# Explain how light moves in relative motion due to special relativity?

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1. Apr 2, 2014

### iliedonUA

I've been told on many occasions that light will always be measured at the same speed regardless of measuring equipment speed as long as there is no acceleration and you are moving the same speed relative to that object you are measuring from

if that is incorrect please ignore what i say below and tell me why because it doesn't make any sense to me either

i know that this isn't an affect of time dilation, but i will include that to justify it

there is no human observer, just robotic, pretend we have a train, the train is moving in outer space at 83% of the speed of light where time dilation is slowing time by half, let's say the train is 1 light hour long, now... this is some extreme rounding here but let's say from still observer outside the train, the train is moving at let's say 5/6th the speed of light, so if from inside the train, light was shined from the back of the train to the front, it would take about 6 total light hours for the light to reach the front of the train (again more rounding)

however there is a detector at the beginning and end of the train that are part of the train, if what i said in the beginning of this thread is true, then to the machine equipment on the train measuring the light, the fact that the train is moving 83% is ignored and the equipment measured the light on the train would measure that the light has went from the back to the front of the train in 1 hour

alright in the image above there are six gaps (i fully realize the image and gaps are inaccurate), each gap is 1 light hour - the length of the train

the blue line is the path of light, starting from the right and moving left

we'll pretend that the green line is a very rough estimation of the light's current position form the perspective of an outside observer, from the outside observers stationary perspective, it will take 6 hours to complete the journey from the front to the end of the train, the train is only 1 light hour in length, but it's moving so from an outside non-moving perspective we'll say 6 hours

okay, so basically the onboard detectors should say the (according to what i said at the very beginning) that the light took 1 hour to reach the front of the train? how is this possible? even at the brown mark, if time was slowed by half for the detector 1.5 hours should've passed EVEN AT THE HALF WAY POINT, so that's already more than 1 hour passed for the onboard detector at the end of the train before the light is even close to reaching it? how is it supposed to reach the back detector in only 1 hour, time dilation is curved, not linear so please someone explain?

2. Apr 2, 2014

### Staff: Mentor

You have forgotten to take the relativity of simultaneity into account. (This is the most subtle of the three main relativistic effects, and the one at the root of most confusions.)

According to the 'stationary' (embankment) observers, the light does take about 6 hours to go from the rear of the train to the front. And according to the train observers, that trip only takes about 1 hour. But realize that according to the stationary observers, those clocks at the front and rear of the train that were used to measure that 1 hour were not only running slowly (time dilation) but were wildly out of sync (relativity of simultaneity). According to the stationary observers, the clock in front of the train is way behind the clock in the rear--that's why the train observers only measure a one hour time of travel.

3. Apr 2, 2014

### Staff: Mentor

A massive thread derailment has been removed.

iliedonUA, as Doc Al mentioned, the key is the relativity of simultaneity.

4. Apr 3, 2014

### ghwellsjr

That is correct.

You've got some misconceptions going on here. First off, at 83%c, time doesn't slow by half, that happens at 86.6%c. You should be able to see that from your graph. At 5/6th c (83.33%c) the time dilation factor is 1.8 (a time slowdown to 0.55). And more significantly, it won't take anywhere near 6 hours for the light to get from the back of the train to the front for an observer outside the train, it's more like 3 1/3 hours (200 minutes). How did you arrive at 6 hours?

That is correct.

Are you thinking that if the light took 6 hours for the outside observer, then if the clock was running at about half speed for the train, it should take 3 hours for the total time and at the midpoint of the train it would take 1.5 hours? Otherwise, I have no idea what your thought process is.

But even if you used the correct factors, that the light takes 3 1/3 hours for the outside observer and the train clocks slow down to 0.55, then you would conclude that it got to 1.83 hours for the full distance and half that at the midpoint is 0.92 hours (55 minutes) which seems way to big and it is, it should be exactly 1/2 hour.

The way to solve a problem like this is to start in the rest frame of the train showing the propagation of the light from back to front and passing a clock at the midpoint. Here is a spacetime diagram depicting the scenario:

The back of the train is shown as the thick blue line, the midpoint as the red line and the front of the train as the black line. The thin blue line shows the propagation of the light going from the back to the front of the train. The dots mark off 10-minute intervals for clocks at the back, midpoint and front of the train. You can see that the light reaches the midpoint at 30 minutes and the front at 60 minutes.

Now we transform the coordinates of all the dots to the frame of the outside observer at a speed of 0.8333c and we get this spacetime diagram:

Now you can see the slowdown of the train's clocks as they take 18 minutes of coordinate time for every 10 minutes of their own time. One hour is stretched out to 108 minutes. But, more importantly, the clocks are offset in this frame. This is the Relativity of Simultaneity that Doc Al mentioned and it provides the means by which the clock at the midpoint reaches 30 minutes even though a naive calculation would have put it at 55 minutes, which, by the way, is the correct answer for the time on the blue clock at the back of the train.

Does this all make sense? Any questions?

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Last edited: Apr 3, 2014
5. Apr 3, 2014

### abitslow

"I've been told on many occasions that light will always be measured at the same speed regardless of measuring equipment speed as long as there is no acceleration and you are moving the same speed relative to that object you are measuring from " Is incorrect.
In ANY inertial frame (non-accelerating, not in a gravitational field) the speed of light (in a vacuum) is constant. Period."Constant" means "the same". C is the same in ALL inertial frames.
Your comment about "you" "moving the same speed relative to that object" is pretty poor English, I am having trouble understanding what you could possibly mean. There is absolutely no requirement that the source and detector must be moving at the "same speed". Even if we give you the great benefit of the doubt and allow your use of "speed" to mean "velocity". (Velocity is a vector. It has magnitude AND direction.)
Others have pointed out some of your other errors. One thing you need to be clear about and which is rarely stated is that in a two dimensional chart of space-time (usually time increasing vertically up, and one dimension of space increasing (distance from the origin) to the right is not like a graph of a Euclidean 2d plane. For one observer you can draw the coordinates (x,t) say orthogonally, (at right angles) for a different observer (at a non-zero velocity relative to the first) the (x',t') coordinates are drawn in that diagram as lines (axes) which meet at less than a 90° angle. The clear implication of this is that you need to be very careful in applying the same concepts you used when you were using plane geometry, and learning how to graph to this situation. On this chart, the shortest distance between two points is NOT necessarily the shortest line between them on the chart. Other errors you made include your assumption that the train HAS a property called "length" and that this property isn't a function of the rest frame used to measure (determine) it. Length and time are relative to the rest frame used to measure them, two different observers in different rest frames (regardless whether they are inertial or accelerating) will be unlikely to agree on length or time.

Last edited: Apr 3, 2014
6. Apr 3, 2014

### iliedonUA

well you've certainly all given me something to think about, if i don't reply to this in a few more days assume i get it, etc, thanks, i didn't really mind the derailment from earlier, and yes i messed up the 83% should've been 86% i went from memory

7. Apr 3, 2014

### ghwellsjr

His statement was very clear to me and there was nothing incorrect about it. He was talking about measuring the speed of light which does not require establishing or knowing anything about inertial frames or knowing anything about Special Relativity. It's just an experiment that anyone can perform and they will all and always get the same answer if performed the way he described it.

Yes, it is, as long as the 2D plane includes time vertically and 1D space going horizontally.

Or you can draw the coordinates for the second observer on a different diagram with orthogonal axes, as I did, nothing wrong with that and it's so much easier to understand. Your way is also valid but I'll bet you won't draw a single diagram showing the coordinates for both observers as you described. Please prove me wrong and show us that diagram and then explain it in a way that the OP can understand it.

This is a very confusing statement because you left out a whole bunch of assumptions. You need to explain it in more detail.

But the train has a property called Proper Length and its clocks have a property call Proper Time which all observers will agree on. You should make it clear that different observers, or rather, the rest frames of different observers will assign different coordinates to the events of the scenario as my two diagrams clearly indicate and that is what you are talking about.

There is nothing wrong with my diagrams or my explanation. Rather than complain about my post, I again invite you to draw the diagram that you think is valid and provide your own explanation. Can you do it?

8. Apr 3, 2014

### ghwellsjr

No matter, it would make only small differences in the explanation and diagrams.

9. Apr 4, 2014

### Simplyh

How can this be compatible with Lorentz deduction of LT; Einstein deduction of LT; the Principle of Relativity and all the logical frame of deduction of SR?

10. Apr 4, 2014

### ghwellsjr

He's talking about the very popular way of drawing a Minkowski diagram such as:

I'm still waiting for him to draw one depicting the OP's scenario.

11. Apr 6, 2014

### Simplyh

However this does not answer my question: how can two equivalent frames of reference be represented by systems of axis with different angles? The mathematics are obvious but do not correspond to the Principle of Relativity from which all the theory began.

I think the explanation on your link to Wikipedia is not well developed and I never found any explanation I could consider as logic. I wish to know if you can explain, on a better way, the logics that allow the compatibility of that geometry with SR principles, in particular, the Principal of Relativity.

Divirtam-se (have fun)

12. Apr 6, 2014

### Staff: Mentor

Why not?

The principle of relativity says only that the laws of physics need to be the same in different frames. In other words, the principle of relativity says that if you start with some physical system A in an inertial frame S and if you transform from S to S' then the transformed system A' will behave according to the same laws of physics.

In other words, if L is a law of physics in S then the corresponding law of physics in S' is L'=L. The principle of relativity places restrictions on the relationship between L and L', not on the relationship between A and A'.

That certainly doesn't seem to restrict the transformations in the way that you seem to think it does. Can you provide some explanation about why you think this?

13. Apr 6, 2014

### Staff: Mentor

That representation is just two lines drawn on a piece of paper as an illustration. The non-square angles between those lines are no more "real" than the non-square angles and intersecting parallel lines that appear in a perspective drawing of a three-dimensional scene on a two-dimensional sheet of paper.

It is very unfortunate that there's no easy way of drawing a picture of both coordinate systems on the same sheet of paper and making the lines that represent the axes intersect at right angles for both sets of axes at once. When we choose (completely arbitrarily) which set of axes will be at right angles in our picture, we also confuse people into thinking that that set is somehow "real" or "correct" or "natural" or "special" - in fact it has no physical significance whatsoever and we could just as easily have drawn the picture so that a different observer's axes intersected at right angles in the picture.

Last edited: Apr 6, 2014
14. Apr 8, 2014

### Simplyh

Dalespan:
You are correct but, in this case, we have two different kinds of angles between the space and time coordinates which, according to Nugatory, is just a problem of our capacity to, lets say, draw it better. If this is the case, then we all agree.

Divirtam-se (Have fun)

15. Apr 8, 2014

### Staff: Mentor

Yes, I agree. We have different kinds of angles between space and time coordinates. Angles involving only space coordinates are measured with protractors and angles involving time coordinates are measured with speedometers. As long as protractors and speedometers work the same in all frames the principle of relativity is satisfied.

This is conceptually similar to how space coordinates are measured with rods and time coordinates are measured with clocks.

16. Apr 8, 2014

### ghwellsjr

If your concern is merely with the angle of the coordinates, maybe this animation might help: