Proper time in moving frame of reference

[happyparticle]

Homework Statement

Calculate the proper time between two events for a moving frame of reference.

Relevant Equations

$(d \tau)^2 = \frac{- (ds)^2}{c^2}$

I try to figure out how to calculate the proper time between two events for a moving frame of reference.

Knowing that the proper time is: $(d \tau)^2 = \frac{- (ds)^2}{c^2}$, and thus it is the same for all frame of reference.

Following This example from wikipedia, we have an observer A who moves between the coordinates (0,0,0,0) and (10 years, 0, 0, 0) and an observer B who travels 4.33 light-years in the x direction for 5 years and then -4.33 light-years for 5 years.

The proper times calculated from the observer A rest frame are:
$d(\tau)_A = 10$ years ,$d(\tau)_B = 2 \sqrt{5^2 - 4.33^2} = 5$ years.

Then the proper times calculated from the observer B. Since in this frame of refence the observer B is at rest and the proper time should be the same for all frames. Knowing that $V = 0.866c$, then $\gamma = 2$ and $t = 5$ years

$d(\tau)_B = \sqrt{5^2 - 0^2} = 5$ years

Here comes my problem...

$x = 2.5 \cdot0.866c = 2.165$ light-years

$d(\tau)_A = 2 (\sqrt{(2.5 - 0)^2 - (-2.165 -0)^2} ) + \sqrt{(5 - 2.5)^2 - (0 - (-2.165)^2)} = 2.5$ years

The proper time for the observer A is wrong and it should be 10 years.

 

[QuarkyMeson]

Homework Statement: Calculate the proper time between two events for a moving frame of reference.
Relevant Equations: $(d \tau)^2 = \frac{- (ds)^2}{c^2}$

Knowing that the proper time is: (dτ)2=−(ds)2c2, and thus it is the same for all frame of reference.

This isn't true, the spacetime interval is frame independent for all observers in inertial reference frames, but proper time is path dependent. Proper time depends on the worldline that an object takes through spacetime. $$ d\tau^2 = \frac{-ds^2}{c^2}$$ $$\tau = \int \sqrt{ \frac{-ds^2}{c^2}}$$

So even if your starting and ending coordinates are the same, proper time still depends on the path taken.

 

[anuttarasammyak]

FoR A is a IFR which you can apply faliliar SR formula, but FoR B is not. Go- FoR B and Return- FoR B are IFRs but synthesized Go&Return FoR B is not a IFR.

 

[PeroK]

Homework Statement: Calculate the proper time between two events for a moving frame of reference.
Relevant Equations: $(d \tau)^2 = \frac{- (ds)^2}{c^2}$

I try to figure out how to calculate the proper time between two events for a moving frame of reference.

Knowing that the proper time is: $(d \tau)^2 = \frac{- (ds)^2}{c^2}$, and thus it is the same for all frame of reference.

Following This example from wikipedia, we have an observer A who moves between the coordinates (0,0,0,0) and (10 years, 0, 0, 0) and an observer B who travels 4.33 light-years in the x direction for 5 years and then -4.33 light-years for 5 years.

The proper times calculated from the observer A rest frame are:
$d(\tau)_A = 10$ years ,$d(\tau)_B = 2 \sqrt{5^2 - 4.33^2} = 5$ years.

Then the proper times calculated from the observer B. Since in this frame of refence the observer B is at rest and the proper time should be the same for all frames. Knowing that $V = 0.866c$, then $\gamma = 2$ and $t = 5$ years

$d(\tau)_B = \sqrt{5^2 - 0^2} = 5$ years

Here comes my problem...

$x = 2.5 \cdot0.866c = 2.165$ light-years

$d(\tau)_A = 2 (\sqrt{(2.5 - 0)^2 - (-2.165 -0)^2} ) + \sqrt{(5 - 2.5)^2 - (0 - (-2.165)^2)} = 2.5$ years

The proper time for the observer A is wrong and it should be 10 years.

This is all completely muddled. There are three inertia reference frames here:

The rest frame of A; the outbound rest frame of B; and, the inbound rest frame of B.

You have a description of the worldlines of A and B in the rest frame of A. So, you can directly calculate the proper times - using the coordinates as specified in this frame. Note that as the wordline of B is not a straight line, you have to calculate the proper time for each leg of the journey separately.

You cannot calculute the proper times easily from the rest frame of B, because it is not inertial. Instead, you have to transform the coordinates of the worldlines of A and B for the first part of the experiment into the outbound frame and calculate the proper time in that frame. Then, you have to transform the coordinates for the remainder of the experiment into the outbound frame and calculate the remaining proper time for A and B.

This may be a useful exercise, but you have to be very clear about the intermediate events that represent the end of the outbound phase and the beginning of the inbound phase.

Alterbatively, you can calculate the proper times of A and B using either the outbound frame throughout or the inbound frame throughout. And, of course, you should get the same answer in both cases.

 

[PeroK]

This isn't true, the spacetime interval is frame independent for all observers in inertial reference frames, but proper time is path dependent. Proper time depends on the worldline that an object takes through spacetime. $$ d\tau^2 = \frac{-ds^2}{c^2}$$ $$\tau = \int \sqrt{ \frac{-ds^2}{c^2}}$$

So even if your starting and ending coordinates are the same, proper time still depends on the path taken.

The starting and ending events, and the proper time along a worldline are invariant. It's the description of those things in terms of coordinates that is frame dependent.

 

[happyparticle]

@PeroK I see. I transformed the coordinates of A and B in the frame of B using the Lorentz transformation matrix and then I calculated the proper time for both A and B and now it works. I have the same proper time for A and B in both reference.

Did I understood correctly?