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To explain the question let me give you a short example. In the scenario there are two references frames A and B.

A consists of a x'=1 Ls (lightsecond) long pole in the positive x direction. At t=0 a flash is generated at its origin. 1s later the flash reaches the end of the pole.

B sees A moving with v=0.866c in the positive x direction. Due to length contraction, A's pole appears to only be 0.5 Ls long.

In B, 3.731s after A generated the flash the flash reaches the end of the pole, because: $$x-ct=0 \quad with \quad x=v \cdot t+x'\sqrt{1-v^2/c^2}$$ $$(v \cdot t+x'\sqrt{1-v^2/c^2})-ct=0$$ $$(0.866c \cdot 3.731s+0.5)-3.731s \cdot c=0$$

So the flash reaches the pole's end after 1s in A. But from B's point of view it takes 3.371s.

Wouldn't this require a time dilation factor of 3.371? But the actual factor is $$\frac{1}{\sqrt{1-v^2/c^2}}=2$$

Based on a suggestion in a comment let me write out the problem more detailed:

For A: $$ct'-x'=0$$ $$t'=1s$$ $$c \cdot 1s - 1 (1c \cdot 1s) = 0$$

For B (for the formula of x see above): $$ct-x=0$$ $$t = \frac{t'}{\sqrt{1-v^2/c^2}}=1s / 0.5 = 2s$$ $$c \cdot 2s - x \neq 0$$

xnor
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3 Answers3

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I think that your calculation is correct in that it would take $3.73$ seconds for the light pulse to reach the end of the pole according to B's perspective. However, that number is not the time dilation factor. For $v=0.866$, the factor should be equal to about $2 (= \frac{1}{\sqrt{1-(v/c)^2}})$.

So where did your reasoning go wrong? I think that the problem is your selection of a pole and then considering a light pulse in the positive x-direction. Why did you choose the positive x-direction? Why not send a light pulse in the negative x-direction instead? It should give the same answer if what you calculated was really the time dilation factor, right? But it doesn't. If you send a light pulse in the negative x-direction, you should that it takes less than 1 second for the light pulse to reach the end of the pole from B's perspective. Does that mean the time dilation factor is both less than 1 (for light pulses in the negative direction) and also greater than one (for light pulses in the positive direction)? Obviously, that's nonsense.

You can't use light pulses in either the positive or negative direction for your purpose because in addition to time dilation effects there is also a time contribution due to the fact that the far end of the pole is moving away (or moving towards) the light pulse. Try setting up your calculation so that instead you consider a light pulse moving in the y-direction (i.e., perpendicular to the direction of the motion of the light pulse and the observer B). That should give you the correct time dilation factor if you repeat your calculation.

SchrodingersCat
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There is a value related to the two events that all observers can agree on. The "interval" $(\Delta s)^2$ computed as $$ (\Delta s)^2 = (c\Delta t)^2 - (\Delta\vec{r})^2 \,,$$ is the same in every frame of reference.

This is generally the first Lorentz scalar introduced in a basic development of special relativity, and you will notice that for time-like separated particles (that is, when $(\Delta s)^2$ is positive) it is equal to the proper time experienced by an inertial observer between the two events.

The second Lorentz scalar introduced is often the mass $$ (mc^2)^2 = E^2 - (\vec{p}c)^2 \,.$$ In modern parlance this "invariant mass" is the only mass of the particle. It corresponds to what was called the "rest mass" in the bad old days.

The squares are important. By way of a metaphor, consider a line segment in $\mathbb{R}^2$ observed in two different reference frames (i.e. two different sets of axis) using the same length units.

Because the line segment is the same it's total length-squared $$ \ell^2 = (\Delta x)^2 + (\Delta y)^2 \,,$$ is the same in both frame, but in general the sum of it's x- and y-projections $s = \Delta x + \Delta y$ is not the same if the primed frame is rotated relative the un-primed frame.

Note that some authors prefer to use the negative of the RHS, this has no effect on the constancy of the value, but does change the sense of comparisons (i.e. using that convention the interval of time-like separated events is negative rather than positive).

dmckee --- ex-moderator kitten
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I think the problem is this:

In A light would move 1s to the end of the 1Ls long pole, e.g. be reflected by a mirror and take 1s back to A.

In B the light catches up with the contracted (0.5 Ls long) pole with a relative speed of 1-0.866c=0.134c, so takes 3.731s. Then it is reflected and approaches A with a relative speed of 1.866c, so that only takes 0.268s back to A.

Sum that up and we get 4s in B, 2s in A, exactly as the Lorentz factor says.

There is also a mistake in my calculations: $$t \neq \frac{t'}{\sqrt{1-v^2/c^2}}$$ instead $$t=\frac{t'+\frac{vx'}{c^2}}{\sqrt{1-v^2/c^2}}=(1s+0.866c \cdot 1(1c \cdot 1s)/c^2)/0.5=3.732s$$

xnor
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