Confusion over Special Relativity - Time and Length

Question

I've just started learning about special relativity. Apart from realising that everything I thought I knew about reality was a lie, I'm trying to wrap my head around a few concepts relating to the fundamentals of spacetime.

My current understanding:

1. The units of spacetime are absolute and are called spacetime intervals, which can be spacelike/timelike

   • Space and time can be considered components of spacetime - therefore we can think of time as a 'temporal distance'

2. Everything is travelling through spacetime at a constant speed: the speed of light

   • Spatial and temporal speed can be considered components of spacetime speed
   • I've found the concept of distributing this constant spacetime speed to spatial and temporal speed helpful, e.g. increasing your spatial speed will decrease your temporal speed

However, I'm still having trouble with a few issues.

1. How exactly do we measure temporal speed?

I often see the language used of 'your clock ticks slower' in relation to time dilation, but I don't understand exactly what this actually means. For instance, when we say that you move through space slower, it is referring to the fact that less spatial distance in e.g. m is covered per unit time. Similarly, what units are being discussed in phrases like 'A's clock ticks once every two seconds'? Does it mean that one unit of proper time passes every two seconds of coordinate time or vice versa? What exactly are the units of temporal speed and how do clocks represent this?

2. How can time dilation be related to length contraction?

When considering time dilation, I've thought of it as decreasing temporal speed in order to increase spatial speed. But when applying the reverse to space, it seems to me that increasing spatial speed should result in some sort of length dilation and not contraction. How can I understand this concept using the approach of distributing speed to time and space respectively?

Geniuses of the internet, please enlighten me.

Answer 1

The idea that everything travels through spacetime at constant speed $c$ can be quite confusing. Firstly, speed is defined as distance divided by time, so it’s inherently confusing to talk about spatial speed vs temporal speed. There is a way to do it, if you must: spatial speed is distance $\Delta x$ divided by proper time $\Delta\tau$ and temporal speed is coordinate time $\Delta t$ divided by proper time $\Delta\tau$. Then, since

$$c^2\Delta t^2 - \Delta x^2 = c^2\Delta\tau^2$$

by definition, you have the “identity”:

$$\sqrt{c^2(\mbox{temporal speed})^2- (\mbox{spatial speed})^2}=c.$$

While you can make this work, it is a contortion of the concept of speed, that in practice does not really help. Case in point: you said “increasing your spatial speed will decrease your temporal speed“, which is wrong. Increasing spatial speed, increases temporal speed, as you can see in the formula.

You ask how one measures temporal speed. It’s in the definition: you need to compare coordinate time change with proper time change. Proper time is measured by a clock that you carry with you, just look at it. Coordinate time is a bit different, because it depends on the reference frame. Einstein imagined having a 3D array of clocks, all at rest with each other, ticking in unison. As you fly by each of them, you take their reading as coordinate time.

Finally, length contraction is totally related to time dilation. Indeed, a lot of books first derive one, and use that to derive the other. My favourite is about thinking about muon decay. Say a muon travels at speed $v$ down a linear accelerator of lengths $L$ in a time $t=L/v$, where $t$ is much greater than the half-life of the muon. How is this possible? Well, time dilation makes it so that the proper time $\tau$ along the trajectory of the muon is much shorter than $t$ and the half life. But in the rest frame of the muon, the accelerator is moving at a speed $v$, and $\tau v \ll t v =L$. So muon only a fraction of the accelerator flies past the muon? No, by assumption the end of the accelerator reaches the muon before the muon decays. The accelerator is shorter in the rest frame of the muon, its length is $L’= v\tau$.

Answer 2

what units are being discussed in phrases like 'A's clock ticks once every two seconds'?

This simply means, that in B's frame , the time interval between 2 ticks of A's clock is 2 seconds.
Similarly, it will also be true that, that in A's frame , the time interval between 2 ticks of B's clock is 2 seconds. ( Assuming the scenario is that A and B are moving past each other at uniform speed )

How can I understand this concept using the approach of distributing speed to time and space respectively?

I do not think the approach you are using to understand this , might be the most useful way. It can lead to the kind of confusion you are facing right now. There are many more useful ways of trying to understand this , atleast at the stage of your learning, that you seem to be at right now. There are some excellent questions and answers on this topic in this site. I would suggest you to read through them.

Also i can suggest a good external link for understanding basics of special relativity. I feel this is a great one for beginners, as it is not too math heavy and explains the concepts and logic behind it more than the maths

https://archive.org/details/EinsteinsRelativityAndTheQuantumRevolution