Time Dilation Explained: SR vs GR | Modern Physics by Taylor and Zalifatos

  • Thread starter Thread starter OS Richert
  • Start date Start date
  • Tags Tags
    Time
Click For Summary
The discussion centers on understanding time dilation as explained in "Modern Physics" by Taylor and Zalfiratos, particularly through the example of a moving train and a plane experiment involving synchronized atomic clocks. Participants express confusion over why the plane's clock shows less elapsed time compared to a ground clock, despite both observers perceiving time dilation in their respective frames. The conversation highlights that the plane's non-inertial motion and the effects of acceleration complicate the application of special relativity (SR) versus general relativity (GR). Key points include the importance of recognizing that simultaneity differs across reference frames and that time dilation results from the paths taken through spacetime, rather than an absolute change in clock rates. Ultimately, the discussion emphasizes the need to abandon the notion of absolute time to resolve these paradoxes.
  • #31
wxrocks said:
If I send a beam of light at someone moving towards or away from me at some speed -- It is going to take a different amount of time for the light to get there depending on if I am looking from my POV or the person on the spacecraft because the person moving towards or away from me sees the length between me and him/her contracted.
That is not correct, both of you will observe the distance contracted not just the one in the spacecraft .

In general terms one can say that the observed distance between two objects that are in relative motion with each other decreases.

That is why it is so easy to reach a far away star system for someone who would be traveling at near light speed towards it, the observed distance simply decreases as soon as he increases his speed.
And by the way, it is not because his time slows down!
:smile:
 
Last edited:
Physics news on Phys.org
  • #32
MeJennifer said:
So I hope it is clear to everyone that the explanation that Hans de Vries gives here is diametrically opposed to the explanation that I gave. :smile:

While he holds the idea that time in some way can be seen as slowing down I hold the opposite opinion. The age of an object depends on the path taken in space-time not on the effects of relativistic measurements.

So pick your model!

:smile:
MeJennifer,

Would you care to carry your computations to the end for us? You start with :

1. "At home twin" ds^2=dt_1^2 because dx_1=0

2. "Travelling twin" ds^2=dt_2^2-dx_2^2

Could you please carry on the calculations from here?
 
Last edited:
  • #33
MeJen --

If I am standing on Earth and you are in a spacecraft moving away from Earth, your spaceship will look shorter, but my observed distance to Pluto is not going to change. You on the other hand will think your spacecraft still has a 3 meter cockpit, but will think Pluto moved a lot closer.
 
  • #34
wxrocks said:
If I am standing on Earth and you are in a spacecraft moving away from Earth, your spaceship will look shorter
Yes and for me in the spaceship the Earth will be observed shorter!

wxrocks said:
, but my observed distance to Pluto is not going to change. You on the other hand will think your spacecraft still has a 3 meter cockpit, but will think Pluto moved a lot closer.
Correct!

With regards to calculating the times in the twin problem.

So we have the Minkowski metric:

ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2

Let's simplify that a bit for the example:

ds^2=t^2 - dx^2

To find the time difference between the twins one has to compare the proper time, which is the time that each person experienced.

Proper time in a Minkowski space is:

<br /> \int d \tau = \int ds<br />

For the person that stayed at home this calculation is trivial since it's velocity was 0, only time increased. So basically we can simply look at dt only.

For the traveler the problem is more complicated because dt and ds become interrelated.
Perhaps this is going a bit to deep into the matter but what happens is that the 3D space of the traveler is tilted proportionally to his velocity but in an opposite way. The extreme being at the speed of light (but of course he cannot reach it), were space is tilted to a maximum. The tilting allows him to take a "shortcut" in space, the more tilt the shorter distances in the direction of travel are observed.

Still the way to calculate his proper time is the same, we take

\int d\tau = \int ds

Now in order to compose the total proper time for him we have to add the following parts:

a) The initial acceleration
b) The travel with a constant speed up to the deceleration near the halfway point
c) The the deceleration to zero
d) The the acceleration for the return
e) The travel with a constant speed up to the deceleration near the return point .
f) The the deceleration to 0

No how do we calculate the proper time?
For the constant speed parts it is simple:

\int d\tau = \int {\sqrt{1 - (dx/dt)^2 } dt

The acceleration and deceleration parts are more complicated, it depends if a is constant or variable. I would have to look it up.

So in summary, the traveler noticed that the distance became shorter the faster he traveled because of the tilt in 3D space. However his clocks were running as fast as before the trip. For the person who stayed behind there were no noticable changes. He did observe that the traveler's clock slowed down, but of course in reality it did not. It would be kind of silly to suggest that physical properties are changed because someone observes it from another frame of reference.

Hope this helps :smile:

This latex feature is nice but it took me quite some edits to get it right since I was not familiar with it.
 
Last edited:
  • #35
MeJennifer said:
Yes and for me in the spaceship the Earth will be observed shorter!Correct!

With regards to calculating the times in the twin problem.

So we have the Minkowski metric:

ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2

Let's simplify that a bit for the example:

ds^2=t^2 - dx^2

OK, so far.
To find the time difference between the twins one has to compare the proper time, which is the time that each person experienced.

This is better than what you were writing earlier. Looks like my question got you looking up the correct solution.

Proper time in a Minkowski space is:

<br /> \int d \tau = \int ds<br />

Would be nice to explain how you arrive to the above, what is the physical explanation?
For the person that stayed at home this calculation is trivial since it's velocity was 0, only time increased. So basically we can simply look at dt only.

Please do, you are now resorting to literature instead of math.

For the traveler the problem is more complicated because dt and ds become interrelated.

More literature instead of math, this is not saying anything of any use.

Perhaps this is going a bit to deep into the matter but what happens is that the 3D space of the traveler is tilted proportionally to his velocity but in an opposite way. The extreme being at the speed of light (but of course he cannot reach it), were space is tilted to a maximum. The tilting allows him to take a "shortcut" in space, the more tilt the shorter distances in the direction of travel are observed.

Now, the above is a pure word salad, devoid of any math.
Still the way to calculate his proper time is the same, we take

\int d\tau = \int ds

You are repeating yourself without giving any real info.

Now in order to compose the total proper time for him we have to add the following parts:

a) The initial acceleration
b) The travel with a constant speed up to the deceleration near the halfway point
c) The the deceleration to zero
d) The the acceleration for the return
e) The travel with a constant speed up to the deceleration near the return point .
f) The the deceleration to 0

Correct but where is the math?
No how do we calculate the proper time?
For the constant speed parts it is simple:

\int d\tau = \int {\sqrt{1 - (dx/dt)^2 } dt

Where did you get this from?
The acceleration and deceleration parts are more complicated, it depends if a is constant or variable. I would have to look it up.

"a" is constant. Please look up a complete solution before confusing people with your posts.
So in summary, the traveler noticed that the distance became shorter the faster he traveled because of the tilt in 3D space.

In essence you are saying "it is the way I told you it is", you haven't proven anything, just added a lot of half baked math creating more confusion than answers.
However his clocks were running as fast as before the trip.

No tangible proof here, just a word salad.

For the person who stayed behind there were no noticable changes. He did observe that the traveler's clock slowed down, but of course in reality it did not.

Physics is about hard, demonstrable facts. Not about word salads.
It would be kind of silly to suggest that physical properties are changed because someone observes it from another frame of reference.

Hope this helps :smile:

No, it didn't. Like many of your other posts, it "helped" confuse things further. It would be good for you to think a little more before posting. The above problem has been explained correctly by many of us in the past, there is no need for confusing people with fudged up answers.
 
Last edited:
  • #36
My motivation for posting here was to help OS Richert with his questions. I think I did help him understand the issue and my objective was to use a minimum of math.

If you think I wrote something incorrect in this topic then demonstrate it and at least provide some supporting arguments for accusing me of fudging up answers. So far you have not contributed anything except for making accusations of me fudging things.

I think what I wrote about the twin problem is correct, please feel free to show where my alledged errors are and where I misled or confused OS Richert.
 
Last edited:
  • #37
MeJennifer said:
My motivation for posting here was to help OS Richert with his questions. I think I did help him understand the issue and my objective was to use a minimum of math.

If you think I wrote something incorrect in this topic then demonstrate it and at least provide some supporting arguments for accusing me of fudging up answers.

I think what I wrote about the twin problem is correct, please feel free to show where my alledged errors are.

Re-read my post. You have "provided" a lot of confusion, I asked you to clear it up by posting the complete and correct math. There are a lot of unanswered questions, can you try to clear them. with math, not with more words.
 
  • #38
clj4 said:
Re-read my post. You have "provided" a lot of confusion, I asked you to clear it up by posting the complete and correct math. There are a lot of unanswered questions, can you try to clear them. with math, not with more words.
So who is confused here? I am not, Richert apparently is not, he got the answers he was looking for. He did not ask for math showing tilts of 3D-planes in space-time, or how to integrate the timelines of accelerated frames, that is not what this topic is about. He simply wanted to understand it in English. :smile:

Now what about you, are you confused about anything in the twin problem? Any problems with the math perhaps? If so perhaps I can help! :smile:

Or is it just that you have an issue with my explanation to Richert?
If so, fine, I have no problem with that.
But then demonstrate where my explanation is wrong or confusing.
 
Last edited:
  • #39
MeJennifer said:
So who is confused here? I am not, Richert apparently is not, he got the answers he was looking for. He did not ask for math showing tilts of 3D-planes in space-time, or how to integrate the timelines of accelerated frames, that is not what this topic is about. He simply wanted to understand it in English. :smile:
Looks like OS Richert never said "ok, MeJennifer, I understand". Instead of explaining things, you confused them.

Now what about you, are you confused about anything in the twin problem? Any problems with the math perhaps? If so perhaps I can help! :smile:
Yes, your "maths" are a collection of incomplete, botched up statements. So, I asked you to fix them. If you can. Please finish your "disertation". With math, not with prose. I understand math. Extremely well.

Or is it just that you have an issue with my explanation to Richert?
If so, fine, I have no problem with that.
Yes, your explanation is mathematically incomplete and incorrect. Please answer the questions and complete the calculations.
But then demonstrate where my explanation is wrong or confusing.

The first post you put up on the calculations were completely bogus. A simple question from me was sufficient for you to see the error and to abandon any folloup. You never answered my request yo complete the calculations. Why? Because you realized that the approach was incorrect.

The second post is a collection of things gathered at random,a little better but with missing explanations. You never completed any calculation, again. Why? Can you complete the calculations? If yes, please do so.
 
Last edited:

Similar threads

Replies
12
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
4K
  • · Replies 9 ·
Replies
9
Views
697
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 88 ·
3
Replies
88
Views
7K
  • · Replies 58 ·
2
Replies
58
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K