# Time in Special Relativity

Discussion in 'Physics & Math' started by Pete, Apr 10, 2010.

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1. ### PeteIt's not rocket surgeryRegistered Senior Member

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Moderator note: This thread has been split from the following thread:

[thread=101005]A thread about Jack's capacity to discuss relativity[/thread]

Please use the current thread only to discuss Jack_'s issues with time in relativity, and not for character analysis. Thankyou.

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Jack, there are a few problems with this question. You'll need to be more precise.
Firstly, a semantic issue. A reference frame doesn't "see" anything. Do you mean to ask when measurements made in the moving reference frame indicate that...?
Secondly, the question is loaded with the assumption of absolute simultaneity. It assumes that a particular instant ("when") in one reference frame universally translates to a single instant in another reference frame.

It is clear that you simply do not understand what SR says about how time transforms between reference frames. You can learn, Jack, it's really not that difficult.
Briefly: The light reaches a distance r in the moving frame at a time t'=r/c after it is emitted. That moving frame instant, when t'=r/c, does not happen at one instant in the stationary frame, but over a period of time.

But, this has already been explained to you, patiently and repeatedly. I fully expect that you will just ignore it once again and repeat your same old boring story. Surprise me, Jack. Take off that steel skullcap and consider a new idea. You don't need to accept it, but we do expect you to honestly consider it. That's how we learn.

Last edited by a moderator: Apr 16, 2010
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3. ### Jack_BannedBanned

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Good, I have posted the interval before.

Absolute simultaneity is not included in this logic. Absolute simultaneity implies it will be simultaneous for both frame. I used time dilation to prove that single time is rγ /c in the time of the stationary frame where in that frame it is r/c. Note, they are not simultaneous. You claimed to have learned all about this.

I posted a very clear proof. When light elaspes rγ /c in O, r/c elapses at the O' clock by time dilation.

When r/c elapses at the O' clock, light must be a distance r from O in all directions.

You see, I am not log jammed in my thinking. I am able to apply time dilation and the light sphere and that arrives at a different conclusion.

You are stuck only on LT and yet you also believe time dilation and the light sphere.

But, to refute my logic in favor of LT implies you refute time dilation or the light sphere.

So, you must confess time dilation is false or you must contend when the O' clock elapses r/c, light is not a distance r from O' in all directions.

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5. ### Jack_BannedBanned

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I have decided to post the time interval again using LT as to when in the frame of O that O' sees light a distance r from O'.

t ε [r/c( 1 - v/c)γ, r/c( 1 + v/c)γ].

Now, you claim to understand SR and I am just thick headed.

OK, when t = r/c( 1 - v/c)γ, you claim light is a distance r along the negative x-axis.

Let's now apply time dilation to the clock at O'.

tO' = r/c( 1 - v/c).

That is interesting, then light should only be a distance r( 1 - v/c) from O' by the logic of the light sphere.

Yet, LT claims it is r.

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7. ### PeteIt's not rocket surgeryRegistered Senior Member

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Jack, this paragraph indicates that we're not on the same page when talking about simultaneity. It seems you are thinking something very different to me. Please let me check, to make sure we're talking the same language.

You said: "Absolute simultaneity implies it will be simultaneous for both frames."
By "It", you mean, "the time when the moving frame sees light a distance r in all directions from its moving origin", right?
And by "simultaneous in both frames", you mean "the time at which this occurs in the moving frame is different from the time at which it occurs in the stationary frame", right?

So when you say "they are not simultaneous", you mean "the value of t' when it happens is not the same as the value of t when it happens", right?

Last edited: Apr 11, 2010
8. ### Jack_BannedBanned

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Yes, the logic I provided included time dilation and therefore, both frames would not agree on time.

9. ### PeteIt's not rocket surgeryRegistered Senior Member

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Yes, I thought so. We are talking different languages.

Jack, when we talk about simultaneity being relative, we're not just talking about different values for time. We mean that you can't unambiguously match up a particular instant in one reference frame with a particular instant in another. You can do it for a single location, but that doesn't apply in other locations.

Consider that particular instant in the moving frame, when the light reaches a distance r from the origin at t' = r/c.

You claim that SR says that this occurs in the stationary frame at t = γr/c.
You correctly derived this at a particular location - the location of O'.

But, at other locations in the stationary frame, that particular moving frame instant occurs at different times. Try transforming the location of the light, for example:

In the moving frame, light reaches x'=r at t'=r/c. When does this occur in the stationary frame? Hint: It's not t=γr/c.

In the moving frame, light reaches x'=-r at t'=r/c. When does this occur in the stationary frame? Hint: It's not t=γr/c.

Do you see? SR says that the moving frame instant is not a single instant in the stationary frame.

But like I said, you've made this mistake before, and it's been pointed out and explained patiently and repeatedly. I'm really not confident that this time will be any different, but feel free to surprise me.

10. ### PeteIt's not rocket surgeryRegistered Senior Member

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Jack, did you miss my last post? Have you tried to grasp what relative simultaneity means?

You say you know how to use the lorentz transform, so why not answer your own question properly:
In the moving frame, light reaches a distance r from the origin at t'=r/c.
Let's examine three events that occur at that time:
• at x'=0, the clock at O' reads r/c
• at x'=-r, light arrives
• at x=r, light arrives

So, when in the stationary frame do these events occur, according to SR?
You already know the first one: the O' clock reads r/c at t=γr/c. Right?
You can confirm it by transforming the event coordinates (x',t')=(0,r/c)

What about the other two things?
Their coordinates are (x',t')=(-r,r/c) and (x',t')=(r,r/c)
Hint - they don't happen at t=γr/c.

Last edited: Apr 15, 2010
11. ### Jack_BannedBanned

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OK, this is funny.

12. ### PeteIt's not rocket surgeryRegistered Senior Member

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This is most certainly on task, Jack.
Time in SR is about more than just time dilation. That's why you need to do proper transformations - naively applying time dilation is likely to cause mistakes.

What I want to see you do is realise that the answer is more involved than you think.

Yes, in the moving frame the clock at O' reaches r/c when the clock at O reaches γr/c.
No, this is not true in the stationary frame.
And no, at t'=r/c in the moving frame, the stationary-frame-time at locations other than O is not γr/c. At O', for example, the stationary-frame-time is t=r/γc.

Try it yourself:
Transform the stationary frame location of O' at t=r/γc to the moving frame.
(x,t)=(vr/γc,r/γc)
...transforms to...
(x',t')=(0,r/c)
ie this is the moving-frame location of O' when the O' clock reaches r/c.

So to answer your question:
You'll find that you get different times at different locations.

Try it. The moving frame time is t'=r/c, but there are several x' locations of interest. Transform them to the stationary frame, and see what stationary frame times you get:
(x',t')=(0,r/c) (location of O')
(x',t')=(-r,r/c) (location of light)
(x',t')=(r,r/c) (location of light)
(x',t')=(-vr/c,r/c) (location of O)

13. ### Jack_BannedBanned

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What you have done above is refute SR using circular reciprocol time dilation.

Here is how it works.

Frame A see time t, then moving A' sees r/γ as the time. But, taking that frame as stationary, that means A must see r/γ²

You set up a sequence based on taking the other frame as stationary.

r/γ^n that converges to 0.

What do you think about that?

Anyway, if the O frame shows r/γc at the origin, what does the O' frame show at the origin? r/c.

If a frame showd r/c on the clock at the light emission point, light is r in all directions. You cannot stop any of this logic

Yo beef aint with me, it is with SR.

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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No, Jack, that's not how it works. That's one of the common mistakes with naive time dilation I was talking about.
Do the transforms, post the results. Then we can talk.

15. ### James RJust this guy, you know?Staff Member

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Clearly you haven't understood a word that Pete has said to you.

Nor have you made any effort to follow through on what he asked you to do.

In short, you have made no attempt to understand Lorentz tranformations.

16. ### PeteIt's not rocket surgeryRegistered Senior Member

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As long as it takes for you to understand that time in SR is more than simple time dilation.

Consider these events at time r/c in the moving frame.
Transform them to the stationary frame, and see what stationary frame times you get.
(x',t')=(0,r/c) (location of O')
(x',t')=(-r,r/c) (location of light)
(x',t')=(r,r/c) (location of light)
(x',t')=(-vr/c,r/c) (location of O)

I'll step you through the first one: (x',t')=(0,r/c)

$\large\begin{eqnarray} x &=& \gamma (x' + vt') \\ &=& \gamma vr/c \end{eqnarray*} \\ \begin{eqnarray*} t &=& \gamma (t' + \frac{vx'}{c^2}) \\ &=& \gamma r/c \end{eqnarray*}$

Can you manage the rest? The arithmetic can get tricky. You might need to expand gamma.

Last edited: Apr 15, 2010
17. ### AlphaNumericFully ionizedRegistered Senior Member

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As a vague point of interest the book 'Mathematical Physics' by Robert Geroch is an excellent introduction to the mathematical formalism associated to many things used by physicists. Its quite abstract in places, he goes at everything from the point of view of category theory (or as he calls it 'the mathematics of mathematics') but well worth the effort. I mention it because Jack is complaining about a lack of mathematical consistency and the book goes from about as basic a starting point as you can without having to redo entire 1st year courses (quickly goes through basic definitions of sets, map, function etc but you should already know about them) and explicitly considers the Minkowski vector space in Chapter 15 and then the Lorentz Group in Chapter 16. They are given as the standard example of some interesting results developed before those chapters and both covered in succession because they are linked in precisely the manner I've outlined which Jack refuses to address, namely one is a vector bundle and the other its associated principle G bundle, aka one examples the space, the other its symmetries (respectively).

If you like the more formal side of physics its a great book. If you have graduate level knowledge of the more mathematical notions used in physics its useful. If you have taught vector calculus to undergraduates then it'll be easy to follow (at least in relation to Chapters 15 and 16). If Jack is intellectually honest and interested in expanding his knowledge and understanding he'll take a look. If he's still at UW then he'll have free access to it via its library. Otherwise it's not expensive, its paper back.

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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How are you going with these, Jack?

I'll step you through the second one: (x',t')=(-r,r/c)

$\large\begin{eqnarray} x &=& \gamma (x' + vt') \\ &=& \gamma (-r + vr/c) \\ &=& \gamma vr/c - \gamma r \end{eqnarray*} \\ \begin{eqnarray*} t &=& \gamma (t' + \frac{vx'}{c^2}) \\ &=& \gamma (r/c - \frac{vr}{c^2}) \\ &=& \gamma r/c - \gamma vr/c^2 \end{eqnarray*}$

Note carefully the value of t. Do you see that there is more that just time dilation involved?

19. ### Jack_BannedBanned

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Yes, I see that.

That is one calculated and I am glad to see everyone now meeting the same calculation I had under LT.

That can be a self check to know you are correct.

Now, we have an issue with time dialtion and the light sphere.

But, we are going to apply logic.

20. ### PeteIt's not rocket surgeryRegistered Senior Member

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Show us, Jack.
Show us that you can transform these events at t'=r/c to the stationary frame, and explain the stationary frame times you get:
I've done the first two. Can you do the others?
Yes, we do have an issue with time dilation, and I think we should explore it.
Do those transforms, and we'll examine how the results conflict with simple time dilation.

Last edited: Apr 15, 2010
21. ### Jack_BannedBanned

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OK, now that everyone is up to speed on the LT equations, let do the time dilation equations.

What is the time at O'?

22. ### PeteIt's not rocket surgeryRegistered Senior Member

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Are you, Jack?
Show us.
Transform these events to the stationary frame:
a) (x',t')=(0,r/c) (location of O')
b) (x',t')=(-r,r/c) (location of light)
c) (x',t')=(r,r/c) (location of light)
d)(x',t')=(-vr/c,r/c) (location of O)

Please, Jack, you need to make your questions more precise, like this:
What is the time in the rest frame of O at O' when t' = r/c?
Transforming event a), we find the answer:
t=γr/c

Your turn now.
What is the time in the rest frame of O at O when t' = r/c?
Transform event d) to find the answer.

Last edited: Apr 15, 2010
23. ### Jack_BannedBanned

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Heck, that would be r/(γc).

Then calculate the time at O' after that.

Pretty funny no?

Basides, this does not matter. Einstein said the moving clock beats slower by a factor of 1/γ.

GPS says the moving satellites beat slower by a factor of 1/γ.

So, the moving O' will see r/c on its clock.

When r/c is on the O' clock, light must be r in all directions from O' by the light sphere.

Last edited: Apr 16, 2010
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