# WIKI and Time Dilation

1. Jan 2, 2011

2. Jan 2, 2011

### NobodySpecial

Seems like the standard explanation and the simplest way of deriving the lorentz-fitzgerald formula, whats the problem ?

3. Jan 2, 2011

### chinglu1998

It states t' = t γ and t is from the rest frame. That means if t is 1s and since γ > 1, then t' > t which means more time on primed clock which is moving.

4. Jan 2, 2011

### Dr Lots-o'watts

Why do you write t? It's delta t. You're looking for trouble.

5. Jan 2, 2011

### chinglu1998

What?

Delta t then.

Do you have a comment on the OP? Is WIKI correct?

6. Jan 2, 2011

### JesseM

Yes, it's correct. In the primed frame where the clock is moving, the clock is running slow so it takes more time to tick forward by a given amount. For example, if a clock is moving at 0.6c in my (primed) frame, it takes 25 seconds in my frame for the clock to tick forward by 20 seconds of its own time.

Also, I don't understand your use of the phrase "more time on primed clock which is moving". Motion is relative, there is no objective truth about which is "moving". The time dilation formula assumes you have a clock at rest relative to the unprimed frame, and which is moving relative to the primed frame, and you want to know how much time goes by in the unprimed frame when the clock ticks forward by a given amount.

7. Jan 2, 2011

### Staff: Mentor

I strongly and consistently recommend against ever using either the time dilation or length contraction formulas. It is too easy to use it incorrectly, and it doesn't save a significant amount of effort.

I recommend always using the complete Lorentz transform. This avoids any risk of using the simplified formulas when they do not apply, and it automatically simplifies when appropriate.

8. Jan 2, 2011

### chinglu1998

Well Einstein wrote the following
τ = t √( 1 - v²/c² )
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Section 4 halfway down.

t is in the rest frame time and this is the same as WIKI, the unprimed frame is the rest frame.

Both WIKI and Einstein used the unprimed frame as the rest frame.

So, let say the rest frame (unprimed frame) ticked 20 seconds.

Einstein said τ = 20 √( 1 - (.6c)²/c² ) = 20 √( 1 - (.6c)²/c² ) = 20 √( .64) = 20 (.8) = 16.

Thus, the moving frame ticked 16 seconds or slower than the rest unprimed frame.

This is not what WIKI says.

WIKI claims the moving frame ticks off more time.

9. Jan 2, 2011

### chinglu1998

I agree, but, if you look at WIKI, it has the rest frame beating slower or the moving frame clock beating faster. The unprimed frame is the rest frame.

10. Jan 2, 2011

### chinglu1998

The example said the primed frame was the moving frame. And the unprimed frame is taken at the stationary system of coordinates. So, that means the clocks in the primed frame are moving relative to the unprimed frame.

11. Jan 2, 2011

### JesseM

I think you're confused about how relativity works, no frame is declared "moving" or "stationary" in any absolute sense, you can only talk about relative motion. The primed frame is moving relative to the unprimed frame, and the unprimed frame is moving relative to the primed frame. Clocks at rest in the unprimed frame are running slow in the primed frame (and the wiki article is specifically considering a clock at rest in the unprimed frame), just as clocks at rest in the primed frame would be running slow in the unprimed frame (though the time dilation equation in the wiki article doesn't deal with this case).

12. Jan 2, 2011

### chinglu1998

13. Jan 2, 2011

### JesseM

What do you mean by "rest frame"? Do you understand my point that no frame is declared to be "at rest" in any absolute sense, you can only talk about a frame as the "rest frame" of some particular object? In the wiki, the unprimed frame is the rest frame of the clock, and the primed frame is the rest frame of the observer.
No, it says the unprimed frame is the rest frame of the clock being considered in the time dilation equation. Some of the links you gave do use a different convention, but in any case it's purely a matter of convention which coordinates we use to label time in the clock's own rest frame, it's usually the primed frame but it could be the unprimed frame.

14. Jan 2, 2011

### chinglu1998

I can't see how you conclude I said any frame is at absolute rest.

I said when the frame is taken as stationary.

So, I think you are claiming, if the unprimed frame is taken as stationary, the moving clocks in the primed frame are beating slower.

You can tell the unprimed frame is at rest with the light source because there is no light abberation (angle to the light beam).

So, the unprimed frame is considered stationary and the primd frame is considered the moving frame. Clocks therefore, when viewed from the stationary frame are time dilated.

For example, when do you say you are in the moving frame and the clocks in the other frame are at rest?

Do you see this makes no sense what you say?

15. Jan 2, 2011

### chinglu1998

Then is it possible you compare the equations?

WIKI says t' = t γ

The others say t' = t / γ.

Can you see the difference?

16. Jan 2, 2011

### JesseM

While you were posting this I was editing my post to read:

17. Jan 2, 2011

### chinglu1998

I agree with you the convention is important.

But, the fact that the unprimed frame has no light abberation clearly indicates that frame is at rest with the light source which is being used to determine time dilation and that plus the fact WIKI said the unprimed frame is at rest, I am not sure what other evidence I could produce that the unprimed frame is not moving and any clock in the primed frame is.

Moving clocks beat time dilated and that is not what the equation on WIKI shows.

18. Jan 2, 2011

### JesseM

"Taken as stationary" relative to what? Better to say it's the rest frame of the observer, I think.
The phrase "taken as stationary" doesn't seem to add any meaning here. I would just say that from the perspective of the unprimed frame, the clocks in the primed frame are moving and beating slower.
Or you could just note that the position of the source in the unprimed frame doesn't change with time.
Again the words "stationary" and "moving" are completely unnecessary and confusing. I would just say "clocks at rest in the unprimed frame, when viewed from the primed frame are time dilated"
"At rest" is meaningless unless you specify what frame something is at rest relative to. If I am at rest relative to the primed frame, then clocks at rest relative to the unprimed frame are moving relative to me.

19. Jan 2, 2011

### chinglu1998

Here count the number of times Einstein uses "stationary". 62 times.
I think it has a clear meaning.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

20. Jan 2, 2011

### JesseM

It is the rest frame of the light clock whose time dilation we want to determine, as seen in the frame of an observer (the primed frame). Regardless of what symbol convention is used, the time dilation equation always has the following form:

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma
No it didn't, it only said the unprimed frame is the one where the clock is at rest. Likewise, the primed frame is the frame of the observer, so it is obviously the frame where the observer is at rest. When the article refers to the "moving observer", they only mean the observer is moving relative to the clock.
As usual, it is physically meaningless to say a given object is "moving" or "at rest" unless you are really saying it is moving or at rest relative to something. It's true that some authors just talk about a "stationary frame" and a "moving frame" but it is always clear from the context that they are implicitly defining motion relative to a particular object or observer. Obviously this convention is getting you all confused because you fail to understand the unspoken "relative to", so I'd suggest that from now on you never use phrases like "stationary" or "moving" unless they are followed by "relative to X", where X is some object or observer.

Last edited: Jan 2, 2011