Proper Time and Time Dilatation

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SUMMARY

The discussion clarifies the concept of proper time in the context of special relativity, specifically using the Minkowski space framework. Proper time, denoted as ##d\tau##, is defined as the time measured by a clock moving with an observer, and is calculated using the formula ##d\tau = \frac{dt}{\gamma}##, where ##\gamma## is the Lorentz factor. The confusion arises from the relationship between proper time and coordinate time, as proper time is always less than coordinate time due to time dilation effects. The discussion emphasizes that proper time is frame-dependent and is only equal to coordinate time when the observer is at rest.

PREREQUISITES
  • Understanding of special relativity concepts
  • Familiarity with Minkowski space
  • Knowledge of the Lorentz factor (##\gamma##)
  • Basic grasp of time dilation principles
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  • Study the derivation of the Lorentz transformations
  • Explore the implications of time dilation in practical scenarios
  • Learn about the twin paradox and its resolution
  • Investigate the role of proper time in general relativity
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Students of physics, particularly those studying special relativity, educators explaining time dilation concepts, and anyone interested in the mathematical foundations of relativistic physics.

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Homework Statement
I am a little confused with the concept of proper time
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I am a little confused with the concept of proper time: Using the invariance of the distance square in the Minkowski space, we can get the expression ##d \tau = \frac{d t}{ \gamma}## Now the problem:

Aren't the proper time the time measured by a moving clock? That is, if i am moving with speed v and carries with me a clock, "my" proper time is the time elapsed in my clock, right?

But, to the best of my knowledge, the time elapsed in a moving frame shouldn't be dilated? The so called time dilatation, so why the proper time in the expression above is lesser than the coordinate time?

The expression above has a special frame?
 
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In your rest frame, you are not moving, and thus ##\gamma = 1## and ##d\tau = dt##. There is no such thing as an inherently "moving frame", frames can only move relative to each other and you have to be careful in how you reference this.

In the frame where you move with speed ##v##, you indeed have ##d\tau = dt/\gamma## with ##dt## being the coordinate time differential in that frame.
 

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