Time dilation . . . again!

  • Thread starter OS Richert
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  • #26
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Hans de Vries said:
Do not jump to conclusions. Many things look strange and non consistent when studying Special Relativity, but believe me, there are all good physical explanations and with time comes understanding.
I understand it very well thank you very much. It seems you feel a need to patronize.

Hans de Vries said:
Then you start learning what physics actually causes the speed of light to be constant and what causes the actual laws of Special Relativity.
So to you the constancy of the speed of light is no longer a postulate but something that can be derived?

Hans de Vries said:
I will have written a small illustrated booklet in a few months or so just for this purpose, sparing the reader for example from the need to study complicated graduate text on Relativistic Quantum Mechanics.
Well good for you.

I am sorry I would not be able to recommend it if it includes absurdities like that the speed of time slows down.
 
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  • #27
Hans de Vries
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MeJennifer said:
I am sorry I would not be able to recommend it if it includes absurdities like that the speed of time slows down.
These are your own words, not mine: "the speed of time slows down"
It is not at all clear to me what you exactly mean here nor what your
objections are.

It is mainstream physics to say that physical processes proceed slower
(take more time) when moving, like the standard example of muons
moving around in a cyclotron which take longer to decay.

Again as I mentioned earlier. Using the time as defined in the restframe
one would say that: Moving physical processes proceed slower while
in the moving frame one would say that "They pass through less time"
Both remarks are equally correct according to Special Relativity and you
are familiar enough with SR to know this.


Regards, Hans.
 
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  • #28
Garth
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Time can only ever 'pass' at the tautological 'one second per second'.

What time dilation is all about is the observation of one clock by another. Physical processes in an object are observed to proceed more slowly when observed from a frame of reference moving relative to that object, or when the object and the observer are at different gravitational potentials.

Garth
 
  • #29
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Garth said:
What time dilation is all about is the observation of one clock by another. Physical processes in an object are observed to proceed more slowly when observed from a frame of reference moving relative to that object, or when the object and the observer are at different gravitational potentials.
Indeed.

But just because something is observed as such does not make it so for the object in question.
The reality of an object or process is what is or would be observed at the locality, in the rest frame, not what we observe due to relativistic coordinate transformations.

Would we claim that for instance an object approaching the event horizon of a black hole will never enter it because we observe it as such?
 
  • #30
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To me, whats missing is the concept of length contraction (or at least it is not mentioned directly). 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.
 
  • #31
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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:
 
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  • #32
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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" [tex]ds^2=dt_1^2[/tex] because [tex]dx_1=0[/tex]

2. "Travelling twin" [tex]ds^2=dt_2^2-dx_2^2[/tex]

Could you please carry on the calculations from here?
 
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  • #33
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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 alot closer.
 
  • #34
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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 alot closer.
Correct!

With regards to calculating the times in the twin problem.

So we have the Minkowski metric:

[tex]ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]

Let's simplify that a bit for the example:

[tex]ds^2=t^2 - dx^2[/tex]

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:

[tex]
\int d \tau = \int ds
[/tex]

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 [tex]dt[/tex] only.

For the traveler the problem is more complicated because [tex]dt[/tex] and [tex]ds[/tex] 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

[tex]\int d\tau = \int ds[/tex]

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:

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

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.
 
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  • #35
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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:

[tex]ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]

Let's simplify that a bit for the example:

[tex]ds^2=t^2 - dx^2[/tex]
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:

[tex]
\int d \tau = \int ds
[/tex]
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 [tex]dt[/tex] only.
Please do, you are now resorting to literature instead of math.

For the traveler the problem is more complicated because [tex]dt[/tex] and [tex]ds[/tex] 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

[tex]\int d\tau = \int ds[/tex]
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:

[tex]\int d\tau = \int {\sqrt{1 - (dx/dt)^2 } dt[/tex]
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.
 
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  • #36
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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.
 
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  • #37
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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
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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.
 
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  • #39
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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 realised 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.
 
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