Special relativity -- Proper time

In summary: The difference between the coordinate times for B and C/D will be larger than the difference between the proper times for B and C/D.Using the metric equation is equivalent. You can say$$\Delta s_{ED-CD} = 0$$and calculate the corresponding Δt. You'll get the same result.In summary, Alice is driving a race car at a constant speed of 60 m/s on an essentially circular track. Brian measures the time it takes for her to complete a lap by starting his watch when she passes by and stopping it when she passes by again. This is also observed by Cara and Dave on a train passing close to Brian. The clocks used by all three are considered to be "present" at both
  • #1
Lito
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Homework Statement



Alice is driving a race car around an essentially circular track at a constant speed of 60 m/s. Brian, who is sitting at a fixed position at the edge of the track, measures the time that Alice takes to complete a lap by starting his watch when Alice passes by his position (call this event E) and stopping it when Alice passes his position again (call this event F). This situation is also observed by Cara and Dave, who are passengers in a train that passes very close to Brian. Cara happens to be passing Brian just when Alice passes Brian the second time (see Figure). Assume that the clocks used by Alice, Brian and Cara are close enough together that we can consider them all to be “present” at event E; similarly, that those used by Alice, Brian and Dave are “present” at event F. Assume that the ground frame is an inertial reference frame.

(a) Who measures the shortest time between these events? Who measures the longest?
(b) If Brian measures 100 s between the events, how much less time does Alice measure between the events?
(c) If the train carrying Cara and Dave moves at a speed of 30 m/s, how much larger or smaller is the time that they measure compared to Brian’s time?
2q9ypol.jpg


Homework Equations



Coordinate time ≥ Space-time interval ≥ Proper time
$$ Δt≥Δs≥Δr $$

Measuring proper time in an inertial reference frame between events A and B, and V is constant:
$$\Delta r_{AB} = \sqrt{1-v^2}\Delta t_{AB} \approx (1-0.5 v^2) \Delta t_{AB}$$ (Using Binomial approx.)

The Attempt at a Solution


my answers:
A- Alice, B- Brian, CD- Cara and Dave

(a)
The Shortest time-
A and B measures Δr (which is defferent in each RF)
Because B measures Δs also, and A is moving (therefore time moves slower for her) – Alice measures the shortest time.

The longest time- B and CD measures Δt
B measures Δs which is the shortest possible Δt.
Since CD don’t measure Δs they measure the longest time interval.

*** Question: Is it correct to state that CD don't measure Δs ? and therefore they measure the longest time interval? ***

(b)
Converting the speed the SR units:
$$ 60 m/s = 20*10^{-8} (SR) $$
$$ Δt_{AB-Brian} = 100_s $$
Manipulating the equation by subtracting Δt in order to get the answer:
$$\Delta r_{AB- Alice} - Δt_{AB} = (1-0.5 v^2) \Delta t_{AB} - Δt_{AB} = (1-0.5 v^2-1) \Delta t_{AB} = -0.5 Δt_{AB} $$
$$ -0.5 * (20*10^{-8})^2 *100_s = -2*10^{-12}_s $$
*** No questions about this section ***

(c)
Train speed $$ 30 m/s = 10^{-7} (SR) $$
Meaning that in CD reference frame B is moving in a constant speed of 10^{-7}

*** Question: I thought I need to use the same equation like in section b, but CD don't measure proper time, or maybe they are indeed measuring proper time ?
I'm not sure how to combine the given data into the equation ****


Thanks so much in advance !
 
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  • #2
Lito said:
*** Question: Is it correct to state that CD don't measure Δs ? and therefore they measure the longest time interval? ***
Sure, they have different points on the train for the two events.

Why do you write seconds as index?

You can find the time C/D measure by considering the reference frame of the train, for example. B is moving from C to D with a known velocity and for a known proper time.
 
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  • #3
mfb thank you so much !
The second's index was a mistake.

mfb said:
You can find the time C/D measure by considering the reference frame of the train, for example. B is moving from C to D with a known velocity and for a known proper time.

I made the train RF:
qrhroi.jpg


so is it correct to use the equation as follows:

$$ Δr_{EF-Brian} = (1-0.5 v^2) \Delta t_{ED-CD}$$
$$ \Delta t_{ED-CD} = \frac{100 }{(1-0.5*(10^{-7})^2)} = \frac {100}{(1-5*10^{-15})} $$

And therefore CD Coordinate time is bigger.

Or maybe should I use the metric equation (but I am not sure its possible due to the data) ?
 
  • #4
Lito said:
And therefore CD Coordinate time is bigger.
Correct.
 
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Related to Special relativity -- Proper time

What is special relativity?

Special relativity is a theory in physics that explains how objects move at high speeds and how time and space are affected by these movements. It was developed by Albert Einstein in the early 20th century.

What is proper time in special relativity?

Proper time is the time measured by an observer who is in the same frame of reference as the moving object. It is the time experienced by the object itself and is always shorter than the time measured by an observer in a different frame of reference.

How is proper time related to time dilation?

Proper time is closely related to time dilation, which is the phenomenon where time passes slower for objects that are moving at high speeds. The faster an object is moving, the more time dilation occurs, and the shorter its proper time becomes.

Can proper time be measured?

Yes, proper time can be measured, but it requires precise instruments and accurate measurements. It is also important to take into account the relative velocities and frames of reference of the observer and the object being measured.

How does proper time affect our daily lives?

Proper time and the principles of special relativity are most noticeable in extreme conditions, such as objects moving at very high speeds or in space. However, even in our daily lives, the effects of time dilation can be observed in technologies such as GPS systems which need to account for the time difference between satellites and receivers due to their relative speeds.

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