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In his book Simply Einstein Richard Wolfson describes time dilation this way: "The time between two events is shortest when measured in a reference frame where the two events occur at the same place."

I think I have a counterexample to this but I am not sure. Suppose Battlestar Galactica is flying over the earth. In the front of BG a flash of light is emitted. Then, a short time later, at the back of BG a flash of light is emitted. As it happens, BG has moved exactly one length in the interval, so, from our standpoint on Earth, those flashes of light have occurred in the same place. Doesn't special relativity say that, if the time interval between the two events measured on BG is one second, the time interval between those events measured on Earth will be longer than one second?

My original example involved two clocks that are (at least from the standpoint of BG) synchronized with one another, one in the front, one in the back. I have no idea if it makes any difference.

Anyway, is Wolfson's formulation correct? Is my example a counterexample to his statement?

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Wolfson is correct, but the flaw in your example is a bit subtle. Let me recap first.

In the ship's frame, there are synchronized clocks at the front and back. A light flash is emitted at the front when the front clock reads $t = 0\ \mathrm{s}$, and a light flash is emitted at the back when the back block reads $t = 1\ \mathrm{s}$. So in the ship frame, $\Delta t = 1\ \mathrm{s}$.

In the Earth's frame, let the ship have length $L$ and velocity $v$, so $L/v = 1\ \mathrm{s}$. You argue that by time dilation, the clocks will run slower by a factor of $\gamma$. But by the relativity of simultaneity, the clocks are not synchronized in the Earth's frame; instead, the clock in the back is ahead by $Lv/c^2$. Then the total time interval is $$\Delta t_{E} = \gamma(1 - Lv/c^2) = \frac{1-v^2/c^2}{\sqrt{1-v^2/c^2}} = \sqrt{1-v^2/c^2}\ \mathrm{s}.$$ This is less than one second, as expected.

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  • $\begingroup$ Thank you, that explains it very well, much appreciated. $\endgroup$ – Mark D Jun 9 '16 at 0:18
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Doesn't special relativity say that, if the time interval between the two events measured on BG is one second, the time interval between those events measured on Earth will be longer than one second?

No, SR doesn't say that; SR says that moving clocks run slower than stationary clocks (relative to some reference frame).

Is my example a counterexample to his statement?

In your thought experiment, the times of emission of the two light pulses can be measured by one clock stationary relative to the (unprimed) Earth's reference frame and co-located with the emission events.

According to the Lorentz transformation, the elapsed coordinate time in the (primed) BG reference frame is

$$\Delta t' ~=~ \gamma\left(\Delta t - \frac{v}{c^2}\Delta x\right) ~=~ \gamma \Delta t $$

since $\Delta x = 0$ (the events are co-located in the unprimed coordinates).

Since $\gamma \ge 1$, it follows that $\Delta t$, "the time interval between those events measured on Earth" is the shortest.

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  • $\begingroup$ Thank you, I see how the result follows from the equation; I appreciate your helpful answer. $\endgroup$ – Mark D Jun 9 '16 at 0:20

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