A thought experiment regarding special relativity

[KyriakosTsp]

Hello everyone,

I am currently delving into the intricacies of Einstein's theory of relativity and striving to grasp its fundamental essence and implications.

Your assistance would be greatly appreciated.

Here's my thought experiment concerning special relativity:
Suppose my friend and I are moving toward each other at a relatively high speed, V, each in our own spaceship. We have agreed to measure the time intervals between the same two events and then erect a vertical rod in our spaceship, with the length corresponding to our time measurement. We assume the rods are perpendicular to the direction of our motion and are equipped with markers at their top ends. As we pass each other at a constant speed, V, the shorter rod will naturally scratch a mark on the longer rod. After this encounter, we check our rods. The question is, who will have the mark?

Note 1: Clearly, I have designed this theoretical experiment to maintain symmetry between my friend and me throughout the two events.

Note 2: We assume the rods are perfectly perpendicular to our motion (and sufficiently lengthy) so that no length contraction will be observed.

 

[Nugatory]

We have agreed to measure the time intervals between the same two events

Which two events? We will get different answers for different choices.

The vertical rods are an unnecessary complication. We’re using them only to compare our time measurements; it would be as easy to have you and your friend just radio your measurements to one another, or write them on pieces of paper that we collect and compare afterwards.

 

[Sagittarius A-Star]

The question is, who will have the mark?

Nobody, because the rods are equally long, reason see below.

Note 1: Clearly, I have designed this theoretical experiment to maintain symmetry between my friend and me throughout the two events.

That means the two events shall be i.e. considered to be two ticks of a clock at rest in the "middle-frame" between the two frames. This clock is, with respect to the two other frames, equally time-dilated.

 

[Dale]

my friend and I are moving toward each other … measure the time intervals between the same two events

This is not possible. If you and your friend are moving at constant velocity relative to each other then there is at most 1 event that is on both of your worldlines.

 

[Ibix]

the same two events

Here's the problem. An event is a time and a place, and if you are both moving then you can only both be present at one event, the one when you pass each other. So at least one of you cannot start your clock at the other event, but instead has to start their clock at the same time as the other. And the two frames don't agree on what "at the same time" means.

As others have noted, your requirement of symmetry implies that the simultaneity condition being used is not the one used by either you or your friend, but that of the frame where you are both moving in opposite directions at equal speeds. Thus both of you will say that the other started their clock early by your own definition of "at the same time", in such a way that the early start and the slow ticking cancel out and the clocks read the same when you pass.

 

[PeroK]

Here's my thought experiment concerning special relativity:
Suppose my friend and I are moving toward each other at a relatively high speed, V, each in our own spaceship. We have agreed to measure the time intervals between the same two events and then erect a vertical rod in our spaceship, with the length corresponding to our time measurement. We assume the rods are perpendicular to the direction of our motion and are equipped with markers at their top ends. As we pass each other at a constant speed, V, the shorter rod will naturally scratch a mark on the longer rod. After this encounter, we check our rods. The question is, who will have the mark?

That's actually an important thought experiment to justify that there is no length contraction in a direction perpendicular to motion.

In terms of the elapsed times between the two events, why do they need the rods? They could simply exchange that information as they pass each other.

 

[Hill]

Let's call them, event A and event B. In your frame, they have coordinates $t_A, x_A, y_A, z_A$ and $t_B, x_B, y_B, z_B$. In your friend's frame, their coordinates are $t'_A, x'_A, y'_A, z'_A$ and $t'_B, x'_B, y'_B, z'_B$.

You and your friend will agree on the invariant interval:
$(t_A-t_B)^2-(x_A-x_B)^2-(y_A-y_B)^2-(z_A-z_B)^2=$
$(t'_A-t'_B)^2-(x'_A-x'_B)^2-(y'_A-y'_B)^2-(z'_A-z'_B)^2$.

Thus, the difference in your time measurements will be
$(t_A-t_B)^2-(t'_A-t'_B)^2=$
$(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2-(x'_A-x'_B)^2-(y'_A-y'_B)^2-(z'_A-z'_B)^2$.

(I take $c=1$)

 

[FactChecker]

Note 1: Clearly, I have designed this theoretical experiment to maintain symmetry between my friend and me throughout the two events.

Is it symmetric? Consider the simplifying coincidence where you think that your two events, separated in the direction of relative motion, are simultaneous. Then your friend can not think that those events were simultaneous. You cut your rod down to nothing and he has a remaining rod of some length. There can be some symmetry in that there can be two OTHER events that your friend thinks are simultaneous, but you do not. But those are different events. So the rod-length result depends on which events you are talking about.

 

[Mister T]

Suppose my friend and I are moving toward each other

Then you can't both be present at either of your two events. Thus you'll need distant clocks synchronized, and the answer depends on where the two events occur. If, for example, both events occur at the same place in your rest frame, then your friend's rod will be longer. But if the two events occur in the same place in your friends rest frame then yours will be longer. There are lots of other possibilities, too.

It would be worthwhile for you to consider examples of specific events. Specify where they occur, and at what time according to both observers (both you and your friend).

 

[pervect]

Hello everyone,

I am currently delving into the intricacies of Einstein's theory of relativity and striving to grasp its fundamental essence and implications.

Your assistance would be greatly appreciated.

Here's my thought experiment concerning special relativity:
Suppose my friend and I are moving toward each other at a relatively high speed, V, each in our own spaceship. We have agreed to measure the time intervals between the same two events and then erect a vertical rod in our spaceship, with the length corresponding to our time measurement. We assume the rods are perpendicular to the direction of our motion and are equipped with markers at their top ends. As we pass each other at a constant speed, V, the shorter rod will naturally scratch a mark on the longer rod. After this encounter, we check our rods. The question is, who will have the mark?

Note 1: Clearly, I have designed this theoretical experiment to maintain symmetry between my friend and me throughout the two events.

Note 2: We assume the rods are perfectly perpendicular to our motion (and sufficiently lengthy) so that no length contraction will be observed.

This seems a bit overly complicated, so I am concerned that I might be misunderstanding the details of the question. A space-time diagram would help clear up any ambiguities.

In general, for the question as stated (as I interpret it), there is insufficient information to give an answer. If both events happen to be on the worldline of one of the travelers (i.e. that observer is physically present at zero distance at both events), that traveller will measure the longest time interval, and hence have the longest rod.

The reason the question has no answer is "the relativity of simultaneity". A lot has been written about it, the only specific source I have that I'd recommend is aimed at teachers and not students, however.

"The challenge of changing deeply-held student beliefs about the relativity of simultaneity", Scherr, et al, available at https://arxiv.org/abs/physics/0207081. Also of some interest is "Student understanding of time in special relativity: simultaneity and reference frames", by the same authors.

Here is a quote from the second paper. Sherr asked several versions of the same basic question, I am only going to give one of the four versions of the question.

... two volcanoes,
Mt. Rainier and Mt. Hood, that erupt
simultaneously according to an observer at
rest on the ground, midway between the
volcanoes. A spacecraft moves at a given
relativistic velocity from Mt. Rainier to
Mt. Hood.

....

Spacecraft Question: Undirected version
The students were asked to draw
spacetime diagrams for both the ground and
spacecraft frames. They were told to show
the volcanoes, the spacecraft, and the
eruption events. They were not asked
explicitly whether the eruption events are
simultaneous in the spacecraft frame.
Rather, we inferred their ideas indirectly
from their diagrams.

Following are correct and "typical" incorrect space-time diagrams for this question from Scherr's paper. Of some concern is whether or not you are familiar with space-time diagrams. They're not hard, if you're not familiar I'd encourage you to read about them. They illustrate some important points about special relativity that are hard to explain otherwise.

a) is the correct space-time diagram for the ground observer, b) is the correct space-time diagram for the spacecraft observer, and c) is a "typical" incorrect space-time diagram.

 

[Mister T]

If both events happen to be on the worldline of one of the travelers (i.e. that observer is physically present at zero distance at both events), that traveller will measure the longest time interval, and hence have the longest rod.

Shouldn't that be the shorter time period, and hence the shorter rod?

 

[Nugatory]

Shouldn't that be the shorter time period, and hence the shorter rod?

Longer - zero $\Delta x$ maximizes $\Delta t$ between two timelike-separated events.

 

[Mister T]

Longer - zero $\Delta x$ maximizes $\Delta t$ between two timelike-separated events.

Okay, I think I see where I went wrong with my analysis of @pervect 's scenario. But I don't see why you're saying that $\Delta x = 0$.

 

[Nugatory]

Okay, I think I see where I went wrong with my analysis of @pervect 's scenario. But I don't see why you're saying that $\Delta x = 0$.

If both events are on one traveller’s worldline then $\Delta x$ will be zero using coordinates in which that traveller is at rest.

 

[Mister T]

If both events are on one traveller’s worldline then $\Delta x$ will be zero using coordinates in which that traveller is at rest.

Setting the interval between the events for each observer equal we have
$$(\Delta t)^2-(\Delta x)^2=(\Delta t')^2-(\Delta x')^2$$ $$(\Delta t)^2=(\Delta t')^2-(\Delta x')^2.$$ Thus $\Delta t'>\Delta t$ for all nonzero $\Delta x'$.

Isn't this consistent with what I wrote in Post #9?

 

[pervect]

A few comments. (Note that I am assuming special relativity throughout, avoiding general relativity).

1) The particular example I gave (about both events being on the worldline of one observer) is simply an example of the principle of "maximal aging", an important principle popularized by authors such as Taylor. It doesn't generalize to GR however, where one needs the principle of extremal aging instead.

2) The example of the explosion of the two volcanoes IS a particular example which illustrates the problem. If one is not familiar with space-time diagrams (I am sure most responders are, but I'm not so sure if the OP is), one may not realize that this example shows that the "length of the rod", given by the time difference between the two volcano events, is zero the ground frame, and non-zero in the spaceship frame.

To make another general observation of a general principle rather than focusing on a specific case, for any two events that are "space-like separated", which operationally means any two events in which are separated by enough distance so that light cannot propagate from one event to the other, there will be some frame of reference in which both events are simultaneous. This of course makes the "length of the rod" zero in that frame. And it will be non-zero in any other frame.