Exploring the Speed of Light in Deep Space: A Scenario on a Space Station

[zoobyshoe]

OK. My next step, I believe, should be to look at the distance L. In the inertial frame of the station the distance L = 300,000 km. It will not be the same distance to the ship by virtue of the ships speed. To the ship the length L will be contracted. The ship will find L to measure something less than 300,000 km.

jcsd has provided the formula $$\gamma L$$ to use to determine what the ship will find L to be from its perspective.

He has provided this formula:

$$\gamma = \frac{1}{\sqrt{1 - \beta^2}}$$

to describe "gamma"($$\gamma$$).

The value I got when I solved for gamma earlier using .5 c for the relative velocity of the ship and the station was
1.1547005.

1.1547005 times 300,000 = 346,410.15 km

346,410.15 km > 300,000, not < 300,000.

Instead of contracting, the length has dilated!


jcsd has given us the wrong formula. He has given us the formula for time dilation, not length contraction.

The formula for length contraction is simply:

$$\sqrt {1 - \frac {v^2}{c^2}}$$

the result is the percentage of the length that the observer will see that length contracted to.

In our example, the ship will see the 300,000 km to have contracted to .8660254 % of its original length.

That comes out to be 300,000 times .8660254 = 259,807.62 km.

 

[plover]

Actually jcsd is correct.

When he writes

$$M=(0,0)$$
$$A=(L,0)$$
$$B=(-L,0)$$

He is stating that the distance to either A or B from M is L and thus each half of the train measured in the embankment frame is also length L (so total length 2L). However, the train is moving in the embankment frame and thus appears contracted.

So the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed $\gamma L$ (an un-contraction, if you will). This is the first value you calculated (~346k).

However, in the train reference frame the track is moving and thus the distance from M to A appears contracted. This length is usually expressed as $\frac{L}{\gamma}$. This is the second value you calculated (~260k).

 

[Doc Al]

OK. My next step, I believe, should be to look at the distance L. In the inertial frame of the station the distance L = 300,000 km. It will not be the same distance to the ship by virtue of the ships speed. To the ship the length L will be contracted. The ship will find L to measure something less than 300,000 km.

Two comments:

(1) Unless I haven't been paying close enough attention, it looks like you have changed the scenario a bit from your original one. I thought that your original scenario had the distance = 300,000 km as measured in the rocket frame, not the station frame. But to get a close analogy to Einstein's train problem, you should have the distance measured in the station frame. That way the station frame corresponds to the embankment frame. (But realize the answer to your original question will be different now.)

(2) Yes, if the distance is L in the station frame, the rocket will measure that distance to be shorter.​

jcsd has provided the formula $$\gamma L$$ to use to determine what the ship will find L to be from its perspective.

I believe you are misinterpreting what jcsd provided. jcsd analyzed the Einstein train example from both the embankment and the train frames. He gave the space-time coordinates of three events: A (the light flash at A), M (the passing of M' by M), and B (the light flash at B). So, when jcsd gave these equations:

$$M = (0,0)$$
$$A = (L,0)$$
$$B = (-L,0)$$​

he was giving the location and time coordinates of these events according to the embankment frame. Thus the three events happen at the same time (T=0) and A and B happen at positions -L and +L from the midpoint.

When he gave these equations:

$$M\'\; = (0,0)$$
$$A\'\; = (\gamma L,\frac{-\gamma\beta L}{c})$$
$$B\'\; = (-\gamma L,\frac{\gamma\beta L}{c})$$​

he was providing the location and time coordinates of these same events, but according to the train frame. Note that the times are different.

$\gamma L$ is not the distance L as measured by the train! It is the position coordinate of an event.

 

[zoobyshoe]

Two comments:

(1) Unless I haven't been paying close enough attention, it looks like you have changed the scenario a bit from your original one. I thought that your original scenario had the distance = 300,000 km as measured in the rocket frame, not the station frame.​

If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.

(2) Yes, if the distance is L in the station frame, the rocket will measure that distance to be shorter.

Kinda goes without saying.

I believe you are misinterpreting what jcsd provided.

What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)

He says: "x is the distance from...", so, the L given by jcsd represents the lengths M--->A and M---B respectively.

jcsd analyzed the Einstein train example from both the embankment and the train frames. He gave the space-time coordinates of three events: A (the light flash at A), M (the passing of M' by M), and B (the light flash at B). So, when jcsd gave these equations:

$$M = (0,0)$$
$$A = (L,0)$$
$$B = (-L,0)$$​

he was giving the location and time coordinates of these events according to the embankment frame. Thus the three events happen at the same time (T=0) and A and B happen at positions -L and +L from the midpoint.

I would buy what you say about -L and +L as positions, except you can't apply a Lorentz transformation to a point. A point has no dimensions.

When he gave these equations:

$$M\'\; = (0,0)$$
$$A\'\; = (\gamma L,\frac{-\gamma\beta L}{c})$$
$$B\'\; = (-\gamma L,\frac{\gamma\beta L}{c})$$​

he was providing the location and time coordinates of these same events, but according to the train frame. Note that the times are different.

Yes, I noticed that the second term in the parentheses for A' and B' were different. I found that mighty peculiar also, since no time has passed yet. We are still at T=0 for both reference frames.

$\gamma L$ is not the distance L as measured by the train! It is the position coordinate of an event.

Whatever it is, he tried to apply the Lorentz tranformation for time dilation to it. There's no good reason to do that to a length or a point in a coordinate system.​

 

[zoobyshoe]

Actually jcsd is correct.

When he writes

$$M=(0,0)$$
$$A=(L,0)$$
$$B=(-L,0)$$

He is stating that the distance to either A or B from M is L and thus each half of the train measured in the embankment frame is also length L (so total length 2L). However, the train is moving in the embankment frame and thus appears contracted.

So the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed $\gamma L$ (an un-contraction, if you will). This is the first value you calculated (~346k).

However, in the train reference frame the track is moving and thus the distance from M to A appears contracted. This length is usually expressed as $\frac{L}{\gamma}$. This is the second value you calculated (~260k).

Sophist.

Any length that one observer measures to be contracted will be its original length when "uncontracted, not greater than its original length.

 

[Physicsfreek]

Below is an excerpt (Chapter 12) from my upcoming book, "Space, Time and Elementary Particles. (Proof in chapers 13 and 14, not attached)

Chapter 12 The Speed of Light and Space-Time

c = d / [(t2-t0) / 2]
The speed of light is measured with the Fizoau procedure (for Armand Hippolyte Louis Fizoau (1819-1896) a French physicist who, as one of his many accomplishments, was the first to measure accurately the velocity of light (on Earth) in 1849. The following is a description of the Fizoau procedure.
Definition: The speed of light is the speed of propagation of electromagnetic waves in a vacuum, which is a physical constant equal to exactly 299 792.458 km/s.
McGraw-Hill Dictionary of Scientific & Technical Terms
The Fizoau procedure takes a beam of light, bounces it off a mirror a measured distance away and then divides that distance by one-half of the measured time for the round trip: d / [(t2 - t0) / 2]. A photon is emitted from RSX at time t0 toward the mirror M that reflects the photon at time t1. RSX sees the photon at time t2. Since the speed of light is a constant, the two world lines of the photon (RSXt0 to Mt1 and Mt1 to RSXt2) are equal.
Fizoau set up an apparatus, which consisted of a rapidly spinning wheel with tooth gaps evenly spaced around the edge of the wheel. At a point (RSX), Fizoau directed a beam of light through the gaps at the edge of the rapidly rotating wheel towards a mirror eight kilometers away. The speed of the wheel was adjusted until the returning light could be seen through the next gap. Fizoau, by measuring exactly how fast the wheel was turning, could calculate an exact time for one gap on the edge of the wheel to be replaced at a specific location by the next gap around the edge of the wheel. After taking many readings, Fizoau calculated the speed of light to be 315,000 kilometers per second. Leon Foucault improved on that measurement a year later by using rotating mirrors and attained a speed of 298,000 kilometers per second. Foucault’s technique was accurate enough to confirm that light travels slower in water than in air.
Different forms of Fizoau’s experiment have been carried out many times over the years with a longer or shorter distance d, using a different frequency of photon, at different locations on the earth, at different times of day, with different positions for the apparatus and with different reference systems in motion relative to each other. The answer is always 299,792.458 kilometers per second exactly (in a vacuum). The speed of light is a constant for all observers that are in motion with a constant velocity.
The speed of light is an exact measurement: 299,792,458 meters per second exactly (in a vacuum) but that exactness is not by accident. Galileo has been credited with being the first scientist to try to measure the speed of light. Galileo used a lamp that he could uncover and when his assistant, at some distance away, saw the light from Galileo’s lamp, the assistant uncovered his own lamp. By knowing the distance between himself and his assistant and by measuring the elapsed time (Galileo had to use a clepsydra, a water clock, to measure the time interval) until he saw his assistant’s lamp, Galileo tried to calculate the speed of the light beams. Galileo concluded that if light is not “instantaneous, it is extraordinarily rapid.” This was in 1667 and Galileo figured that the speed of light was at least ten times faster than the speed of sound. (Actually, the speed of light is almost nine-hundred thousand times faster than the speed of sound.)
In 1675, the Danish astronomer Ole Roemer measured the speed of light. The astronomer noticed that the eclipses of the moons of Jupiter changed with the relative positions of Jupiter and Earth. When Earth was closer to Jupiter, the eclipses happened sooner than when Jupiter was far from Earth. The time difference between when the moons of Jupiter were expected to eclipse and when they did eclipse was up to sixteen minutes. Roemer reasoned that the change was caused by the greater distance that light had to travel when Earth was farther from Jupiter. Using the then accepted distances between the orbits, he calculated that the speed of light to be 225,000 kilometers per second.
After Fizoau made his measurement in 1849, Foucault and other experimenters began measuring the speed of light with greater precision. As the speed of light was measured with increased precision, the definition of a meter and of a second had to be defined with greater precision.
Historically, the meter was defined by the French Academy of Sciences as 1/10,000,000 of the distance from Earth’s North Pole to the equator on a line that ran through the city of Paris. This definition was replaced 1874 and refined in 1889 by an exact measurement between two marks on a bar of 90% platinum/10 % iridium that was kept in a vault in Paris. Then in 1960, the meter was re-defined as 1,650,763.73 wavelengths of the emissions of the gas Krypton. In 1983, a meter was re-defined again as the distance that light travels in a vacuum in 1/299,792,458th of a second. Then, to define a second, there had to be very accurate clocks. In 1955, the first atomic clock was built and refined in 1958. 1967, the 13th General Conference on Weights and Measures first defined the International System (SI) and a second was defined by 9,192,631,770 periods of the radiation of the cesium-133 atom. These clocks provide an accuracy of 2 nanoseconds per day or one second in 1,400,000 years.
The meter went from being defined as a distance in space (1/10,000,000 of the distance from the North Pole to the equator) to being defined as a measure of time (1/299,792,458th of a second) which can be equated, with the speed of light, into a distance in space.

c = d / [(t2-t0) / 2] When t2 – t0 is 2/299,792,458 seconds, d is defined as 1 meter.
The exact length of a meter is a fraction of space-time.
By international standards, a second is defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between two hyperfine levels of the ground state of the cesium-133 atom.
Over a distance in time of one meter, the radiation of the cesium-133 atom oscillates 30.663319 times (1/299,792,458 of a second, times 9,192,631,770 Hz/sec. = 30.663319 Hz).
The oscillation of one wavelength equates to a distance of 1.087827757*10-10 seconds on Tk.
Or, one wavelength as measured as a metric distance on Tk is 3.261225571 centimeters.


Measuring Time in Units of Distance
Minkowski described the space-time continuum as an inseperate union and he used the complex configuration (√-1) for time measurement to allow time to be considered as the forth dimension equal to the three spatial dimensions: (ds)2 = x2 + y2 + z2 + tk2. However, space and time are measured in different units. By international convention, the units that are used around the world for scientific discussion are standardized: the meter/kilogram/second standard is used to measure length, mass and time. However, for any unit of distance that is chosen to measure space, a corresponding unit of distance can be used to measured time.
Thus, to change seconds into meters, the time interval in seconds is multiplied by the speed of light measured in meters per second.

c = 299,792,458 meters per second.
▲t = 1 second. c * ▲t = 299,792,458 m/s * 1 second. Then ▲t equals 299,792,458 meters.

▲t = 1/299,792,458th of a second.
c * ▲t = 299,792,458 m/s * 1/299 792 458 second. Then ▲t equals 1 meter.


When t is measured in seconds, a formula to measure a distance, ds, in space-time may be written as ds2 = x2 + y2 + z2 + c2t2 and the time interval conversion is positioned in the formula. In a space-time diagram, when using equal scales of space and time, one unit of space equaling one unit of time, the time units are directly convertible into spatial units. When one unit on the spatial axes is measured as one meter, then ▲t is equal to one meter on the Tk axis (time axis in the forth dimension). Whatever units the speed of light is measured in; meters, miles or cubits, the forth dimension can be measured in exactly the same units.
When ▲t on Tk, is measured as one second, then the passage of space through time, N to N+1sec., has progressed through 299,792.458 kilometers of the forth-dimension.
https://www.physicsforums.com/attachment.php?attachmentid=1165&stc=1
With distance in the forth dimension measured the same as distance is measured in space, a time interval that is measured as one second is equal to 299,792.458 kilometers on Tk. The corollary is that three-dimensional space, N, progresses to N+▲t through the forth-dimension at 299,792.458 km per second measured on Tk. Space, all of space, is physically moving through the forth dimension of time. The speed of light reflects the velocity of space through the forth dimension. Any observer in any reference system is measuring the velocity of N as N progresses to N+▲t when they measure the speed of light.
Space travels through the forth dimension in one direction. This forward direction creates the progression of our existence from the present to the future. As three-dimensional space progresses through the forth dimension from N to N+299 792 458 m, the clocks in RSS (Reference System Space)register one second. Other reference systems that have a velocity reference to RSS will measure space and time differently than RSS (the Lorentz transformations), but every reference system that maintains a continuous existence in space progresses through time with RSS into the future in N to N+▲t.

 

[zoobyshoe]

Nay, I think what we are all amazed at is the fact that the moving observer will be gullible enough to think receiving the signal at a different time than the stationary observer is equivalent to the source emitting the signal at a different time.

The observer on the train is meant to be ignorant of the actual timing of the flashes: he doesn't know when the man at the mid point receives them. All we want to know from him is: "Did you see them at the same time, or one before the other?"

The question, more or less, that Einstein has posed in a previous chapter is "When something seems simultaneous, how do we know if it really is or not?" (Not an exact quote, just a characterization.) He is pointing out in this chapter that you can't tell from the relative timing of the arrival of the flashes, if they were emitted simultaneously in their own rest frame.

 

[Doc Al]

If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.

So now you have two lengths? One is 300,000 km measured by the rocket? And the other is 300,000 km measured by the station? What are these two lengths? They aren't the same thing, obviously.

Kinda goes without saying.

I would have thought... but here we are.

He says: "x is the distance from...", so, the L given by jcsd represents the lengths M--->A and M---B respectively.

L is the distance between A and M, and B and M, measured in the embankment frame.

I would buy what you say about -L and +L as positions, except you can't apply a Lorentz transformation to a point. A point has no dimensions.

Lorentz transformations are used to convert space-time coordinates of an event (a point in space time) from one frame's measurements to another's.

Yes, I noticed that the second term in the parentheses for A' and B' were different. I found that mighty peculiar also, since no time has passed yet. We are still at T=0 for both reference frames.

You are missing the entire point. The events (1) Light A flashes and (2) Light B flashes do not occur at the same time according to the train frame!

Whatever it is, he tried to apply the Lorentz tranformation for time dilation to it. There's no good reason to do that to a length or a point in a coordinate system.

I suggest that you learn a bit about the Lorentz transformation.

 

[Doc Al]

If A' coincides with A at the moment of the flashes and B coincides with B' at the moment of the flashes and M' coincides with M at the moment of the flashes, how can "the moment of the flashes" be different for either the designated stationary or moving observer? That's like saying given 1=1 and 2=2 and 3 = 3 and 4 = 4, 1 is not 1 and 2 is not 2 and 3 is not 3 and 4 is not 4.

While A' (a point on the train) coincides with A (a point on the embankment) at the moment that A flashes, this does not mean that the train clock at A' agrees with the embankment clock at A. In fact, they obviously don't.

Obviously the the events are absolute in time and place for both inertial observers and the relative simultaneity being argued in that booklet is that of the arrival time of the light reflected from the events.

If that's what you think, then you'd better read it over. The two frames disagree as to when the flashes occured, not just when the light is detected by the observers.

 

[Eyesaw]

While A' (a point on the train) coincides with A (a point on the embankment) at the moment that A flashes, this does not mean that the train clock at A' agrees with the embankment clock at A. In fact, they obviously don't.


If that's what you think, then you'd better read it over. The two frames disagree as to when the flashes occured, not just when the light is detected by the observers.

I don't quite understand this are you saying the moment when A coincide with A' coincide with A flash are not simultaneous because someone on the embankment is wearing a watch synchronized with the time in Bei Jing while the gal on the train is keeping the time in France?

 

[Doc Al]

I don't quite understand this are you saying the moment when A coincide with A' coincide with A flash are not simultaneous because someone on the embankment is wearing a watch synchronized with the time in Bei Jing while the gal on the train is keeping the time in France?

:rofl: Good one, Eyesaw.

When A and A' coincide is a single event in space-time (something that happens at particular place and time). But the time that this event occurs will be measured differently in different frames.

The issue for simultaneity is not: Do A and A' pass each other at "the same instant"? Of course they do! The issue is: does the passing of A and A' happen at the same time as the passing of B and B'? And this question is answered differently in different frames: these two events are observed to happen at different times in the train frame.

 

[plover]

This post started out as a simple comment on the last sentence of thread post #73 by Doc Al. The actual result is more of a digression on some of the formal aspects of Special Relativity that, in the context of this discussion, either have only been sketched or have been described piecemeal.

$\gamma L$ is not the distance L as measured by the train! It is the position coordinate of an event.

I agree with what you say here, but as stated (and presuming it is meant as a warning about my previous post [#72]), it may obscure what is correct and what might be misleading about my post.

1) (To clarify for readers unfamiliar with the terminology)
Event is the term used for a given point in the mathematical space used to describe relativity. Specifying an event requires both space coordinate(s) and a time coordinate. The definition covering the current situation, as given by jcsd, is:

"What I've done is basically create a co-ordinate susyetm the co-ordinates for each event are (x,t) where x is the distance from the midpoint of the space station (i.e. x = 0 at M) and t is the difference in time from when the spaceship and M occupy the same spot (i.e. t = 0 as the spaceship arrives at M)."​

Each reference frame describes the same set of events. However, different reference frames overlay different systems of coordinates onto these. The differing measurements determined by these different coordinate systems reflect the space and time distortions that occur when measurements are made in different reference frames.

2) The Lorentz transformation is the mathematical operation that maps the coordinates of events in one reference frame to the coordinates in a second frame.

The ordinary Lorentz transformation maps event coordinates in the rest frame to event coordinates in some other inertial frame.

If an event has coordinates (x, t) in the rest frame, then the general transform $$\Lambda$$ to coordinates (x', t') in a frame with relative velocity u is:

$$(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)$$​

If, however, what is known is the coordinates (x', t') in the moving frame, then to find the coordinates (x, t) in the rest frame what is required is the inverse Lorentz transformation $$\Lambda^{-1}$$. This has the general form:

$$(x, t) = \Lambda^{-1} (x', t') = \left( \gamma (x' + ut'), \gamma(t' + \frac{ux'}{c^2}) \right)$$​

3) Ok, I screwed up. I just realized that jcsd was talking about the spaceship scenario not the train scenario, so the way my previous post (#72) compares what I say to what jcsd said makes no sense, and I should probably edit it. (Without the comparison, the logic still holds - though with the caveats given below.)

4) Just to be clear: I'm discussing the train scenario.

In the embankment frame, coordinates are denoted:

$$(x_m, t_m)$$​

where x_m is a spatial coordinate along an axis defined by the track, and t_m is a time coordinate.

For any time t_m in the embankment frame, the observer M has coordinates:

$$M(t_m) = (x_M, t_M) = (0, t_m)$$​

We also have two events - flash at A, and flash at B - that happen at t_m = 0. Coordinates for these are:

$$\textrm{flash}_A = (x_A, t_A) = (-L, 0)$$
$$\textrm{flash}_B = (x_B, t_B) = (L, 0)$$​

(I've switched the signs so that the train moves toward positive x_m.)

The flashes occur at the same moment that the rear and front of the train pass (respectively) A and B. In other words, in the embankment frame the length of the train equals the distance from A to B.

In the train frame, coordinates are denoted:

$$(x_{m'}, t_{m'})$$​

The axis along which the spatial coordinate x_m' is measured is defined by the track and thus coincides with spatial axis used in the embankment frame (it's coordinate system, however, is, of course, separate).

For any time t_m' in the train frame, the observer M' has coordinates:

$$M'(t_{m'}) = (x_{M'}, t_{M'}) = (0, t_{m'})$$​

The coordinates of flash_A and flash_B transformed into the train frame (as given by jcsd, but with signs reversed) are:

$$\textrm{flash}_{A'} = (x_{A'}, t_{A'}) = (- \gamma L, \frac{\gamma\beta L}{c})$$
$$\textrm{flash}_{B'} = (x_{B'}, t_{B'}) = (\gamma L, \frac{-\gamma\beta L}{c})$$​

This is an example of the inverse Lorentz transformation as defined in item 2) above. (We consider the train to be at rest so we are going from a moving frame to a rest frame.) Calculating using the formula from item 2) we get:

This has been corrected. See post #99 below for the reasoning behind the use of -u.

$$\begin{equation*} \begin{split} (x_{A'}, t_{A'}) &= \Lambda^{-1}(x_A, t_A) = \Lambda^{-1}(-L, 0) \\ &= \left( \gamma (-L + 0(-u)), \gamma (0 + \frac{(-u)(-L)}{c^2}) \right) \\ &= ( -\gamma L, \frac{\gamma\beta L}{c} ) \end{split} \end{equation*}$$​

The last point necessary for the set up is that both the observer M and the observer M' set the origin of their respective time axes to the moment that M' passes M.

5) So far quantities such as L and $\gamma L$ are, as Doc Al said, "position coordinate of an event".

To get the spatial distance in one dimension we can use the formula:

$$\Delta x = |x_1 - x_0|$$​

Thus the distances from the observer M to the points A and B are (in the embankment frame):

$$\Delta x_A = |x_A - x_M| = |-L - 0| =\ L$$
$$\Delta x_B = |x_B - x_M| = |L - 0| =\ L$$​

and thus as I said in the previous post, "each half of the train measured in the embankment frame is also length L (so total length 2L)"

And the distances from the observer M' to the points A' and B' are (in the train frame):

$$\Delta x_{A'} = |x_{A'} - x_{M'}| = |-\gamma L - 0| = \gamma L$$
$$\Delta x_{B'} = |x_{B'} - x_{M'}| = |\gamma L - 0| = \gamma L$$​

and so, again as in the previous post, "the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed $\gamma L$"

I'll leave off for now, the third assertion of my previous post will have to wait for next time.

 

[zoobyshoe]

So now you have two lengths? One is 300,000 km measured by the rocket? And the other is 300,000 km measured by the station? What are these two lengths? They aren't the same thing, obviously.

They are the lengths from the midpoint to the ends. If we're talking about the rocket scenario we can use Janus' rod attached to the rocket, which measures 300,000 km. This will give the rocket a length to measure in its frame if it needs one. The train is already this length from midpoint to the end.

L is the distance between A and M, and B and M, measured in the embankment frame.

Yes.

Lorentz transformations are used to convert space-time coordinates of an event (a point in space time) from one frame's measurements to another's.

Not exactly right. They can't be applied to a single point. In the case of length contraction you need the beginning and ending coordinates of the length to be contracted. Einstein gives the beginning coordinate as 0 for simplicity in the example he gives. For time dilation you also need two times to compare. T=0 and t=>0. A single point doesn't contain enough information to do anything with.

You are missing the entire point. The events (1) Light A flashes and (2) Light B flashes do not occur at the same time according to the train frame!

I don't know why you think I have missed that point. Jcsd set up coordinates for the M and M' systems at t=0 for both. Nothing has happened yet. I thought it was peculiar that he indicated that something had happened already in the M' system when no time has elapsed yet.

I suggest that you learn a bit about the Lorentz transformation.

Well, there is no doubt that I have more to learn about them.

 

[Doc Al]

This post started out as a simple comment on the last sentence of thread post #73 by Doc Al. The actual result is more of a digression on some of the formal aspects of Special Relativity that, in the context of this discussion, either have only been sketched or have been described piecemeal.


I agree with what you say here, but as stated (and presuming it is meant as a warning about my previous post [#72]), it may obscure what is correct and what might be misleading about my post.

Actually I was just pointing out that I thought zoobyshoe was interpreting the $\gamma L$ in jcsd's equations as the length L as measured by the train. This is not so.

5) So far quantities such as L and $\gamma L$ are, as Doc Al said, "position coordinate of an event".

To get the spatial distance in one dimension we can use the formula:

$$\Delta x = |x_1 - x_0|$$​

Thus the distances from the observer M to the points A and B are (in the embankment frame):

$$\Delta x_A = |x_A - x_M| = |-L - 0| = L$$
$$\Delta x_B = |x_B - x_M| = |L - 0| = L$$​

and thus as I said in the previous post, "each half of the train measured in the embankment frame is also length L (so total length 2L)"

Maybe I'm misinterpreting your notation, but $\Delta x_A = |x_A - x_M| = |-L - 0| = L$ is the distance--as measured in the embankment frame-- between the events (A) A flashes and (M) M' passes M. Of course, since these events are simultaneous in the embankment frame, then this is a length. It's the length between point A and point M in the embankment frame, which was of course given as L. I don't see this having anything to do with the length of the train. (Was that even specified?)

And the distances from the observer M' to the points A' and B' are (in the train frame):

$$\Delta x_{A'} = |x_{A'} - x_{M'}| = |-\gamma L - 0| = \gamma L$$
$$\Delta x_{B'} = |x_{B'} - x_{M'}| = |\gamma L - 0| = \gamma L$$​

and so, again as in the previous post, "the rest length of each half of the train, i.e. the length measured in the train reference frame is indeed $\gamma L$"

Now you are measuring the distance between those events (A) & (B) and (M) according to the train frame. $\Delta x_{B'} = |x_{B'} - x_{M'}| = \gamma L$ is not the length L (between A and M) as measured by the train, since these events are not simultaneous in the train frame. I suppose that if (for some unknown reason) the ends of the train happen to coincide with locations A and B when the lights flash, then $\gamma L$ would be the proper length of half the train. (Is that what you are assuming?)

 

[zoobyshoe]

So, I just looked through Plover's post about the Lorentz Transforms, and he makes a case I can't follow for jcsd having used the correct formula under the circumstances. Not having the patience to try and inform myself about that particular application of the transforms at this point, I will simply stipulate he is correct, apologise to jcsd for having asserted he used the wrong formula, and try and get back to the question that started this.

Employing jcsds method, and given a speed of .5c for the ship or train, what will be the relative times and distances from t=0 for the detection of the light signals by the observer on the train or ship?

 

[Doc Al]

They are the lengths from the midpoint to the ends. If we're talking about the rocket scenario we can use Janus' rod attached to the rocket, which measures 300,000 km. This will give the rocket a length to measure in its frame if it needs one. The train is already this length from midpoint to the end.

In the train gedanken, L is the length from one light to the point M in the embankment. It is not the length of the train. (I guess it could be, but that would be just an arbitrary coincidence.)

Not exactly right. They can't be applied to a single point. In the case of length contraction you need the beginning and ending coordinates of the length to be contracted. Einstein gives the beginning coordinate as 0 for simplicity in the example he gives. For time dilation you also need two times to compare. T=0 and t=>0. A single point doesn't contain enough information to do anything with.

Thanks for the lecture, but once a common origin is defined the LT can certainly be applied to a point (an event in space-time).

I don't know why you think I have missed that point. Jcsd set up coordinates for the M and M' systems at t=0 for both. Nothing has happened yet. I thought it was peculiar that he indicated that something had happened already in the M' system when no time has elapsed yet.

The two systems agree that the M clock and the M' clock both read t = 0 at the instant that M' passes M. That's all. They certainly do not agree that the lights flashed at the same time. At time t=0, the train frame says light B already flashed and light A didn't flash yet. But the embankment insists that both lights flashed at t = 0.

 

[zoobyshoe]

In the train gedanken, L is the length from one light to the point M in the embankment. It is not the length of the train. (I guess it could be, but that would be just an arbitrary coincidence.)

The length of the train is 2L. Always has been.

Thanks for the lecture, but once a common origin is defined the LT can certainly be applied to a point (an event in space-time).

Point, as defined by plover, yes, you may well be right. However, he/she may simply have made that whole post up. I certainly don't trust anything he/she says farther than I can throw it. The "uncontraction" post where the length "uncontracts" to one larger than it had before has made me suspicious.

The two systems agree that the M clock and the M' clock both read t = 0 at the instant that M' passes M. That's all. They certainly do not agree that the lights flashed at the same time. At time t=0, the train frame says light B already flashed and light A didn't flash yet. But the embankment insists that both lights flashed at t = 0.

Yes. This is absolutely what Einstein has said. However, I still don't know by what math we discover the difference in the reception time of the flashes by M'.

 

[plover]

Maybe I'm misinterpreting your notation, but $\Delta x_A = |x_A - x_M| = |-L - 0| = L$ is the distance--as measured in the embankment frame-- between the events (A) A flashes and (M) M' passes M. Of course, since these events are simultaneous in the embankment frame, then this is a length. It's the length between point A and point M in the embankment frame, which was of course given as L.

Yes, this was my intention. Since L was introduced as a coordinate, I was just making the derivation of how it also ends up as a length.

I don't see this having anything to do with the length of the train. (Was that even specified?)

Over the course of the thread, the assumption that the length of the train as measured in the embankment frame is the same as the distance from A to B has drifted in and out of use. The assumption is not present in Einstein's original set up, and in fact is implicitly denied by both Einstein's diagram and the one I gave in an earlier post. I should have been explicit that I was using it here.

Now you are measuring the distance between those events (A) & (B) and (M) according to the train frame.

If it appears that way, I expect its because I forgot to make the assumption about the length of the train explicit.

$\Delta x_{B'} = |x_{B'} - x_{M'}| = \gamma L$ is not the length L (between A and M) as measured by the train, since these events are not simultaneous in the train frame.

Yes.

I suppose that if (for some unknown reason) the ends of the train happen to coincide with locations A and B when the lights flash, then $\gamma L$ would be the proper length of half the train. (Is that what you are assuming?)

Well, there's someone standing at each end of the train holding a pointy metal pole, and as it's the middle of an electrical storm, these someones must be fairly gullible, and, well, being one of those ubiquitous "assistants" that show up in gedanken-experiments always has been a thankless job...

 

[Doc Al]

When do the photons arrive at M'? Here's when.

However, I still don't know by what math we discover the difference in the reception time of the flashes by M'.

Ask nicely and we'll tell you.

Let's be clear. I'll define two events:

(1) photons from light B arrive at M'
(2) photons from light A arrive at M'​

According the to the train, these events occur at:
$$(x_1', t_1') = (0,\frac{\gamma L}{c} - \frac{\gamma v L}{c^2})$$
$$(x_2', t_2') = (0,\frac{\gamma L}{c} + \frac{\gamma v L}{c^2})$$

Note added: The order of the lights in my set up is A, M, B. M' moves towards B. So event #1 occurs before event #2 in both frames.

 

[zoobyshoe]

Incidently, when we multiply gamma times length and get 346,410.15 and -346,410.15, what do these numbers tell us about A and B in relation to M, other than that they are still equidistant from M?

 

[Janus]

If the rocket measures it to be 300,000 km in its own frame, the station will also measure it to be 300,000 in its own frame. Each will measure it to be shorter in the other's frame, however.
Kinda goes without saying.

No, That's not right.

Let's try this again with the following attachment.

Assume two very long trains. each car is 20,000 km long, as measured from the train to which it belongs.
Train 1 has three observers, M, A and B, A and B are 300,000 km from M, according to M, which puts them 15 cars away from M.
Train 2 is moving at .5c relative to train 1. It has observers M', A' and B'.
A" and B' are each placed an equal distance from M' such that from the frame of M, A and B, when M and M' pass each other, A' passes A and B' passes B. since train 2 is length contracted according to train 1, this places A' and B' about 17 cars away from M' on train 2. This means that the distance between A' and M' (or B' and M') is 346420 km according to train 2.

Now look at what happens according to train 2. A' and B' are a little over 17 cars from M' (this has not changed) putting them 346420 km from M'
A and B are still 15 cars from M, but since train 1 is length contracted according to train 2, each of these cars is only 17320 km long, meaning that A and B are, according to train 2, only 259800 km from M.
When M passes M' A' does not pass A and B' does not pass B.

so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.

 

[Doc Al]

Incidently, when we multiply gamma times length and get 346,410.15 and -346,410.15, what do these numbers tell us about A and B in relation to M, other than that they are still equidistant from M?

I assume you are talking about the $\gamma L$ that appears in these equations:

$$B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})$$
$$A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})$$​

(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)

If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:

- when light A flashed it was at position $x' = -\gamma L$ in the train frame
- when light B flashed it was at postion $x' = \gamma L$ in the train frame​

Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).

 

[zoobyshoe]

so if the rocket measures the distance as 300,000km in its frame, the station does not measure it to be 300,000km in its frame.

Somewhere back there I gave the distance A---->B as 600,000 km. That's in the embankment frame.

I have also given the train a length of 600,000 km in its own frame wherever I mentioned a length for the train.

This means that when the train is at rest in the embankment frame they both measure 600,000 km.

When the train is moving at .5 c, it still measures itself to be 600,000 km long. The distance A--->B in the embankment frame, is still measured to be 600,000 km long in its frame.

If the train measures the embankment while moving at .5 c, it is, of course a different story. The moving train will see the distance A--->B contracted from 600,000 to .8660254% of that which is 5196152.4 km.

If an observer in the embankment frame measures the train, it will find the train to have contracted to 519,615.4 km, as well, from the embankment viewpoint.

If we need to demonstrate the same for the ship example for any reason we can do so by attaching your rods, provided the rods give it a length of 600,000 km in its own frame.

 

[zoobyshoe]

I assume you are talking about the $\gamma L$ that appears in these equations:

$$B\'\; = (\gamma L,\frac{-\gamma\beta L}{c})$$
$$A\'\; = (-\gamma L,\frac{\gamma\beta L}{c})$$​

That's right.

(I just noticed that jcsd defined the lights opposite to the way I would. In my set up, the lights are arranged in order A, M, and B. M' moves towards B. I hope this doesn't add any confusion.)

Nope.

If so, then you must understand what they mean. These are the space-time coordianates according to the train frame of the events (A) light flashes at A and (B) light flashes at B. They tell us that:

- when light A flashed it was at position $x' = -\gamma L$ in the train frame
- when light B flashed it was at postion $x' = \gamma L$ in the train frame​

OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened. Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?

Yes, those flashes occur at postions that are equidistant from the midpoint M' of the train (but they occur at different times).

 

[Doc Al]

OK. This is exactly what I thought. However, what confuses me is that 346,410.15 km is a greater distance than 300,000 km. I would expect it to be contracted, not lengthened.

You are still thinking that $\gamma L$ is the length L as measured by the train. It's not. According to the train, things go like this: Light B flashes. Then, some time later, light A flashes. During that time the train is moving (of course, in the train frame, the train sees the platform move).

Of what use is it to apply the time dilation equation to a length? What does this help us understand about the situation?

I'm not sure what you mean by "apply the time dilation equation" to a length, as I did no such thing. I applied the Lorentz transformation to the space-time coordinates of an event.

If you wish to think of sticks shrinking and clocks moving slowly, then you apply the LT to sticks (lengths) and clocks. For example, a length of L in the embankment is measured to be $L/\gamma$ from the train (yes, the train measures the lights A and B to be only $2L/\gamma$ apart); and the train frame will measure a time of $\gamma \Delta t$ when an embankment clock measures only a time of $\Delta t$.

To analyze a problem correctly you need to apply length contraction, time dilation, and the relativity of simultaneity all together. That's tricky. The Lorentz transformation takes account of all of that for you.

 

[zoobyshoe]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:

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

$$y'=y$$

$$z'=z$$

$$t'=\frac{t-\frac{v}{c^2}•x}{\sqrt{1-\frac{v^2}{c^2}}}$$

The two interesting ones, of course, are those for x' and t'.

In the next chapter, Einstein gives a simpler version of x':

$$^x$$(beginning of rod)$$^=0\sqrt{1-\frac{v^2}{c^2}}$$

$$^x$$(end of rod)$$^1•\sqrt{1-\frac{v^2}{c^2}}$$

"The distance between the two points being
$$\sqrt{1-\frac{v^2}{c^2}}$$

If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.

Instead, jcsd used the next one referred to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:

$$t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

If the 1 in the numerator of the equation given by jcsd for $$\gamma$$ is not the one used by Einstein in the time dilation version, I am quite confused about how jcsd arrived at the 1 from x-vt, which is the numerator in the more general equation for x'.

 

[zoobyshoe]

Those are from this chapter:

Chapter 11. The Lorentz Transformation. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/11.html

And the following one:

Chapter 12. The Behaviour of Measuring-Rods and Clocks in Motion. Einstein, Albert. 1920. Relativity: The Special and General Theory
Address:http://www.bartleby.com/173/12.html

 

[Doc Al]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation as:

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

$$y'=y$$

$$z'=z$$

$$t'=\frac{t-\frac{v}{c^2} x}{\sqrt{1-\frac{v^2}{c^2}}}$$

The two interesting ones, of course, are those for x' and t'.

Yes, those are the same Lorentz transformations that jcsd and I both used.

In the next chapter, Einstein gives a simpler version of x':

$$^x$$(beginning of rod)$$^=0\sqrt{1-\frac{v^2}{c^2}}$$

$$^x$$(end of rod)$$^1 \sqrt{1-\frac{v^2}{c^2}}$$

"The distance between the two points being
$$\sqrt{1-\frac{v^2}{c^2}}$$

Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame. Note that each end must be located at the same time.

If we want to express A or B as A' or B', it seems to me this is the one to use, multiplying it by the length, L, which is the distance between the two points (M--->B, or M--->A) in the embankment frame that we want to locate in the train frame.

If the problem is: Find out the distance from A to B as seen by the train frame, then that would be correct. But that's not what jcsd was calculating. Instead he (and I) calculated when and where those two flashes occurred according to the moving frame. That's a different problem. For one thing, those flashes occur at different times--so that length contraction formula doesn't apply.

jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame. (I'm not crazy about this notation, but that's what he used.)

How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!

Instead, jcsd used the next one referred to by Einstein; the one he uses to find the time dilation between two successive ticks of a clock. The first tick is t = 0. The second tick is:

$$t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

This is not what jcsd did. This particular equation describes the time dilation of a moving clock.

 

[plover]

In the same online version of Relativity from which I've been quoting parts of chapter IX, Einstein gives the Lorentz transformation

This is a four dimensional version of the transformation I gave above. As it shows, spatial distortions happen only along the direction of motion (in this case along the x-axis). This is why the two dimensional system suggested by jcsd is useful.

Note: When Doc Al says: "jcsd defined the lights opposite to the way I would", he makes precisely the same change as I did where I said "I've switched the signs so that the train moves toward positive x_m".​

The equation I gave was:

$$(x', t') = \Lambda (x, t) = \left( \gamma (x - ut), \gamma(t - \frac{ux}{c^2}) \right)$$
​

This is entirely equivalent to the separate equations:

$$x' = \gamma (x - ut)$$

$$t' = \gamma \left(t - \frac{ux}{c^2} \right)$$
​

(I expect this goes without saying -- I'm just ensuring the intention is clear.)

The notations introduced by jcsd (and which I should perhaps have reiterated):

$$\beta = \frac{u}{c}$$
$$\gamma = \frac {1} { \sqrt{1 - \beta^2} } = \frac {1} { \sqrt{1 - \frac{u^2}{c^2}} }$$​

These notations are quite standard. Using u for the velocity in SR calculations also seems to be a current standard practice.

Again just to be clear, I'll apply these definitions to the separated equations above:

$$x'\ =\ \gamma (x - ut) \ =\ \frac{1}{\sqrt{1 - \beta^2}} (x - ut) \ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}}$$

$$t'\ =\ \gamma \left( t - \frac{ux}{c^2} \right) \ =\ \frac{1}{\sqrt{1 - \beta^2}} \left(t - \frac{ux}{c^2} \right) \ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}}$$​

These are obviously identical to the equations given by Einstein.

Note: There is a typo in Ch 12 of the Einstein. Near the end of paragraph 1 a sentence reads:

"For the velocity v = 0 we should have

$$\sqrt{1 - v^2/c^2} = 0$$​

and for still greater velocities the square-root becomes imaginary."
​

This sentence only makes sense if it is changed to read: "For the velocity v = c we should have...". (Someone should probably back me up on this.)
​

There is one issue that everyone has neglected so far (I think). Einstein's coordinate system diagram (in Ch 11) shows that the k system views the k' system as moving to the right, i.e. toward positive x. From the viewpoint of k', however, k is moving to the left, i.e. toward negative x. Thus if k measures the velocity of k' to be u then k' will measure the velocity of k to be -u. Similarly, if the x axes of the train and embankment frames are oriented to be positive in the same direction, then if an embankment observer measures the velocity of the train as u, an observer on the train measures the velocity of the embankment to be -u.

If I'm reading everyone correctly, jcsd and Doc Al took this into account (though I don't think they mentioned it). I, on the other hand, um, managed to make the issue disappear by making a calculation error when I applied the inverse Lorentz transform in my previous post, which I should fix (and which is probably why zoobyshoe found my argument for jcsd's values ambiguous...).

I assume it is agreed that the coordinates in the embankment frame for the event "flash at B" (which we are also taking to be the coordinates of "front of train passes B") are:

x = L
t = 0​

If we insert these values into Einstein's equations (and reapply the standard notations), we get:

$$x'\ =\ \frac{x - ut}{\sqrt{1 - \frac{u^2}{c^2}}} \ =\ \frac{L - u \cdot 0}{\sqrt{1 - \frac{u^2}{c^2}}} \ =\ \frac{1}{\sqrt{1 - \beta^2}}\ \cdot\ L \ =\ \gamma L$$

$$t'\ =\ \frac{\left(t - \frac{ux}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}} \ =\ \frac{\left(0 - \frac{uL}{c^2} \right)}{\sqrt{1 - \frac{u^2}{c^2}}} \ =\ \frac{-1}{\sqrt{1 - \beta^2}}\ \cdot\ \frac{u}{c}\ \cdot\ \frac{L}{c} \ =\ \frac{-\gamma\beta L}{c}$$​

So if we set u to .5c our value for x' is:

$$\beta\ =\ .5c/c\ =\ .5$$

$$\gamma\ =\ \frac{1}{\sqrt{1 - (.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15$$

$$x'\ =\ \gamma L\ =\ 1.15L$$​

Now since:

• the origin of the x' axis of the train frame is the center of the train, and
• the front of the train is at the point (x, t) that we transformed from the embankment frame

we know that the value of x' also equals half the length of the train. So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?

The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.

 

[plover]

Sophist.

Any length that one observer measures to be contracted will be its original length when "uncontracted, not greater than its original length.
_______________________________________________

he/she may simply have made that whole post up. I certainly don't trust anything he/she says farther than I can throw it. The "uncontraction" post where the length "uncontracts" to one larger than it had before has made me suspicious.

:uhh: :uhh:

What are you talking about?

I used the word "uncontraction" to refer to the purely mathematical operation of finding the original length given a contracted length. When did I ever claim that something would be greater than its original length?

And what have I done that would lead you to countenance that I might intend such a reading? That makes a suspicion of willful deceit or "making whole posts up" more plausible than a mistake on one or both of our parts?

What's really going on here? (If you've got some beef, my pm box is open...)

 

[zoobyshoe]

Careful. This is not a "simpler version of x' ", it is a specific application to find how a length gets transformed when observed from a moving frame.

Good point. "Simpler version" is not particularly accurate. "Specific application" is much more to the point, much more accurate. I appreciate the distinction.

Note that each end must be located at the same time.

By "at the same time" do you mean at a specific time coordinate such as T=0? Or something else?

Instead he (and I) calculated when and where those two flashes occurred according to the moving frame.

This is what I have understood you to be saying about this since I raised my objection. However, I haven't been able to understand from you how the answer that is produced: -346, 410.16 and + 346,410.16, contains any useful information about the when and where according to the moving frame. If I observe that the A or B event took place in the embankment frame at t=0, x=+/-300,000 km. What does a value of -/+346,410.16 tell me about the moving frame?

According to you, this number says something useful about the when and where of event A' or B' in the moving frame. If this number actually has any use, I can't see it. In fact, to me it looks like it is saying something both incorrect and useless about these events in the A' and B' frame. These numbers seem to me to say nothing at all about the when in the A' and B' frame, and to say something outright incorrect about the position A' and B' will have in the moving frame.

From the general direction and tone of your answers, it seems clear that you understand the difficulty I am having with these numbers. I am trying to phrase my difficulty more specifically in the hope you'll see how to solve it.

jcsd used A to represent the coordinates of the event "Light A flashes" in the embankment frame, and A' to represent the coordinates of that same event in the train frame.

Yes, this is what I have understood him to be doing.

How did we get the answers we did? It's easy. We know the coordinates of the events in the embankment frame, that's a given: x = -L, t = 0 (for A flashing) and x = +L, t = 0 (for B flashing). Now to find out when and where the train observers determine these flashes took place: apply the LT. Try this for yourself!

Specify which of the four you wish me to try for myself. I can use the one for x' and I can use the one for t' but there is no one of the four that gives an answer for both x' and t'. You are giving me one equation and claiming the answer is telling me both the x' and t' coordinates with one number.

This is not what jcsd did. This particular equation describes the time dilation of a moving clock.

jcsd gave:

$$\gamma=\frac {1}{\sqrt{1-\beta^2}}$$

where:

$$\beta = \frac{u}{c}$$

This is exactly the same equation as:

$$t=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

solving for u/c separately and then squaring the result
$$\beta^2$$

gives the same answer as: the square root of u²/c².

I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.

 

[jcsd]

Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.

One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.

It is worth learning 4-vector algebra as it makes problems like these extremely simple to solve and also helps to eliminate errors.

 

[Azriel]

This speed of light do-hickey

I think the answer is A.c . This is because the speed of light is measured as constant in all reference frames. you can't add a sublight speed to a light, or translight speed!

 

[jcsd]

I need for you and jcsd to justify giving me the special application of the Lorentz Transformation for time dilation and telling me it has something useful to say about the x' coordinate of either A' or B'. Here's the primary problem I have with it: it doesn't address position, it only addresses time. The result, when it is applied to position, strikes me as downright useless.

What we've done is perform the Lorentz transformation completely, to do this you must consider all four dimensions (though saying that we've cut out two because they don't change as the direction of motion is orthogonal to them, for confirmation of this in the method examine the Lorentz transformation in full), the metod I used as outlined in my last post you don't even need to consider each directrion independently.

t doesn't equal gamma, gamma is just a unitless quantity that is useful in special relativity and it comes into the equation for length contraction too.

 

[zoobyshoe]

Zoobyshoe:
I'll explain the exact method I used: I treated the Lorentz tensor and 4-vector postions as matrices and then simply used matrix mutiplication to perform the Lorentz transformation.

Fascinating!

One worth pointing out s that we've all apporached the problem in slightly different ways, but still obtained equivalent results which is a strong indication in itself that we are correct.

Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.