I really do not understand.
If we assume c '= c-u then c' = c?
I am even more confused.
Now assuming c = c' then $$\quad C' = \gamma (u) C$$
Light at Light Speed
- Thread starter Bowser
- Start date
If we assume c '= c-u then c' = c?
Now assuming c = c' then $$\quad C' = \gamma (u) C$$
I think you have still more to learn.
Emil is another one, along with Motor Daddy and Farsight, who do not understand and accept relativity.
Pointing out that relativity has been experimentally verified to a very high degree of accuracy never seems to make any impact in their belief.
Pointing out that relativity has been experimentally verified to a very high degree of accuracy never seems to make any impact in their belief.
Peoples personal belief systems can make them blind to even the most blatant of facts. After all if someone believes that aliens can pop up for UFO picture shoots on planet earth from light years away, their belief requires a spaceship that can apparently go faster than light.
Mentioning verified experiments is not going to undermine their belief, since they have already made their decision beforehand.
(I'm just using the UFO/spacecraft analogy because it's an extreme.)
I have to admit giving a decent analogy to express what the Lightspeed constant actually means to the layperson is difficult
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The definitions of A and B as physical bodies, C as the distance between them, and D as the distance traveled by light moving between them is in post 76.
The definition of u and c as the speed of the physical bodies and the speed of light, respectively, as seen by the implied observer's rulers and clock, follows from the Cartesian equations for uniform motion in space and time.
The definition of u' and c' as the speed of the physical bodies and the speed of light, respectively, as seen by A, given a test theory of space and time transformations.
That transformation had four parameters, K (with units of square seconds per square meter), v (with units of meters per second), X' (with units of meters), and T' (with units of seconds). v is fixed by the situation to be equal to -u. X' and T' turn out to irrelevant when space-time has transitional symmetry and any number of experiments are consistent with one specific, non-zero value of K.
Rather than address this, you pretend inability to distinguish c (ASCII/Unicode
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Motor Daddy:
It's one point in time (in ONE reference frame), but TWO points in space, which makes TWO points in spacetime. Spacetime events have both spatial and time coordinates. Change either the space or the time coordinate and you have a different event.
Examples:
1. I eat breakfast today. I eat breakfast at the same place tomorrow. Clearly, two different spacetime events.
2. I eat breakfast today at my house. You eat breakfast today at your house. Again, quite clearly two different spacetime events.
3. One strike of lightning hits the front of the train now. A different strike hits the rear of the train now. Clearly, two separate spacetime events.
No, because as you say the speed of light is the same in every frame. Not that you have any actual appreciation of what that means.
For the speed of light to be the same in all frames, space and time MUST be different in different frames. Anything else is impossible.
If you're on the train when you do that, you won't record different times, but the same time in both cases. That's the difference between the real world and your fantasy world. Asserting something that isn't true over and over again just doesn't make it any truer.
It follows DIRECTLY from the constancy of the speed of light. If you were able to follow any basic introductory relativity text, then you'd be able to follow the simple derivation (even without maths!) of the relativity of simultaneity. Once you truely accept the constancy of the speed of light, you can't avoid the relativity of simultaneity. Your problem is that you think the constancy of the speed of light really only applies in your favorite "absolute" reference frame. You don't really accept that the speed is constant in every frame, despite your statements to the contrary. And that's because, even after a lengthy discussion in which I walked you through what a reference frame is, in the end you still couldn't grasp the concept. And, as I predicted you would, you have forgotten that entire discussion and reverted to your previous tired arguments.
That's fine for the track's reference frame. What are the numbers in the train's reference frame?
It's one point in time (in ONE reference frame), but TWO points in space, which makes TWO points in spacetime. Spacetime events have both spatial and time coordinates. Change either the space or the time coordinate and you have a different event.
Examples:
1. I eat breakfast today. I eat breakfast at the same place tomorrow. Clearly, two different spacetime events.
2. I eat breakfast today at my house. You eat breakfast today at your house. Again, quite clearly two different spacetime events.
3. One strike of lightning hits the front of the train now. A different strike hits the rear of the train now. Clearly, two separate spacetime events.
No, because as you say the speed of light is the same in every frame. Not that you have any actual appreciation of what that means.
For the speed of light to be the same in all frames, space and time MUST be different in different frames. Anything else is impossible.
If you're on the train when you do that, you won't record different times, but the same time in both cases. That's the difference between the real world and your fantasy world. Asserting something that isn't true over and over again just doesn't make it any truer.
It follows DIRECTLY from the constancy of the speed of light. If you were able to follow any basic introductory relativity text, then you'd be able to follow the simple derivation (even without maths!) of the relativity of simultaneity. Once you truely accept the constancy of the speed of light, you can't avoid the relativity of simultaneity. Your problem is that you think the constancy of the speed of light really only applies in your favorite "absolute" reference frame. You don't really accept that the speed is constant in every frame, despite your statements to the contrary. And that's because, even after a lengthy discussion in which I walked you through what a reference frame is, in the end you still couldn't grasp the concept. And, as I predicted you would, you have forgotten that entire discussion and reverted to your previous tired arguments.
That's fine for the track's reference frame. What are the numbers in the train's reference frame?
Yes, you're right. It is my negligence.
But then it raises another problem.
Length contraction is not only for observer located in the direction of travel?
Or is it also for the observer which is perpendicular to the direction of travel?
That's entirely the point - when he tests for the train's velocity, as far as he can tell the train is at rest.
You know, I'm pretty sure that you've been shown the numbers more than once.
But yes, we have a deal. We will continue in the thread you abandoned: [post=2662612]The Relativity of Simultaneity[/post]
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Eh? For any observer, length contraction of a physical object is along the axis of the physical object's movement relative to that observer, and is independent of the position of the observer.
Only relative velocity matters for computing the axis and amount of length contraction. Position doesn't matter when computing length contraction.
Maybe I have not expressed correctly.
You have two objects, and relative speed between them.
One is the object itself and the other "object" is the observer.
The observer "sees" a length contraction depending on the relative speed between itself and the object.
What I said above is correct?
At the moment when the object passes in front of him, at a distance d, the speed of the object relative to the observer is 0. (Velocity vector component which is oriented towards the observer is 0.)
You have two objects, and relative speed between them.
One is the object itself and the other "object" is the observer.
The observer "sees" a length contraction depending on the relative speed between itself and the object.
What I said above is correct?
I understand this observer, which is not along the axis of the physical object's movement.
At the moment when the object passes in front of him, at a distance d, the speed of the object relative to the observer is 0. (Velocity vector component which is oriented towards the observer is 0.)
No, that's wrong.
Let's use the rulers and clocks of the observer.
Then, in vector notation, we may describe the position of the object as a vector function of time pointing from the origin of the coordinate system: $$\vec{r}_{\textrm{object}} (t)$$
In complete generality, the observer also has a vector describing the oberver's own location relative to the origin of the coordinate system: $$\vec{r}_{\textrm{observer}} (t)$$ but it does not vary over time. $$\frac{d \vec{r}_{\textrm{observer}} (t)}{dt} = 0$$.
So we may speak of the vector from the observer to the object, where we drop the subscript because this is the main subject of your previous post: $$\vec{r}(t) = \vec{r}_{\textrm{object}} (t) - \vec{r}_{\textrm{observer}} (t)$$ and we may speak of the relative velocity of object: $$\vec{v}(t) = \frac{d \vec{r} (t)}{dt} = \frac{d \vec{r}_{\textrm{object}} (t)}{dt} $$ Now if we are talking about uniform motion, this velocity is constant over time. $$\frac{d \vec{v}(t)}{dt} = \frac{d^2 \vec{r} (t)}{dt^2} = \frac{d^2 \vec{r}_{\textrm{object}} (t)}{dt^2} =0$$ so $$\vec{v}(t) = \vec{v}$$.
In Cartesian coordinates, we have the equation of a line in space and time by expanding the following equation in x, y, and z. $$\vec{r} - t \vec{v} = \vec{r}_0$$ where $$\vec{r}_0$$ is the value of $$\vec{r}$$ at time t=0.
So we can solve for the time of closest approach, by writing the distance (squared) as a function of time. $$f(t) = \left( \vec{r}_0 + t \vec{v} \right)^2$$, taking the derivative: $$f'(t) = 2 \vec{r}_0 \cdot \vec{v} + 2 t \left| \vec{v} \right| ^2$$ and solve $$f'(t) = 0$$ which gives us $$t_0 = - \frac{ \vec{r}_0 }{\left| \vec{v} \right|} \cos \theta $$ where theta is the angle between r_0 and v.
So when the observer is not on the axis of movement, $$f(t_0) = d^2 > 0$$ as you say. And the component of the velocity of in the direction of the observer is zero, or $$\vec{r}(t_0) \cdot \vec{v} = \left( \vec{r}_0 + t_0 \vec{v} \right) \cdot \vec{v} = \vec{r}_0 \cdot \vec{v} + \left( - \frac{ \vec{r}_0 \cdot \vec{v} }{\left| \vec{v} \right|^2} \right) \left| \vec{v} \right|^2 = \vec{r}_0 \cdot \vec{v} - \vec{r}_0 \cdot \vec{v} = 0 $$.
But that has nothing to do with length contraction, which depends on the velocity, not the position or the component of velocity in the direction of the observer. It's the the observer's clocks and rulers which matter, not the observer itself.
Let's use the rulers and clocks of the observer.
Then, in vector notation, we may describe the position of the object as a vector function of time pointing from the origin of the coordinate system: $$\vec{r}_{\textrm{object}} (t)$$
In complete generality, the observer also has a vector describing the oberver's own location relative to the origin of the coordinate system: $$\vec{r}_{\textrm{observer}} (t)$$ but it does not vary over time. $$\frac{d \vec{r}_{\textrm{observer}} (t)}{dt} = 0$$.
So we may speak of the vector from the observer to the object, where we drop the subscript because this is the main subject of your previous post: $$\vec{r}(t) = \vec{r}_{\textrm{object}} (t) - \vec{r}_{\textrm{observer}} (t)$$ and we may speak of the relative velocity of object: $$\vec{v}(t) = \frac{d \vec{r} (t)}{dt} = \frac{d \vec{r}_{\textrm{object}} (t)}{dt} $$ Now if we are talking about uniform motion, this velocity is constant over time. $$\frac{d \vec{v}(t)}{dt} = \frac{d^2 \vec{r} (t)}{dt^2} = \frac{d^2 \vec{r}_{\textrm{object}} (t)}{dt^2} =0$$ so $$\vec{v}(t) = \vec{v}$$.
In Cartesian coordinates, we have the equation of a line in space and time by expanding the following equation in x, y, and z. $$\vec{r} - t \vec{v} = \vec{r}_0$$ where $$\vec{r}_0$$ is the value of $$\vec{r}$$ at time t=0.
So we can solve for the time of closest approach, by writing the distance (squared) as a function of time. $$f(t) = \left( \vec{r}_0 + t \vec{v} \right)^2$$, taking the derivative: $$f'(t) = 2 \vec{r}_0 \cdot \vec{v} + 2 t \left| \vec{v} \right| ^2$$ and solve $$f'(t) = 0$$ which gives us $$t_0 = - \frac{ \vec{r}_0 }{\left| \vec{v} \right|} \cos \theta $$ where theta is the angle between r_0 and v.
So when the observer is not on the axis of movement, $$f(t_0) = d^2 > 0$$ as you say. And the component of the velocity of in the direction of the observer is zero, or $$\vec{r}(t_0) \cdot \vec{v} = \left( \vec{r}_0 + t_0 \vec{v} \right) \cdot \vec{v} = \vec{r}_0 \cdot \vec{v} + \left( - \frac{ \vec{r}_0 \cdot \vec{v} }{\left| \vec{v} \right|^2} \right) \left| \vec{v} \right|^2 = \vec{r}_0 \cdot \vec{v} - \vec{r}_0 \cdot \vec{v} = 0 $$.
But that has nothing to do with length contraction, which depends on the velocity, not the position or the component of velocity in the direction of the observer. It's the the observer's clocks and rulers which matter, not the observer itself.
In the real world seldom happens that an object's velocity vector to coincide with the direction of the observer. Usually there is a degree of deviation, A degrees.
In this case how it applies time dilation and length contraction?
As I said, the amount of length contraction depends only on $$\left| \vec{v} \right|$$ not on $$\vec{r}$$, $$\vec{r} \cdot \vec{v}$$ or $$\theta$$. The amount of length contraction is the same for an observer on the line of motion as for an observer not on the line of motion.
Length contraction depends on the rulers and clocks of the observer, not the observer's position.
Length contraction depends on the rulers and clocks of the observer, not the observer's position.
Motor Daddy
Valued Senior Member
Pete, rpenner, and James R., you continue to evade the task at hand which is showing your numbers for Einstein's Chapter 9. You need to show your numbers before we can continue this discussion. Posting a bunch of word salad is worthless. Show me your numbers! Can't do it, can you? Why not, because there exists no relativity of simultaneity?
I want to see the following, as I've shown you in my numbers, which are 100% facts:
1. What time was it at A and B when the lightening struck?
2. What is the velocity of the tracks/embankment?
3. What is the velocity of the train?
4. What time(s) did the light from A and B hit the embankment observer?
5. What time(s) did the light from A and B hit the train observer?
6. How much time did it take for light to travel from A and B to impact each observer?
7. What distance did the light travel from A and B to reach each observer?
8. What is the distance between A and B in the train frame?
9. What is the distance between A and B in the embankment frame?
I maintain the following:
Show me your numbers and stop beating around the bush!!!
I want to see the following, as I've shown you in my numbers, which are 100% facts:
1. What time was it at A and B when the lightening struck?
2. What is the velocity of the tracks/embankment?
3. What is the velocity of the train?
4. What time(s) did the light from A and B hit the embankment observer?
5. What time(s) did the light from A and B hit the train observer?
6. How much time did it take for light to travel from A and B to impact each observer?
7. What distance did the light travel from A and B to reach each observer?
8. What is the distance between A and B in the train frame?
9. What is the distance between A and B in the embankment frame?
I maintain the following:
Show me your numbers and stop beating around the bush!!!
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I have to admit, those numbers certainly look impressive when they're all lined up, like trophies on a shelf.
That .000000016678480590418212900804736688488 is looking a bit worn though. On the left. Can you see the scratches?
That .000000016678480590418212900804736688488 is looking a bit worn though. On the left. Can you see the scratches?
You put me in a serious problem because I do not know what is mean to you, the relative speed between two objects when velocity direction does not correspond with the axis between the two objects. (The "observer" is just one of the objects.)
Almost one year ago, I wrote:
http://sciforums.com/showthread.php?p=2540723#post2540723
Then, as now, Motor Daddy does not argue from physical observation or internally consistent logic -- he merely asserts.
Four months ago, on a thread where I was not active, Motor Daddy inflicted his tortured logic and superfluous assertions on us again.
http://www.sciforums.com/showthread.php?p=2670829#post2670829
(I would eventually put the last post on the thread because of my concern about facts and the nature of time.)
So where do these numbers come from?
299792458 meters per second = 0.299792458 meters per nanosecond is exactly the speed of light in vacuum, c.
.00000004 seconds = 40 nanoseconds. Why is this number chosen? Probably because it is a "nice" combination of powers of 2 and 5 and so later gives an exact integer number. It is based on nothing on Chapter 9 of Einstein's 1920 pop-physics book.
40 light-nanoseconds = 11.99169832 meters exactly.
Motor Daddy asserts that the length of the train is 11.9915 meters "in the frame of the train" but has not properly justified this with any argument or math. He has just chosen a number here, but that does not match with any observation.
Then the velocity is calculated from the elementary rate equation: $$\ell = (c - v) \Delta t$$ which Motor Daddy rearranges as $$v = \frac{c \Delta t - \ell}{\Delta t}$$ and tells himself that this is the secret formula to determine absolute velocity despite the implications this would have to the historical determinations of the speed of light. But this would not be the last time Motor Daddy saw fit to contradict empirical fact by naked assertion and hubris.
So v = 4958 meters per second exactly.
Replacing Motor Daddy's fooking insane assertion that 11.9915 meters is the length of the train "in the frame of the train" we have a simple rate-time problem in the frame of the embankment.
So $$\ell$$ (= 11.9915 meters) is the length of the train "in the frame of the embankment".
v = 4958 meters per second is exactly the speed of the train relative to that embankment.
c = 299792458 meters per second is exactly the speed of light in vacuum, everywhere, anyhow (in SR or in GR measured locally)
c - v = 299787500 is the speed, as seen from the embankment of a object moving at speed c catching up with an object at speed v.
$$\Delta t$$ = 40 nanoseconds = 0.00000004 seconds is the elapsed time, as seen from the embankment, between the time light started at the rear of the train until the time it reached the front of the train.
$$\Delta x = c \Delta t$$ = 11.99169832 meters is the total distance traveled by light during the time.
And so $$\ell = (c - v) \Delta t$$ makes sense from the point of view of the embankment. (Note: $$\Delta x = \ell + v \Delta t$$ )
So if that's the view from the embankment, what the hell is the view from the train. Well we have to put ourselves in another's shoes and the universe still needs to make sense from that point of view. In other words both $$\Delta x' = c' \Delta t'$$ and $$\ell ' = (c' - v') \Delta t'$$ must apply to the same setup from the train's point of view, with perhaps different numbers from the train's point of view.
Using the "agnostic" version of the space-time transformation, we have:
$$\begin{eqnarray} \Delta x ' & = & \frac{\Delta x - v \Delta t}{\sqrt{1 - K v^2}} \\ \Delta t ' & = & \frac{\Delta t - K v \Delta x}{\sqrt{1 - K v^2}} \\ u ' & = & \frac{u - v}{1 - K u v} \end{eqnarray}$$
So $$v' = \frac{v - v}{1 - K v^2} = 0$$,
$$c' = \frac{c - v}{1 - K c v} $$,
$$\Delta x' = \ell ' = \frac{\ell + v \Delta t - v \Delta t}{\sqrt{1 - K v^2}}$$, and
$$\Delta t ' = \frac{\Delta t - K v \ell + K v^2 \Delta t}{\sqrt{1 - K v^2}}$$ and so $$\ell ' = (c' - v') \Delta t'$$ requires $$\frac{\ell + v \Delta t - v \Delta t}{\sqrt{1 - K v^2}} = \frac{c - v}{1 - K c v} \frac{\Delta t - K v \ell - K v^2 \Delta t}{\sqrt{1 - K v^2}}$$. But this expression does not constrain K or c. So we must turn to physical experiment to find out that within experimental precision $$K = c^{-2}$$ and that beyond the shadow of a doubt the empirical evidence requires that $$K \neq 0$$.
So $$v'$$ = 0, $$c' = c$$, $$\ell' = \gamma(v) \ell = \frac{149896229 \sqrt{23983}}{300000 \sqrt{41638530}} \quad \textrm{meters} \approx 11.99150000164- \quad \textrm{meters} $$, $$\Delta t' = \frac{23983}{600000 \sqrt{998616864990}} \quad \textrm{seconds} \approx 39.99933848+ \quad \textrm{nanoseconds}$$ all as seen by the embankment.
This is the essential disconnect with Motor Daddy -- he believes himself entitled to assert facts in contradiction with reality. If you compare with Chapter 9 of Einstein's 1920 pop physics book, you will see that none of these numbers comes from that chapter and are so only shed light on Motor Daddy's willingness to assert falsehoods.
With that background out of the way, we may finally address the malformed questions posed by Motor Daddy from his uninformed and counter-factual mental picture.
A and B are not places, but events. Einstein does not establish any global standard for time, but only defines them in terms of relative time with respect to each other in a particular frame ("simultaneous with reference to the railway embankment").
Thus $$t_A = t_B$$ and Einstein in that same paragraph says that he will show that in the framework of the train $$t'_A \neq t'_B$$.
From the table below, $$t_A - t_B = 0$$ but $$t'_A - t'_B = \frac{v \ell '}{c^2} \neq 0$$.
In the frame of the embankment, 0.
In the frame of the train, -v. Einstein was talking to people interested in physics circa 1920, so sees no need to specify a figure for v, which he defines as the movement of the train along the rails from left to right. This is the same as the Cartesian convention for positive horizontal movement. Naturally, with such conventions in place, the velocity of the tracks as viewed from the train must be of equal magnitude but opposite direction, so we supply a minus sign.
In the frame of the embankment, v.
In the frame of the train, 0.
That time would be $$t_A + \frac{\ell}{2 c} = t_B + \frac{\ell}{2 c}$$ where $$\ell$$ is the distance (as measured by rulers on the embankment) between the places as seen on the embankment where lightning strikes A and B occurred. Thus, $$\ell = x_B - x_A$$.
From the table below, this is $$t_M$$.
From the table below, these are $$t'_1 \quad \textrm{and} \quad t'_2$$. (They differ from $$t'_B \quad \textrm{and} \quad t'_A$$, respectively, by the above indicated amount.
From the table below, the times are $$t_M - t_A = t_M - t_B$$ and $$t'_2 - t'_A = t'_1 - t'_B$$.
As before, the distance traveled depends on the rulers being used. $$\frac{\ell}{2}$$ for the embankment, $$\frac{\ell '}{2}$$ for the train.
From the table below, the distances are $$x_M - x_A = x_B - x_M$$ and $$x'_2 - x'_A = x'_B - x'_1$$.
From the table below, the distance is $$x'_B - x'_A$$.
From the table below, the distance is $$x_B - x_A$$.
But what Motor Daddy don't ask is how these quantities are related.
For convenience's sake, define $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \quad > \quad 1$$. Then $$\ell \gamma = \ell '$$ (length contraction).
Event A: $$(x_A, t_A, x'_A, t'_A)$$
Event B: $$(x_B = x_A + \ell, t_B = t_A, x'_B = x'_A + \ell ', t'_B = t'_A - \frac{v \ell '}{c^2} )$$
Event 1: $$(x_1 = x_A + \frac{c + 2v}{c + v} \frac{\ell}{2}, t_1 = t_A + \frac{\ell}{2c + 2v}, x'_1 = x'_A + \frac{\ell '}{2}, t'_1 = t'_A + \frac{\ell '}{2 c} \left( 1 - \frac{2v}{c} \right) )$$
Event M: $$(x_M = x_A + \frac{\ell}{2}, t_M = t_A + \frac{\ell}{2c}, x'_M = x'_A + \frac{\ell '}{2} \left( 1 - \frac{v}{c} \right), t'_M = t'_A + \frac{\ell '}{2 c} \left( 1 - \frac{v}{c} \right) )$$
Event 2: $$(x_2 = x_A + \frac{c}{c - v} \frac{\ell}{2}, t_2 = t_A + \frac{\ell}{2c - 2v}, x'_2 = x'_A + \frac{\ell '}{2}, t'_2 = t'_A + \frac{\ell '}{2 c} )$$
Using geometry or algebra is a stronger demonstration of correctness than the numbers Motor Daddy requests for the same reason that a stopped clock is right twice a day without being useful. Better to have the working clock and learn to read the moving hands. (AlphaNumeric's right -- I need to take a class on better analogies.)
http://sciforums.com/showthread.php?p=2540723#post2540723
Resorting to trolling from thread to thread and argumentum ad font are merely the expected symptoms.
Then, as now, Motor Daddy does not argue from physical observation or internally consistent logic -- he merely asserts.
Four months ago, on a thread where I was not active, Motor Daddy inflicted his tortured logic and superfluous assertions on us again.
http://www.sciforums.com/showthread.php?p=2670829#post2670829
Already he has assumed a Galilean universe, so Motor Daddy's emphasis on "numbers" is spurious and non-fact-based.
(I would eventually put the last post on the thread because of my concern about facts and the nature of time.)
So where do these numbers come from?
299792458 meters per second = 0.299792458 meters per nanosecond is exactly the speed of light in vacuum, c.
.00000004 seconds = 40 nanoseconds. Why is this number chosen? Probably because it is a "nice" combination of powers of 2 and 5 and so later gives an exact integer number. It is based on nothing on Chapter 9 of Einstein's 1920 pop-physics book.
40 light-nanoseconds = 11.99169832 meters exactly.
Motor Daddy asserts that the length of the train is 11.9915 meters "in the frame of the train" but has not properly justified this with any argument or math. He has just chosen a number here, but that does not match with any observation.
Then the velocity is calculated from the elementary rate equation: $$\ell = (c - v) \Delta t$$ which Motor Daddy rearranges as $$v = \frac{c \Delta t - \ell}{\Delta t}$$ and tells himself that this is the secret formula to determine absolute velocity despite the implications this would have to the historical determinations of the speed of light. But this would not be the last time Motor Daddy saw fit to contradict empirical fact by naked assertion and hubris.
So v = 4958 meters per second exactly.
Replacing Motor Daddy's fooking insane assertion that 11.9915 meters is the length of the train "in the frame of the train" we have a simple rate-time problem in the frame of the embankment.
So $$\ell$$ (= 11.9915 meters) is the length of the train "in the frame of the embankment".
v = 4958 meters per second is exactly the speed of the train relative to that embankment.
c = 299792458 meters per second is exactly the speed of light in vacuum, everywhere, anyhow (in SR or in GR measured locally)
c - v = 299787500 is the speed, as seen from the embankment of a object moving at speed c catching up with an object at speed v.
$$\Delta t$$ = 40 nanoseconds = 0.00000004 seconds is the elapsed time, as seen from the embankment, between the time light started at the rear of the train until the time it reached the front of the train.
$$\Delta x = c \Delta t$$ = 11.99169832 meters is the total distance traveled by light during the time.
And so $$\ell = (c - v) \Delta t$$ makes sense from the point of view of the embankment. (Note: $$\Delta x = \ell + v \Delta t$$ )
So if that's the view from the embankment, what the hell is the view from the train. Well we have to put ourselves in another's shoes and the universe still needs to make sense from that point of view. In other words both $$\Delta x' = c' \Delta t'$$ and $$\ell ' = (c' - v') \Delta t'$$ must apply to the same setup from the train's point of view, with perhaps different numbers from the train's point of view.
Using the "agnostic" version of the space-time transformation, we have:
$$\begin{eqnarray} \Delta x ' & = & \frac{\Delta x - v \Delta t}{\sqrt{1 - K v^2}} \\ \Delta t ' & = & \frac{\Delta t - K v \Delta x}{\sqrt{1 - K v^2}} \\ u ' & = & \frac{u - v}{1 - K u v} \end{eqnarray}$$
So $$v' = \frac{v - v}{1 - K v^2} = 0$$,
$$c' = \frac{c - v}{1 - K c v} $$,
$$\Delta x' = \ell ' = \frac{\ell + v \Delta t - v \Delta t}{\sqrt{1 - K v^2}}$$, and
$$\Delta t ' = \frac{\Delta t - K v \ell + K v^2 \Delta t}{\sqrt{1 - K v^2}}$$ and so $$\ell ' = (c' - v') \Delta t'$$ requires $$\frac{\ell + v \Delta t - v \Delta t}{\sqrt{1 - K v^2}} = \frac{c - v}{1 - K c v} \frac{\Delta t - K v \ell - K v^2 \Delta t}{\sqrt{1 - K v^2}}$$. But this expression does not constrain K or c. So we must turn to physical experiment to find out that within experimental precision $$K = c^{-2}$$ and that beyond the shadow of a doubt the empirical evidence requires that $$K \neq 0$$.
So $$v'$$ = 0, $$c' = c$$, $$\ell' = \gamma(v) \ell = \frac{149896229 \sqrt{23983}}{300000 \sqrt{41638530}} \quad \textrm{meters} \approx 11.99150000164- \quad \textrm{meters} $$, $$\Delta t' = \frac{23983}{600000 \sqrt{998616864990}} \quad \textrm{seconds} \approx 39.99933848+ \quad \textrm{nanoseconds}$$ all as seen by the embankment.
This is the essential disconnect with Motor Daddy -- he believes himself entitled to assert facts in contradiction with reality. If you compare with Chapter 9 of Einstein's 1920 pop physics book, you will see that none of these numbers comes from that chapter and are so only shed light on Motor Daddy's willingness to assert falsehoods.
With that background out of the way, we may finally address the malformed questions posed by Motor Daddy from his uninformed and counter-factual mental picture.
The question is not properly worded and assumes absolute time. As chapter nine is cited as the background material, the question proves Motor Daddy failed to understand the material.
A and B are not places, but events. Einstein does not establish any global standard for time, but only defines them in terms of relative time with respect to each other in a particular frame ("simultaneous with reference to the railway embankment").
Thus $$t_A = t_B$$ and Einstein in that same paragraph says that he will show that in the framework of the train $$t'_A \neq t'_B$$.
From the table below, $$t_A - t_B = 0$$ but $$t'_A - t'_B = \frac{v \ell '}{c^2} \neq 0$$.
(Here again Motor Daddy asks for a numerical answer against some invisible absolute framework of space and time. That defeats the purpose of citing Einstein, so I ignore it.)
In the frame of the embankment, 0.
In the frame of the train, -v. Einstein was talking to people interested in physics circa 1920, so sees no need to specify a figure for v, which he defines as the movement of the train along the rails from left to right. This is the same as the Cartesian convention for positive horizontal movement. Naturally, with such conventions in place, the velocity of the tracks as viewed from the train must be of equal magnitude but opposite direction, so we supply a minus sign.
(Here again Motor Daddy asks for a numerical answer against some invisible absolute framework of space and time. That defeats the purpose of citing Einstein, so I ignore it.)
In the frame of the embankment, v.
In the frame of the train, 0.
There is no embankment observer. A careful read of the paragraph following Figure 1 shows that Einstein placed the observer on the train at position M' (defined in terms of the train) and counter-factually assumed that the observer would stay fixed at position M (defined in terms of the embankment) to see the flashes of light from lightning strokes A and B arrive at the same time at M.
That time would be $$t_A + \frac{\ell}{2 c} = t_B + \frac{\ell}{2 c}$$ where $$\ell$$ is the distance (as measured by rulers on the embankment) between the places as seen on the embankment where lightning strikes A and B occurred. Thus, $$\ell = x_B - x_A$$.
From the table below, this is $$t_M$$.
Two different times. $$t'_A + \frac{\ell'}{2c}$$ and $$t'_B + \frac{\ell'}{2c}$$. Here, $$\ell ' = x'_B - x'_A$$ where as we use the primed coordinates to talk about the train's clocks and rulers.
From the table below, these are $$t'_1 \quad \textrm{and} \quad t'_2$$. (They differ from $$t'_B \quad \textrm{and} \quad t'_A$$, respectively, by the above indicated amount.
Once again, times are measured by different clocks for different observers, and since $$\ell \ne \ell '$$ the times seen by a hypothetical observer on the embankment and the observer on the train are just gotten by subtraction. $$\frac{\ell}{2c}$$ for the embankment at position M, $$\frac{\ell '}{2c}$$ for the train at position M'.
From the table below, the times are $$t_M - t_A = t_M - t_B$$ and $$t'_2 - t'_A = t'_1 - t'_B$$.
Do you get the sense of the low level of the questions?
As before, the distance traveled depends on the rulers being used. $$\frac{\ell}{2}$$ for the embankment, $$\frac{\ell '}{2}$$ for the train.
From the table below, the distances are $$x_M - x_A = x_B - x_M$$ and $$x'_2 - x'_A = x'_B - x'_1$$.
Correctly worded at last. $$\ell '$$.
From the table below, the distance is $$x'_B - x'_A$$.
$$\ell$$
From the table below, the distance is $$x_B - x_A$$.
But what Motor Daddy don't ask is how these quantities are related.
For convenience's sake, define $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \quad > \quad 1$$. Then $$\ell \gamma = \ell '$$ (length contraction).
Event A: $$(x_A, t_A, x'_A, t'_A)$$
Event B: $$(x_B = x_A + \ell, t_B = t_A, x'_B = x'_A + \ell ', t'_B = t'_A - \frac{v \ell '}{c^2} )$$
Event 1: $$(x_1 = x_A + \frac{c + 2v}{c + v} \frac{\ell}{2}, t_1 = t_A + \frac{\ell}{2c + 2v}, x'_1 = x'_A + \frac{\ell '}{2}, t'_1 = t'_A + \frac{\ell '}{2 c} \left( 1 - \frac{2v}{c} \right) )$$
Event M: $$(x_M = x_A + \frac{\ell}{2}, t_M = t_A + \frac{\ell}{2c}, x'_M = x'_A + \frac{\ell '}{2} \left( 1 - \frac{v}{c} \right), t'_M = t'_A + \frac{\ell '}{2 c} \left( 1 - \frac{v}{c} \right) )$$
Event 2: $$(x_2 = x_A + \frac{c}{c - v} \frac{\ell}{2}, t_2 = t_A + \frac{\ell}{2c - 2v}, x'_2 = x'_A + \frac{\ell '}{2}, t'_2 = t'_A + \frac{\ell '}{2 c} )$$
Using geometry or algebra is a stronger demonstration of correctness than the numbers Motor Daddy requests for the same reason that a stopped clock is right twice a day without being useful. Better to have the working clock and learn to read the moving hands. (AlphaNumeric's right -- I need to take a class on better analogies.)
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I don't know about space changing, but we do know that time changes and I'm pretty sure that is all that's needed. If anyone knows better I'm listening.
