If length contraction is just apparent (not real), then why is time dilation real? [duplicate]

Question

Length contraction and time dilation are said to be the two sides of the same coin (the same phenomenon as viewed from different frames).

That is to say, the time dilation and length contraction are symmetrical.

Length contraction, like time dilation, exists when there is relative motion and goes away when there is no relative motion, but there isn't any "accumulation" with length contraction, so there is nothing to be "left over".

Why isn't length contraction permanent even though time dilation is?

Now in my thought experiment, if I have two meter rods (each of them has an attached photon clock, so that the mirrors are at the end of the two rods), and initially synchronize the clocks, and send one rod (with the clock attached) to a journey at relativistic speeds, and bring it back , the rods won't change length, but the clocks will read different ticks (lets say the clock is able to register how many it ticked).

Now when the traveling rod (with a clock attached) travels, from the stationary frame (relative to Earth) you can see that the traveling rod undergoes length contraction because of its relativistic speeds.

Now when the the traveling rod (photon clock) arrives back:

1. you can see that the two rods are still the same length, so length contraction was just an apparent phenomenon, it is not real

2. you can see that the two clocks are now showing different ticks, because of time dilation one of the clocks ticked less (the traveling one) then the other clock, so time dilation is a real effect

Now if length contraction and time dilation are just the same phenomenon viewed from different frames, then how is that time dilation was real (the clocks), but length contraction was just apparent?

Question:

1. If length contraction is just apparent (not real), then why is time dilation real?

Answer 1

Length-contraction is not just "apparent" in the sense that it is not an illusion (say, unlike a mirage). In particular, the length of a moving rod is genuinely different from the length of the same rod in its rest frame. But, I don't think the main point here is to parse the Talmudic difference between real and apparent. The point that you raise is correct, it just needs to be turned on its head. Time dilation and length contraction are not symmetrical. Here is the disanalogy:

• Time dilation corresponds to the difference in the time interval between the same pair of events as observed by two different observers.
  • For example, let's consider a light beam being emitted from one point and then arriving back at that point after being reflected off of a mirror (the events must have the same location in space to avoid confounding time dilation with simultaneity differences). These events have coordinates in spacetime of $(0, x)$ and $(t, x)$ in the rest frame. Applying the Lorentz transformation to these gives coordinates of $\gamma (vx, x)$ and $\gamma (t + vx, x + vt)$ in the moving frame. So the dilated time, i.e., the time between the events in the moving frame, is $\gamma (t + vx) - \gamma vx = \gamma t$.
• Length contraction corresponds to the difference in the spatial interval between two different pairs of events as observed by two different observers.
  • In particular, let's consider the spacetime trajectory of the end-points of a rod in the rest frame. The coordinates of these events are given by $(t,0)$ and $(t,l)$ respectively. Now, the Lorentz transformation of the coordinates of the same set of events in a moving frame gives us $\gamma(t, vt)$ and $\gamma(t+vl,l+vt)$. However, as you can see, this pairing of events is not simultaneous in the moving frame. So, they're not the end-points of the rod at a given time in the moving frame. What would be the end-points of the rod at a given time in the moving frame are such events whose coordinates in the rest-frame are $(t,0)$ and $(t+\Delta t, l)$ where $\Delta t$ is such that $\Delta t=-vl$. Now, the coordinates of such events in the moving frame are $\gamma(t, vt)$ and $\gamma(t,l+vt-v^2l)$, which gives you the length of the rod in the moving frame as the spatial interval between these simultaneous events, i.e., $\gamma(1-v^2)l=l/\gamma$.
  • So, to recap, the length of the rod in the rest-frame is the spatial interval, as observed by the rest-frame, between such a pair of events whose coordinates are $(t,0)$ and $(t,l)$ in the rest frame. Whereas, the length of the rod in the moving frame is the spatial interval, as observed by the moving frame, between such a pair of events whose coordinates in the rest-frame are $(t,0)$ and $(t-vl,l)$. In other words, it's not the same pair of events between whom the spatial interval is being measured by the two frames.

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To re-emphasize the point that we are dealing with two different pairs of events when we consider length-contraction, I will quote here a part of the discussion from the comments (now deleted):

• The length of a rod in its rest frame is the spatial interval (as observed by the rest frame of the rod) between a pair of events whose coordinates are $(0,0),(0,𝑙)$ in the rest frame of the rod.
• On the other hand, the length of the rod in the ground frame (that is moving at a velocity −𝑣 w.r.t. the rest frame of the rod) is the spatial interval (as observed by the ground frame) between events whose coordinates are $(0,0),(−𝑣𝑙,𝑙)$ in the rest frame of the rod.

Since I am keeping the frame in which I am writing the coordinates the same, you can see directly that the pairs of events that we are using to calculate the length in the two different frames are different.