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A question about stationary reference frame 
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#19
Dec2511, 04:40 AM

Sci Advisor
P: 8,799

Coordinate time is not clock time.
Clock time is "proper time", and is the same in every reference frame. It is a property of the spacetime trajectory of a clock. Coordinate time is the "time" coordinate assigned when "space" and "time" coordinates are assigned within one reference frame to all conceivable events in spacetime. Proper time coincides with the coordinate time of a Lorentz inertial frame for a clock that is stationary with respect to that Lorentz inertial frame. For clocks that are moving relative to a Lorentz inertial frame, the proper time can be calculated from the coordinate time of that Lorentz inertial frame and the "Lorentz factor". 


#20
Dec2511, 10:37 AM

PF Gold
P: 4,792

γ = 1/√(1β^{2}) And beta, β, is the ratio of the speed that the clock is traveling at divided by the speed of light: β = v/c So the other way to express gamma is: γ = c/√(c^{2}v^{2}) But I like to use the first formula for gamma since it uses speeds as a ratio of the speed of light which is what you used also. You said the Lorentz Factor is equal to the change in coordinate time/change in proper time. So that means the change in proper time is equal to the change in coordinate time divided by the Lorentz Factor, correct? So we could express this as: Δτ = Δt/γ Δτ = Δt * √(1β^{2}) where Δτ (delta tau) is the change in proper time and Δt is the change in coordinate time. So given a speed as a factor of the speed of light as in your example of β=.5, did you actually do this calculation for the Lorentz Factor? γ = 1/√(1β^{2}) γ = 1/√(10.5^{2}) γ = 1/√(10.25) γ = 1/√(0.75) γ = 1/0.866 γ = 1.1547 Now you said there was a 5year change in the coordinate time so the change in the proper time for the train is equal to 5/1.1547 which equals 4.33 years. Now since you got the correct answer, how did you get it if you didn't understand how to do the calculations? Where are you still confused? It looks to me like you have perfect understanding (except for the incomplete quotes from wikipedia). Also, have you studied the calculations I gave you for doing the Lorentz Transform in my previous post? Does it all make sense? Did I answer your question about how to do the Lorentz Transform? 


#21
Dec2511, 04:15 PM

P: 127

Anybody can plug numbers into a calculation. This has nothing to do with understanding what they actually mean.



#22
Dec2611, 04:03 AM

P: 127

Just to clarify.
I am the stationary observer. There is a moving ship going at .8c relative to me. I experience one year on my clock. This 1 year is equal to my coordinate time and proper time, correct? Now the Coordinate time for the moving ship would be equal to 1 year as well, correct? The proper time would be equal to .6 years, correct? Is that what I call t'? Would this be symmetrical? I read .6 years on his clock but he reads 1 year on his clock and .6 on mine? 


#23
Dec2611, 05:07 AM

PF Gold
P: 4,792

Δτ = Δt * √(1β^{2}) where Δτ (delta tau) is the change in proper time and Δt is the change in coordinate time. So because your speed in your FoR is 0, β=0 and Δτ = Δt. Δτ = Δt * √(1β^{2}) Δτ = 1 * √(10.8^{2}) Δτ = 1 * √(10.64) Δτ = 1 * √0.36 Δτ = 1 * 0.6 Δτ = 0.6 years Now if you want to know how much time he reads on your clock during the same time interval that you read 1 year on your clock and .6 years on his clock, it would be .36 years. If you want to talk about a different time interval of 1 year on his clock, then, yes, he will read .6 years on your clock. For both of these situations we would switch to the ship's FoR and specify whatever time interval we want to consider, in the first instance, a proper time and a coordinate time of .6 years and we would calculate a proper time for you of .36 years using the same formula above. For a proper time and coordinate time of 1 year for the ship, the proper time for you would be .6 years. Keep in mind that we are not using the Lorentz Transform to do these calculations where t' is applied. 


#24
Dec2611, 02:41 PM

P: 127

http://www.trell.org/div/minkowski.html
Using this interactive minkowski diagram I plugged in a relative velocity of .8c, and for event B I plugged in a (1,.8). This means in my reference frame one year has passed and event B is at location of .8light years. It tells me that t' = .6 and x' = 0. This would mean in his reference frame he is only at .6 years and he is at his origin. Now when I plug in 2,0 as in for the time the ship comes back to Earth I get t' = 3.33 and x' = 2.67. So what exactly is this saying? It's saying that his time is 3.33 when he is at location 2.67. Surely I don't think this is right. I had thought he only measured 1.2 years. This must not be indicating the twin paradox correctly. Now I plug in for the stationary frame (2,1.6) and I get t' =1.2 and x' =0. Explain to me the differences. Why do I plug in the total distance he traveled relative to me even though he is now back at location 0 for me? 


#25
Dec2611, 04:52 PM

PF Gold
P: 4,792




#26
Dec2611, 11:46 PM

P: 140

I think you are missing what people are having a problem with. Lets take the earth and a ship moving away from earth at .8c.
from the earth's FOR the earth is stationary and the ship is moving at .8c. also the ship's clocks are ticking slower than the earth's. from the the ship's FOR the ship is stationary and the earth is moving at .8c. also the earth's clocks are ticking slower than the ships. 


#27
Dec2711, 12:15 AM

P: 127

So doesn't that mean if I am in the FOR of the earth and 1 year passes by for me while .6 years passes by for the ship? What I want to know is what is going on in the FOR of the ship? Does he see his clock at one year and my clock at .6? 


#28
Dec2711, 02:05 AM

PF Gold
P: 4,792




#29
Dec2711, 02:50 AM

P: 127

So it's only on the turn around, or when the clock enters a new FOR, that the symmetry breaks?



#30
Dec2711, 03:10 AM

PF Gold
P: 4,792




#31
Dec2711, 09:33 PM

P: 127

Let's say I am the stationary observer and there is space ship moving at .8c relative to me.
I see one year on my clock and I see .6 years on his clock. What time does he see on his own clock? I want to say he sees 1 year on his clock and .6 on mine, but I'm not sure. 


#32
Dec2711, 09:43 PM

P: 126

In either inertial frame, the spaceship twin or the earth twin, it is valid for each to consider themselves at rest and the other as moving.
However, in order to compare the clocks, one, or the other, or both frameworks must undergo acceleration to bring them into a common frame. It is this acceleration which differentiates one inertial frame from the other. 


#33
Dec2811, 11:37 AM

PF Gold
P: 4,792

But you should be aware that neither of you can actually see the others clock as you are asking about. When you see 1 year pass on your own clock, you will actually see 4 months (1/3 year) year pass on the ship's clock and in the same way when the ship see's 1 year pass on its own clock, it will see 4 months (1/3 year) pass on your clock. This is called the Relativistic Doppler effect and is a result of the time dilation of .6 years plus the time it takes for the image of the ship's clock to propagate across space to your clock and vice versa. 


#34
Dec2911, 02:16 AM

P: 127

so all moving bodies that are inertial are symmetrical to whatever inertial frame you choose, correct?



#35
Dec2911, 05:11 AM

PF Gold
P: 4,792

No, they are symmetrical to each other but you can use any inertial frame you choose to define, analyze and demonstrate what is going on.



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