$dt/ d\tau $ = Lorentz factor

Imagine inertial frame A is situated on a spaceship moving near the speed of light relative to inertial frame B, which is an astronaut in space. The Lorentz factor = $\frac {1 }{\sqrt{(1 - (v/c)^2)}}$ = some number greater than 1 (let's say it happens to be 5 in this case for concreteness).

If the astronaut peers into the window of the spacecraft zooming by him, he will observe clocks in the spaceship to be ticking at a factor of 5 slower than his own wrist watch. So, it makes intuitive sense that $dt/ d\tau $ is 5. The rate that the astronaut appears to be measuring his clock to tick is 5 times faster than the rate he observes the clock ticking on the spacecraft.

Is this a correct interpretation?

  • $\begingroup$ Yes, with the caveat that you have to be careful about what you mean by "observe". Each tick of the passing ship's clock comes to our astronaut's ship from a different distance, and hence with a different time delay due to the fact that information travels at a finite speed. The correct statement is that after correcting for these time delays, the astronaut will say that his own clock ticks (say) five times for each tick of the passing clock. $\endgroup$ – WillO Mar 13 '17 at 2:15
  • $\begingroup$ Yes, it's correct. WillO was talking about retarded positions. If we disregard that your interpretation is correct. $d\tau$ is the proper time (the time for the moving object), $dt$ is the observer's time, so $d\tau = dt \sqrt{1 - \frac{v^2}{c^2}}$. This is for $v$ when the object is moving on the same axis with the observer's axis. $\endgroup$ – Mihai B. Mar 14 '17 at 21:28

By my reading of your question, I would say no. With the caveats suggested earlier, I would agree and say yes.

When describing what one "sees" with one's eyes, the Doppler factor $k$ is the important factor that captures the key features (not the Lorentz factor $\gamma$). The Lorentz factor $\gamma$ is only part of the story.

The following spacetime diagram may help

spacetime diagram

Two inertial observers, Alice (along AD) and Bob (along HL), will meet momentarily at event O. In this diagram, Alice is at rest and Bob's velocity is PJ/OP=3/5 according to Alice. ( 3/5 is chosen to make the arithmetic easier and to make things easier to draw on rotated graph paper. )

The Lorentz factor (according to Alice) is the ratio OP/OJ=5/4, where Alice regards P and J as simultaneous. (By a radar experiment to measure J, Alice must send a signal at S and receive at C. Then the midpoint-event P is simultaneous with J.) You can check that for $v=3/5$, then $\gamma=1/\sqrt{1-v^2}=5/4$.

[Trigonometrically, the Lorentz factor $\gamma$ is the hyperbolic cosine of the rapidity angle $\theta$ between the worldlines: (ADJACENT/HYPOTENUSE)=OP/OJ=5/4. The velocity is the hyperbolic tangent: (OPPOSITE/ADJACENT)=PJ/OP=3/5.]

What Alice "sees" (with her eyes by looking out the window) are the images of Bob's clock, as marked by the dashed lightlike lines sent from Bob's periodically ticking clock [at events H, I, O, J, K]. When Bob is approaching, Bob's clock visually appears to be running fast according to Alice: HI/AB=4/2=2. Think of Alice watching the incoming images of Bob's clock while watching her own clock. Here's a TV transmission interpretation. If each diamond on a worldline represents 15-minutes, a 60-minute broadcast by Bob is viewed by Alice [once received] in 30 minutes. After the momentary meeting and Bob recedes, Bob's clock visually appears to be running slowly according to Alice: JK/CD=4/8=(1/2). Bob's 60-minute broadcast now takes Alice 120 minutes to view. These ratios HI/AB and JK/CD are associated with the Doppler factor $k=\sqrt{\frac{1+v}{1-v}}=2$.

[Trigonometrically, $k=\exp\theta=(\cosh\theta+\sinh\theta)=\cosh\theta(1+\tanh\theta)=\gamma(1+v)$].

(The same diagram can show that Bob would observe the same thing about Alice by drawing a different set of lightlike signals sent from a periodic sequence of Alice's ticks.)


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