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Notes on Relativity and Cosmology - Physics Department, UCSB

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70 CHAPTER 3. EINSTEIN AND INERTIAL FRAMES<br />

t = us<br />

1<br />

1 - v 2/c2<br />

x =0<br />

us<br />

A<br />

x = 0<br />

f<br />

t = f<br />

1<br />

1 - v 2/c2<br />

t = 1 sec<br />

f<br />

B<br />

t<br />

us<br />

= 1 sec<br />

Us<br />

t =0 us<br />

Note that I have marked two special events <strong>on</strong> the diagram. The dot labeled A<br />

is the event where the moving (friend’s) clock ticks t = 1 sec<strong>on</strong>d. It is an event<br />

<strong>on</strong> the friend’s worldline. The dot labeled B is the event where our clock ticks<br />

t = 1 sec<strong>on</strong>d. It is an event <strong>on</strong> our worldline.<br />

3.5.3 Proper Time<br />

We have seen that 3 different observers in different inertial frames measure different<br />

amounts of time to pass between two given events. We might ask if any<br />

<strong>on</strong>e of these is a “better” answer than another? Well, in some sense the answer<br />

must be ‘no,’ since the principle of relativity tells us that all inertial frames<br />

are equally valid. However, there can be a distinguished answer. Note that, if<br />

<strong>on</strong>e inertial observer actually experiences both events, then inertial observers in<br />

other frames have different worldlines <strong>and</strong> so cannot pass through both of these<br />

events. It is useful to use the term proper time between two events to refer to<br />

the time measured by an inertial observer who actually moves between the two<br />

events. Note that this c<strong>on</strong>cept exists <strong>on</strong>ly for timelike separated events.<br />

Let’s work through at a few cases to make sure that we underst<strong>and</strong> what is<br />

going <strong>on</strong>. C<strong>on</strong>sider two observers, red <strong>and</strong> blue. The worldlines of the two<br />

observers intersect at an event, where both set their clocks to read t = 0.<br />

1) Suppose that red sets a firecracker to go off <strong>on</strong> red’s worldline at t red = 1.<br />

At what time does blue find it to go off? Our result (3.3) tells us that<br />

t blue = 1/ √ 1 − (v/c) 2 .<br />

2) Suppose now that blue sets a firecracker to go off <strong>on</strong> blue’s worldline at<br />

t blue = 1. At what time does red find it to go off? From (3.3) we now have<br />

t red = 1/ √ 1 − (v/c) 2 .<br />

3 Still assuming that Einstein’s idea is right.

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