Do relativistic events need to match if accounted for time dilation and length contraction?

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): $$ct-x=0$$ $$t = \frac{t'}{\sqrt{1-v^2/c^2}}=1s / 0.5 = 2s$$ $$c \cdot 2s - x \neq 0$$

• Comments are not for extended discussion; this conversation has been moved to chat. – David Z Sep 22 '15 at 7:21

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.

• If the pole and flash is in the negative x direction then it is the same 1s in A, but of course a fraction of a second in B. So the answer is NO, events do not match even if we account for time dilation and length contraction? – xnor Sep 20 '15 at 19:20
• My point is that your choice of choosing a positive-x direction for your calculation of time dilation was an arbitrary choice. Why not choose the negative-x direction or a direction at a 30-degree angle off of the positive-x direction? What's so special about the positive-x direction that you would expect the total transit time of the light pulse as seen by B would directly give you the time dilation factor? As for your question, if you take everything into account - including time dilation and length contraction - then the observed events should agree from both perspectives. – Samuel Weir Sep 20 '15 at 19:28
• If the choice is arbitrary then it doesn't matter. You said the observed events should agree ... so where did I go wrong given my arbitrary choices? – xnor Sep 20 '15 at 19:37
• I did a quick calculation of your setup and found that the time interval observed by B is gamma*(1+v) where gamma is the time dilation factor that you want. If you send a light pulse in the negative-x direction instead, the time interval observed by B is gamma*(1-v). The "+v" and "-v" terms are due to the fact that the end of the pole is moving away or towards the light pulse, respectively. If you send the light pulse in the y-direction, you should find that the time interval observed by B is just gamma, which is the number you wanted. – Samuel Weir Sep 20 '15 at 19:44
• The relativity of simultaniety is a big deal here, it means that the observers in different frames are in general not comparing the same things. One thing using invariants like the interval does is force all observers to agree on what events they are comparing. – dmckee Sep 20 '15 at 19:45

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).

• With s=0 isn't this exactly 0=ct-x like in my example? – xnor Sep 20 '15 at 19:51
• Sure $0=(ct)^2 - (x)^2$ is satisfied by any $c$ and $t$ satisfying $0=ct-x$, but that is only one frame. After transformation to the primed frame the two conditions are different. Try writing them out explicitly. – dmckee Sep 20 '15 at 19:57
• I edited my question and added more detailed calculations at the end. It does not match. Where is the mistake? – xnor Sep 20 '15 at 20:22
• dmckee: "There is a value related to the two events that all observers can agree on. The "interval" $(\Delta s)^2$ [...]" -- I have two questions regarding your suggested notation for the "interval". 1. If the value is related to (or depending on) two specific events, then should its notation not include the names of these two events explicitly as two arguments? 2. By using the pair of parentheses in your notation you're suggesting that "$\Delta s$" denotes by itself a meaningful quantity whose square is the "interval". How do you call this quantity "$\Delta s$" itself? – user12262 Sep 26 '15 at 13:43
• @user12262 For time-like separated event $\Delta s$ is the proper time (if you use $c=1$ units). Indeed Golsdtein denotes the equivalent differential quantity $\mathrm{d}\tau$. At least some authors (Landau and Lifshitz for instance) call $\mathrm{d}s$ the interval; see physics.stackexchange.com/questions/114958/…. The choice to use parenthesis here is merely for consistent appearance on both side of the expression when I want to emphasize that the RHS involves squares of differences rather than differences of squares. – dmckee Sep 26 '15 at 20:58

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 $$t=\frac{t'+\frac{vx'}{c^2}}{\sqrt{1-v^2/c^2}}=(1s+0.866c \cdot 1(1c \cdot 1s)/c^2)/0.5=3.732s$$

• "In B the light catches up with the contracted (0.5 Ls long) pole with a relative speed of 1-0.866c=0.134c" Light always travels at $c$. – bright magus Sep 23 '15 at 1:51
• As to the one-way or two frame lightspeed measurement you might want to read this and this. – bright magus Sep 23 '15 at 2:35
• @brightmagus "Light always travels at c." I didn't say otherwise. In B it is A that moves at 0.866c. Light moves at 1c. So B would see the light catching up with the end of A's pole with 0.134c. – xnor Sep 23 '15 at 15:44