Time dilation: speed relative to what?

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Time dilation is fundamentally about how speed is measured relative to a frame of reference, with significant implications for how clocks are perceived to run. A spaceship traveling at 0.866c experiences time slowing by a factor of two, but this effect is dependent on the observer's frame of reference. When two spaceships move in opposite directions at 0.433c, each measures the other's clock as running slow, despite both clocks running at 0.901 of a standard clock as observed from a stationary midpoint. The discussion emphasizes that time dilation is a measurement effect rather than a change in the clock's actual rate, highlighting the importance of specifying the frame of reference. Ultimately, relativity asserts that all observers in inertial frames have equally valid perspectives on time, reinforcing that measurements are influenced by relative motion.
  • #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.
 
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  • #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.
 
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  • #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.
 
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  • #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.
 
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  • #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 s^2=x^2+y^2+z^2-(ct)^2, but your expression in the primed frame includes ##\gamma##s, which it should not. The correct expression is s^2=x'^2+y'^2+z'^2-(ct')^2.

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.
 
  • #71
Grimble said:
from the embankment, where the two events are simultaneous, the interval between the two lightning strikes would be s2 = 0 + x2 + y2 + z2
OK
Grimble said:
in the trains frame it would be -(cγt')2 + (γx')2 + (y')2 + (z')2
No, in the train's frame it is ##s^2 = -c^2 t'^2 + x'^2 + y'^2 + z'^2##. It most definitely does not equal what you have written.
Grimble said:
but as y' = y and z' = z this reduces to
x2 = s2 = -(cγt')2 + (γx')2
In the train's frame ##ct'=-vx \gamma/c##, ##x'=\gamma x##, ##y'=y##, and ##z'=z##.

So ##s^2 = -c^2 t'^2 + x'^2 + y'^2 + z'^2 = - v^2 x^2 \gamma^2/c^2 + x^2 \gamma^2 + y^2 + z^2 = x^2 + y^2 + z^2##
 
  • #72
harrylin said:
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.
Yes indeed I apolodise for my wording being a little slack there.
 

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