6
$\begingroup$

I'm trying to understand the first postulate in special relativity. My take on that is that the outcome of an experiment does not depend on the observer (inertial frame of reference).

However, since simultaneity is relative, such that the order of two events can be different in different frames of reference, it can affect the outcome of an experiment, or can it?

A simple example is that one measures the light emitted from point A and point B at the same distance with regard to the observer. If the light from A arrives first then he will detonate a bomb, otherwise, he'll do nothing. As the order of the light emitted from the points changes from frame to frame, it is possible to find a pair of frames where the outcome of the measurement is different (in one frame, the bomb explodes while nothing happens in the other).

From my understanding of SR, what I describe in my example can happen, but I don't know if I can consider it as an experiment. If not, then how can I define an experiment.


Here I describe a thought experiment to illustrate my question:

Let there be a cart of length $L_1 = L$ when at rest to the observer. At each end of the cart (A and B), a light source is install. The cart can slide on a rail.

On the same rail, we install a "tray", we name the ends of the tray A' and B' such that vector AB and A'B' point in the same direction. There is a mechanism on the cart, so that when A contacts A', the light source at A will flash, and when B contacts B', the light source at B will flash. Otherwise, no light flashes.

The experiment starts by letting the cart slide toward the tray at a constant velocity with magnitude $|V|$. An observer will determine the order of light flashing when the cart pass through the tray. If the light at A is observed first, then a harmless bomb is detonated. Otherwise, nothing happens.

As length contracts, the observer standing at the center of the cart will see the light at B flash first, while the observer standing at the center of the tray will see the light at A flash first. Thus the bomb is detonated in the frame of the tray.

Below is a sketch of the experiment:

1


My question is about understanding the first postulate of special relativity, why is it considered as homework question? It doesn't make sense.


I want to update that from the posts, I assume the answer to my question is that an "experiment" that measures the order of two events from different frames is not the type of experiment implied in the first postulate.

Although time order does not change the equations of motion, it still seems odd to me because, in the end, the ordering still affects the decisions of the observers, thus making different outcomes. So I still need time to think and digest. Thank you very much for all the comments.

$\endgroup$
16
  • 2
    $\begingroup$ It's best to keep relativity questions within "tradition", and in this case, it's not a tray and cart, it's a train and a train station. It just makes it easier to understand the question, since we all know the train is "moving" and station is "at rest". Moreover, you introduced the stationary frame first and it is unprimed, while the moving system is primed. It's also confusing that the tray/cart has a length with both subscripts 1 and 2, and then you have 2 separate cases in which the observer is moving around, but it looks equivalent to one system from two frames. $\endgroup$
    – JEB
    Commented May 23 at 19:50
  • 2
    $\begingroup$ If you rephrase the question with "Alice at the station of length $L$" with flashes $F_1$ and $F_2$, and Bob is on a train of length $L$ with triggers $T_1$ and $T_2$, and light is emitted when $T_i$ touches $F_i$, and Alice has two events $E_i$ which is the detection of light from $F_i$, with $F_2$ detonating the bomb, while $F_1$ disables it. It would be a lot easier to follow. $\endgroup$
    – JEB
    Commented May 23 at 19:56
  • 2
    $\begingroup$ There is no possible physical mechanism to accomplish this: “An observer will determine the order of light flashing when the cart pass through the tray. If the light at A is observed first, then a harmless bomb is detonated. Otherwise nothing happens.” You can select one specific observer and design a mechanism that works for them, but there is no physical mechanism that will do this for all observers. Thought experiments are not limited by engineering or economic issues, but they must still follow the laws of physics and be logically self consistent $\endgroup$
    – Dale
    Commented May 23 at 21:48
  • 3
    $\begingroup$ A light sensor would be located at some specific place and would only be able to detect when the light arrives to the detector simultaneously. It could not detect anything about whether it was emitted simultaneously or not. $\endgroup$
    – Dale
    Commented May 23 at 23:30
  • 1
    $\begingroup$ You would do better with a discussion forum like physicsforums.com rather than a Q&A forum like this one $\endgroup$
    – Dale
    Commented May 24 at 17:48

8 Answers 8

14
$\begingroup$

In special relativity, an event is a single point in spacetime, i.e. a particular spatial position and a particular moment in time. The two emission events are spacelike separated so there exist inertial frames in which they have different order in time. However, the two arrival events at the cart-observer occur at the same location so they are timelike separated. Therefore, their order must be the same in all inertial frames. The following is a detailed analysis of your situation.

Let $S$ and $S'$ denote the frames of the tray and cart respectively, i.e. $S'$ moves at $v$ with respect to $S$. The two events are the contact of the two ends $A$ and $B$ respectively. So we have $$\Delta x = L \\ \Delta t = \frac{L}{v}\left(1-\frac{1}{\gamma}\right) = T$$ and using the Lorentz transformation, $$\Delta x' = L \\ \Delta t' = \frac{L}{v}\left(\frac{1}{\gamma}-1\right) = -T.$$ So the two contacts indeed have their order reversed as you claimed. (You can confirm this by showing that $\Delta x^2 \gt c^2\Delta t^2$ i.e. they are spacelike separated.) However, this is different from the two events where the light from A and B are respectively received by the cart-observer. In $S'$, the spatial separation between these two events is zero since they are received at the same spatial location. Since, in $S'$, the cart-observer is at rest at the midpoint and the sources are also at rest, the time between these two events is simply the time difference between the times of emission. Therefore, we have $$\Delta x'=0 \\ \Delta t' = -T$$ and using the Lorentz transformation from $S'$ to $S$, $$\Delta x = \gamma v\Delta t' = -\gamma vT \\ \Delta t = \gamma\Delta t' = -\gamma T$$ so the orders in the two frames are the same since $\Delta t$ and $\Delta t'$ have the same sign. This confirms the observations in the first paragraph. The order in which the lights reach the cart-observer is the same in both frames.

A similar analysis can be performed for the reception by the tray-observer, from which we find that the tray-observer will indeed receive the lights in the opposite order from the cart-observer. There is no contradiction because the emission, reception by the cart-observer and reception by the tray observer are three different pairs of events.

To be completely explicit, the receptions of light by the cart-observer is a different pair of events from the receptions of light by the tray-observer. They are not the Lorentz transformation of each other. Indeed, since the two pairs of receptions are both timelike separated with opposite order, they cannot be related by a Lorentz transformation. The Lorentz transformation deals with the coordinates of the same event(s) in different frames. You have two different pairs of reception events. Each pair has its own respective coordinates in $S$ and $S'$. You have confused the physical observer (which is a spatial point i.e. worldline in a frame) with the frame itself.

$\endgroup$
19
  • $\begingroup$ my original question may be unclear, I added a more detail example in my post. $\endgroup$
    – Rekkhan
    Commented May 23 at 12:52
  • 1
    $\begingroup$ @Rekkhan I have updated the answer. $\endgroup$ Commented May 23 at 13:34
  • 1
    $\begingroup$ @Rekkhan The light emitted from B (with velocity $c$ to the left) travels less in $S$ precisely because the observer on the cart is moving to the right with $v$. So the time it takes is $(L/2\gamma)/(c+v)$. And similarly for the light emitted from A, we get $(L/2\gamma)/(c-v)$. This is basic kinematics. The light is traveling at $c$ but its speed relative to the observer, in $S$, is $c\pm v$. You are simply wrong when you assert that the times and distances are the same in $S$. $\endgroup$ Commented May 23 at 23:03
  • 1
    $\begingroup$ @Rekkhan I have already told you three times that it is not true in the frame of the tray (the observer is moving). However you refuse to believe basic kinematics and keep repeating the same wrong things again and again. As such there is no point in continuing this discussion given your level of receptivity. If you insist on using incorrect premises you will very well derive contradictions from them. $\endgroup$ Commented May 24 at 8:42
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Commented May 24 at 9:52
9
$\begingroup$

since simultaneity is relative, such that the order of two events can be different in different frame of reference, it can affect the outcome of an experiment, or can it?

It cannot.

Although the ordering of some events depends on the reference frame, the ordering of causally related events is frame invariant. If $A$ causes $B$ then $A$ will precede $B$ in all frames.

Experiments can depend on the ordering of causally related events, which are timelike or null separated. They cannot depend on the ordering of non-causally related events, which are spacelike separated.

A simple example is that one measure the light emitted from point A and point B at the same distance with regard to the observer. If the light from A arrive first then he will detonate a bomb, otherwise he'll do nothing. As the order of the light emitted from the points change from frame to frame, it is possible to find a pair of frame where the outcome of the measurement is different (in one frame, the bomb explode while nothing happen in the other).

The result of this does not depend on the reference frame. The order of the light emission changes from frame to frame, but not the order of arrival.

$\endgroup$
14
  • $\begingroup$ if the distance between each light source to the observer is the same, then how can the order of emission does not affect the order of arrival? $\endgroup$
    – Rekkhan
    Commented May 23 at 11:51
  • 1
    $\begingroup$ The distance that the light travels is not the same. Don’t confuse the distance between the light source and the detector with the distance the light travels. They are only the same in a reference frame where both emitter and detector are at rest $\endgroup$
    – Dale
    Commented May 23 at 12:03
  • $\begingroup$ Sorry but I'm a bit confused. If a light source emit a photon toward the observer, then the time it take the photon to travel to the observer should not depend on the speed of the light source, which mean if an observer is at the center of AB, then the light from A and the light from B should take the same time to reach the observer, regardless of the relative speed of A and B to the observer. Thus I assume that the order of receiving light and emission should be the same. $\endgroup$
    – Rekkhan
    Commented May 23 at 12:38
  • 1
    $\begingroup$ The distance the light travels is the difference in the position of the emitter at the time of emission vs the position of the receiver at the time of reception $\endgroup$
    – Dale
    Commented May 23 at 16:13
  • 1
    $\begingroup$ So then what is the problem? You can certainly have one bomb explode and another not explode. Two different devices are not required to do the same thing. $\endgroup$
    – Dale
    Commented May 24 at 13:22
8
$\begingroup$

To understand relativity, learn how to draw and interpret spacetime diagrams.

The spacetime diagram is like a position-vs-time graph, only we usually draw time going vertically on the diagram, and space going horizontally. To interpret the diagram in full you need to learn how to find lines of simultaneity, but for this question it is not needed. The central point is that there is just one spacetime and one set of events in spacetime.

Here is a diagram suited to the scenario described in the question: spacetime diagram showing A,B,A',B' and light pulses

Lines A and B show the two ends of the cart going from left to right. Lines A' and B' show the two ends of the tray going from right to left. The orange lines show light pulses. One pair is emitted when A meets A', the other when B meets B'. The person standing in the middle of AB receives the pulse from B before the one from A. You can tell this simply by looking where the line (called worldline) of each light pulse intersects the line (called worldline) of the person. The person standing in the middle of A'B' receives the pulse from A before the one from B.

Ok, that's everything. All we need to do now is state who is carrying the bomb (or the trigger for the bomb). If it is the person on the cart then they will not detonate the bomb. If it is the person on the tray then they will detonate the bomb. If there are two people and they both have a trigger then one will pull the trigger, the other will not. To resolve your puzzle, you see, you have to be very physical at every stage. Do not try to imagine some abstract idea of observer and some abstract bomb. Rather, put everything right there in spacetime where it has its physical existence.

In this example, the surprise from relativity is that the person on the cart deduces that event (B meets B') happens before event (A meets A'), while the person on the tray deduces that these events happen in the other order. But these deductions have no effect whatsoever on the events themselves.

(For completeness, I should mention that the term 'observer', in this area of physics, refers to the whole set of conclusions about all of spacetime which a person in a given state of motion would infer from observations such as the ones we have been discussing.)

$\endgroup$
6
  • 1
    $\begingroup$ I see your point. So an "experiment" to observe time order of 2 events is not a real experiment. $\endgroup$
    – Rekkhan
    Commented May 23 at 23:43
  • 1
    $\begingroup$ Yes that's exactly right. It could be compared to an "experiment" designed to determine whether a chair is to the left or the right of a table. $\endgroup$ Commented May 24 at 13:27
  • $\begingroup$ I get 80% of the analogy. The rest 20% was cause by the arrow of time. Left and right are allowed in space as one can move in any direction in space. But no one move back ward in time. $\endgroup$
    – Rekkhan
    Commented May 24 at 15:13
  • $\begingroup$ @Rekkhan I don't understand your objection - one object can be to the left or right of another, just as one event can precede or follow another in time. Motion is not required for the analogy. $\endgroup$
    – J. Murray
    Commented May 24 at 16:28
  • $\begingroup$ my point is, in real life, one can change his perspective to change the direction of the motion, let's say 180 rotation on the XY plane can change the direction of motion from $v_{x} > 0$ to $v_{x}<0$, but no such transformation can cause an object to go back in time. That is why I'm not very comfortable with the analogy. $\endgroup$
    – Rekkhan
    Commented May 24 at 17:13
6
$\begingroup$

As soon as you write "The bomb is detonated in the frame of the tray", you know you've made a mistake. It's entirely akin to concluding that "The bomb is detonated in Spanish (but not in Portuguese)". Bombs detonate or they don't, entirely independent of what frames (or what languages) you and I choose to use when we talk about them.

So the next step is to find the mistake. My understanding of your thought experiment is that an observer notes the order of two events and then does or does not detonate the bomb depending on that order. Obviously the outcome can depend on the observer.

Here's another version of essentially the same thought experiment: A bomb sits in Chicago. An observer in another city looks at the sky, and if there are rain clouds, he pushes a button that detonates the bomb. Obviously the outcome of the experiment can depend on whether the observer happens to be in Seattle (where the sky is very cloudy) or in Tuscon (where the sky is entirely clear). You would not want to conclude that "the bomb explodes in Seattle but not in Tuscon".

$\endgroup$
4
  • $\begingroup$ hmm, in your case, the observer look at different place (and possibly different time too), then it make no sense to talk about the outcome, as it's not even the same experiment. $\endgroup$
    – Rekkhan
    Commented May 23 at 15:03
  • 2
    $\begingroup$ @Rekkhan : Exactly. The experiment that depends on what one observer sees is not the same as the experiment that depends on what the other observer sees. That is the whole point. $\endgroup$
    – WillO
    Commented May 23 at 15:21
  • $\begingroup$ but what does it have anything to do with my experiment? $\endgroup$
    – Rekkhan
    Commented May 23 at 15:45
  • 2
    $\begingroup$ @Rekkhan "but what does it have anything to do with my experiment?" - there's a subtle way in which what you're thinking of as the "same experiment" is not the same experiment. When you change the observer, the situation should be symmetrical, everything is mirrored, except the labels for the lights - and this changes the experimental procedure in the mirrored version. It's like saying: if I raise the hand on the east side of my body (which happens to be your left hand) detonate the bomb, then turning around 180 degrees and raising your left hand expecting the detonation to happen. $\endgroup$ Commented May 23 at 21:55
5
$\begingroup$

You have two different observers in this experiment, and they will naturally observe different things. Whether the bomb detonates depends on which observer controls the bomb. It does not, however, depend on which frame you calculate in.

To illustrate this, consider not only what each observer will observe, but also what each observer can calculate that the other observer will observe.

Observer O is in the cart and sees the tray moving with length contraction. In the frame of observer O, light B flashes first, followed by light A. The two light flashes have equal distance to cover to reach observer O, so O observes them in that same order - B, then A. Observer O will not trigger the bomb.

Now consider, in the frame of observer O, what observer O' will see. Light B flashes first, since we're calculating this in the frame of observer O. However, observer O' is moving rapidly to the left in that frame, and is already left of light B when it flashes, so observer O' is moving rapidly away from light B when it flashes. The light from source B has to chase after observer O' to catch up. Observer O' is to the right of light A when light A flashes, on the other hand, so observer O' and the light from source A are moving toward each other. As a result, the light from source B has to cover more distance than the light from source A to reach observer O', and the difference in travel time due to this is greater than the difference in time between when the light sources flashed.

Observer O calculates, in their own frame, that observer O' will see the light from source A first, followed by the light from source B, due to the motion of observer O' changing the travel distance of the light to reach observer O'. Consequently, observer O' will trigger the bomb.

Now to repeat this in the frame of observer O'. The tray is stationary, and the cart is moving with length contraction. In the frame of observer O', light A flashes first followed by light B. The two light flashes have equal distance to cover to reach observer O', so O' observes them in that same order - A, then B. Observer O' will trigger the bomb.

Now consider, in the frame of observer O', what observer O will see. Light A flashes first, since we're calculating this in the frame of observer O'. However, observer O is moving rapidly to the right in that frame, and is already right of light A when it flashes, so observer O is moving rapidly away from light A when it flashes. The light from source A has to chase after observer O to catch up. Observer O is to the left of light B when light B flashes, on the other hand, so observer O and the light from source B are moving toward each other. As a result, the light from source A has to cover more distance than the light from source B to reach observer O, and the difference in travel time due to this is greater than the difference in time between when the light sources flashed.

Observer O' calculates, in their own frame, that observer O will see the light from source B first, followed by the light from source A, due to the motion of observer O changing the travel distance of the light to reach observer O. Consequently, observer O will not trigger the bomb.

Both frames agree that observer O will not trigger the bomb, and that observer O' will trigger the bomb. They just have different reasons for their conclusions.

This principle is consistent and reliable: No matter what frames you choose, and no matter how you design your experiment, any difference in the order of events will be compensated for by a difference in the distances and times required for signals from each event to reach the observer. The outcome of the experiment will always be the same in all reference frames.

$\endgroup$
0
5
$\begingroup$

If the light at A is observed first, then a harmless bomb is detonated. Otherwise nothing happens.

Lets say both observers have a detonator switch that is connected by very long wires to the same bomb that is at rest in the rail frame. If either observer sees light A come on before light B, they press their detonator switch and the bomb explodes. The end result is that the observer on the cart always sees light B come on first and does not press his switch, from the point of view of any reference frame. The observer on the tray always sees light A come on first, in any reference frame, so the tray observer always presses the switch and the bomb detonates. Everyone agrees, no matter what reference frame they were in, that the bomb detonates and it was as a result of the tray observer pressing his detonator switch. The tangible results are the same in all reference frames.

It is not immediately obvious why the cart rider does not see light A come on first, when looked at from the point of view of an observer at rest with the tray. The reason is because when light A is activated, the light from the bulb has to catch up with the observer at the centre of the cart before he sees it. Before that signal arrives, the front of the cart has arrived at B on the tray and the light from that bulb is much quicker arriving at the centre of the cart because the cart is heading towards where the light came from.

Likewise, in the rest frame for the cart, the B light comes on first but it does not arrive at the centre of the tray until after the light from the A light has arrived at the centre of the tray, because the light from B has to catch up with centre of the tray that is moving away from the source.

All reference frames agree that:

  1. The cart rider sees light B come on before light A.
  2. The tray rider sees light A come on before light B.
  3. The bomb exploded.
  4. The cart rider did not detonate the bomb.
  5. The tray rider detonated the bomb.

There is nothing to disagree about, except the apparent simultaneity or otherwise of spatially separated events, that has no physical consequences.

Events are always connected by light cones. In any reference frame, the order of cause and effect of related but spatially separated events, is always the same and all observers agree on this order.

$\endgroup$
4
$\begingroup$

Thus the bomb is detonated in the frame of the tray.

This sentence seems like it might be a clue to where the misconception lies, so I'll base my answer on the following interpretation of what you're saying.

From what I inferred, your scenario involves each observer having a separate detonator, where each will decide whether or not to activate their own bomb based on what they see.

If so, then this is what's causing the confusion.

Suppose the observer in the tray frame (the T-observer) detonates the bomb, as you described. Then the cart frame observer (the C-observer) can either: (1) watch the same experiment in the literal sense (i.e., C can observe T perform the experiment and detonate the bomb), or (2) replicate the experiment in their own frame.

If (1), then C must observe the bomb explode (even though the lights flash in a different order for C). The reason for this is that when viewed in the context of spacetime, the 4D arrangement of events is what it is, and does not depend on observer's frames in any way. It's just that each observer sees/experiences different 3D slices of these 4D objects (these slices represent each observer's space "now", and are at an angle to each other, "mixing" time and space directions). This is the sense in which outcomes are observer-independent.

If (2), then when you replicate the experiment with separate (or partially separate) equipment, to call it "the same experiment" you must exactly reproduce all the relevant aspects (things that affect the outcome of the experiment), and make sure that you're observing/measuring analogous things.

In your description, you say the bomb is detonated if A and A' meet first. But when you change the observer, you effectively mirror the entire setup: the velocities change direction, the moving object is coming from the other side, the whole thing is "backwards" in a way. Everything is mirrored, except your labels - which changes which light is the relevant one.

The consequence of this is that each observer performs their experiment under different rules. Suppose the observers are facing each other (this is consistent with the mirrored setup - each is facing what is to them the incoming object). Then for T, the rule is to detonate the bomb if the light in front of them flashes first, while for C the rule is to detonate the bomb if the light behind them flashes first.

Thus, it is not the same experiment, they are following different instructions.

$\endgroup$
1
  • $\begingroup$ if I understand you correctly, you mean my original experiment to observe time order of 2 events is in fact not an experiment. (the bomb is just a way to dramatically emphasise the result) $\endgroup$
    – Rekkhan
    Commented May 23 at 23:32
2
$\begingroup$

You have made a basic mistake. If two events are simultaneous in the frame of the cart, they won't be simultaneous in the frame of the tray, and likewise if the events are simultaneous in the frame of the tray, they won't be simultaneous in the frame of the cart. So when you say that the bomb explodes only if the two flashes are simultaneous, then that statement can only be true in one frame. If you create an experiment in which the bomb goes off only when there are two simultaneous flashes in the frame of the cart, then you have created an experiment in which the bomb is triggered by two non-simultaneous flashes in the frame of the tray.

$\endgroup$
2
  • $\begingroup$ Althought you made a small mistake in your answer, I get the main idea. It is similar to the other answers stating that a "experiment" that measure the other of the events is not a real experiment, or more precise, not a experiment implied by the first postulate. But I have to think more about that. Thank you for your explanation. $\endgroup$
    – Rekkhan
    Commented May 24 at 12:46
  • $\begingroup$ Cheers! Mistake corrected. $\endgroup$ Commented May 24 at 16:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.