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I know similar questions have been asked, but I've had trouble finding an answer I understand as this is all pretty new to me. I'm trying to get a better understanding of the Special Relativity, with the goal of understanding how causality is preserved during time dilation.

I'm starting with the example of a photon clock, where a photon bounces between a defined distance. Photon Clock A is not moving in my frame of reference, and Photon Clock B is moving. If this explanation is unclear, it is based upon this short video:

Clock A is not moving in my frame of reference, so the photon only travels the defined distance, which we will call 100 meters. Clock B does appear moving to me, so the distance that photon travels is, say, 1000 meters.

This lets me create two equations for the speed of light, based upon s=d/t:

   c=100/tA for Clock A
   c=1000/tB for Clock B.

Now, for the simplicity of the math, lets pretend the speed of light is actually 10m/s, those equations resolve to:

10=100/tA -> tA=10 for Clock A
10=1000/tB - tB=100 for Clock B

Based upon all that, I would expect that clocks that appear to be moving would look faster to me. However, every resources says clocks observed in motion appear slower, and I'm struggling to understand why. Any help resolving this would be greatly appreciated.

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  • $\begingroup$ Based on your equations, your A clock takes 10 seconds to do 1 tick in your frame, but the B clock (measured in your frame) takes 100 (of your) seconds to do 1 tick. So it's going slower. $\endgroup$ – PM 2Ring Dec 7 '20 at 20:45
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Special Relativity can be summed up in two postulates; everything else from the theory follows directly from them.

1: The laws of physics are the same for all inertial reference frames.

2: The speed of light, c, is the same for all observers.

You correctly identified that, for the observer in the stationary reference frame of Clock A, they would measure the distance of the photon of clock B to be moving a larger distance per clock tick than for Clock A. Time as measured by Clock A from the observer's perspective is called the proper time; time as measured in the rest frame of an event. Any time measured from something moving with respect to that frame will show a GREATER time; i.e. time dilation. To see why in this example, note from your observation that Photon B travels farther per tick from the observers frame because Clock B is moving with respect to the observer's reference frame. But, since the clocks measure time by the "ticks" of the photons, which move at the speed of light, by Einstein's second postulate (the speed of light is measured the same for observers of any reference frame), both photons move with the same speed between ticks. Thus:

$$ c\, =\, \frac{s_A}{t_A}$$ $$ c\, =\, \frac{s_B}{t_B}$$

where

$$s_B > s_A$$

Thus:

$$\frac{s_B}{c} > \frac{s_A}{c}$$

and

$$t_B > t_A$$

This comes about because the photons both travel at the same speed, but photon B travels more distance than photon A in one tick in the reference frame of the observer. The only way they could travel the same speed but B could cover more distance than A is if it took longer for photon B to make one "tick". A good way to think about these situations is to remember that the shortest time that can be measured in Special Relativity is proper time, which is the time measured in the rest frame of an event (an event is something that occurs at a given position and time; i.e. a point in spacetime). Any time interval measured for events in a reference frame that is moving with respect to a reference frame will have time that moves slower for an observer; a dilated time. How much time dilation is given by the Lorentz transformations. Something important to note that many people don't realize at first is the symmetry between frames. While what I wrote holds true for observer A in reference frame A, by the same reasoning from an observer B in reference frame B, Clock A would be moving slowly in reference frame B by the same amount that A sees Clock B moving slowly in reference frame A. Which is correct? BOTH! By Einstein's first postulate, the laws of physics (and their results) are the same for all observers in inertial reference frames.

As far as your question about causality, this is best explained by Minkowski diagrams. On these diagrams are geometric regions called light cones. The specific event the diagram refers to is the event in spacetime at the apexes of the cones. Any events within these cones are timelike separated, meaning these events occur within spacetime in which light can travel between these events. A more accurate name for the speed of light is the speed of causality; the maximum speed which which interactions can be connected within spacetime. Light happens to move at the speed of causality, and it was discovered first, so we usually refer to this speed as the speed of light (you may hear things such as, "Gravitational waves travel at the speed of light.") Any events on a Minkowski diagram that fall outside of the lightcone of an event means that, at a given time, even something moving at the speed of light (and more importantly the speed of causality) could not traverse in spacetime to connect these events. These events outside of the event's lightcone is called spacelike separated. You may wonder what would happen to the Minkowski diagram for an event if we transformed by the Lorentz transformations into another interial reference frame. Would events look different in that reference frame??? YES! however, the light cones shift under such a transformation. However, there is a special property of events withing the lightcone. These events NEVER change in such a way where events become causally flipped. If event A takes place before event B in the lightcone, it will for all inertial reference frames. This is NOT true for spacelike separated events. There are different references frames where event A and event C's order switches based on reference frame. Does this break causality? NO! Because these events are spacelike separated; i.e. they were NEVER causally connected. So, it doesn't matter if an observer sees one event before another, vice versa, or even simultaneously.

For reference, here is a 3D Minkowski diagram:

enter image description here

Notice the axes. The xy plane is 2D space. The other axis is time multiplied by the speed of light so that the units match. The worldline is the the path through spacetime that an object takes; its evolution through spacetime.

This video is EXCELLENT at explaining these concepts visually.

https://www.youtube.com/watch?v=FdWMM6aXpYE

Hope this helped!

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  • $\begingroup$ Thanks Brian. Based on my example, I was thinking that since tB is 10x tA, Observer A would see 10x as many ticks on Clock B. But it actually means the opposite. For one tick of Clock A, they would only see 1/10th a tick of Clock B. This is because if they observed Clock B having 10x as many ticks, Clock B's photon would exceed the speed of light. Thus, to keep the photon's at consistent speed, events appear slower. Would that be an accurate recap? Trying to make sure my understanding is right. I appreciate the resources! $\endgroup$ – DoubtingThomas3005 Dec 7 '20 at 23:24
  • $\begingroup$ That's exactly right! You're welcome; glad I could help. $\endgroup$ – Brian Opatosky Dec 7 '20 at 23:43
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Let's say that the clock is 50 meters tall, so that the time interval you're describing is one round trip up and down. An observer at rest relative to the clock can call such a round trip one "tick." An observer at rest relative to clock B says that clock B made one tick. Observer A, who sees clock B moving, says that 10 ticks have gone by on clock A. Therefore they say clock B is slow.

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    $\begingroup$ I'm still a bit foggy at the end. Observer A says 10 ticks have gone by on Clock A, which is not in motion relative to them. Why would would they see 10 ticks on a stationary clock, wouldn't they see just one tick on Clock A, just as Observer B sees one tick for Clock B? $\endgroup$ – DoubtingThomas3005 Dec 7 '20 at 20:18

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