# Special relativity in layman's terms

So I'm learning special relativity in high school right now, and I'm having some difficulty understanding some of the relations.

Right now, we're learning about time dilation and length contraction. What we've been told is that proper time is measured between two events upon the frame of reference for which both events occur at the same (relative) point in space. This is understandable.

However, we were introduced to length contraction today. What I understand, from the information we are given, is this: relativistic time passes slower than proper time, and proper length is longer than relativistic length. From this, based on a possibly incomplete equation we are given from which I must assume the relation between time and distance is proportional, it must be concluded that
proper length/relativistic time = relativistic length/proper time = speed

And yet the given definition of proper length (a word-based definition) leads me to believe that proper length is in fact over proper time.

Can you please provide as simple and rudimentary and intuitive explanation of how to find which of two frames of reference has proper length?

• The easiest way to understand SR is to discover it yourself. See youtube.com/… – Sean Jan 13 '17 at 6:37

Length's are measured by determining the locations of two points of an object (the ends of a rod, the corners of a rectangle, etc) simultaneously. You, thus, have two spacetime events which are simultaneous in your reference frame, but are at different locations.

If you are at rest with respect to the object, i.e., as short time after your measurement, the positions of those points haven't changed, then you have measured the proper length.

If you are moving with respect to the object, you still make simultaneous measurements in your reference frame, but they will not be simultaneous in other frames. Neither will the proper length measurement be simultaneous events in your frame.

Edit to explain non-simultaneity: Special relativity is actually a transformation of coordinates of both space and time. The time value of an event in one frame depends on both the time value and the location in the other frame. For example, in a primed frame moving at $\beta$ with respect to unprimed, with overlapping origins at $t'=t=0$, an event at (x,t) will have coordinates at $x'=\gamma x - \gamma\beta c t$ and $t'=\gamma t - \gamma\beta x/c$.

Two simultaneous events occur in the unprimed frame at different locations: $(x_1,t)\text{ and }(x_2,t).$ Transforming to the primed frame we get $$x_1'=\gamma x_1 - \gamma\beta c t \text{ and } t_1'=\gamma t - \gamma\beta x_1/c$$ $$x_2'=\gamma x_2 - \gamma\beta c t\text{ and } t_2'=\gamma t - \gamma\beta x_2/c$$ Notice that the times in the primed frame are different! And this means that we can't take $x_2'-x_1'$ as the length in the primed frame.

• So, if I'm on a spaceship travelling really fast and make a length observation, since I'm at rest with respect to the spaceship, I am measuring the proper length? I'm also a little confused as to what you mean in the last paragraph by "you still make simultaneous measurements in your reference frame, but they will not be simultaneous in other frames." Why wouldn't they? If the location of the two relevant points are determined simultaneously, how does it become desynchronized in another frame of reference? – YogrtMan Jan 12 '17 at 2:51
• Position and time are coordinates which are rotated by a coordinate transformation between reference frames which move relative to each other. That transformation is called a Lorentz transformation. See my edit for a "breaking" of simultaneous events. – Bill N Jan 12 '17 at 3:21
• And yes, if you are at rest with respect to the object you're measuring, that is the proper length. Your movement relative to other things is irrelevant for that first object. – Bill N Jan 12 '17 at 3:30