Since this question is causing all kinds of problems in a number of threads discussing relativity, I thought I would post an explanation of the concept, here. You can think of a reference frame as follows: Take an infinite number of rods of equal length, and an infinite number of identical clocks. Connect these up in a cubical lattice, so that each cube is constructed from 12 of the rods. At each corner of each cube, place one of the clocks. The resulting (imaginary) structure stretches over the whole universe. Now, suppose that each rod is L metres long. Follow the procedure below to synchronise the clocks. 1. Designate 1 particular clock as the coordinate (x,y,z) = (0,0,0) The "nearest neighbour clocks" of this clock exist at the following coordinates: (L,0,0) , (0,L,0), (0,0,L), (-L,0,0), (0,-L,0), (0,0,-L). 2. At time T on the (0,0,0) clock, send a light signal from that clock outwards in all directions. 3. When the light signal is received by a clock at distance nL from the (0,0,0) clock, set the clock receiving the signal to read T + nL/c, where c is the speed of light. For example, when the light signal reaches the clock at (L,0,0), the time on that clock should be set to T + L/c. This is the same time as is displayed on the (0,0,0) clock at the time the (L,0,0) clock receives the signal, since the time taken for the light to travel distance L is L/c. 4. Once the synchronisation procedure has been carried out, we have an infinite set of clocks spread throughout space, all of which display the same time. This system of synchronised clocks and rods is a reference frame. Suppose an explosion occurs somewhere and at some time in space. To specify the location of the explosion, we count the number of rods from the (0,0,0) clock to the position where the explosion occurred, in each of the three coordinate directions. That gives the spatial coordinates of the explosion (x,y,z). To get the time coordinate, t, of the explosion, we simply read off the time displayed on the clock sitting at the point where the explosion occurred. Thus, every event in spacetime has a unique set of spacetime coordinates (x,y,z,t). Some points to note: A reference frame covers the whole of space and time. All objects in spacetime exist in all reference frames. The spacetime coordinates of a single event are unique for a particular reference frame. There are an infinite number of possible reference frames. For example, set up another lattice of clocks and rods, and have it move in the x direction at some speed v. This new set of clocks and rods needs a new (0,0,0) clock, which can be any chosen clock in the set. It also needs its own synchronisation. Coordinates in this new frame must be given new labels, such as (x',y',z',t'). It is obvious that events which occur in one reference frame will, in general, have different coordinates (x,y,z,t) than the coordinates in some other reference frame (x',y',z',t'). Often, we are concerned with how to translate coordinates from one reference frame to another. 3 examples are given. Coordinates in frame 1 are given without the prime (') symbol. Coordinates in frame 2 have the prime (') symbol. Example 1 - rotated coordinate system at rest Reference frame 1 is set up. Reference frame 2 is chosen such that the (0,0,0)' clock is at exactly the same location as the (0,0,0) clock. The z' direction is chosen to be the same as the z direction. The x' direction is chosen to be rotated by an angle of q from the x axis. The coordinate tranformation is: x' = x cos(q) + y sin(q) y' = -x sin(q) + y cos(q) z' = z t' = t Example 2 - Galilean tranformation Reference frame 1 is set up. Reference frame 2 is chosen so that, in frame 1, the (0,0,0)' clock is located at position (vt,0,0) at any time t. In other words, frame 2 moves in the positive x direction at speed v relative to frame 1, and is not rotated in any way relative to frame 1. Universal time is assumed. This results in Galilean transformation: x' = x - vt y' = y z' = z t ' = t Example 3 - Lorentz transformation Reference frame 1 is set up. Reference frame 2 is chosen so that, in frame 1, the (0,0,0)' clock is located at position (vt,0,0) at any time t. In other words, frame 2 moves in the positive x direction at speed v relative to frame 1, and is not rotated in any way relative to frame 1. The postulates of Special Relativity are assumed. This results in Lorentz transformation: x' = g(x - vt) y' = y z' = z t ' = g(t - xv/c<sup>2</sup>) where g is the Lorentz factor 1/sqrt(1-(v/c)<sup>2</sup>) and c is the speed of light.