How can moving observer explain non-simultaneity?

Question

This question comes from the classic "passing train" scenario you see in almost all textbooks on relativity.

Summary: A guy on a train passes a guy on a platform. The moment they pass two bombs (or lightning bolts) explode at the same distance from the platform.

The guy on the platform says they exploded simultaneously, but the traveling guy says no -> both are right

Just to clarify the question we give the person on the platform (Alice) an optical trigger (a lamp), and fiber optic cables of length L connected to the two bombs, rigged as before.

We get the same result: Alice says the bombs exploded simultaneously, and Bob on the train says no. This is all in accordance with standard physics.

Now we ask Bob: How is it physically possible for Alice to detonate the two bombs at different times?

How can Bob explain the non-simultaneity in his reference frame?

Of course he can't say light traveled at different speeds to the left and right of Alice.

Notes:

Bob sees a shorter L than Alice, but both agree the two bombs are equidistant from the platform.

I'm aware of similar posts, but none ask Bob to explain himself, which is the point here.

Answer 1

Bob knows the light travels toward the two bombs at the same speed. He also knows the bombs are equidistant from the detonator when it is pressed. However, in his frame of reference, the bombs are moving. One bomb is moving toward the pulse of light, while the other one is moving away from the pulse, requiring the light to "chase" it. Therefore, the bomb moving toward the pulse detonates before the one that is running away from the pulse.

Answer 2

Introducing all the complexity of the detonator etc is an unnecessary elaboration. Try this instead. At the moment Bob passes Alice, one or other of them flashes a light- it doesn't matter which. The light heads off in all directions. After a second, Bob says the light is 'now' a light second away from him to either side, while Alice says the light is 'now' a light second away from her to either side. Since Bob and Alice are no longer standing in the same place, they can only both be correct if they disagree on what time is 'now' a light second away.

If you apply that same thinking to the set-up you described, in Bob's frame, the light pulse heads off along the two lengths of optical fibre in different directions. After some time, the light must have gone equal distances along the fibre from Bob. But equal distances along the fibre from Bob are not the same points as equal distances along the fibre from Alice, since Bob and Alice are some distance apart. Bob therefore sees the bomb he is approaching detonate earlier than the bomb he is leaving further behind.

I think your mistake is to assume that Bob sees the light moving at equal speeds along the optical fibre from Alice- he doesn't, as she's moving. Bob sees the light moving at equal speeds along the fibre from him. Given that, the light has to travel further from him to reach one bomb than the other, so in his frame they go off at different times.

Answer 3

This is a good question, and part of the answer is that Relativity is not obliged to provide these details. In particle accelerators we see particles decay at far slower rates than the laws of quantum mechanics dictate. We also observe the line spectra of distant stars that is redshifted, whose wavelengths correspond to none of the known elements in existence – but they do correspond to a shifted version of hydrogen, or shifted helium, or shifted carbon. Thus the observed spectrum does not demand new physics or new elements, but rather the physics we know in a moving reference frame.

So while Alice's and Bob's measurements may be equally valid, and equally real, they do not necessarily have different but equally valid physical explanations for the cause of the measurements. Both will use relativistic physics to deduce and agree on what happened.

Answer 4

Number one rule: if you're still confused: do the Lorentz transform.

Rule No 2: is pick good coordinates. Start in $S$, at rest with Alice. Introduce a moving ($\beta$) Bob in $S'$.

Make sure their origins overlap, or else you can do LTs. We have Alice's world-line as $(t, 0)_S$ and Bob's is $(t', 0)_{S'}$.

Now $t=t'=0$ occur when Alice and Bob are at the same event. This is when Alice sees the bombs explode at:

$$ E_{\pm} = (0, \pm \frac L{2\gamma})_S $$

Note we cheated a bit, because we knew length contraction was a thing.

This means she fires her light-trigger at:

$$E_0 = (-\frac L{2\gamma}, 0)_S $$

So for her: she hits the button, two equidistant bombs get the signal simultaneously--done. There just happens to be a train passing by such that the bombs explode at the front and the back, exactly.

Now we LT from $S\rightarrow S'$, and figure out Bob's 3+1.

Note: we do, a priori, not need to thing about Alice moving, nor the bombs moving. We are just looking at events (this is why I prefer lightning strikes--as they have no time duration. That the bombs exist in a frame, it makes our Galilean sensibilities think that the frame is special--it's not. Events don't move, they just exists at single time and point).

Anyway, for Bob:

$$t'_0 = \gamma\big(-\frac L{2\gamma} - 0\big) = -\frac L{2\gamma}$$

$$x'_0 = \gamma\big(0-\beta\frac{-L}{2\gamma}\big)=\beta \frac L{2\gamma}$$

So:

$$ E_0 = \Big( -\frac L{2\gamma}, \beta \frac L{2\gamma} \Big)_{S'}$$

Meanwhile, the explosions transform to:

$$ E_{\pm} = \Big(\mp \frac{\beta L} 2, \pm \frac L 2 \Big )_{S'} $$

So the forward explosions occurs 1st, at the front of the train, and then later, the caboose blows up.

You can verify that in all frames:

$$(E_{\pm}-E_0)^2 = 0 $$

(they are light-like) and that:

$$(E_+-E_-)^2 = -\big(\frac L{\gamma} \big)^2$$

Final Comment: note that in the OP, you introduced fiber optics, and then complained that the speeds were different. One answer pointed out that this was superfluous--and it is: UNLESS $v \lt c$.

You should redo this problem where the propagation speed of the Alice's signal is sub-light: $v = c/n$. With that, Alice triggers earlier, so that:

$$E_0 = (-\frac {nL}{2\gamma}, 0)_S $$

while $E_{\pm}$ are unchanged. In $S'$, Bob should see the two signals propagating at different speeds:

$$ v_{\pm} = \frac{\beta \mp \frac 1 n}{1 \mp \frac \beta n}$$

Answer 5

Can I get concrete?

Suppose Bob is moving at 0.5 c relative to Alice. The train goes West to East.

Alice, in her frame of reference, is 1 light second away from each Bomb, due West and East.

Alice pushes the button, and 1 light second later the Bombs go off. Alice sees the light from the Bombs arrive 2 seconds after she pushes the button, at the same time.

As Alice pushes the button, Bob's train drives through Alice in the direction of the Bombs. (Ie, Bob's position is identical to Alice's at the moment the button is pushed).

The distance between Alice and the Bombs is length contracted from Bob's perspective down to $\frac{\sqrt{3}}{2}$ light-seconds.

From Bob's perspective, the Bombs are moving at 0.5 c East to West. The light travels at c, but the distance between the light pulse and the East Bomb decreases at a rate of 1.5c and the West Bomb at 0.5c; light takes time to move, and in that time the Bombs move.

It hits the East Bomb after $\frac{1}{\sqrt{3}}$ seconds, the Bomb goes off, and Bob sees the East Bomb explode at $\frac{2}{\sqrt{3}}$ seconds after the button is pushed (about 1.15 seconds).

The signal hits the West Bomb after $\sqrt{3}$ seconds, and Bob sees the Bomb go off $2\sqrt{3}$ seconds after the button is pushed; 3 times slower, about 3.46 seconds after the button is pushed.

If the signal wire "lights up" as it transmits the signal, Bob can watch the light move at c along the wire (with it taking distance/c/2 seconds to see it at a specific distance away, as it is moving away from Bob). This distance value isn't the distance along the wire in Alice's frame; it is the location in Bob's frame, where the wire is flying at 0.5 c as well.

To Bob, the signal travels away from him at the same speed in both directions. But while the signal travels, the Bombs on one side move closer while the Bombs on the other move further away. The one that moves closer as the signal travels explode when the signal reaches them.

Here is Bob and two bombs marked X. Each character is a fraction of a light second. Each line is a fraction of a second. The signal, visible along the cord, is the .; when it reaches a bomb it produces an explosion *.

This represents what Bob sees at each second.

X          B          X
 X       . B .         X
  X    .   B   .        X
   X .     B     .       X
    *      B       .      X
    ^!^    B         .     X
     _!_   B           .    X
       !   B             .   X
        _  B               .  X
         _ B                 . X
          _B                   .X
           B                     *

this image isn't quite right, because Bob actually sees the bomb "moving towards" him as being bit closer than the one "moving away". But I drew it with the two Bombs appear into be equal distance.

At a given time, the light coming from the Bomb "moving towards" Bob that hits Bob is from further in the past than the one "moving away" in a sense.

Answer 6

For any two space-like non simultaneous events in a frame S, it is possible to find a frame S' such that they are simultaneous for S'.

For example: now on Earth and 1 min from now on Mars. For a ship at a suitable speed passing by here going to Mars these 2 events are simultaneous.

Let's examine Bob's frame. It happens that Alice's frame has the precise relative speed such that these non simultaneous events (in his frame), are simultaneous on her frame.

Answer 7

The problem of information propagation can be misleading. In order to get a better understanding, you can remove it by using two infinite trains with an infinity of regularly spaced clones of Alice and Bob on each.

All Alice clones being unmoving with one another, you can synchronize them all by sending photons back and forth between them, calculating the correction as half the time of all round travel. You can do as well with bob clones.

This makes as if you had two infinite rulers watching each other, and they both not only have distance ticks but also clock ticks for which they can compare their position and their time with the position and time of the tick in front of them in the other train.

Then, you can apply the Lorentz transform to see exactly what each clone of Alice will see in front of her and vice versa. You will see that All clone of alice will watch a different delta t between her clock and the clock of the clone of Bob in front of her, But Also the distance between de Clones of Bob are not the same as the distance between her clones. This is explains why they both have the impression than the time is slower on the other train.

In fact, one Alice going to the past of Bob's train, she has the feeling that the bob's clone passing in front of her have a slower time, but when all the alice's clones compare the time the recorded from the same clone of Bob, they will see that his own time was in reality faster because The same Bob Also went to Alice from the past giving them the illusion that his own time went faster.

Answer 8

I appreciate the input & thanks for correcting me.

I see a lot of new answers and will study them in turn. Pretty sure there are multiple valid ones. Also very happy about this post being re-opened, which lets me add my own answer, for completeness.

To be overly picky the question was "What would Bob say?"

which is tricky, because it depends on how much Bob knows.

I like the idea (from answer #2) that Bob knows math and can work it out with measurements.

He can tell when the light from the triggers/bombs are shifted unequally (red vs blue).

He knows there is only one unique frame of reference that sees the explosions simultaneously, and with identical wavelengths (no doppler).

Bob knows he is not in that special frame.

Bob can now give the right answer:

"It is physically impossible to trigger the bombs at different times in Alice's/bombs' rest-frame. The bombs did indeed explode simultaneously when I was midway between them, but my speed carried me away from the mid-point, which caused (doppler shifted) light to hit me at different times."