The Patterns of Spacetime

Philosophy Forums Science, Technology, and Math
281

You don’t need to worry about the practicalities of things like air molecules and exact photon to clock sizes. We are discussing the ideals involved, not a practical model.

Huh? How did you determine that? You seem to be thinking in terms of a special reference frame for the light that might be different from other objects. SRT doesn’t allow the light to have its own special reference frame.

If I had said that the station was traveling at 0.4c and the train was traveling at 0.9c during all of this, would you still say that the station stayed centered with respect to the flash points?

SRT specifies that from every frame of reference, light must be measured to travel at the same velocity within that frame. That means that if any clock, moving or not, is centered between any flash points, from that clocks pov, the light from both flashers must travel to it in the same amount of time because as far as that clock’s pov, the clock is not moving relative to the flash points whether the flashers were moving or not.

See above;

Hi Ed,

Wait. “t0” is one fixed point in time when the clock reads 0, not a variable in that equation.
k1 is a variable, t.
t varies from its initial t0. t0 doesn’t vary.

t0 is the one instant when all clocks are initialized/reset. Perhaps think of it as the trains are coming and at one instant, all clocks are reset to 0. At that moment, the front train is closer to the station than the rear train and both clocks have been reset to 0. That means that the x0 for each clock is different with respect to the station. Thus the x distance from the station clock cannot determine any value at all because at that one moment, every clock, at every x distance, would read zero.

The x distance is NOT the distance from the station clock, but the distance from whenever the clock in question was reset or initialized, its x0 and t0. Each clock must have its own “x0” and “x”; “xr” and “xf” for the rear and front clocks measured from xr0 and xf0 (or P1 and P3 in that diagram). The station’s “x” is always zero because the station isn’t moving relative to the station frame’s “x0”.

You have to calculate each clock separately. Each clock has a different x0.

… but for each clock separately because the “x0” from which x is measured is different for each clock. And xr is always equal to xf (D1 and D2 in the diagram).

282

Hi James

The only reason that I can think of for you to think that t0 is 0 is that you may have used t0 somewhere in your analysis. Therefore I will replace the label t0 with ta.

ta represents any arbitrary time, as viewed from the station clock, after the train reaches its maximum velocity. You should consider ta fixed, or constant which is why I used the term k1. (A reasonably common term for constants).

The reason that I did this was so that we could focus on what happens to t’ as x, as viewed from the station, varies.

Psychologically, this should be similar to how one deals with Partial Differential Equations.

Ed

283

As long as you understand that “x” represents the distance from each clock’s reset location (or synchronization point; P1 and P3) to the location it is at when t is reached (xr and xf or D1 and D2 in the diagram).

284

Hi James,

In the Lorentz transform x measures the distance from the origin of the preferred reference frame which in this case would be the center of the station.

Ed

285

Well, no. If you want to use the station for x0, you will have to ensure that your “x” is actually a “Δx”, “change in x” from t0 to t. And for each clock separately, Δxr and Δxf.

Look at the situation from the very beginning when the clocks are first reset. All clocks read zero, yet one clock is perhaps 100 meters away while the other is 120 meters away. What does that do to your equation? Within merely 1 picosecond from reset, suddenly there is a vast difference between the two clock’s t’ readings.

Lorentz never intended that time be location dependent (interpretation 2). We had a tread about this long ago. Lorentz synchronizing ensures that any and all clocks at any distance read identically regardless of location. That means that immediately after reset, all clocks in all frames would still be very close to the reset reading even though one clock might be 1000 miles from x0. You can’t use Lorentz’ equation with interpretation 2, else things will not stay coherent. If you remember the first debate with Carleas years ago concerning this issue, once I saw that is what the confusion was, we finally resolved that you simply cannot use that interpretation or the numbers just won’t add up. You’ll get obvious non-sense numbers. And that is what ended that debate.

286

Hi James,

The following sketch of the Lorentz transforms derivation is provided to help you and perhaps others see why x is likely to be the distance to the center of the Station. A complete derivation is found at:

viewtopic.php?f=1&t=173286&p=2175197&hilit=Derivation+Lorentz+Transform#p2175197

The Algorithm:

Step 1:
Establish a standard coordinate system centered around the Station.

Step 2:
Rotate that system in a manner such that its x axis corresponds to the train track

Step 3:
Assume that S’ is any inertial reference system moving at velocity v and coincidental to that track.

Step 4 through Step 6:
Reduce the general relationships

x’ = f1(x, y, z, t)
y’ = f2(x, y, z, t)
z’ = f3(x, y, z, t)
t’ = f4(x, y, z, t)

to

x’ = a11(x –vt)
y’ = y
z’ = z
t’ = a41x + a44t,

using the assumptions that space is homogeneous, isotropic and that the physical laws are the same in all inertial reference frames.

From here there are a couple of paths to obtaining the Lorentz equations. The classical path assumes that the speed of light is constant.

Step 8

Substitute the right side of the above 4 equations into the equation:

x’^2 + y’^2 + z’^2 = (ct’)^2 and solve for the coefficients of the equation x^2 + y^2 +z^2 = (ct)^2

The results gives you the Lorentz equations.

X throughout is the distance to the origin.

There are only 3 assumptions, the rest is just mathematical calculations. (A good high school student math student can do it).

This is all very straight forward and unless you disagree with the assumptions (an important point and one which any person with a strong sense of intellectual curiosity should try to do – But after trying to find flaws myself -):

Any rational person would have to conclude that the Lorentz transforms are valid.

Tinkering with the actual thought experiment:

Throughout the following x and t are assumed to be variables in the Station’s Reference frame.

In the case of the accelerated train there will be a lag to get that train up to the maximum velocity.

To determine t’ after the train hits its maximum velocity, vmax, I will consider an inertial frame S’ that is moving continuously at the velocity of vmax. (Since the train’s velocity is always less than vmax until it reaches vmax, we know that S’ must be located, at t = 0, some distance less than 0, in order to coincide with the train when the train reaches vmax.

Let xvmax be the distance at which the train reaches its maximum velocity.
Let txvmax be the time at which the train reaches its maximum velocity.

We know that S’ will travel a distance of (vmax)(txvmax) during the time period of tvmax. This means that the offset distance is [(vmax)(txvmax)] – xvmax], we will call it Dx, is the offset from the station’s origin on the x axis.

Now let’s look at the equation:

t’ = (t – (v/c^2)x)L where L is the Lorentz factor.

At time t = txvmax

t’ = (txvmax – (v/c^2)( xvmax +Dx))*L. There needs to be an offset in x because we are measuring from the starting point of S’.

As we go forward from txvmax, t = txvmax + (t – txvmax) and x = xvmax +Dx + (x – xvmax) and by substituting we get:

t’ = (t – (v/c^2)(x +Dx))L

While there is an offset of Dx, again x is the distance to the origin at the center of the Station.

In these types of “phase offsets” Dx is typically negitive and I would like someone who is competent, like Phyllo, to review this.

Some comments on Lorentz:

Lorentz was a very brilliant Physicist and Mathematician (if I recall correctly) – one of only 5 or 6 people to understand General Relativity early on.

I would certainly be foolish to claim to know more than he. But I would like to know more about your reference in order to try to reconcile these statements.

Selected Wiki commentary on Lorentz:

The bold and brackets are my modifications to the Wiki article.

Ed

287

Re’s Ed,

I have noticed something about you. As reflected by your Twin Paradox argument, you run off into extremely detailed explanations (which I normally appreciate) before you have ensured that your assumptions are applicable. With the Twin Paradox, you assumed it all too complicated. With the SCP, you assume it all too simple.

The issue is NOT the accuracy of Lorentz. If we had only one S clock and one S’ clock, his equations would do fine.
The issue is the number of clocks/Timers.

What does t and t’ represent when you have two station frame clocks, perhaps separated by 20 meters (SC1 & SC2, where SC1 is further from the train), and two train clocks also separated by 20 meters (TC1 & TC2, where TC1 = lead train)?

At t = 0 and t’ = 0 (initialization), and the leading train is 100 meters to the station, what are the following values?
SC1 = ?
SC2 = ?
TC1 =?
TC2 = ?

And then immediately after initialization at t = 0.0001, assuming constant velocity v = 0.8c;
SC1 = ?
SC2 = ?
TC1 =?
TC2 = ?

Additionally, do you understand the following Clock Synchronization?

Such “One-way” synchronization is sufficient since we are assuming no aether existence and it yields consistent time readings (“simultaneity slices”) throughout any chosen inertial frame. So if one were to freeze the universe when SC1 came to read exactly 4:00, all S frame clocks would read exactly 4:00. And if the train had not moved yet, its clocks would also read exactly 4:00 regardless of how far down the track it happened to be. At that one instant SC1, SC2, TC1, and TC2 all = 0.

But now look at your equations;
x’ = a11(x –vt)
y’ = y
z’ = z
t’ = a41x + a44t,

At that one instant, they become;
x’ = a11(x - 0), or x’ = x
t’ = a41x + 0, or t’ = a41x
And that condition could only be true if x = x’ = 0. In other words, doing it the way that you are suggesting (interpretation 2) at reset or initialization, the train absolutely must be at the station. If the train is anywhere else in the entire universe, t’ can never equal zero.

Again, let me remind you of this scenario;

If you had a clock on each of the blue and green trains, which one would represent t’ ?

288

You guys carry on your conversation around me–I just want to post my next thought on spacetime coordinate systems:

I’m wondering if it’s essential to a spacetime coordinate system to have a single origin. I’ve imagine a 3D spatial coordinate system in which there is no origin (or in which there are three distinct origins depending on how you think about it). Usually, when we picture 3D space to ourselves, injecting a coordinate system into the image, we picture three axes all at 90 degrees from each other and crossing at a single point–we call that point the origin. But now imagine that this coordinate system is representing by a line drawn in the sand, a rope crossing that line at 90 degrees to it, and a flag pole staked into the sand right at where the line and the rope cross. In other words, the line is the x-axis, the rope is the y-axis, and the flag pole is the z-axis. Now take the pole out of the ground and fork it back into the ground somewhere away from where the line and the rope cross (it would be in one of the quadrants marked out by the line and the rope). Now grab the rope at both ends and raise it up (say to chest level). You have just created a 3D coordinate system with no origin (or 3 distinct origins if you want to consider each of the axes to have their own origin, but each one can be anywhere along the axis).

Now according to my earlier definition of what makes for a valid coordinate system, this one passes: for any point in space you choose, it still has a definitive and unambiguous value in relation to the three axes of this coordinate system. It still works. Spacetime coordinate systems don’t necessarily need to have a single origin.

289

One chooses a coordinate system in the same way as an ontology or truth system … by its usefulness/rationality.

290

Actually, this is all wrong.

I suddenly realized that you can’t get rid of the origin. Even with my 3D spacetime coordinate system in which none of the axes are touching, you can still have the point (0,0,0). In fact, this ought to be another characteristic of any coordinate system worth its salt: not only should any point have a definitive and unambiguous value according to the coordinate system, but any value possible should have a point. So if you want to ask: does the point (0,0,0) exist in your coordinate system, the answer must be yes. But that just is the origin of the coordinate system. All I did with my “axes” was that I picked three arbitrary lines, all orthogonal to each other and none of them touching, and drew them out in my mind a bit thicker than all other lines. I called them the main “axes” of my coordinate system, and maybe I can say that they are the main axes of my coordinate system, but I haven’t thereby gotten rid of point (0,0,0). It’s still there somewhere in the coordinate system–just not necessarily on any of the axes.

291

Hi James,

You wrote:

I do not understand what you are saying.

If 4:00 is the time at which t = 0, then v = 0 and a11, which will turn out to be the Lorentz factor, = 1.
Here you are right: x’ = x.

At t = 0, t’ = a11((a41/a11)x + (a41x/a11)t) (I used the distributive law to pull out the Lorentz factor. This will put the equations in their proper form). Since a41/a11 will equal -v/c^2, the coefficient of x will be 0.
Since a44/a11 will be 1, t’ = t; and at t = 0 t’ =0.

All the clocks at t = 0 will coincide. If you are interested, all of these functions are continuous, as long as v is continuous.

I no longer consider the Stopped Clock Paradox a viable paradox in The Special Theory of Relativity.

You have not accounted for the Lorentz time transform. There is only a picture of clocks in the train where all of the times look the same. The concept that “Equal actions imply equal results” is misapplied in this case. The actions in the station’s reference frame do not require the exact same actions in the trains reference frame.

Ed

P.S. Just a quick word about pictures.

The French group of mathematicians named Nicolas Bourbaki refused to allow pictures in their text books. The reason that they did not want pictures was because they considered pictures to be analogies. A generous assessment is that while pictures can carry valuable insights, they can also carry miscellaneous unrelated information. A less generous statement is that many pictures carry out right lies.

A picture may be worth a thousand words, but 127 of them are wrong. :smiley:

Most people consider the Bourbaki position to be radical and that pictures are very useful. Benoit Mandelbrot was a nephew of one of the Bourbaki and he absolutely repudiated the Bourbaki.

My own position is that pictures can give an intuitive feel for many things including mathematics, but after you have had a reasonable introduction, you should scrap your background and learn the material all over. This time rigorously. Occasionally this means that you (this is true for me anyway) have to give up some closely held beliefs.

292

Hi Gib,

I would like to spend some time some explaining why your concept of space, which is consistent with Kant and the Stanford Encyclopedia of Philosophy, is wrong.

But I am old and senile and need some time off.

Thanks Ed

293

Emmm… a41 is the coefficient for x. How can a41/a11 = -v/c^2 and a41 = 0 except at zero velocity, in which case it is the trivial case, no motion, only?

For v = non-zero (the only case when you need any transform), a41 cannot be zero and thus x must be zero => only on top of the station.

You keep switching your time reading convention. Are you getting lost in the math? You can’t have both conventions.

And you didn’t answer the questions for multiple clocks/timers. Those bring out the whole problem with your proposed equations. The train represents two clocks in different locations that each read the same thing at reset or initialization.

The pictures are to more ensure that the words are understood.
If you don’t like the pictures, don’t look at them. But if they are showing something different than you think I am saying, then you are probably misunderstanding what I am saying.
:text-worthless:

294

Hi James,

Recapping:

After applying the assumptions that space is homogeneous, isotropic, and that the laws of physics are the same in all inertial reference frames, we arrive at:

x’ = a11(x - vt)
y’ = y
z’ = z, and
t’ = a41x + a44t

We know that the Lorentz transforms, which are mathematically derived from above equations, have the following form:

x’ = (x – vt)*L
y’ = y
z’ = z, and
t’ = (t – (v/c^2)x)L

By writing x’ = (x – vt)L as x’ = L((x – vt) and comparing it to x’ = a11(x – vt), we get:

a11 = L. Equation 1

Now, in order to get t’ = a41x + a44t to look like t’ = (t – (v/c^2)x)L, we need to introduce the a11 term.
Multiplying the right side of the equation t’ = a41x +a44t by a11/a11 we get:

t’ = (a41x +a44t)(a11/a11). Distributing the denominator a11 we get:
t’ = ((a41/a11)x + (a44/a11)t)a11.

Since a11 = L, from equation 1 we can write:
t’ = ((a41/a11)x + (a44/a11)t)L.

Using the commutative law of addition we can write:
t’ = ((a44/a11)t + (a41/a11)x)L.

Comparing the above equation, t’ = ((a44/a11)t + (a41/a11)x)L, with the Lorentz transform for time, t’ = (t – (v/c^2)x)L, we get:

A44/a11 = 1 and
A41/a11 = -(v/c^2)

Hopefully this explains the three lines beneath your first quote.

Now, to answer your questions:

At t = 0, (the station’s clock),

SC1 = 0, because of the setup.
SC2 = 0, because this clock is in the same reference frame as SC1
TC1 =0, because, at v =0, TC1 is also in the same reference frame
TC2 = 0, for the same reason as above.

At t = .0001 and v = .8c we get:

SC1 = .0001, from above
SC2 = .0001, because SC2 is still in the station’s reference frame

Before giving the answers for TC1 and TC2, I should note that in this setup the trains are moving from right to left and v needs to be changed to –v.

TC1 =(.0001 + (.8/c)100)5/3
TC2 = (.0001 + (.8/c)120)5/3

Ed

295

I wasn’t questioning any of that. I was referring to your assertion that a41, the coefficient of x, = 0. If you hold that to be true, the velocity, v, must = 0 and nothing can work out from there, as can be seen below;

Your use of the equations with your time reading convention (with which in the past you have always disagreed) leaves you saying that as the station clock counted 1/10,000 of a second, one train clock counted 6.11 times that and the other train clock, an even 7 times that. You now have both train clocks running faster than the station clock. Does that make sense to you?

The error is in your use of “x”. The variable “x” must represent the amount of distance traveled during the period of t. And it is the same for both train clocks. It can NOT be used as the distance from the station clock.

Realize that you have only one moving frame and that means that you have only one t’ and it is derived by how fast that entire frame moved during t.

296

…and if the train was 1000m away from the train, you would have the train clock running 46 times faster than the station train.

This is the equation you have to use;

except that, as you know, Wiki posts t and t’ backwards
So at 0.8c, you get;
t’ = TC1 = TC2 = 0.00006

The train’s clock is 60% of the station’s no matter how far away it is. And it doesn’t matter how many train clocks there are. They all read the same because it is only their velocity that matters, not how far away they are.

297

Hi James,

Length contraction and time contraction both come from the inverse transforms:

x = (x’ + vt’)L
y = y’
z = z’
t = (t’+ (v/c^2)x’)L

This is straight out of Resnick’s text book.

Basically, when you are comparing ticks of the clock you looking at time intervals, and need to be comparing data in the unprimed or preferred reference frame.

The derivations are fairly straightforward.

Ed

298

So are you then agreeing that if the train was initialized 1000m away, its clocks would be running 46 times faster than the stations? And if it was initialize merely 120m away, it would be running 7 times faster?

The problem is with how you are interpreting “x”. The x in those equations needs to be a “Δx” if the clocks are not starting off from the same location (such as a rocket taking off from Earth vs a train leaving from far away and arriving at the station). Those equations were written for initially collocated clocks.

299

Hi James,

To answer your first question:

Your idea is approximately correct, but your math is wrong.

The derivation of the Lorentz transform assumes that, when x = 0 and t = 0, x’ = 0 and t’ = 0. I am not certain how to deal with the offsets (though I have been tinkering with them).

Also I have discovered that when I initially tried to calculate TC1 and TC2, I forgot to take into account the fact that in .0001 seconds the train will have moved 23,983.4 meters away from 100 meter mark. This means that I should have evaluated (x’, t’) at (-23834.4, .0001). But due to the offsets in time and space, I cannot use the transforms in a straight forward manner.

Just for fun:

Let’s assume that we can evaluate the train using the standard transform. (i.e. when the train is at the origin (Station) x= 0 and t = 0, which implies that x’ = 0 and t’ = 0.
We know that x = vt, where x and t are measured in the station’s reference frame, so t = x/v. We have assumed that v = -.8c. Therefore at x = 100 meters t = 100/-.8c. Since c = 299,792,458 meters per second, -.8c = -239,833,966.4 meters per second. This means that t = -.000000417 seconds.

(It would be common practice to write this as -4.17 X 10^-7 but I will use this method of representation for now).

Similarly, if x = 120 meters, t = 120/-.8c or t = -.0000005 seconds.

Now to get t’ at (100, -.000000417) we would evaluate t’ = (t – (v/c^2))L
t’ = (t - (v/c^2)x)L or t’ = (-.000000417 – (-.8c/c^2)(100))L = (.000000417-(-239,833,966.4/c)(100))L.
Thus t’ = (-.000000417 + (.8/299,792,458)(100))L = (-.000000417 + .000000267)L =- .000000150L.

Since L = 5/3 we get:

TC1’ = -.000000250

Substituting 120 for 100, we would get at (120, -.000000417):

t’ = (-.000000417 – (-.8c/c^2)(120))L.

Completing the evaluation as before we get:
t’ = (-.000000417 + .000000320)L = -.000000097

This means that TC2 = - .000000097. Recapping:

TC1 = -.0000025
TC2 = -.000000097

End of Just for fun.

I will probably keep tinkering. The main idea that I am working on is to set up a new coordinate system at (100, -.000000417) whose coordinates are (0, 0). This would be a simple translation relative to the train station and it would not be moving relative to the station. v again would be -.8c. Then I would explore the results for (x’, t’) in the new reference frame. (Unfortunately, x’, an t’ in this new system may not equal x’ and t’ for the standard system).

I have no idea where your interpretation 1 comes from. Can you give me references or derive it yourself from the Lorentz transform?

Ed

300

Whereas your idea is wrong and your math is approximately correct. I am just using your equations to minimize argumentation.

The way that you are doing it has a different time dilation at every location as though the train’s clocks got faster and faster as the train got closer to the station. You are applying the equations incorrectly, misunderstanding the meaning of x and x’.

You are applying the math to the wrong scenario. The equations that you are using, in the way that you are using them, are only for clocks that are initialized in the same location (the station clock location => everything =0). Imagine that the train is at the station when the clocks are initialized. If the train then goes backwards at 0.8c, in -100 meters, you will get the t’ that you have been calculating.

The issue has been that the train clocks are NOT initialized while the train is at the station, but rather (in this case) at -100m. Thus at t0, we have;
t0 = 0
t0’ = 0
x0 = -100
x0’ = -100

And because of that, the rest of what you calculated is irrelevant to anything.

Dealing with the offsets is easy;
(x - x0) = ((x’ - x0’) + v(t’ - t0’))L
(t - t0)= ((t’ - t0’)+ (v/c^2)(x’ - x0’))L

That would give you the time reading for each clock for every location, assuming that each clock had its own x0.

But easier than that is to not worry about x at all and use what Wiki provided for you;

Because it is the time dilation that you want, not the time reading at any one point. The time dilation has nothing to do with the location of the train, merely its velocity.

The way that you are doing it has a different time dilation at every location as though the train’s clocks got faster and faster as the train got closer to the station merely because they were closer (or vsvrsa).

After you have done it correctly, you will see that the time dilation for both train clocks are equal and where the train is located is irrelevant until you want a particular time reading at a particular location. And you find that using the constant time dilation, velocity, and distance traveled by the clock from that clock’s x0’ (not from the station clock’s location).