# Lorentz Transformation in Different Reference Frames

TO ALL: Thanks to the help of many dedicated forum members, I have learned that this problem can be explained by understanding Minkowski diagrams. Here is a consolidated list of helpful links: https://www.khanacademy.org/science/physics/special-relativity/minkowski-spacetime-2016-01-18T22:56:14.718Z/v/starting-to-set-up-a-newtonian-space-time-diagram https://www.physics.byu.edu/faculty/allred/222%2011/minkowski%2011.pdf

How can the speed of light remain constant in reference frames that experience time differently?

Maybe some context would be helpful, so let’s do a quick thought experiment: Imagine that you are standing on the platform of a railway station. A train approaches you at 99% the speed of light. It’s headlight flashes when the locomotive is directly in front of you*. At that instant, what do you see?

• Note: The fact that the headlight is directly in front of the observer when it flashes is crucial to this version of the thought experiment. The light source is not moving away from the observer in any direction, thus simplifying things by eliminating the relativistic doppler effect.

Although an overall analysis of this thought experiment would be greatly appreciated in the comments, there is a specific question that I am trying to answer: Is the absolute, (as opposed to observed) speed of light truly constant? Let’s take two perspectives to examine this scenario from. An engineer in the locomotive would observe that the photons emitted from the headlight are traveling at the speed of light, C, relative to him. He is also experiencing time at the same rate, or is in the same “time bubble” so to speak, as the light source because they are both moving at the same velocity. On the other hand, at what speed would the photons appear to be traveling from the perspective of the (relatively) stationary observer on the platform? This person is in a different time bubble than the engineer and the light source, since he is traveling at a different velocity. If the observed speed of light is constant, then the person on the platform would observe that the photons are traveling at C, relative to him. However, wouldn’t this mean that light was traveling at two different absolute speeds? In fact, it seems as though the absolute velocity of the photons would be greater for the observer on the platform than for the engineer, since the person on the platform observes the photons traveling away from the moving train at the same velocity that the engineer observers them when moving along with the train.

Essentially, how can time dilation explain why the observed speed of light remains constant in a scenario where the observer experiences time at a different rate than the light source which is being observed?

• According to special relativity (SR), motion is relative and so there is no notion of absolute speed (although there is an invariant speed, $c$). What do you mean by absolute when you write "However, wouldn't this mean that light was traveling at two different absolute speeds?"? – Hal Hollis Jun 12 '17 at 15:47
• I understand that all motion is relative and C has been observed to be invariant. To my understanding, Lorentz transformations adjust all variables in order to maintain a constant ratio of Distance/Time. However, how can these variables be adjusted proportionally if the observer is not experiencing the same Lorentz transformations as the light source? Perhaps an expansion of the original question, (in my response to Willy Billy Williams below) would also be helpful in understanding the problem. – Joe Smith Jun 12 '17 at 21:03

(From a comment by the OP)

To my understanding, Lorentz transformations adjust all variables in order to maintain a constant ratio of Distance/Time. However, how can these variables be adjusted proportionally if the observer is not experiencing the same Lorentz transformations as the light source?

First, observers don't experience Lorentz transformations.

The Lorentz transformation is a coordinate transformation that relates the spacetime coordinates in one inertial reference frame (IRF) to the spacetime coordinates of another, relatively moving IRF.

To be clear, there is no preferred IRF from which we 'Lorentz transform' from. For an inertial observer, all other relatively moving IRFs that are moving while she is at rest.

Second, if an entity has speed $c$ in one IRF, that same entity will have speed $c$ in all IRFs; this is easy to show using the Lorentz transformations.

Third, the Lorentz transformation preserves the spacetime interval, a kind of 'distance' through spacetime. Two events, A and B have an invariant interval $\Delta s^2_{AB}$; all inertial observers find the same interval for events A and B.

Fourth, in special relativity (SR), there is a distinction between observing (recording the spacetime coordinates of an event) and seeing (taking a photograph).

It’s headlight flashes when the locomotive is directly in front of you*. At that instant, what do you see?

You observe (record the coordinates with your rods and synchronized clocks at rest) that the headlight has flashed at that instant and at that position but you don't see the light until later because it must first propagate (at speed $c$) the distance between the headlight and your eyes (or camera).

If event A is the headlight flashing and event B is the arrival of the light at your eyes (camera), you will find that the interval $\Delta s^2_{AB}$ is null (light-like).

However, other observers will disagree on the spatial distance $\Delta x_{AB}$ and the temporal distance $c\Delta t_{AB}$ but they all agree that $\Delta x_{AB}^2 = (c\Delta t_{AB})^2$ since the interval is null.

• Thank you for this explanation. Combined with a new and clearer understanding of Minkowski diagrams, (thanks to Willy Billy Williams) it all makes sense. This was a problem that I had difficulty understanding for years, but have only now had the courage to post on a forum. However, I am sure that there will be many more questions and thought experiments to come. – Joe Smith Jun 12 '17 at 23:10

They both measure the same speed, which is independent of the speed of the source. The two observers not only see that the other observer's clock run slower, they also see length contraction and clock desynchronization. Suppose the guy on the train synchronize clocks with the guy on the platform at the time the pass each other and the light is emitted. Both observers measure the speed of light by dividing the distance to the next station to the time it takes the light to reach the station. The observer in the train will see that the distance is smaller. He will also see that the clock at the next station is out of synchrony with the one at the original station. If, instead of a clock at the next station, the clock at the emitting station were used, then the observer on the train will see that the person will stop the clock after the light, according to him, has reached the second station.

• I understand that length contraction is a factor in this thought experiment. However, though I might be misunderstanding you explanation, I do not think that it fully explains what the observer on the platform will see. If photons always travel at the same velocity, then how can they appear to be "outrunning" the train (projecting away from the headlight) at C if the train has an initial velocity from the perspective of a (relatively) stationary observer? – Joe Smith Jun 12 '17 at 20:38
• (Cont.) If the speed of light is constant, why wouldn't the observer on the platform see the light traveling away from the train at (C - Velocity of the Train)? However, I must admit that my understanding of relativity is far from that of an actual physicist. Please excuse my elementary misunderstandings of this topic if any arise. – Joe Smith Jun 12 '17 at 20:53
• You are thinking in a galilean way. The second intuition is correct, the guy in the platform sees that the difference in velocity between the light and the train is c-v, He will not see the light running at c+v, will see that light travels at c. Even if the train were moving at 0.99999999% c, both observers will see the light moving at c. That is why space and time get mixed, it would make no sense if time were absolute. – user126422 Jun 12 '17 at 21:16
• I cannot make an easy explanation unless you are familiar with minkowski diagrams. If you could learn that, everything will make sense en.wikipedia.org/wiki/Minkowski_diagram – user126422 Jun 12 '17 at 21:18
• Thanks for the link! I will do more reading on Minkowski diagrams in the coming days. To clarify a previous point though, are you saying that the speed of light is constant by the speed at which light is observed to be traveling away from a moving object is variable? – Joe Smith Jun 12 '17 at 21:52

I think you are getting confused about the simultaneity of events.

The person on the platform observes the train passing and the headlight as being turned on simultaneously.

From the point of view of the train driver the two events do not happen simultaneously.

Both see the light travelling at c.