The Lorenz Kinematic Transforms (truly) Explained

In the last post, I discussed relativistic time dilation, length contraction, and the twin paradox. Unlike that post, however, the following is not just conceptual; tricky algebraic and kinematic calculations are required.

 

The Lorenz kinematic transforms are a collection of general formulae to transform kinematical calculations from one frame to another: the distance transforms, the time transforms, the velocity/velocity-addition transforms, and the acceleration transforms. They are the relativistic equivalent of the Gallian kinematic transforms, the technical term for the formulas that convert kinematic measurements from one frame to another in simple distance-rate and (constant) acceleration problems. Older than relativity itself, the transforms were originally—and ironically—derived to account for the antiquated “ether wind.”  

 

This post consists of semi-original derivations of the inertial formulas (excluding acceleration), each detailing the steps of their respective events. Most relativity techniques are “analog methods,” similar to graphical techniques but, in this case, even more analogic. None of them made any sense to me, whatsoever--and I doubt they could have satisfied me even if they did. Actually wanting to understand what’s “really happening” during these processes, I was forced, yet again, to discover them myself. This is the result of those inquiries.

 

 

Although the transforms could have been used, events in the previous post occurred at the same place within moving frames, such as the clock in the front of the spaceship at the beginning and end of its trip across the ruler. If an event, such as the previous, occurs simultaneously in both frames, the ground frame can know what time the ship frame’s clock read at that moment, even if they disagree on the accuracy of the clock, and the same would be true when it reaches the end of the ruler. In these cases, the general equations would have reduced to the formulae that were used. Otherwise, there is no “actual” time in which something occurred in relativity. Things get trickier when events in the moving frame occur at different— that is, TWO--locations within that frame, such as when one occurs in the front of the ship and the other in back. 

 

The most illustrative example of the application of the Lorenz transforms is when a projectile is thrown across a moving vessel and the relativist wants to know the ground frame’s calculations as a function of the other frame’s variables. Such a scenario would be analogous to a frame, such as a spaceship, calculating the velocity of another spaceship as a function of a third. But it doesn’t have to involve a nonrelative velocity: Lightning could strike both ends of a train and the time transform equation can transform the time between the events on the train frame to the ground and vis versa. You could also use the velocity-addition transforms to measure a HYPOTHETICAL velocity--how fast would a stream of fire-retarding fluid have to travel across the train to prevent a fire from starting upon the second strike if it were released immediately following the first, as would be measured by the ground?    

 

First, the clocks must be synchronized in the ship frame, one where the ball will be thrown (the “origin clock”), the second where it will land (the “destination clock”). For simplicity, I’ll say the ball will be thrown at the back of the ship, which is moving right, and lands at the front. To sync the clocks, the ship frame places a light projector at equal distance between the clocks; once the pulses initiate the clocks, the ball will be thrown. But the clocks can’t be synchronized from a second frame if the synchronizing light pulses are projected in the plane of relative motion. 

Questions: Viewed from the ground looking at the ship’s measurements: What is the CLOSING SPEED (the speed the light pulse and the front/back toward each other) at the front of the ship? What of the back? What distance will they be “closing in” on? In other words, what distance will they have to travel as a function of the moving frame? Calculate the “lag time” between the clocks, or how much ahead one will be than the other? And when calculating time once the ball is thrown, does it matter which clock is used.

 

While the light pulses still move at C, the motions of the clocks alter the CLOSING SPEED between them and the pulses: C+V in the case of the pulse and the origin clock, since it moves TOWARD the light, and C-V in the case of the destination clock, which moves AWAY from the other pulse. And they will be “closing in” on the contracted length of the ship. Thus, the delay between them is ∆X’/2y/C-V -- ∆X’/2y/C+V (y=(1-V^2/C^2)^1/2; see the last post). When the calculations are made, either clock can be used to find T as a function of T’, which I’ll explain. 

 

The algebra works out to be yV’∆X/C^2. I call this “the lag time” between the clocks. If the ship frame where instructed to throw the ball at the moment the start clock is activated and to stop both clocks when it reaches the destination clock, they will both have to calculate the same change in time (note that since I’m dealing with what I’ve called the frame measures, this disregards nonrelativistic delays such as the delay, in both frames, for the light of the event of the ball reaching the destination clock to reach the origin or start clock.) 

 

Questions: Will the origin clock start on time, early, or late? Will it end on time, early, or late? What about the destination clock? 

 

The origin clock will BEGIN on time but will stop yV∆X’/C^2 EARLY, since it can’t calculate more time than the other clock and must stop ticking before it; the destination clock will start that amount LATE but END on time. Each clock measured too LITTLE time by an amount equal to the lag time. Whichever clock you use, you, therefore, must add the lag time above to ∆T’y to get the time from the ground. Total time equals y(∆T’ + V∆X’/C^2).

 

Questions: In the case when the ball moves left as the ship moves right, will the destination clock, previously the origin clock, start on time, early, or late? Will it end on time, early, or late? What about the other clock? 

 

The destination clock, previously the origin clock, will BEGIN early, being the one the ground perceives to be ahead, but will END on time, given its arrival and the ceasing of the clock happen simultaneously from both frames. The origin clock, previously the destination clock, must BEGIN on time, given that event happens simultaneously with the throwing of the ball from both frames, but will END LATE, since it’s behind. Each clock measured too MUCH time by an amount equal to the lag time. Either way, you must SUBTRACT the lag time. Next,

I will calculate change in position from the ground’s view as a function of the ship frame in the first scenario above, where the ball moves in the direction of ship’s motion. From the ground frame, the ball will travel the contracted length of the ship plus whatever distance the ship traveled during that time. This equals ∆X’1/y Vy(∆T’ + V∆X’/C^2). This comes out to be y(∆X’ +V∆T’). In the case of the ball moving in the opposite direction: the ball travels the contracted length of the ship minus the distance the ship traveled, ∆X’1/y – yV (∆T’ – V∆X’/C^2). This comes out to y(∆X’-V∆T’).

 

Now I’m ready to calculate the ball’s velocity, U’, from the point of view of the ground, U. 

Setting ∆X’ to U∆T’, this comes out to U plus or minus V/ 1 plus or minus V∆X’/C^2.

 

 

The Z and Y Axis’:

 

Although a projectile can be thrown in the Y and Z plane WITHIN the moving frame, since the frame measurer is always perpendicular to motion, a frame, by convention, always moves exclusively along the X axis.

 

Questions: Will there be length contraction in the Y and Z planes? If two clocks are on the same point on the X plane, will synchronizing pulses have to travel more, less, or equal distance in the frame that see the clocks in motion? What does this mean as far as whether those clocks can be synced? Will the clocks be dilated? What is Z as a function of the moving frame? What is T as a function of moving frame? And, finally, what is Uz?

 

The Y and Z lengths/distances are always agreed upon, and, from the point of view of the frame measurer, the ship doesn’t move vertically or laterally. Because the distances synchronizing light pulses will travel are, therefore, always equal between moving clocks on the same point on the X axis regardless of their location on the Z and Y, all clocks at the same point on the X will be synchronized. 

 

Questions: What is ∆X as a function of the moving frame? What is ∆T[T’]? What is U as function of the moving frame?

 

Distance in the Z plane as a function of the moving frame will always be Uz ∆T’, since it’s agreed upon. The clocks will move slow but no more so than usual, making ∆T[T’] ∆T’y. Thus, U in the Z as a function of the moving frame will be Uz∆T’/∆T’y=Uz /y.     

 

Because the transforms are invariant, whatever they reduce to is also invariant. If you plug the time and distance transforms into (∆CT)^2 -∆X^2, they, conveniently, reduce back to (∆CT)^2 -∆X^2. Thus, (∆CT’)^2 -∆X’^2 always equals (∆CT)^2 -∆X^2. Hint: Multiply C and T before squaring and group the ∆X’ terms with each other and the ∆T’ terms with each other.  

 

 

Hopefully, you successfully answered the foregoing questions and are satisfied with your resulting understanding. Relativistic kinematics is a peculiar subject. It’s effects on reality make it as titillating and important as anything in physics, prompting a great deal of interest, but it’s exploited as much for entertainment and the promotion of science as the advancement of science itself. On the one hand, it’s very basic and could in theory be understood by anyone skilled in naught but basic kinematics and algebra. On the other hand, it’s the most basic branch of theoretical physics, and its courses are taken by those who’ve already advanced beyond freshman physics and calculus; most of its pedagogy and techniques are designed to accommodate the language, needs, and inclinations of those trained/being trained in far more advanced subjects. Though I’m more disinclined toward graphical and analogical techniques than most, special relativity, nonetheless, seems devoid of an accessible and effective curriculum for the “serious laymen,” such as engineers and philosophers with an avocational interest in physics.

 

This and the other relativity posts are an attempt to create such curriculum. Feedback is greatly desired.