated by a time interval ##T_0## in your rest frame). Then use the Lorentz transformation to transform those coordinates into the ship's rest frame. Looking at both the time and space coordinates, not just the time coordinate, should help to resolve your confusion.

(Similar remarks apply to length contraction.)
 
  • #3
Thanks. This bothers me though, because the textbook completely ignores relativity of simultaneity at this point. Not even a single mention. It merely creates the lorenz transformation for time dillation from the postulates of SR, then uses that to deduce time slowing down for moving observers. Any clue as to why the book wants me to make this logical step, even though I cannot (according to your answer).
 
  • #5
29,604
5,902
In this equation, T0 is measured by a person at rest (in his own frame) while T is measured by a person moving (in the rest frame of the first observer.
It is a little more subtle than that. T and T0 are measured between two given events. T is in a reference frame for which the two events are co-located. T0 is in a frame which is moving at velocity v wrt the first. In this frame the two events are not co-located.

That said, I agree that the Lorentz transformation is the way to go. It will automatically simplify when appropriate.
 
Last edited:
  • #6
russ_watters
Mentor
19,662
5,946
Thanks. This bothers me though, because the textbook completely ignores relativity of simultaneity at this point. Not even a single mention. It merely creates the lorenz transformation for time dillation from the postulates of SR, then uses that to deduce time slowing down for moving observers. Any clue as to why the book wants me to make this logical step, even though I cannot (according to your answer).
As someone who also has only a thin knowledge of Relativity, I can guess:

When being introduced to a subject, it is often better to keep things simple and tackle only one issue at a time. Bringing together both time dilation and relativity of simultenaity adds unnecessary complexity (IMO) and potential to mistake one issue for the other or think they are interdependent when in reality they are completely separate issues.

For that reason I prefer starting with simpler scenarios where only time dilation is needed, such as a simple version of the twins paradox.
 
  • Like
Likes Fervent Freyja
  • #7
446
159
Everything is easier to keep track of if you specify reference events in spacetime.

Event A: the nose of the spaceship crosses a "starting line" to your left

Event B: the nose of the spaceship crosses a "finishing line" to your right

A single wristwatch worn by a passenger at the nose of the spaceship logs the spaceship's proper time. (Let's not worry about YOUR proper time right now.) Between Events A and B, that wristwatch measures a proper time interval ##\Delta t_0##.

For you on the ground, the coordinate time interval between Events A and B is measured by two different synchronized clocks: one at the starting line, which takes a reading when the nose of the spaceship arrives (marking your coordinate time for Event A); and the other at the finishing line, which records your coordinate time for Event B. After the readings are taken, you compare the clocks and find that Event B happened a time ##\Delta t## after Event A.

The time dilation formula tells you that ##\Delta t = \gamma \Delta t_0##, guaranteeing that ##\Delta t > \Delta t_0##. So you would say that the passenger ages less than you do during the spaceship's journey from starting line to finishing line.



Now, if you wanted to bring YOUR proper time into this, then you'd need to specify events that both happen at YOUR location. For example:

Event C: someone at the NOSE of the spaceship gives you a high-five while passing by

Event D: someone at the TAIL of the spaceship gives you a high-five while passing by

Because you're physically present at both events, YOUR wristwatch logs the inertial proper time interval between Events C and D. From the spaceship's perspective, it is YOU who is "moving" while the spaceship is "stationary." Two different synchronized clocks on the spaceship measure the coordinate time interval between Events C and D: one at the nose of the ship, which takes a reading when you high-five the nose-passenger; and the other at the tail of the ship, which takes a reading when you high-five the tail-passenger.

In this scenario, the passengers on the ship would say that you age less than they do during "your journey" from nose to tail.
 
  • #8
Mister T
Science Advisor
Gold Member
2,564
820
I've studied SR previously and, after a lot of work, grew to understand it. Now I have to re-work through it using a different textbook and can't figure out the deal with time dillation.

I derived the equation

T = T0γ

With relative ease. In this equation, T0 is measured by a person at rest (in his own frame) while T is measured by a person moving (in the rest frame of the first observer. Due to the nature of γ, T will always be larger than T0.
##T_o## is the time that elapses between two events that occur at the same place. This is called the proper time. Note that this can be the time that elapses between two events in only one of your two frames.

So, in the example of you watching a ship pass across the sky, we can choose two events such as "ship is located on the western horizon" and "ship is located on the eastern horizon". For you those two events occur at different places so the time that elapses between them, as measured by say your wristwatch, is not a proper time. For a person aboard the ship they do occur at the same place, say his chair, so the time that elapses between them as measured on his watch is a proper time.
 
  • #10
1,420
112
I've studied SR previously and, after a lot of work, grew to understand it. Now I have to re-work through it using a different textbook and can't figure out the deal with time dillation.

I derived the equation

T = T0γ

With relative ease. In this equation, T0 is measured by a person at rest (in his own frame) while T is measured by a person moving (in the rest frame of the first observer. Due to the nature of γ, T will always be larger than T0.

At this point, my textbook states that this means that the observer at rest observes time slowing down for the moving person. I don't understand this.

Say a spaceship was passing by Earth. I'm laying in the grass, looking up. In my frame, I'm at rest. The spaceship passes over, quickly. I measure the time I can see it. Since I'm at rest in my own frame, this time is T0.

The same chain of events could be measured from the spaceship. Staying in my frame, my equation tells me that the spaceship measures a higher time. Perhaps I measured 4 seconds, while the spaceship measured 8!

In other words, more time passed for the ship. Doesn't that mean that I observe a spaceship where time is moving much faster! After all, I "saw" twice as much as time passing on their clocks than on mine!



Moving clock runs slowly. Mathematically:

Tnot_moving = Tmovingγ

if we change that to your notation we get this:

T0 = Tγ

Not this: T = T0γ
 
  • #11
Mister T
Science Advisor
Gold Member
2,564
820
Moving clock runs slowly. Mathematically:

Tnot_moving = Tmovingγ
That's a simplification that's often presented in introductory discussions of relativity, but it leads to the very type of learner confusion that caused the OP to start this thread. Here's why. If an observer is moving in such a way that he co-moves with a clock, and that motion is relative to a stationary (that is, not-moving) clock, then he will observe that the stationary clock is running slow.

Now I realize that you can get around this pedagogical difficulty by emphasizing that by "moving clock" you mean moving relative to the observer. And if that works for the learners you're trying to educate, then that's what you can do.

For me, though, I find that he following scheme works better because it forces the learner to think about a pair of events, and the time that elapses between them. So rather than using "not moving" and "moving" to distinguish between the clocks, one can use "proper time" and "dilated time" to distinguish between the things that the clocks are measuring. The dilated time is always larger than the proper time, whether it's the moving clock or the not-moving clock that's measuring it. And for a time to be a proper time it must be the time that elapses between two events that occur in the same place, whether that's the same place for the moving frame or the same place for the stationary frame.
 

Related Threads on How does this equation imply that time slows down?

  • Last Post
Replies
10
Views
5K
  • Last Post
Replies
22
Views
5K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
3
Views
3K
Replies
98
Views
15K
Replies
36
Views
7K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
16
Views
3K
  • Last Post
2
Replies
34
Views
4K
Top