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The following statement combined with the application of time dilation principle confused me on how proper length and time are measured and what they represent.

"The speed at which electrons traverse the 3-km beam line of the SLAC National Accelerator Laboratory is slower than c by less than 1 cm/s. As measured in the reference frame of such an electron, the beam line (which extends from the top to the bottom of this photograph) is only about 15 cm long!"

According to the definition of proper length, we know that it is the length measured from the reference frame in which the object is at rest. In this case it makes sense to observe the beam line as shorter from the electron's reference frame, as the electron is at rest in its reference frame, and according to it, the beam line moves with the speed of the electron. On the other hand, we can apply the same reasoning when making observations from the reference frame of the beam line.

However, when a person measures the time it takes an electron to move a certain distance, this is not considered as proper time. Based on the given definitions, proper time is the time between two events measured in a frame where the event occur at the same location. When length is predicted to be relative in Special Relativity, how do we determine the frame where two events occur at the same location?

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  • $\begingroup$ Your confusion comes from two sources. One is using a poor definition of proper time. The other is misinterpreting this definition. According to this definition, proper time is the time between two events happening at the same location in that frame of reference (in which the events are at the same location). You are correct that length is relative and this is exactly the reason for the second part of the definition to specify relative to what the distance is measured. Just use a better definition, such as your proper time is the time measured by your own wristwatch, no matter how you move. $\endgroup$ – safesphere Nov 18 '17 at 0:21
  • $\begingroup$ Why not? Can you provide a reference or logical argument to the contrary? The time measured using your own wristwatch is indeed your proper time not only in SR, but in GR as well. $\endgroup$ – safesphere Nov 18 '17 at 23:44
  • $\begingroup$ According to your reasoning time measured in all inertial reference frames is proper time. Let's say you measure time in the inertial frame of reference on Earth and that of a moving spaceship. If proper time (to) was the time you measured on Earth, according to the time dilation equation t > to , therefore internal clock of a person moving with the spaceship ticks faster than that of a person on Earth, and you would have a contradiction @safesphere $\endgroup$ – user171347 Nov 18 '17 at 23:52
  • $\begingroup$ No, as I mentioned and highlighted above, time measured in any frame (inertial or not) is the proper time of the observer in this frame only. In your example, time measured on the Earth is the proper time of the Earth. The proper time of the spaceship is the time measured by the wristwatch of the astronaut in this spaceship. $\endgroup$ – safesphere Nov 19 '17 at 0:23
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A spacetime diagram might be the clearest way to answer the question. I've drawn it on rotated graph paper so that the intervals ("tickmarks") are easier to see.

For my example, I am using v=(4/5)c (so that the arithmetic is easier).
So, $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=5/3$.

Suppose an electron travels from one side of the beam line to the other. (Here, it's 20 units long in the lab frame. Since the beam line is at rest in the lab frame, the proper length of the beam line is 20 units.)

So, there are two events O (leave the starting wall) and P (arrive at the ending wall).

Join these [timelike-related] events by a line-segment. That segment is the worldline of the inertial observer Elle who experienced both O and P, and that observer would assign position coordinates [in her frame] $x^{Elle}_O=0$ and $x^{Elle}_P=0$.

According to the lab frame,
the velocity of that frame of reference is the slope $v^{Lab}_{OP}=\displaystyle\frac{x^{Lab}_P-x^{Lab}_O}{t^{Lab}_P-t^{Lab}_O}=\frac{20}{25}=4/5$.

Rotated Graph paper

From Elle's frame of reference,...
when she starts at event O, the end of the beam line is at $x^{Elle}_{O'}=12$ units.
(Note: $12=20/\gamma=20/(5/3)$... length contraction)
When she arrives at the end of the beam at event P, the start of the beam line is at $x^{Elle}_{P'}=-12$ units.
Elle says the trip took $15$ units of time [the proper time* along her worldline]
(Note: $25=\gamma * 15= (5/3)15$... time dilation),
and she concludes
the beam line had velocity $v^{Elle}_{OP'}= \displaystyle\frac{x^{Elle}_{P'}-x^{Lab}_O}{t^{Lab}_{P'}-t^{Lab}_O}=\frac{-12}{15}=-4/5$.

(*Note: "proper time" is associated with a worldline [not just a pair of events]; it is like the arc-length along a curve from one point to another.)

UPDATE:

Using WolframAlpha with
(1-tanh(arccosh(3000/0.15)))*(speed of light) ,
I get (c - 37.4 cm/s) as the velocity to get 15 cm.

WolframAlpha: 3km*sqrt(1-((speed+of+light-37.4*cm/s)/(speed+of+light))^2) gives 15 cm.
With (c - (1 cm/s) ), I get 2.45 cm.
With (c - (1 ft/s) ), I get 13.53.cm. They probably meant "1 ft/s slower than c".

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If a spaceship moves from star A just at the moment when the star explodes in a supernova and reaches star B just at the moment when it explodes as well, both explosions happened at the same location in the spaceship's frame (namely it its very position).

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how do we determine the frame where two events occur at the same location?

The two events have the same spatial coordinates.

For example, consider a clock at rest in an inertial coordinate system. The event that the clock reads $t=0$ and the event that the clock reads $t=1$ are two events that have the same spatial coordinates in this system. Thus, the proper time between the two events is just the elapsed coordinate time as given by the stationary clock: $\tau = \Delta t = 1$.

From any other inertial coordinate system, relatively moving with respect to this one, the elapsed coordinate time between the two events is greater: $\Delta t' > \Delta t$

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