# Thread: Tutorial: Relativity - what is a reference frame?

1. ## Tutorial: Relativity - what is a reference frame?

In many discussions of Einstein's theory of Relativity on sciforums, I have found that people do not have a clear concept of what a Reference Frame is. So, here is a short primer on frames of reference. I am hoping we can link back to this thread when this kind of confusion crops up in future. Comments and discussion are welcome, of course.
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Here's what a reference frame is, approximately. We'll work in one spatial dimension for simplicity. Imagine you have constructed 100 identical straight sticks of equal length. The exact method of construction and the exact length is not important, as long as the lengths are the same when you lay one stick on top of another.

Now, to constuct a reference frame, you lay out your 100 sticks end-to-end along a straight line. You now have a distance scale. If you want the distance from the end of the row of sticks to some other point on the line, you just count how many sticks there are between the end and the point you're interested in.

Now, as well as a distance scale, we need some clocks. At every intersection of two sticks along the line, we place a clock. All of the clocks along the line are adjusted so that they read the same time always. Again, we won't worry about the exact method used to synchronise them for now; we just assume they are synchronised.

So, a reference frame is this set of sticks and clocks.

Now, suppose we number the intersections between any two sticks. We start at the end of the row (let's say) and call that position zero. The next intersection is position 1, the one after that position 2 etc. We now have a definition of position relative to our reference frame.

Imagine a ball located at some particular intersection, say number 57. Then we say the position of the ball is at x=57 (sticks).

Next we define velocity. If an object moves from one position to another along our line of sticks, then while it is moving it has a velocity equal to the number of intersections crossed divided by the time taken (as measured on the clocks). For example, if the ball moves from location x=57 to x=58 in 2 seconds, then it has a velocity of 1 stick per 2 seconds or 0.5 sticks per second.

Finally, we have acceleration. Acceleration is the rate of change of velocity. For example, if the ball changes speed from 3 sticks per second to 5 sticks per second in a time of 1 second, then its acceleration is 2 sticks per second per second, or 2 sticks per second squared.

Now, let's be clear about what is meant by the term "at rest". An object, such as the ball, is at rest in our reference frame if its position is not changing with time. In other words, if it stays at x=57 then it is at rest with respect to this reference frame. Its velocity in this case is the change in position over time, and since its position is not changing, the velocity is zero when it is at rest. "At rest" says nothing about acceleration. "At rest" simply tells you velocity is zero.

Now, consider a different type of motion. Imagine the ball is "stationary" (whatever that means), and we move our whole assembly of sticks and clocks past the ball. Can you see that, in the reference frame we have constructed, the ball now has a velocity? The reason is that the stick intersections (57,56, 55, 54...) are moving past the ball, and so its position in the reference frame is changing over time. Therefore, in the reference frame of the sticks/clocks, the ball has a velocity, because of the way we defined velocity.

We could achieve the same velocity for the ball in a different way, of course. We could hold the stick and clocks "stationary" and move the ball along the sticks, instead of holding the ball stationary and moving the reference frame. In either case, the speed of the ball would be the same in the reference frame.

Now, suppose we put our whole assemblage of sticks and clocks in an aeroplane which is flying above the Earth. And suppose we place our ball at x=57 and leave it there.

Now, relative to our reference frame of sticks and clocks, the ball is at rest. Recall that we defined "at rest" to mean that an object does not change position, and position is defined by the intersections of the sticks. So, our ball is at rest in this reference frame. Note that the motion of the plane is completely irrelevant to whether the ball is at rest in the reference frame. As long as the ball stays at x=57 it is at rest, by definition.

Ok, so now we have a definition of "reference frame", "position", "velocity" and "acceleration".

So, let's now consider TWO reference frames, by which I mean two identical sets of sticks and clocks laid out parallel to each other. Obviously, we can slide one entire set of sticks along the other set of sticks, so that each set of sticks has a velocity relative to the other set. For definiteness, let's assume on set of sticks is laid out on the Earth (attached to it) and the other set of sticks is laid out in our aeroplane, which is flying parallel to the Earth's surface, in the direction that both sets of sticks are laid out.

Now consider the ball on the plane. Suppose it just sits in the aisle at location x=57 relative to the sticks on the plane. Then, the ball is at rest relative to the plane's reference frame. But the ball obviously has some velocity relative to the ground's set of sticks, its position on those sticks is changing as the plane flies past.

So, is the velocity of the ball "really" zero, as indicated by the plane's reference frame, or is it "really" 800 km/hr (say), as indicated by the Earth's reference frame? Answer: it is BOTH.

Or, to put it another way, it makes no sense to say "The ball's velocity is zero" unless you specify which reference frame you're using. You ought to say "The ball's velocity is zero in the reference frame of the plane", or equivalently "The ball's velocity is zero relative to the plane." Obviously, at the same time the ball's velocity is NOT zero relative to the ground.

More points to appreciate: EVERY reference frame - every set of sticks and clocks - is attached somewhere. The position of any given object may be different in different reference frame, as may its velocity and acceleration. Also, it makes no sense to define "position", "velocity" or "acceleration" in the absence of a reference frame.

When it comes to light, we can consider a light photon as just another ball moving along our lines of sticks. If light covers a certain number of sticks in a certain amount of time (as measured by the clocks at the intersections) then light has a certain speed determined by the reference frame. Up to this point I have talked about lengths in terms of stick lengths. Obviously, if we can somehow ensure that each stick has length 1 metre, then we can get velocities in metres per second.

Ok, that's enough background.

Clearly, it is a precursor to any discussion of relativity to agree with the above description of what a reference frame is, as well as definitions of position and velocity. Because if we can't agree on these basics then we're talking at cross purposes right from the start.

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Reference: Most of this post was extracted from a long discussion about relativity. The corresponding post is here:

2. Good explanation, and good of you to help the non-technical people understand this. It's free tutoring--who can beat that?

3. Excellent tutorial James. Good work!

4. James, Can you elaborate a little more on how you would go about determining that a stick in your frame is 1 meter, using light (since a meter is defined by light travel time), while keeping in mind that light travels independently of objects? Also, speaking strictly about light and time, in what reference frame are the sticks laid out for the light sphere??

James, It's clear that you can be stationary compared to the bus while remaining seated on the bus, and also be traveling with respect to the road when the bus is in motion compared to the road. Nobody disputes that!! Get a grip on yourself.

James, Can you elaborate a little more on how you would go about determining that a stick in your frame is 1 meter, using light (since a meter is defined by light travel time), while keeping in mind that light travels independently of objects?
The length we call 1 metre is defined to be the distance that light travels in 1/299792458 seconds. The second is defined independently of the speed of light and the metre.

So, to determine that a stick is 1 metre, you set off a light pulse from one end of the stick and let it travel for 1 second. Then, you arrange (somehow!) to mark off how far it has gone after 1 second on your stick. The distance between the start point and the time that the light reached after 1 second is 1 metre. In practical terms, this isn't quite how you'd do it, but that's the general principle. I'm sure you can fill in the technical details.

Also, speaking strictly about light and time, in what reference frame are the sticks laid out for the light sphere??
The sticks are the reference frame, as explained in the opening post. (This is a conceptual picture, of course.) They aren't "laid out" in some superior reference frame or something.

If you want a reference frame, just follow the instructions about laying out sticks and clocks, as given in the opening post. You needn't worry about anything else.

James, It's clear that you can be stationary compared to the bus while remaining seated on the bus, and also be traveling with respect to the road when the bus is in motion compared to the road. Nobody disputes that!! Get a grip on yourself.
I haven't talked about buses at all in this thread. What are you talking about? Get a grip on yourself.

6. Originally Posted by James R

The length we call 1 metre is defined to be the distance that light travels in 1/299792458 seconds. The second is defined independently of the speed of light and the metre.

So, to determine that a stick is 1 metre, you set off a light pulse from one end of the stick and let it travel for 1 second. Then, you arrange (somehow!) to mark off how far it has gone after 1 second on your stick. The distance between the start point and the time that the light reached after 1 second is 1 metre. In practical terms, this isn't quite how you'd do it, but that's the general principle. I'm sure you can fill in the technical details.
K. So you are at a constant velocity in a ship, not accelerating. You send a light signal from the rear of the ship and find where the light hits after 1/299792458 of a second. The distance the light traveled in 1/299792458 of a second in that frame is a meter, correct? Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?

7. Originally Posted by Motor Daddy
K. So you are at a constant velocity in a ship, not accelerating. You send a light signal from the rear of the ship and find where the light hits after 1/299792458 of a second. The distance the light traveled in 1/299792458 of a second in that frame is a meter, correct? Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?
Which sticks?

8. Originally Posted by James R
Which sticks?
The set of sticks nailed to the floor of the ship. The same set of sticks in both tests in which you marked on the sticks where light was after 1/299792458 of a second. You know, THE STICKS!

9. Originally Posted by Motor Daddy
K. So you are at a constant velocity in a ship, not accelerating. You send a light signal from the rear of the ship and find where the light hits after 1/299792458 of a second. The distance the light traveled in 1/299792458 of a second in that frame is a meter, correct? Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?
Yes, it will always be the same. That's just a small part of the beauty of Relativity.

James, It's clear that you can be stationary compared to the bus while remaining seated on the bus, and also be traveling with respect to the road when the bus is in motion compared to the road. Nobody disputes that!! Get a grip on yourself.
Then you do understand the meaning of "Frame of Reference"? In this case the frame of reference is the bus, not the road. If the bus was traveling very near lightspeed you would still measure the speed of light and the distance it travels/time it takes exactly the same as if the bus was stationary to the road(paint over the windows and you could not tell whether you were stationary or moving very fast), but the effects you experience(speed of time, length in the direction of travel etc.)would be different from effects the people standing beside the road measure IN YOUR FRAME, and that is Relativity. IE what you experience within your frame is always the same as measured within that frame, but is relative to all other frames. This is a simple fact, get used to it.

Grumpy

11. Originally Posted by Aethelwulf
now $x' = 0$ actually corresponded to $x = vt$ where $x'=0$ described the origin of the moving observer.
You know, I have a nagging suspicion you only know that because I've written almost exactly the same thing many times in various relativity related threads here over the last few years.

12. Originally Posted by Aethelwulf
I have no idea what you are talking about.
If you set the sign convention on the velocity such that

$
\begin{eqnarray}
t' &=& \gamma (t \,-\, \frac{v}{c^{2}} x) \\
x' &=& \gamma (x \,-\, vt) \,,
\end{eqnarray}
$

then the inverse equations are

$
\begin{eqnarray}
t &=& \gamma (t' \,+\, \frac{v}{c^{2}} x') \\
x &=& \gamma (x' \,+\, vt') \,,
\end{eqnarray}
$

Also it's a bit odd that you write

Originally Posted by Aethelwulf
$t' = \frac{t - vx}{\sqrt{1 - \frac{v^2}{c^2}$
because the numerator $t \,-\, vx$ is only correct if you apply the common convention of setting $c \,=\, 1$, which makes it rather odd that you leave $c$ explicitly in the denominator.

Also,

$x'' = \frac{x' - ut'}{\sqrt{1 - \frac{v^2}{c^2}}$

$= \frac{x-vt}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{u(t-vx)}{\sqrt{1-\frac{v^2}{c^2}}}$
actually, if I ignore the weird $c \,=\, 1$? thing I mentioned just above, that should be

$
\frac{x \,-\, vt}{1 \,-\, \frac{v^2}{c^2}} \,-\, \frac{u(t\,-\,vx)}{1\,-\,\frac{v^2}{c^2}}
$

(without the square roots).

Obviously anyone can make silly mistakes, but it doesn't look much like you really understand what you're doing.

13. So Motor Daddy, a reference frame is simply ''another point of view.'' The only thing in this universe that does not vary from one reference frame to another, is of course the speed of light - always gets measured moving at the speed of light.

14. Originally Posted by Aethelwulf
Don't flatter yourself.
I'm just observing that your post seems to follow a format that I've used many times (the Lorentz transformation is..., notice that $x' \,=\, 0 \,\Rightarrow\, x \,=\, vt$..., prove something that should be obvious to anyone who already understands relativity...).

I'm not flattered so much as amused. I wouldn't normally bother pointing something like this out (if people are reading and learning from my posts, then great!), except that you have something of a history of writing posts that consist mostly of a collage of stuff you read and didn't understand elsewhere, and you've even been caught out in blatant plagiarism before.

Your last few posts roughly look like you're trying to show that the composition of two Lorentz boosts is still a Lorentz boost and that the velocity of the resulting boost is given by the Lorentz velocity addition formula ($w \,=\, \frac{u \,+\, v}{1 \,+\, \frac{uv}{c^{2}}}$), except you got that wrong too and it doesn't seem to be a useful addition to this thread or have any relation to the comment you were replying to.

15. Originally Posted by Aethelwulf
Please don't talk about me like you know me, because you don't.
I don't think I know you. I have no idea what you'd be like in person. But many people here are familiar with your online persona and patterns in your posting behaviour.

And whether or not you're the latest incarnation of reiku, you should keep in mind it is damn obvious to everyone when you use well known equations that you don't understand the meaning of.

16. If you assume the laws of physics are translationally invariant in space and time, and assume the laws of physics are rotationally invariant so that the directions we call X, Y and Z are arbitrary choices, and you assume the laws of physics are invariant with respect to inertial frame such that any inertial body is just as good as any other inertial body as the standard of "motionless," then one is compelled to deal with the geometry of space and time. It is then a matter of experiment, not opinion, as to what geometry our universe has.
The possibilities (that don't involve rotation or motion other than in the X direction) are parametrized in a generalization of the Galilean transform.
$\begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex}\Delta x' \\ \rule{0pt}{20ex}\Delta y' \\ \rule{0pt}{20ex}\Delta z' \end{pmatrix} = \begin{pmatrix}\rule{0pt}{20ex}\frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex}\frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex}0 & 0 & 1 & 0 \\ \rule{0pt}{20ex}0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix}$

This transform converts coordinate differences on a motionlessness world-line in one coordinate system to motion in another:
$\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} v \Delta t' \\ \rule{0pt}{20ex}0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$
This transform also converts motion in one coordinate system to motionless in this specific case:
$\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \Delta t\\ \rule{0pt}{20ex} -v \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} \Delta t - \frac{K v^2}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} \Delta t - \frac{v}{\sqrt{1 - K v^2}} \Delta t \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} 0 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex}0 \\ \rule{0pt}{20ex} 0 \end{pmatrix}$
And we see it is self-consistent if we compose two different transforms and see that the result is also a transform:
$\begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K u^2}} & \frac{K u}{\sqrt{1 - K u^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{u}{\sqrt{1 - K u^2}} & \frac{1}{\sqrt{1 - K u^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & \frac{K u + K v}{1 + K u v} \times \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{u + v}{1 + K u v} \times \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K w^2}} & \frac{K w}{\sqrt{1 - K w^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{w}{\sqrt{1 - K w^2}} & \frac{1}{\sqrt{1 - K w^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix}$
Since $w = \frac{u + v}{1 + K u v} \, \wedge \, \left| K u^2 \right| < 1 \, \wedge \, \left| K v^2 \right| < 1 \; \Rightarrow \; \frac{1}{\sqrt{1 - K w^2}} = \frac{1 + K u v}{\sqrt{1 - K u^2}\sqrt{1 - K v^2}}$

So the generalization of the Galilean transformation gives us the self-consistent generalization of the velocity composition law. $w = \frac{u + v}{1 + K u v}$. We also have three different conservation laws (isometries) depending on the sign on K.
If K < 0, then $\left( \Delta t \right)^2 + \left| K \right| \left( \left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 \right) = \left( \Delta t' \right)^2 + \left| K \right| \left( \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2 \right)$. This is elliptical geometry.
If K > 0, then $\left( \Delta t \right)^2 - \left| K \right| \left( \left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 \right) = \left( \Delta t' \right)^2 - \left| K \right| \left( \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2 \right)$. This is hyperbolic geometry.
If K = 0, then $\left( \Delta x \right)^2 + \left( \Delta y \right)^2 + \left( \Delta z \right)^2 = \left( \Delta x' \right)^2 + \left( \Delta y' \right)^2 + \left( \Delta z' \right)^2$ and $\Delta t = \Delta t'$ This is Euclidean geometry of space and an unconnected absolute time.

All physical experiment (since 1859 when testing became precise enough) favors $K = c^{\tiny -2}$ over $K = 0$ which is why we study the Lorentz transformations, Einstein velocity composition laws, special relativistic interval $c^2 \left( \Delta t \right)^2 - \left( \Delta x \right)^2 - \left( \Delta y \right)^2 - \left( \Delta z \right)^2$ and hyperbolic geometry in modern physics classes. It also unified the mechanics of material bodies with the phenomena of electromagnetism.

17. Originally Posted by Aethelwulf
So Motor Daddy, a reference frame is simply ''another point of view.'' The only thing in this universe that does not vary from one reference frame to another, is of course the speed of light - always gets measured moving at the speed of light.
I know what a frame of reference is, so stop pretending you are teaching me something, because you're not.

I am still waiting on an answer from James. If you care to answer the question then take a stab at it. You have sticks nailed to the floor in your ship. What is your process of determining a "meter?"

The set of sticks nailed to the floor of the ship. The same set of sticks in both tests in which you marked on the sticks where light was after 1/299792458 of a second. You know, THE STICKS!
Fine. It's just that there was the potential in your question (quoted below) to use a set of sticks nailed to the floor of the ship, or one nailed to the floor of a room or space in which the ship is moving. We'll go with the ones nailed to the floor of the ship, then.

K. So you are at a constant velocity in a ship, not accelerating.
We don't really care about "you" here. Let's just talk about those sticks nailed to the floor, because they are the reference frame.

You send a light signal from the rear of the ship and find where the light hits after 1/299792458 of a second. The distance the light traveled in 1/299792458 of a second in that frame is a meter, correct?
Yes.

Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?
Yes.

19. Moderator note: 38 off-topic posts have been deleted.

20. Originally Posted by James R

Now, you fire your rocket engine and accelerate for a duration of time and then turn off the engine and are once again at a constant velocity. You then perform the test again, sending a signal from the rear of the ship and find where the light is after 1/299792458 of a second. Are you saying that in both tests the light will be at the same point along the sticks after 1/299792458 of a second has elapsed?
Yes.
Impossible, James. Don't you know what you are saying?

The more you make increases or decreases in the ship's velocity by accelerating for a duration of time, and then retesting, the more your point of impact of light is going to change where you find it along your sticks 1/299792458 of a second later.

If the signal is sent from the rear of the ship, the greater the ship's velocity in the forward direction (x direction, we'll say) the closer to the rear of the ship along the sticks the light will impact, if tested for the same 1/299792458 of a second each test.