Does time dilation work in 1d space?

[jbriggs444]

So back to the original idea that a spaceship is traveling back to Earth in view of a video projection from Earth of a clock, does the spaceship see the clock tick faster on the return journey?

Yes. That's the Doppler effect. The space ship sees the clock tick fast. Not as fast as in a Newtonian universe, but still faster than the on board clock.

The faster you return, the more time you get to spend with them. You see the clock speed up, yes. But you see it run faster for a shorter time. Less total elapsed time shown on the video.

 

[Dale]

Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as viewed by B.

I thought this was illegal. I thought if we measure the speed of light in the positive x direction relative to the speed of light in the negative x direction we are not allowed to attain a velocity of 2c.

It is a valid mathematical operation. However, I would not call it “the relative speed as viewed by B” precisely to avoid the confusion you are feeling.

I would reserve the term “relative speed” of two objects for the speed of one object measured in the frame of the other. I would use the term “separation speed” to describe the difference in velocity in a frame where neither is at rest.

 

[Mister T]

With the triangles you have shown, where is distance, time and velocity?

Nowhere. It's a drawing of a pair of wedges. A wedge is something that you might use, for example, to keep a door open.

 

[robphy]

Here is the spacetime-diagram example of the Euclidean "slope composition" diagram given by @Mister T and elaborated on by @jbriggs444 .
(Time runs upwards.)

Below, $\bot$ indicates that the two lines meeting there are Minkowski-perpendicular (that observer's spaceline (their x-axis) is Minkowski-perpendicular to that observer's timeline (their t-axis, their worldline)).

Choosing arithmetically-convenient values...

By counting diamonds, the velocity of Bob with respect to Alice [using Alice's diamonds for her time and space units]
$v_{BA} =\frac{6}{10}$, and so forth.

$\begin{align*}v_{CA} &=\frac{v_{CB}+v_{BA}}{1+v_{CB}v_{BA}}\\ &=\frac{(\frac{5}{13})+(\frac{6}{10})}{1+(\frac{5}{13})(\frac{6}{10})}=(\frac{4}{5})=\frac{16}{20} \end{align*}$

 

[YouAreAwesome]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

 

[Nugatory]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

If the traveller is watching through a telescope, yes.
However, when we allow for the light travel time (we're getting closer to the clock so with every tick the light reaches us sooner) we will calculate that the clock is ticking slower than ours. We see the ticks happening faster because of the Doppler effect, but not as much faster as we'd expect from a non-relativistic calculation.

And note that we will get the same result if we say that the "traveller" is not moving while the clock is rushing towards them at close to teh speed of light.

 

[Ibix]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

@Nugatory already answered this, noting that there's a distinction between what you literally see (clocks apparently running faster or slower as you approach or recede from them) and what you calculate is happening once you correct for the changing light travel time (moving clocks run slowly). Just to add to that, you need to be very wary whenever you come across the word "see" in a discussion of relativity, because it is often used sloppily to mean the calculated effect that you don't literally see (!). For example, it's quite common to come across statements like "a moving observer sees my clock tick slowly just as I see hers tick slowly". That's a correct statement about time dilation if I take "see" to mean "measure the apparent clock rate and correct for the changing distance", but that's not what "see" normally means. Such usage is, therefore, potentially very confusing and unfortunately rather common. Most people here will try to be clear about the distinction, and say things like "you literally see..." and "once you've corrected for the changing distance you'll find...", because we've corrected the misunderstandings so many times that being careful about it is second nature, but many sources are nowhere near as careful.

 

[PeroK]

Again thanks for all the replies. So if a person is traveling at close to the speed of light towards a clock, does it appear the person traveling that the clock is ticking faster?

Yes, but that has nothing to do with special relativity. That's simply the varying travel time of the signal from a sequence of events to an observer. In fact, the first good estimate of the speed of light was made using this idea back in 1676 by measuring the amount by which the orbit of one of Jupiter's moons got out of sync as Jupiter got further or closer to the Earth. You can read about that here, for example:

https://en.wikipedia.org/wiki/Speed_of_light#First_measurement_attempts

Note that SR is not about the finiteness of the speed of light but about the invariance of the speed of light across all inertial frames of reference.