Time dilation: speed relative to what?

In summary: It only applies if the observer is moving relative to some standard clock.But differential ageing is also relative, right?Differential ageing is not coordinate-dependent, but time dilation is.
  • #36
OK, let us say that the relative speed of C to A and to B is v.

I can't follow your reasoning, but perhaps that is due to a lack of definition of what you mean with "running at the same rate". According to all Galilean reference frames except the one in which the clock is not moving, the clock is "running at a slower rate".

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/ [Broken]

So clocks, A & B, are running at the same rate as measured by C,

Reference https://www.physicsforums.com/threads/do-clocks-really-run-slow.810605/ [Broken]

What is your problem here? If C is permanently at the mid point of the line AB, then AC = BC and the speed of each relative to C will be the same - or the distances AC and BC would not continue to be equal.
If the speed of each relative to c is identical then their Lorentz factors will be the same and the slowing of those clocks, relative to C's clock will be the same, therefore clock A must read the same as clock B as measured by observer C. How is that not symmetry?

So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...
 
Last edited by a moderator:
Physics news on Phys.org
  • #37
Grimble said:
It seems like a reciprocal arrangement to me...

That's seems like a "privileged" perspective right? Analysis of all the results at the same time is different from when it is playing out in a continuum. This is the same perspective where you mentioned something happening "at the same time".
 
  • #38
All ideal clocks run at a rate of one second per second of proper time (trivially ##d\tau/d\tau = 1## or barely less than trivially ##c\,d\tau/ds=1##). Ideal clocks run at a rate of less than one second of proper time per second of coordinate time in any inertial frame in which they are moving (##d\tau/dt \le 1 ##).
 
Last edited:
  • #39
From the initial post in this thread:
Grimble said:
Extending this line of logic we can prove that all clocks must be running at the same rate.

I must confess that I'm not seeing the proof here. If you could spell it out in a bit more detail how to extend that line of logic to the conclusion that "all clocks must be running at the same rate?

I suspect that when you do, you will find that you are making an assumption:
If an observer finds that clocks A and B are both in motion relative to him and running at the same rate; and a second observer finds that clocks B and C are both in motion relative to him and running at the same rate then there exists some observer for whom all three clocks are running at the same rate.​
This is not correct, except in the uninteresting special case in which at least two of the three clocks are at rest relative to one another. However, every time that you add another clock "in between" (BTW, have you noticed that the physical positions of the clocks is irrelevant? They don't need to go "in between", they can go anywhere as long as they have the right relative velocities) that clock is in motion relative to the clocks on either side of it as well as all the other clocks, so this special case doesn't apply.
 
Last edited:
  • #40
Grimble, I expect this is going to go in the exact same direction as all your other threads on this. Making increasingly complicated scenarios, guessing at the solution, and then debating it until the thread gets locked hasn't helped in the past, and is unlikely to help if we try it one more time.

A different approach is needed - taking advantage of physics being a quantitative science. You need to explicitly state what calculation would convince you that clocks run slow, and then we can do it together. For example, X observes Y's clock to read t3 when X's reads t1, and observes Y's clock to read t4 when X's clock reads t2. X will say Y's clock runs slow if (t4-t3)/(t2 - t1) < 1.

Issues about symmetries need to be applied in the solution of the problem, not the posing of the problem. As this (and the other threads) show, starting on the solution while still posing the question only adds to the confusion. Give us a clearly posed question.
 
  • Like
Likes Dale
  • #41
Grimble said:
A very basic question, but how does one explain it to someone with this train of thought?

Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of
So how is it that we talk of clocks physically slowing? Of two clocks physically slowing each with respect to the other at the same time?

I would recommend the thread "Symmetrical time dilation implies the relativity of simultaneity", https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/

Symmetrical time dilation is what I call the situation where A says B's clock is slow, and B says A's clock is slow. This notion is not consistent with the notion of absolute time.

I rather strongly suspect here that the issue is that Grimble is implicitly assuming the existence of absolute time. This thread is my best attempt to show why absolute time is incompatible with symmetrical time dilation, and how what's known as "the relativity of simultaneity" fixes the issue.
 
  • #42
Grimble said:
[..] if the 'stationary' observer changed his speed there is no connection to the 'moving' clock [... omit part that I cannot follow...] It can only be that the MEASUREMENT BY the remote observer is affected.
The clocks don't change only the measurements? That can't be right surely?
I agree with that part of your reasoning: the measurements are done with clocks and rulers.
Grimble said:
[..]
So although A and B both show 1pm observed by C, they show different times observed from each other. That is due to relativity of Simultaneity. Can you explain how that works differently for one than for the other? It seems like a reciprocal arrangement to me...
Explaining such a complex arrangement with mere words is complicated and will be difficult to do without ambiguity. The best explication is with numbers - and it avoids wasting time on long discussions. Please prepare an example calculation. Then if you do not already find the answer from it yourself, you can present it here.
 
  • #43
Grimble said:
Two observers, A & B, each with standard clocks are traveling at 0.866c relative to one another, then each will read the others clock as running at half the speed of their own (lorentz factor 2)
If a neutral observer, C, is permanently mid-way between them then each will me moving at 0.433c (lorentz factor 1.1) and their clocks will each, therefore, be running at 0.9 the rate of the neutral observers clock.
So clocks, A & B, are running at the same rate as measured by C, yet each is running at half the rate of the other when measured by that other.
As has been noted before, the observer C would not be traveling at ##\sqrt{3}c/4## because velocities do not add linearly in a relativistic universe. It turns out that, if A is stationary in some frame in which B is approaching at ##\sqrt{3}c/2##, then the observer C must be approaching A at ##c/\sqrt{3}## so that in C's rest frame, both A and B are approaching at the same speed in opposite directions.

So, here is a Minkowski diagram of that setup, as seen by C:
minkowski1.png

Apologies for the lack of labelling. A Minkowski diagram is basically a displacement-time graph, except that time runs vertically upwards and x position is shown horizontally. The vertical gridlines are one light second apart; the horizontal ones are one second apart.

The world-line of C is the purple line right up the middle. C is at rest in this frame, so the x position never changes. It starts at the origin at time zero (marked by a solid purple square) and stays right there.

The world-line of A is represented by the red line. It starts at x=-8ls at time zero (marked by a solid red square), and moves with velocity ##c/\sqrt{3}## - so as time goes on (up the page) it gets closer to C, and eventually meets at the top of the diagram.

The world-line of B is in blue, and is a mirror image of the world-line of A. It starts at x=8ls and moves with velocity ##-c/\sqrt{3}## and reaches C at the same time as A does.

As you noted, the situation is symmetrical as viewed from his frame. I've also added markers to each line showing when, according to C, clocks moving with A, B and C would tick. The purple ticks are every second; the red and blue ticks occur at the same time as each other, but you can see that they are spaced out more - their clocks tick slower in this frame.

But what does this look like according to A? We can use the Lorentz transforms to calculate the coordinates of each of the points in the Minkowski diagram as seen from a frame in which A is at rest and re-draw the diagram. That turns out to look like:
minkowski2.png

You can see that this time A (in red) is stationary, so the position is always the same, giving a vertical line. You can see that the red clock ticks are spaced exactly the same as the grid - they are 1s apart. You can see that there are the same number of ticks on each line as there were in the first diagram, and you can see that this time the purple clock is running slow and the blue clock is running really slow. You can also see why this isn't a problem - in this frame the three observers did not start their clocks at the same time.

So, yes, the clocks are running slow - it's not an illusion (although it is only an effect of co-ordinate choice). The different observers disagree about the time the clocks were started so that there is no contradiction when they meet up and can unambiguously compare clocks.
 
  • #44
To be sure that I can follow your explanations, that with the number of replies can become a little confusing with so many different replies to consider, let me check a few basic premises to be sure that I am correctly appreciating the concepts.
1. The view of space from any Inertial Frame of Reference is at rest; i.e. there are a set of axes that give a fixed set of coordinates for any point in space.
2. A full set of four coordinates specifying a fixed point in space, at a specific point in time, constitutes an event.
2. Every event can be mapped, that is given a unique set of coordinates, in any Inertial Frame of Reference.
3. Events, being fixed in time cannot move.
4. There must be a fixed point with a set of unique spatial coordinates, midway between any two events, in every Inertial Frame of Reference.
5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous. (Einsteins definition of simultaneity).
 
  • #45
Probably you meant it like that, but just to make sure I added some words:

5. If light from two events, arrive at the fixed point midway between them at the same time, (which constitutes a single event), then those events were simultaneous according to that inertial reference system. (Einsteins definition of simultaneity).

PS. this is just the example of Einstein here: http://www.bartleby.com/173/9.html

Personally I find it easier to consider the inverse: a light bulb in the middle of a train gives off a flash of light in both directions. The times that the light rays are thought to reach the ends of the train depend on the used reference system.
 
Last edited:
  • #46
Thank you, yes that is correct. I just wanted to be sure that I was understanding this.

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
And there would be a set of spatial coordinates that would define a fixed point in each and every inertial frame of reference that was spatially mid-way between points A and B?
And that the lights emitted at event A and event B would meet at that said mid-point?
If the lights from A and B arrive together, as a single event and be so measured by an observer at rest at the mid points between event A and event B, would that not imply that they could be measured to be simultaneous, in any, indeed in all, inertial frames of reference?
However it must be at a different mid-point in each frame, as the stationary mid-point in one frame would be moving away in any other frame.
So the observer in any frame would declare that he was the only one who could determine simultaneity...

Now, this is my problem: what is it I am getting wrong here? That all seems so very clear and logical and no matter how many times I have gone over it, I cannot see it!
Please help me and explain it?
 
  • #47
Grimble said:
So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
(my emphasis)
Events are events. They exist, therefore they exist in all frames. If I meet someone at the station at 8pm that is an unchangeable physical fact. My worldline intersected with the other partys worldline. Everyone will agree that we met, furthermore everyone will agree that our clocks read 8pm when we met.
 
Last edited:
  • #48
Grimble said:
those two events A and B would occur in every inertial frame of reference?

Some of your confusion may be that you're thinking that things happen "in" frames. They don't. They happen, and then if it is convenient we assign them time and space coordinates, and a frame is just an arbitrarily chosen convention for assigning these coordinates. When you hear someone saying something like "this event happened in frame F at time t and position x", that's a convenient but sloppy shorthand for the more precise "There's this event. We can choose any frame we want to assign time and space coordinates to that event, and if we choose frame F, we'll end up assigning time coordinate t and space coordinate x to it, but of course if we had chosen a different frame we would have assigned different x and t values".

An event is like a pencil mark on a piece of paper - it's there whether we draw a set of coordinate axes on the paper or not. If it's convenient to describe the location of the point using x and y coordinates, we can draw an x-axis and a y-axis on the sheet of paper as well to create a "frame", and we can choose to draw the axes anywhere we please.
 
  • Like
Likes ComplexVar89 and phinds
  • #49
Grimble said:
And that the lights emitted at event A and event B would meet at that said mid-point?
This can only be true in one frame. In all other frames the light emitted at A and B would not meet at the midpoint.
 
  • #50
Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.
 
  • #51
Ibix said:
Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.

But the flashes of light are events and therefore are fixed. As for timing, if they are triggered by light from a point in the centre of your rod, then they must be simultaneous?
For even measured from a frame that is moving the distances traveled by the light are the same, the movement of the lights are the same, or are we saying that if a light is shone at the centre of the rod and reflected back by two equidistant mirrors, that in one frame alone (where the rod is stationary) will the reflections arrive simultaneously?
And, if the simultaneous arrival of the reflected lights set of a signal, that would only occur in one frame?
Mentz114 said:
Events are events. They exist, therefore they exist in all frames. If I meet someone at the station at 8pm that is an unchangeable physical fact. My worldline intersected with the other partys worldline. Everyone will agree that we met, furthermore everyone will agree that our clocks read 8pm when we met.
Nugatory said:
Some of your confusion may be that you're thinking that things happen "in" frames. They don't. They happen, and then if it is convenient we assign them time and space coordinates, and a frame is just an arbitrarily chosen convention for assigning these coordinates. When you hear someone saying something like "this event happened in frame F at time t and position x", that's a convenient but sloppy shorthand for the more precise "There's this event. We can choose any frame we want to assign time and space coordinates to that event, and if we choose frame F, we'll end up assigning time coordinate t and space coordinate x to it, but of course if we had chosen a different frame we would have assigned different x and t values".
That is how I see it. A Frame of Reference is merely one particular view of Spacetime. So what is in Spacetime is there in all Frames of Reference, it is only the coordinates that will differ.

DaleSpam said:
This can only be true in one frame.
So that would have to be the 'Privileged Frame' that was so abhorred by Einstein?
You see those two events, the emission of the flashes of lightning, being events cannot be moving. They are moments/points in time whose positions are fixed in every frame - only the coordinates are unique. So in any frame there may be an observer positioned midway between those events, who is STATIONARY in that frame of reference and who therefore may measure simultaneity.
The only question then is whether those two events happen at the same time. But if they were triggered by light traveling equal distances from a single event, and in any frame the distances will be equal and the speed of light c, how can they not be at the same time?
 
  • #52
Grimble said:
[..] if they are triggered by light from a point in the centre of your rod [..] or are we saying that if a light is shone at the centre of the rod and reflected back by two equidistant mirrors, that in one frame alone (where the rod is stationary) will the reflections arrive simultaneously? [..]
Yes that is exactly what Einstein's example shows; it directly follows from the assumption (in fact it's a convention) that the speed of light is c relative to the chosen reference reference system. With that assumption, if the rod is moving then already the trigger light cannot reach both ends at the same time - it's as simple as that. :oldsmile:

PS Note that it is quite different if you choose the opposite and simpler example of a light source in the middle of the rod, with mirrors at each end.
Then the depart of the light pulse in both directions is a single event, that is, we designate to it a single (x,y,z,t) and therefore also a single (x',y',z',t'). It is a useful exercise to verify that according to any inertial reference system the reflected rays will arrive simultaneously back at the centre. That is necessary, for that is also a single event.
 
Last edited:
  • #53
Grimble said:
So that would have to be the 'Privileged Frame' that was so abhorred by Einstein?
No. The laws of physics are the same. Nothing priveliged except that A and B happen to be simultaneous.

In spacetime A and B are events, and the light from A and B are worldlines. The light worldlines meet at an event C. The midpoint between A and B is a coordinate line (that is, a line containing a different set of events in each frame), and there is no guarantee that it intersects with event C. That only happens in the frame where A and B are simultaneous.
Grimble said:
But if they were triggered by light traveling equal distances from a single event,
This is never true. A and B are two separate events, even in the frame where they are simultaneous.
 
Last edited:
  • #54
harrylin said:
PS Note that it is quite different if you choose the opposite and simpler example of a light source in the middle of the rod, with mirrors at each end.
Then the depart of the light pulse in both directions is a single event, that is, we designate to it a single (x,y,z,t) and therefore also a single (x',y',z',t'). It is a useful exercise to verify that according to any inertial reference system the reflected rays will arrive simultaneously back at the centre. That is necessary, for that is also a single event.

But that IS exactly what I described!
 
  • #55
Grimble said:
But that IS exactly what I described!
Then I misunderstood what you meant with "shone at"; I misunderstood it in the sense of "firing at" while you meant "at the position of". In a second reading I see that it could not have meant what I first thought.
Glad to see that one solved. :smile:
 
Last edited:
  • Like
Likes Grimble
  • #56
In harrylin's example, all frames agree that the pulses meet again at the center of the rod. They do not agree on where the center of the rod is with respect to where it was when the pulses were emitted or reflected.
 
  • #57
There remains something funny with "observer who is STATIONARY in that frame of reference and who therefore may measure simultaneity." Observers (sensors) can be moving or not, that doesn't really matter; being stationary in the used reference system merely simplifies things. For example GPS satellites do a great job without ever being stationary in the ECI frame which they relate to.
 
Last edited:
  • #58
Ibix said:
Bear in mind that in at least one frame, the light sources are moving. What does "half way between" the sources mean when they don't emit simultaneously and have moved between one emission event and the other? Do you mean that (assuming that the sources are at opposite ends of a rod) the midpoint of that rod? In that case the pulses will cross there but, because the rod is moving, that's not the same as the point half way between the spatial locations of the emission events.
Yes, the light sources will be moving, but the events where the light is emitted from those source, being events are fixed, regardless of where those sources go subsequently.
'Halfway between' means the midpoint of a line drawn between those events.

Ibix said:
In harrylin's example, all frames agree that the pulses meet again at the center of the rod. They do not agree on where the center of the rod is with respect to where it was when the pulses were emitted or reflected.
But it is not the centre of the rod that it is observers who are at rest at the midpoint between the Events where the lights were emitted, the centre of the rod may no longer be at that point, it will have moved away.
 
  • #59
Grimble said:
Thank you, yes that is correct. I just wanted to be sure that I was understanding this.

So following point 2, the two events, let's all them A and B, from which light was emitted that met midway at, let's call it point C; those two events A and B would occur in every inertial frame of reference?
And there would be a set of spatial coordinates that would define a fixed point in each and every inertial frame of reference that was spatially mid-way between points A and B?
And that the lights emitted at event A and event B would meet at that said mid-point?
If the lights from A and B arrive together, as a single event and be so measured by an observer at rest at the mid points between event A and event B, would that not imply that they could be measured to be simultaneous, in any, indeed in all, inertial frames of reference?
However it must be at a different mid-point in each frame, as the stationary mid-point in one frame would be moving away in any other frame.
So the observer in any frame would declare that he was the only one who could determine simultaneity...

Now, this is my problem: what is it I am getting wrong here? That all seems so very clear and logical and no matter how many times I have gone over it, I cannot see it!
Please help me and explain it?

Maybe this will help.

Consider the following scenario: You have an observer standing along a track and an observer on a moving train car. Two flashes of light originate at points an equal distance from the track observer. The flashes arrive at the track observer at the same instant that the train car passes him. Thus both observers see the flashes simultaneously. these events look like this according to anyone at rest with respect to the tracks.

train1.gif


Now we consider the same events according to the observer on the car. Keeping in mind, the postulate that the speed of light is invariant, meaning that he must measure the speed of each light flash relative to himself as being the same. In other words IF he considered the sources of the flashes as being an equal distance from him at the moment the flashes originated and and he sees the flashes simultaneously, then he can conclude that the flashes originated simultaneously.
However, while he does see the flashes simultaneously, he cannot say that he was an equal distance from the points of origins when the flashes started. He is only at the midpoint between the origins when he sees the flashes. At any moment prior to this he is closer to the left flash's origin point than he is to the right flash's. Since the flashes have to have originated at some time prior to his seeing them, he cannot be at the midpoint when either of them originated, and the flashes, according to him could not have originated simultaneously. Thus event would occur like this according to anyone at rest with respect to the train car.

train2.gif


The flashes still meet at the point where the two observers pass each other, but the flashes do not originate at the same time. If we carry this a bit further we can apply it to the Einstein train example. Here we have observers at the midpoint of a train and on the embankment. Unlike the above example, the flashes originate when the observers are next to each other according to anyone at rest with respect to the embankment. The flashes also originate where the ends of the train and the red dots meet. Thus in the following animation, when the front of the train reaches the left dot a flash is produced and when the rear of the train reaches the right dot a flash is produced.

Thus from the embankment, events occur like this:

trainsimul1.gif


Note that the train observer runs into the left flash before the right flash catches up to him.

Now we consider the same events as they occur in according to the train. The first thing to note, is that as measured from the embankment frame, the train is moving and thus has undergone length contraction, in that its measured length is shorter than what it would be as measured by the train itself. It is this length contracted train that fits between the red dots and allows the ends of the train to hit the red dots simultaneously in this frame.

In the train's frame, the train is not moving an not length contracted, but instead, it is the tracks and embankment that is moving and length contracted. So not only does the train measure its own length as being longer than that measured by the embankment frame, but it also measures the distance between the red dots as being shorter than that as measured by the frame. As a result, the train does not, in this frame fit between the two red dots. the front of the train reaches the left dot before the rear reaches the right dot.

For events to be the same in both frames, (such as the flashes originating when the dots and ends of the train align.) the flashes cannot originate at the same time in the train frame and we get the following sequence of the events according to anyone at rest with respect to the train.

trainsimul2.gif


The flashes originate at different times, but still when the train ends and red dots meet. The train observer sees the flashes at different times, since he is an equal distance from the train ends and they originated at different times. The embankment observer see the flashes at the same time (just like he did according to his own rest frame).

In fact, every event in both frames matches up perfectly. For example, the same railway car is next to the embankment observer when he sees the flashes in both animations and the train observer is next to the same point of the tracks when he sees each flash in both animations. All the events are the same, it is just that the frames don't agree as to the timing of these events.
 
  • Like
Likes Mentz114
  • #60
Grimble said:
'Halfway between' means the midpoint of a line drawn between those events.
But that is a different set of events for every reference frame. And there is only one frame where the event C is at the midpoint.
 
  • #61
Grimble said:
Yes, the light sources will be moving, but the events where the light is emitted from those source, being events are fixed, regardless of where those sources go subsequently.
The events are fixed, but observers do not in general agree on the place or time at which those events occur. This has consequences for their interpretation of what follows.
Grimble said:
'Halfway between' means the midpoint of a line drawn between those events.
By that definition, the pulses do not meet halfway between, in general. Why would they? They were emitted at different times in most frames.
Grimble said:
But it is not the centre of the rod that it is observers who are at rest at the midpoint between the Events where the lights were emitted, the centre of the rod may no longer be at that point, it will have moved away.
I really don't understand this sentence.

Let's draw this out, working with the setup where a light pulse is emitted from the center of the rod, reflected from mirrors at each end, and returned to the middle. Here is a Minkowski diagram of the setup in its rest frame:
minkowski1.png

This is a graph of the position of the various parts of the experiment as a function of time, with some key points marked. The red line represents the middle of the rod. It is not moving, so its position (place on the horizontal axis) is the same at all times. The two blue lines represent the mirrors. Again, these are not moving, so their position is also the same at all times. Note that the middle of the rod and the two mirrors are not events (which are points in 4-d space-time). Rather, they are lines which are what points in 3-d space look like in 4d space-time.

At the bottom of the diagram, a red square marks the event where the light pulses are emitted. I have drawn the lines traced out by the light pulses in yellow. Since the pulses are moving (their position is changing with time), their lines are sloped. They are sloped in opposite directions because they are moving in opposite directions. The pulses strike the mirrors at the events marked in blue squares and return, meeting again at the middle of the rod at the event marked with an open square.

I want to draw your attention to something - the mirrors are not at the reflection events most of the time. They do always share a spatial coordinate with the reflection event, but only share a time coordinate at the instants that their respective light pulses strike.

Now let's look at this from a different frame. In this one, the rod is doing 0.6c, which gives a nice round γ factor of 1.25:
minkowski2.png

The symbols on the diagram are the same as before. In this frame, however, all of the lines are sloped since the rod is moving - changing its position with time. You can see that the reflection events are not simultaneous, and you can see why - one of the pulses meets a mirror coming towards it while the other has to chase after its mirror. The first one has a shorter distance to travel. You can also see that the pulses meet again at the center of the rod (which is in a different place from where it was when it emitted the pulses). You can also see why they meet there - the pulse that met the mirror coming towards it (has a short journey) has to chase after the center of the rod (and has a long journey), while the opposite is true of the other pulse. This is how simultaneous emission and simultaneous reception do not imply simultaneous reflection - the total journey time for each pulse is the same, but the lengths of the legs are swapped round.

Finally, you can see what is wrong with your "'Halfway between' means the midpoint of a line drawn between those events". The midpoint of the line is an event on the worldline of the center of the rod, but it is not (in either frame) the same event as the reception of the reflected pulses. Interpreting your definition a bit more loosely as "the line of constant position through the event half way between the reflection events", then this is a vertical line through the event half way between the reflection events. You can see that this is still wrong - only in the special case of the rest frame of the rod is this where the pulses meet up.

The reflected pulses meet up at the center of the rod in all frames, because there would be self-contradiction if they did not. However, the center of the rod does not, in general, occupy the same spatial coordinates at the time of emission and the time of reception.
 
  • #62
And it is only in that frame where the lights meet at point C. In every other frame point C is moving and is no longer necessarily midway.
BUT in every frame the events cannot be moving they are fixed in time and space. And therefore there will be a fixed point midway between them where an observer that is at rest in that frame will measure the events as simultaneous. Let us call that point Cf (where f is an identifier for that individual frame.)
 
  • #63
Now if spacetime interval S2 = -(ct)2 + x2 + y2 + z2 and that is the invariant spacetime interval,
then from the embankment, where the two events are simultaneous, the interval between the two lightning strikes would be s2 = 0 + x2 + y2 + z2
while in the trains frame it would be -(cγt')2 + (γx')2 + (y')2 + (z')2 but as y' = y and z' = z this reduces to
x2 = s2 = -(cγt')2 + (γx')2
yet x = γx'
therefore
x2 - (γx')2 = -(cγt')2 = 0
 
  • #64
harrylin said:
There remains something funny with "observer who is STATIONARY in that frame of reference and who therefore may measure simultaneity." Observers (sensors) can be moving or not, that doesn't really matter; being stationary in the used reference system merely simplifies things. For example GPS satellites do a great job without ever being stationary in the ECI frame which they relate to.
In Einstein's measure of simultaneity the observer is midway between the two light events when the lights reaching him event occurs. I agree it does not really matter where he was either side of that event. Yet saying he was stationary at that point negates having to prove he was there at that moment as he would be there both before and after.
 
  • Like
Likes harrylin
  • #65
Janus said:
Consider the following scenario: You have an observer standing along a track and an observer on a moving train car. Two flashes of light originate at points an equal distance from the track observer. The flashes arrive at the track observer at the same instant that the train car passes him. Thus both observers see the flashes simultaneously. these events look like this according to anyone at rest with respect to the tracks.

http%3A%2F%2Fhome.earthlink.net%2F%7Ejparvey%2Fsitebuildercontent%2Fsitebuilderpictures%2Ftrain1.png


Now we consider the same events according to the observer on the car. Keeping in mind, the postulate that the speed of light is invariant, meaning that he must measure the speed of each light flash relative to himself as being the same. In other words IF he considered the sources of the flashes as being an equal distance from him at the moment the flashes originated and and he sees the flashes simultaneously, then he can conclude that the flashes originated simultaneously.

Yes, as shown in your diagram.

Note too that according to Einstein's first postulate, the laws of science are the same in any inertial FoR and that the distance measured between the flashes (L) will be the same measured locally in one FoR as they will be measured in the other FoR.
It is only when one observer (eg on the Embankment) transforms that same distance (L) measured at speed v that he will calculate that the distance L measured in the moving FoR, would be L/γ.
The distance L is the same measured in one frame as in the other, by the respective observers. Length contraction only happens when the measurement from the moving FoR is transformed to make it relative to the stationary FoR.
Remember, point A on the train and point A on the track coincided at Event A, point B on the train and point B on the track were together at event B.
In the FoR of the train M is mid way between A and B permanently. A and B are the front and back of the train and M is its midpoint. AM and BM are equal distances and the speed of light is c relative to the train, in the train's FoR. Therefore the light from A and the Light from B will arrive at M simultaneously in the FoR of the train observer sat at M.
It is measuring how the light is traveling within the train from point A and Point B which were adjacent to the light sources when they flashed.
This has nothing to do with the speed of the light along the track in the Embankment frame.
 
Last edited:
  • #66
Grimble said:
[..] the distance measured between the flashes (L) will be the same measured locally in one FoR as they will be measured in the other FoR.[..]
The distance L is the same measured in one frame as in the other, by the respective observers. Length contraction only happens when the measurement from the moving FoR is transformed to make it relative to the stationary FoR.
Regretfully I'm not sure to understand the meaning of those sentences. Perhaps you are thinking of a train in rest in one inertial reference system, compared with a train in rest in another inertial reference system? Indeed the physics should be the same.
 
Last edited:
  • #67
Grimble said:
And it is only in that frame where the lights meet at point C. In every other frame point C is moving and is no longer necessarily midway.
BUT in every frame the events cannot be moving they are fixed in time and space. And therefore there will be a fixed point midway between them where an observer that is at rest in that frame will measure the events as simultaneous. Let us call that point Cf (where f is an identifier for that individual frame.)
You are mixing up 3d and 4d concepts, I think. Certainly there is an event midway between the reflection events, but an event is a point at a specific time. There are infinitely many inertial observers who pass through that event. Each of them will consider herself at rest at a point, but only one of them also passes through the event where the light pulses return to the center of the rod.

I think what you are trying to say is that, for any two space-like separated events A and B, there exists a frame in which they occur simultaneously. In that frame, the event half way between A and B and the event where light from
A and B cross have the same spatial position. This is a backwards statement of the Einstein simultaneity convention, I think.
 
  • #68
Grimble said:
Now if spacetime interval S2 = -(ct)2 + x2 + y2 + z2 and that is the invariant spacetime interval,
then from the embankment, where the two events are simultaneous, the interval between the two lightning strikes would be s2 = 0 + x2 + y2 + z2
while in the trains frame it would be -(cγt')2 + (γx')2 + (y')2 + (z')2 but as y' = y and z' = z this reduces to
x2 = s2 = -(cγt')2 + (γx')2
yet x = γx'
therefore
x2 - (γx')2 = -(cγt')2 = 0
I think you are mixing frames. You are correct that [itex]s^2=x^2+y^2+z^2-(ct)^2[/itex], but your expression in the primed frame includes ##\gamma##s, which it should not. The correct expression is [itex]s^2=x'^2+y'^2+z'^2-(ct')^2[/itex].

Also, you will need to use the full Lorentz transforms to get expressions for x' and t' in terms of x and t. You are trying to use the length contraction and time dilation formulae, which are not applicable to the separation between events.
 
  • #69
Grimble said:
Note too that according to Einstein's first postulate, the laws of science are the same in any inertial FoR and that the distance measured between the flashes (L) will be the same measured locally in one FoR as they will be measured in the other FoR.
It is only when one observer (eg on the Embankment) transforms that same distance (L) measured at speed v that he will calculate that the distance L measured in the moving FoR, would be L/γ.
The principle of relativity says that the laws of physics are the same, not that all measurements are the same. The distance between the flashes as measured by one observer, and the distance between them as measured by another in motion relative to the first will be different and related by the Lorentz transforms.
 
  • #70
Grimble said:
And it is only in that frame where the lights meet at point C. In every other frame point C is moving and is no longer necessarily midway.
C is an event, the event where the light from A and B meet. As such it is not moving in any frame and it is only on the midpoint (which is a coordinate line in spacetime) in one frame.

I agree with Ibix that you are mixing 4D and 3D concepts.

Grimble said:
And therefore there will be a fixed point midway between them where an observer that is at rest in that frame will measure the events as simultaneous. Let us call that point Cf (where f is an identifier for that individual frame.)
This is simply false. The determination of simultaneity depends on the reference frame, not the specific location within that reference frame.
 
<h2>1. What is time dilation?</h2><p>Time dilation is a phenomenon in which time appears to pass slower for an object or person that is moving at high speeds relative to an observer. This is a prediction of Einstein's theory of relativity and has been confirmed through experiments and observations.</p><h2>2. How does time dilation occur?</h2><p>Time dilation occurs due to the relationship between time and space. According to Einstein's theory of relativity, time and space are not absolute, but rather are relative to the observer's frame of reference. As an object or person moves at high speeds, their frame of reference changes, causing time to appear to pass slower for them compared to an observer at rest.</p><h2>3. Is time dilation a real phenomenon?</h2><p>Yes, time dilation is a real phenomenon that has been observed and confirmed through experiments and observations. It is a fundamental aspect of Einstein's theory of relativity and has been used in various applications, such as GPS technology.</p><h2>4. How does the speed of an object affect time dilation?</h2><p>The speed of an object has a direct relationship with time dilation. As an object's speed increases, time dilation also increases. This means that time will appear to pass slower for the object moving at high speeds compared to an observer at rest.</p><h2>5. Is time dilation the same as time travel?</h2><p>No, time dilation is not the same as time travel. Time dilation refers to the difference in the passage of time for an object or person moving at high speeds compared to an observer at rest. Time travel, on the other hand, involves the ability to move backwards or forwards in time, which is currently only theoretical and has not been proven to be possible.</p>

1. What is time dilation?

Time dilation is a phenomenon in which time appears to pass slower for an object or person that is moving at high speeds relative to an observer. This is a prediction of Einstein's theory of relativity and has been confirmed through experiments and observations.

2. How does time dilation occur?

Time dilation occurs due to the relationship between time and space. According to Einstein's theory of relativity, time and space are not absolute, but rather are relative to the observer's frame of reference. As an object or person moves at high speeds, their frame of reference changes, causing time to appear to pass slower for them compared to an observer at rest.

3. Is time dilation a real phenomenon?

Yes, time dilation is a real phenomenon that has been observed and confirmed through experiments and observations. It is a fundamental aspect of Einstein's theory of relativity and has been used in various applications, such as GPS technology.

4. How does the speed of an object affect time dilation?

The speed of an object has a direct relationship with time dilation. As an object's speed increases, time dilation also increases. This means that time will appear to pass slower for the object moving at high speeds compared to an observer at rest.

5. Is time dilation the same as time travel?

No, time dilation is not the same as time travel. Time dilation refers to the difference in the passage of time for an object or person moving at high speeds compared to an observer at rest. Time travel, on the other hand, involves the ability to move backwards or forwards in time, which is currently only theoretical and has not been proven to be possible.

Similar threads

  • Special and General Relativity
Replies
14
Views
576
  • Special and General Relativity
2
Replies
58
Views
2K
  • Special and General Relativity
Replies
25
Views
533
  • Special and General Relativity
2
Replies
65
Views
4K
  • Special and General Relativity
Replies
29
Views
1K
  • Special and General Relativity
Replies
4
Views
839
  • Special and General Relativity
Replies
16
Views
597
  • Special and General Relativity
Replies
18
Views
1K
  • Special and General Relativity
3
Replies
88
Views
3K
  • Special and General Relativity
Replies
10
Views
1K
Back
Top