What does the Lorentz factor actually mean?

[Dale]

What, no picture?

 

[bobc2]

What, no picture?

I figured the sketches were getting annoying.

 

[phyti]

This post is intended to explain gamma/(the Lorentz factor) in terms of physical
processes and minimal math instead of theoretical statements.
The light clock consists of an integrated light emitter/detector, and a mirror, separated
by a rod of length r. The clock counts a unit of time (t=1 tick) when a photon moves
the length of the rod to the mirror, and returns to the detector.
There are two observers, U who is not moving, and A who is moving at .6c relative to U
on the Ux axis. Each has a copy of the clock with the rod oriented perpendicular to the
x axis.
Since the outbound path equals the inbound path, we only need to consider the first
path.
With U and A at the origin, each clock emits multiple photons (shown as a blue quarter
circle because object motion is restricted to the +x axis.)
For U the photon moves a distance r (.5 tick).
For A the intersection of the circular arc and rod determine which photon becomes part
of the clock. [1]The photon path ct can be resolved into the vt component which
compensates for the motion of A and the st component which is the active part of the
clock. When a photon arrives at the U mirror, the A clock photon has not reached the
mirror because the photon speed relative to the rod is s. If r' equals the path length to
the mirror for the A photon, then t'/t = r'/r = c/s = gamma.

[1] A vector can be expressed in components suitable for the situation.

https://www.physicsforums.com/attachments/52882

 

[ghwellsjr]

This post is intended to explain gamma/(the Lorentz factor) in terms of physical
processes and minimal math instead of theoretical statements.
The light clock consists of an integrated light emitter/detector, and a mirror, separated
by a rod of length r. The clock counts a unit of time (t=1 tick) when a photon moves
the length of the rod to the mirror, and returns to the detector.
There are two observers, U who is not moving, and A who is moving at .6c relative to U
on the Ux axis. Each has a copy of the clock with the rod oriented perpendicular to the
x axis.
Since the outbound path equals the inbound path, we only need to consider the first
path.
With U and A at the origin, each clock emits multiple photons (shown as a blue quarter
circle because object motion is restricted to the +x axis.)
For U the photon moves a distance r (.5 tick).
For A the intersection of the circular arc and rod determine which photon becomes part
of the clock. [1]The photon path ct can be resolved into the vt component which
compensates for the motion of A and the st component which is the active part of the
clock. When a photon arrives at the U mirror, the A clock photon has not reached the
mirror because the photon speed relative to the rod is s. If r' equals the path length to
the mirror for the A photon, then t'/t = r'/r = c/s = gamma.

[1] A vector can be expressed in components suitable for the situation.

Like I said in my previous post #29, you have discovered a graphical relationship that relates speed to the reciprocal of gamma but it has nothing to do with your explanation of a light clock.

Your problem is that you claim that there is something significant when a photon hits the moving rod at the 80% mark. In fact, there are photons hitting both rods all along their trips to their respective mirrors. So what? There is no significance to the fact that a photon hits a rod at any particular time. What matters is when a photon hits the moving mirror, which you don't show. If you would continue the diagonal line up to the location of where the mirror would be when it hits it and then measure the time it takes for the photon to hit the mirror, you would see that it take 1.25 times as long as it takes for the photon to hit the stationary mirror which gives the correct illustration "in terms of physical processes".

 

[phyti]

Like I said in my previous post #29, you have discovered a graphical relationship that relates speed to the reciprocal of gamma but it has nothing to do with your explanation of a light clock.

The hyperbola is the inverse function of the quarter circle, a fact of geometry.

Your problem is that you claim that there is something significant when a photon hits the moving rod at the 80% mark. In fact, there are photons hitting both rods all along their trips to their respective mirrors. So what? There is no significance to the fact that a photon hits a rod at any particular time. What matters is when a photon hits the moving mirror, which you don't show. If you would continue the diagonal line up to the location of where the mirror would be when it hits it and then measure the time it takes for the photon to hit the mirror, you would see that it take 1.25 times as long as it takes for the photon to hit the stationary mirror which gives the correct illustration "in terms of physical processes".

The rod is only a visual aid to indicate a rigid device moving with A, and represents the photon path as imagined by A. It helps with those new to SR. The photon doesn't hit the rod. The successful photon has to have the correct angle to intercept the A mirror. How would it 'know' this? It doesn't. As stated, the successful photon has a horizontal vt component that equals the speed of the clock. The vertical st component moves toward the mirror but at less than c. The intersection of the arc of photons and the path/(rod) to the mirror is determined by the speed of A. The angle of the ct photon relative to vertical is a function of v/c. A simple examination shows, a lesser speed, a smaller angle and a greater s,
a greater speed, a greater angle and a lesser s.

The drawing is intended to show why 1 tick for the A clock is longer than 1 tick for the U clock for the general case, which it does. The reason is the constant and independent speed of light. The degree of dilation is a function of v/c. If the path is extended, it will not show anything new. Gamma is still the ratio c/s or (A time)/(U time).

 

[ghwellsjr]

If the path is extended, it will not show anything new. Gamma is still the ratio c/s or (A time)/(U time).

Exactly and that's because the ratio between ct and st is the same for all t. You just have a different size triangle involving ct, st and vt. I just don't know why you think there is anything significant for the specific value of t corresponding to when the photons hit U's mirror. If you had used the value of t corresponding to when the photons hit A's mirror, then st will be equal to r, the distance to U's mirror from O, and ct would be the distance to A's mirror from O, and ct/st = c/s = gamma. If you would draw that diagram, which is the diagram that everyone else draws, then it seems like it would be more helpful to those new to SR.