Is the Universe travelling or the Space Ship ?

[Johnny5]

 

Originally Posted by swansont

If any observer has to measure the same distance, doesn't that require the spatial tranformation term to be vt? That's the Galilean tranformation.

 

 

I am going to try again to prove this. The goal is to construct the simplest proof possible for it. It has to be made rigorously logical first though.

 

For reference, here are the Galilean transformations:

 

[math] x^\prime = x - vt [/math]

[math] y^\prime = y [/math]

[math] z^\prime = z [/math]

[math] t^\prime = t [/math]

 

Swansont mentions the spatial transformation term vt.

 

Let us consider two arbitrary moments in time t2,t1, with t1 before t2, in frame S.

 

The Galilean transfomations require that the following statements be true at moment in time t1, in frame S:

 

[math] x1^\prime = x1 - vt1 [/math]

[math] y1^\prime = y1 [/math]

[math] z1^\prime = z1 [/math]

[math] t1^\prime = t1 [/math]

 

In the statements above, (x1,y1,z1) denotes the coordinates of an arbitrary point in frame S, at moment in time t1, and (x1`,y1`,z1`) denotes the coordinates of that very same point, but in frame S`, which is moving relative to S.

 

Now, consider things at moment in time t2. Using the same logic as before, the following statements are simultaneously true in both frames (S,S`), at moment in time t2:

 

[math] x2^\prime = x2 - vt2 [/math]

[math] y2^\prime = y2 [/math]

[math] z2^\prime = z2 [/math]

[math] t2^\prime = t2 [/math]

 

In the statements above, (x2,y2,z2) denotes the coordinates of an arbitrary point in frame S, at moment in time t2, and (x2`,y2`,z2`) denotes the coordinates of that very same point, but in frame S`, which is moving relative to S.

 

Now, in frame S, the location where the axes of S intersect is the point (0,0,0), and this statement is always true in frame S.

 

In S`, the location where the axes of S` intersect is the point (0`,0`,0`), and this statement is always true in frame S`.

 

So consider the location of the origin of S in frame S`, at moment in time t1:

 

By the Galilean transformations the following statements are true in both frames:

 

[math] x1^\prime = 0 - vt1=-vt1 [/math]

[math] y1^\prime = 0 [/math]

[math] z1^\prime = 0 [/math]

[math] t1^\prime = t1 [/math]

 

So, the coordinates of the origin of frame S in frame S` are:

 

(-vt1,0,0)

 

 

Now consider the location of the origin of S` in frame S, at moment in time t1:

 

By the Galilean transformations the following statements are true in both frames:

 

[math] 0^\prime = x - vt1=-vt1 [/math]

[math] 0^\prime = y [/math]

[math] 0^\prime = z [/math]

[math] t1^\prime = t1 [/math]

[math] vt1+ 0^\prime = x [/math]

 

 

So, the coordinates of the origin of frame S` in frame S are:

 

(vt1,0,0)

 

 

Now, at moment in time t2, we have the following statements being true in both frames:

 

 

[math] x1^\prime = 0 - vt2=-vt2 [/math]

[math] y1^\prime = 0 [/math]

[math] z1^\prime = 0 [/math]

[math] t1^\prime = t2 [/math]

 

So, the coordinates of the origin of frame S in frame S` are:

 

(-vt2,0,0)

 

 

Now consider the location of the origin of S` in frame S, at moment in time t1:

 

By the Galilean transformations the following statements are true in both frames:

 

[math] 0^\prime = x - vt2=-vt2 [/math]

[math] 0^\prime = y [/math]

[math] 0^\prime = z [/math]

[math] t2^\prime = t2 [/math]

[math] vt2+ 0^\prime = x [/math]

 

 

So, the coordinates of the origin of frame S` in frame S are:

 

(vt2,0,0)

 

Now, focus upon the location of the origin of frame S` in frame S, at both moments in time.

 

At t1, the coordinates of the origin of frame S` in frame S were computed to be: (vt1,0,0)

 

At t2, the coordinates of the origin of frame S` in frame S were computed to be: (vt2,0,0)

 

Using the finite discrete difference calculus, the instantaneous change in any quantity Q, is found by subtracting the earlier value from the latter, over two adjacent moments in time. So if Q initially has value Q1, and then one moment in time later it has value Q2, then the change in Q is defined to be Q2-Q1. That is, we have the following definition:

 

Definition:

 

[math] \Delta Q \equiv Q2-Q1 [/math]

 

So in frame S, we can compute the following differences:

 

The change in the z coordinate of the origin of frame S` in frame S, is by definition:

 

[math] \Delta Z \equiv 0 - 0 = 0 [/math]

 

The change in the y coordinate of the origin of frame S` in frame S, is by definition:

 

[math] \Delta Y \equiv 0 - 0 = 0 [/math]

 

The change in the x coordinate of the origin of frame S` in frame S, is by definition:

 

[math] \Delta X \equiv vt2 - vt1 = v(t2-t1) [/math]

 

Now, focus upon the location of the origin of frame S in frame S`, at both moments in time.

 

At t1, the coordinates of the origin of frame S in frame S` were computed to be: (-vt1,0,0)

 

At t2, the coordinates of the origin of frame S in frame S` were computed to be: (-vt2,0,0)

 

Notice, that we could have also said that:

 

At t1`, the coordinates of the origin of frame S in frame S` were computed to be: (-vt1`,0,0)

 

At t2`, the coordinates of the origin of frame S in frame S` were computed to be: (-vt2`,0,0)

 

And the reason that we can also say this, is because for any moment in time t, t=t`, since we have assumed that the Galilean transformations are true.

 

 

Now, in frame S`, we can compute the following differences:

 

The change in the z` coordinate of the origin of frame S in frame S`, is by definition:

 

[math] \Delta Z^\prime \equiv 0 - 0 = 0 [/math]

 

The change in the y` coordinate of the origin of frame S in frame S`, is by definition:

 

[math] \Delta Y^\prime \equiv 0 - 0 = 0 [/math]

 

The change in the x` coordinate of the origin of frame S in frame S`, is by definition:

 

[math] \Delta X^\prime \equiv vt2^\prime - vt1^\prime = v(t2^\prime-t1^\prime) =v(t2-t1) [/math]

 

Since moment in time t2, is different from moment in time t1, we can divide both sides of the equation above by (t2-t1) or (t2`-t1`), to obtain an expression for v.

 

Incidentally, t1 has to be a number, and t2 has to be a number, therefore, we shall assume that there is a clock/clocks at rest in frame S, which map natural numbers onto moments in time.

 

In the case where the amount of time has to be determined by different clocks at rest in a frame, as long as the clocks are initially in sync, they remain synchronous, since they tick at the same rate.

 

So we have:

 

[math] \frac{\Delta X^\prime}{t2-t1} = \frac{\Delta X^\prime}{t2^\prime -t1^\prime} = v [/math]

 

Similarly we find that:

 

[math] \frac{\Delta X}{t2-t1} = \frac{\Delta X}{t2^\prime -t1^\prime} = v [/math]

 

Now, using the transitive property of equality, and the few relations just given, we have:

 

 

[math] v = \frac{\Delta X}{(t2^\prime -t1^\prime)} = \frac{\Delta X^\prime}{(t2-t1)} [/math]

 

To requote swansont

 

Originally Posted by swansont

If any observer has to measure the same distance, doesn't that require the spatial tranformation term to be vt? That's the Galilean tranformation.

 

 

So we are to focus on the spatial transformation term vt.

 

Multiplying both sides of the equation above by (t2-t1) we obtain:

 

[math] v (t2-t1) = \frac{(t2-t1) }{(t2^\prime -t1^\prime)} \Delta X = \Delta X^\prime [/math]

 

Now, denoting (t2-t1) by Dt we have:

 

[math] v \Delta t = \frac{(t2-t1) }{(t2^\prime -t1^\prime)} \Delta X = \Delta X^\prime [/math]

 

The LHS is not quite the spatial transformation term, but it is close. But the thing that I would point out now, is that to even define "relative speed v" you need two moments in time, not one.

 

Now might be a good time to close the scope of any assumptions.

 

What I am trying to do right now, is determine whether or not what swansont said is true.

 

What was asked was this, if different observers must measure the same distance, mustn't it be the case that the Galilean transformations are true?

 

Symbolically, that means this:

 

? [math] \Delta X = \Delta X^\prime [/math] iff GT are true.

 

I am going to have a look at that in two parts:

 

Part 1: If [math] \Delta X = \Delta X^\prime [/math] then GT are true.

Part 2: If GT are true then [math] \Delta X = \Delta X^\prime [/math]

 

From one of the relations above, and the assumption that t=t`, we can see that it must be the case that delta X = delta X`. So all we have proven thus far, is the following:

 

If the Galilean transformations are true then observers with relative speed v must measure the same amount of change in the distance between them.

 

The argument above did not prove the converse, which is part 1 above, namely:

 

Converse: If two arbitrary frames S,S` with relative speed v must measure the same amount of change in the distance between them, then the Galilean transformations are true.

 

Before trying to tackle the converse, let us be clear as to what is meant by relative speed v.

 

In some frame S, something moves, lets say along the X axis. The distance it moves as measured by rulers at rest in S, is to be denoted by:

 

[math] \Delta X [/math]

 

And the time coordinate of this frame is to be denoted by t. So, for something that moves in this frame, the amount of time of travel, as measured by clocks at rest in frame S, is going to be denoted by:

 

[math] \Delta t [/math]

 

So, for two consecutive moments in time t2,t1, with t1 before 2 in this frame, we have:

 

[math] \Delta t = t2 - t1 [/math]

 

The relative speed is then defined to be the ratio of the distance traveled in the frame, divided by the time of travel in the frame, that is:

 

[math] v \equiv \frac{\Delta X}{\Delta t} [/math]

 

Now, the definition of relative speed must be the same in both frames. So in frame S`, the origin of S moved a distance which is denoted by:

 

[math] \Delta X^\prime [/math]

 

And that distance is to be measured by rulers at rest in S`.

 

And the time coordinate of frame S` is denoted by t, and the amount of time of travel of the origin of S through S` as measured by clocks at rest in S` is denoted by:

 

[math] \Delta t^\prime [/math]

 

And using the same definition of relative speed v that an observer at rest in S must use, we have:

 

[math] v^\prime \equiv \frac{\Delta X^\prime}{\Delta t^\prime} [/math]

 

And the relative speeds must be equal, that is v=v` therefore:

 

[math] v \equiv \frac{\Delta X}{\Delta t} = \frac{\Delta X^\prime}{\Delta t^\prime}[/math]

 

And the preceding equation is true, regardless of whether or not the Galilean transformations are true.

 

From the equation above, we see that if the numerators are equivalent, then the denominators are equivalent, and if the denominators are equivalent, then the numerators are equivalent. That is, measurments of distances are equal if and only if measurements of "amounts of time" are equal.

 

If [math] \Delta X = \Delta X^\prime [/math] then GR are true.

 

Let the relative motion be along coincident X axes of two arbitrary frames S,S`.

 

Denote the relative speed by v.

 

Assumption 1: [math] \Delta X = \Delta X^\prime [/math]

 

The LHS of the equation above, is the change in the X coordinate of the origin of S` in S. Let X1 denote the initial X coordinate of frame S` in frame S, at moment in time t1, and let X2 denote the X coordinate of frame S` in frame S, at the very next moment in time, t2.

 

The RHS of the equation above, is the change in the X coordinate of the origin of S in S`. Let X1` denote the initial X coordinate of frame S in frame S`, at moment in time t1`, and let X2` denote the X coordinate of frame S in frame S`, at the very next moment in time, t2`.

 

Therefore:

 

[math]\Delta X = x2-x1 [/math]

 

[math]\Delta X^\prime = x2^\prime - x1^\prime [/math]

 

Now, the speed of the origin of S` in S is by definition:

 

[math] v = \frac{\Delta X}{\Delta t} = \frac{x2-x1}{\Delta t} =\frac{x2-x1}{t2-t1} [/math]

 

And the speed of the origin of S in S` is by definition:

 

[math] v^\prime = \frac{\Delta X^\prime}{\Delta t^\prime} = \frac{x2^\prime-x1^\prime}{\Delta t^\prime} = \frac{x2^\prime-x1^\prime}{t2^\prime-t1^\prime} [/math]

 

And the relative speeds must be equal, that is v=v`. Therefore:

 

[math] \frac{x2-x1}{t2-t1} = \frac{x2^\prime-x1^\prime}{t2^\prime-t1^\prime} [/math]

 

So, the assumption that both observers must measure the same distance travelled, that means that the numerators must be equal. Therefore, the denominators are equal.

 

Therefore, under the assumption that the distance traveled is the same in both frames, it necessarily follows that:

 

[math] t2-t1 = t2^\prime - t1^\prime [/math]

 

Now, let the two clocks be synchronized at moment in time t1=t1`, and let them read the number 0. Therefore,

[math] t2-0 = t2= t2^\prime - 0^\prime = t2^\prime [/math]

 

Therefore, for any moment in time, which comes after t1=t1`, the clocks will still be synchronous.

 

This is one of the four Galilean relationships. There are three left to verify.

 

Now, the relative motion was stipulated to be along the common X axis of both frames, therefore the following statements were stipulated to be true:

 

[math] y^\prime = y [/math]

[math] z^\prime = z [/math]

 

And the relation

[math] x2-x1 =x2^\prime - x1^\prime [/math]

 

was assumed to be true.

 

So the only Galilean relation which is left to verify is the following one:

 

[math] x^\prime = x - vt [/math]

 

Once the previous relation is verified, we will have proven the following:

 

Theorem: If [math] \Delta X = \Delta X^\prime [/math] then the Galilean transformations are true, in the case where S,S` are in uniform relative (non-rotational) motion, with speed v.

 

Which is what I think swansont asked me about. So, the answer is yes to his question. That is, if two observers in uniform relative motion with constant speed v must agree upon distance measurements, then they must also agree upon time measurments, which is equivalent to saying that the Galilean transformations are true.

 

Since the relative speed computations must be equal we have:

 

[math] v = \frac{x2-x1}{t2-t1} = \frac{x2^\prime -x1^\prime}{t2^\prime-t1^\prime} [/math]

 

Under the assumption that delta X = Delta X` we can reverse the numerators above, as follows, so that the following statements are true under the single assumption:

 

[math] v = \frac{x2^\prime -x1^\prime}{t2-t1} = \frac{x2-x1}{t2^\prime-t1^\prime} [/math]

 

Let the clocks by synchronized at t1=t1`, and let them both read zero. Therfore we have:

 

[math] v = \frac{x2^\prime -x1^\prime}{t2} = \frac{x2-x1}{t2^\prime} [/math]

 

And we have already shown that under the assumption that delta x = delta x`, that we must have t=t` for any moment in time t. Therefore:

 

[math] v = \frac{x2^\prime -x1^\prime}{t2} = \frac{x2-x1}{t2} [/math]

 

We have not caused division by zero error.

 

Multiplying by t2 we have:

 

[math] vt2 = x2^\prime -x1^\prime= x2-x1 [/math]

 

Therefore, if delta X = delta X` then

 

[math] x1^\prime= x2^\prime -vt2[/math]

 

This argument is way too long, and it involved meandering, but I just figured out how to finish it off.

 

We are trying to prove this:

 

[math] x^\prime = x - vt [/math]

 

Now, at moment in time t1=t1`=0, the equation above leads to:

 

[math] x1^\prime = x1 [/math]

 

And at moment moment in time t2, the equation above leads to:

 

[math] x2^\prime = x2 - vt2 [/math]

 

So here are our initial conditions for S,S`:

 

t1=t1`=0

And this

 

For any point X1 on the X axis of S, and any point X1` on the X axis of S`: x1=x1`.

 

Now here is the only assumption to be made during the argument:

x2-x1=x2`-x1`

 

And we have already shown that:

 

If x2-x1=x2`-x1` then for any moment in time t2>t1 it must be true that t2=t2`.

 

All that remains, is to use the information properly, to formulate the first of the four Galilean relations, and then clean up the argument, so that the final argument is in its simplest form. It should minimize the number of mental steps necessary to reach the conclusion.

 

We start off with this:

 

Theorem:

 

Initial conditions:

 

t1=t1`=0

And this

 

For any point X1 on the X axis of S, and any point X1` on the X axis of S`: x1=x1`.

 

(What these initial conditions do, is stipulate that at moment in time t1=t1`, the frames S,S` are equivalent. In other words, their axes not only overlap, but the positive x,y,z axes of each also correspond.

 

If " moment in time t>t1:

 

(y`=y and z`=z and Dx = Dx`

 

then

 

" moment in time t>t1:

 

[math] x^\prime = x - vt [/math]

[math] y^\prime = y [/math]

[math] z^\prime = z [/math]

[math] t^\prime = t [/math]

 

We can now use first order logic to solve the problem.

 

 

Solution.

 

We have the following initial condition:

 

For any point x on the x axis of S, and any point x` on the x axis of S`

 

x`=x

 

Now, suppose the following:

 

x2-x1=x2`-x1`

 

It was previously shown that:

 

If x2-x1=x2`-x1` then " moment in time t>t1: t`=t

 

Therefore:

 

t`=t

 

Therefore

 

vt`=vt

 

Therefore:

 

x2-x1 - vt =x2`-x1` -vt`

 

The initial conditions are such that when t1=t1`= 0

 

x1=x1`

 

but at the very next moment in time, it is no longer the case that x1=x`, because the frames are in relative motion with speed v.

 

At all moments in time y`=y, z`=z, and under the single assumption, we also have t`=t, all that remains to be shown, is that for any moment in time after t1=0, we have

 

x`=x-vt

 

as the coordinate transformation from one frame to another.

 

At moment in time t2, the relation above gives:

 

x2-x1 - vt2 =x2`-x1` -vt2`

 

Now, at moment in time t1=t1`=0, we have from the initial conditions x1=x1`.

 

Now, at moment in time t2, a point on the x axis of frame S` has moved along the x axis of frame S, and the distance traveled is equal to the relative speed times the time of travel.

 

The time of travel is given by t2-t1=t2-0=t2

 

So that the distance travelled in frame S, by a point (x`,y`,z`) that is fixed in S` is given by vt2.

 

 

So consider the fixed point (4,0,0) in frame S`.

 

Initially, its coordinates in frame S were the same, namely (4,0,0).

 

Now, at the very next moment in time, it moved a positive distance vt2, in frame S.

 

Suppose the distance traveled in this time was three units.

 

Then this fixed point in S`, no longer has an x coordinate of 4, its x coordinate is now displaced from that by 3 units.

 

Therefore, either its new location in S is 7, or its new location in S is 1.

 

So now, we need one more initial condition. We need to know the direction of motion of a fixed point in S`, in frame S. Thus, we need to know not just the relative speed, but the relative velocity of the frames.

 

Suppose a fixed point in S` moves in the i^ direction of frame S.

 

Then at moment in time t2, the point (4,0,0) fixed in S`, has had its x coordinate in frame S increased by amount vt2=3.

 

Therefore its new coordinate in frame S is 7.

 

And therefore, in frame S`, a fixed point in S moved with the following velocity:

 

[math] \vec v^\prime = -v \hat i^\prime = -\frac{\Delta x^\prime}{\Delta t^\prime} \hat i^\prime [/math]

 

And of course i^ = i^` therefore we can just write:

 

[math] \vec v^\prime = -v \hat i = -\frac{\Delta x^\prime}{\Delta t^\prime} \hat i[/math]

 

 

And therefore, in frame S, a fixed point in S` moved with the following velocity:

 

[math] \vec v= v \hat i = \frac{\Delta x}{\Delta t} \hat i [/math]

 

So, we have supposed that a fixed point in S` moves in the i^direction of frame S, and that a fixed point in S moves in the -i^` direction of frame S`.

 

So here is the idea behind the Galilean transformation.

 

Consider a fixed point in reference frame S, say (5,0,0).

 

At the moment in time t1=t1`=0, the initial conditions are such that the frames overlap perfectly, in other words S=S` when t=t1=t2=0.

 

So, at that moment in time the coordinate of that point in S` is also (5,0,0)

 

Now, we know the relative velocity, I just wrote it. So as the time variable t increases, the point (5,0,0) which is fixed (at rest) in frame S, has moved in frame S`. And it has moved in the -i^` direction of frame S`.

 

So the x` coordinate of (5,0,0) is no longer 5, it is something less than 5, because it moved in the -i^` direction. The distance it has moved in frame S` is given by vt`, but since t=t` in the Galilean transformations, we can just as well say that the distance it has moved in frame S` is given by vt.

 

So that its new position in frame S` one moment in time in the future is given by:

 

5 - vt

 

Its original position in frame S` was 5.

 

So over the first two consecutive moments in time the following statement is true:

 

x` = 5-vt

 

We can write this a little more explicitely as follows:

 

[math] x^\prime = 5 - \frac{(x2-x1)t}{t2-t1} [/math]

 

Now, lets look at the first few instantiations...

 

when the variable t is at t1=0, we have:

 

[math] x^\prime = 5 - \frac{(x2-x1)t1}{t2-t1} = 5 - 0 [/math]

 

This is correct, because initially the frames are equivalent, that is they are really only one frame, and the fixed point (5,0,0) in frame S, has the following coordinates in frame S` (5,0,0), which is what the statement above says.

 

Now, when the variable t is instantiated by t2, we have:

 

[math] x^\prime = 5 - \frac{(x2-x1)t2}{t2-t1} [/math]

 

And since t1 is stipulated to be zero we have:

 

[math] x^\prime = 5 - \frac{(x2-x1)t2}{t2} [/math]

 

And let us stipulate that the relative velocity doesn't change. We can now instantiate the variable t with t3, and obtain:

 

[math] x^\prime = 5 - \frac{(x2-x1)t3}{t2} [/math]

 

Thus, we can write x` as a function of t, as follows:

 

[math] x^\prime(t) = 5 - vt [/math]

 

Where we understand that the relative velocity is constant in time.

 

And the statement above would be true for any fixed point on the x axis of frame S, not just the point (5,0,0). So for any point on the x axis of frame S (x,0,0) the following statement is true:

 

[math] x^\prime(t) = x - vt [/math]

 

Under the single assumption made. Therefore we are done. We have proven the following fact:

 

If delta x=delta x` then the Galilean transformations are true.

Specifically:

 

[math] x^\prime(t) = x - vt [/math]

 

and

 

y`=y

z`=z

 

and

t`=t

[swansont]

But simultaneity isn't relative.

 

So you keep saying. I understand that Polly wants a cracker, too.

 

Your model has certain physical implications. You need to verify it experimentally. i.e. it's time to put up or shut up.

[Johnny5]

So you keep saying. I understand that Polly wants a cracker' date=' too.

 

Your model has certain physical implications. You need to verify it experimentally. i.e. it's time to put up or shut up.[/quote']

 

I am working on it, I know what I intend to do...

 

I am ultimately going to use a proof by contradiction, but right now I am looking for an if-and-only if statement....

 

Well here have a look at this:

 

XY X iff Y

00 1

01 0

10 0

11 1

 

So think about what that means, in temporal logic.

Well...

 

in TL, the statements X,Y can have a truth value which varies in time, or a truth value which is constant in time, it doesn't matter. So...

 

What the meaning of iff is, is clear...

 

The only way that "X iff Y" can be true is for both statements X,Y to have simultaneous truth values that match. Even if we don't know the truth value of the individual statements, we can rest assured that so long as the truth values are identical, that the truth value of "X iff Y" is true. At any moment in time at which the truth value of X=truth value of Y, the statement "X iff Y" is true.

 

And at any moment in time for which not (|X|=|Y|), the statement "X iff Y" is false.

 

The point is, I have a plan, and I am working towards it.

 

Regards

[swansont]

I am working on it' date=' I know what I intend to do...

 

I am ultimately going to use a proof by contradiction, but right now I am looking for an if-and-only if statement....

[/quote']

 

Not good enough. Science is inductive, not deductive. You always have to make an assumption about how nature behaves, so when you set up a proof by contradiction, you will have a second assumption hidden in there. Your contradiction may be because your hidden assumption is wrong, not because the explicitly stated one is wrong - as evidenced by every one of your attempts so far having some implicit assumption about simultaneity being absolute or Galilean transforms being correct or something similar. So if basically you assume that both Galilean and Lorentz transforms are correct, you are always going to find a contradiction - but that doesn't tell you which one is incorrect, it only tells you you can't do both, and that shouldn't be surprising.

 

You have to make a predicition about the real world, and then find an experiment that will work if you are right and not work if you are wrong. That's the only way to tell if a hypothesis has merit.

 

Of course, anybody who has taken a relativity course already knows the answer, since many experiments have been done over the years. Galilean transforms have been shown not to work. You're betting on a race that's already been run, and wagering on the wrong horse.

[Johnny5]

You're betting on a race that's already been run, and wagering on the wrong horse.

 

Well its my money, and I don't mind losing. But you never know.

 

Actually I already know, but whatever, my proof skills need work.

[quick silver]

to me.......... it's all perspective. two people could say the other answer and both could be right. it's like reality.

 

just an opininon

[swansont]

to me.......... it's all perspective. two people could say the other answer and both could be right. it's like reality.

 

just an opininon

 

Only one can agree with observations, if they are mutually exclusive.

[geistkiesel]

No, he can easily tell that he was the one who actually accelerated. He can feel the acceleration, he can fire a laser perpendicular to his path and notes that it curves, etc. And while the time dilation formula holds when used by either twin while they are at constant velocity, and holds for the twin that does not accelerate (remains in the same reference frame the whole time.),

 

If the acceleration gives information to the 'real moving' twin then would not this acceleration postulate apply as a scientific principal across the SRT board? No one will suggest the acceleration postulate is a mere arbitrary tool to "explain away" apparent contradicting arguments re the integrity of SRT.

 

In other words a train accelerates from a train station. There is no recorded acceleration by the train station. Now besides a common sense approach, why not use the acceleation postulate you used above to distinguish which of the two inertial frames, supposedly equivalent, are actually moving and the other at rest as far as physical descriptions apply? A long sentence, but seriously asked.

 

As a corrolary, if the train observer is theoretically justified to make make any consideration regarding motion (i.e. at rest or moving wrt the station), then he may equivalently consider the train as moving, and the station is at rest; he can then, even, now use the acceleration postulate to flip the physical coin for the observer's consideration --he can make the reasonable assumption he is moving; he can use the acceleration principal to add weight to the choices, one or the other, in order for him to produce a physically more 'real universe'. Rejecting the 'real unuivserse' in thihe context of the preceeding would require the observer to ignore the acceleration postulate in favor of maintaining SRT continuity, only. There is no rational excuse that forbids the application of the acceleration postulate as designed by Janus

 

This brings us back, by calculation, to a nonSRT world does it not?

 

Embankments never accelerate, and claims to the converse are not physically true, in general,at least. Earthquakes after all, are real, also. -- motion claimed attributable to the measured relative motion under scrutiny Is it physically reasonable to conclude the motion of the accelerated embankment is reasonable? -- And a different physical result calculated depending on which frame the train observer concluded his motion to be at rest?

 

The observer will see a different algorithim of equations correct? The physics changes with a "point of view"? Simultaneity? Real space? real time? The final conclusion after the observer finishes scrutinizing all the possibilities is:?Answer:

A. Only the train is measured as moving, the embaankment is eternally at velocity zero. or,

B. the train is at rest wrt the necessarily accelerated embankment?

See?

 

And indeed, Janus has spoken the truth.

 

And her moving finger writes and, having writ, moves on; yet, using all her piety and wit, she may lure it back to cancel, at least, half a line.

 

Geistkiesel

[geistkiesel]

t = t'

 

Johnny5, this is a 'linear Sagnac' arrangement. wrap the left and right legs of teh frame into the same circle and there is the original Sagnac. Here the photons emitted at the physical midpoint of the L and R clock/mirrors, to these L and R positiions on the moving frame (wrt the embankment). The schematic shows the left photon lp arriving at L before the rp arrives at R. In the first ct distance the lp arrives at L , the rp 2vt from R. the rp must cross a distance ct' = 2vt + vt', the latter distance occured when the photon was chasing the R across the 2vt disatnce.

Here t'= t(2v)/(c-v).

This says to me that using the moving frame clocks a measured t' =0 proves no motion; a t' > 0 proves motion, absolutely. Does not the emission point of the photons define an adequate Gallilean coordinate system? Cannot all velocity now be measured wrt the emission point? This point is invarinat in space and time correct? Immutable? for all t = t'.

 

 

I cannot see how an observer placed anywhere on the frame is privvy to other than the emission times and the arrival times of the reflected light at the moved physical midpoint of L and R at the very earliest. He must use only this data to fomulate a physically proper consideration of whether he is moving or not.

The velocity is

 

v = ct'/(2t + t'),

 

is this not correct? If the values of the variables on the RHS are measured, then is this not a measure of absolute velocity wrt c?

 

This is an edit after I posted. In other words, if the photons arrive at the L and R simultaneously the frame is at rest, other wise if the photons arrive sequentially the frame is moving and t' > 0. This is obviously a comparison of zero and nonzero velocity.

 

Geistkiesel

[Johnny5]

Johnny5' date=' this is a 'linear Sagnac' arrangement. wrap the left and right legs of teh frame into the same circle and there is the original Sagnac. Here the photons emitted at the physical midpoint of the L and R clock/mirrors, to these L and R positiions on the moving frame (wrt the embankment). The schematic shows the left photon lp arriving at L before the rp arrives at R. In the first ct distance the lp arrives at L , the rp 2vt from R. the rp must cross a distance ct' = 2vt + vt', the latter distance occured when the photon was chasing the R across the 2vt disatnce.

Here t'= t(2v)/(c-v).

This says to me that using the moving frame clocks a measured t' =0 proves no motion; a t' > 0 proves motion, absolutely. Does not the emission point of the photons define an adequate Gallilean coordinate system? Cannot all velocity now be measured wrt the emission point? This point is invarinat in space and time correct? Immutable? for all t = t'.

 

[img']http://ourworld.cs.com/Sandgeist/simul/whereissrt.GIF[/img]

 

I cannot see how an observer placed anywhere on the frame is privvy to other than the emission times and the arrival times of the reflected light at the moved physical midpoint of L and R at the very earliest. He must use only this data to fomulate a physically proper consideration of whether he is moving or not.

The velocity is

 

v = ct'/(2t + t'),

 

is this not correct? If the values of the variables on the RHS are measured, then is this not a measure of absolute velocity wrt c?

 

This is an edit after I posted. In other words, if the photons arrive at the L and R simultaneously the frame is at rest, other wise if the photons arrive sequentially the frame is moving and t' > 0. This is obviously a comparison of zero and nonzero velocity.

 

Geistkiesel

 

It will take me awhile to analyze this, but I will.

 

In the meantime, what does kiesel mean?

 

Geist means ghost?

 

Regards

[el bastardo]

So what I want to know is when I drop the ball, does the earth come up to meet it? Would this be out of line w/relativity? Can the Earth (and everything else) be expanding (and curving (possibly necessarily) in the case of space, and slowing/speeding in relative (local) perception)?

[Johnny5]

So what I want to know is when I drop the ball' date=' does the earth come up to meet it?

[/quote']

The truthful and correct answer is yes, when what happens is viewed from the center of mass frame.

 

Takes only a second to explain.

 

In order to analyze the relative motion of things, one must first choose a frame to view the motion in.

 

I presume you mean simply dropping a ball, and watching it fall to earth.

 

The earth is very massive, compared to a tiny ball. There is a lot more matter in the earth, than a ball. A whole lot more.

 

In order to really grasp the explanation, it will help you to think about an infinitely massive object. No such object exists, yet for some reason this will help you understand.

 

Somewhere in the universe is the center of mass of the universe.

 

Call this location in the universe W.

 

So Omega denotes the center of the universe.

 

Now, suppose that there was an infinitely massive object, shaped in a sphere, and that it currently is at rest relative to location Omega.

 

By Galileo/Newton's first law of motion, the center of mass of this object will remain at rest in this frame.

 

Or to use the term 'distance' to express this, you can say that the distance from Omega to the center of mass of this infinitely massive object is currently constant in time.

 

Now, suppose that you want to accelerate the center of mass of this object, relative to point in space Omega.

 

By Newton's second law of motion, the force F you must apply to give it acceleration a in this frame is related to the mass M of the object as follows:

 

F = M a

 

Since the mass has been stipulated to be infinite, you must apply an infinite force to get that object to move, relative to point Omega.

 

No infinite force can be applied to any object, hence the object will remain at a fixed location in space (relative to Omega) forever.

 

But of course there are no infinitely massive objects, but that's what the math says.

 

Now, think about what you want, which is dropping a ball in the gravitational field of the earth.

 

The earth is far closer to being an infinitely massive object than the ball is.

 

A small force such as you jumping up and down, does almost nothing to "change the position of earth" relative to Omega. (where I am pretending earth is currently at rest relative to the center of the universe)

 

Now, as you hold the ball at a steady height h above the surface of the earth, there are two forces.

 

The force of the earth pulling the ball down, and the gravitational force of the ball pulling the earth up.

 

Now, in between the center of mass of the earth, and the center of mass of the ball, is the location of the center of mass of the two body system.

 

If the two objects had equal massives, the location of this place would be half the distance between them. But of course, the earth is far far far more massive than a little ball.

 

So you are about to release the ball, and you are going to view the motion in the center of mass frame. That is a reference frame at which the center of mass of the two body system doesn't move.

 

Here is what happens.

 

Call the location of the center of the system mass the origin of the reference frame.

 

Once the ball is released, the ball accelerates towards the origin.

 

Also

 

Once the ball is released, the earth accelerates towards the origin.

 

The magnitudes of these accelerations are not equal.

 

But, in this frame, the answer to your question is yes, the earth does "come up to meet the ball."

 

By Newtons third law, the force of the ball on the earth, is equal to the force of the earth upon the ball, but in the opposite direction therefore:

 

F12=-F21

 

Let M denote the mass of the earth, let A denote the acceleration of the earth, let m denote the mass of the ball, and let a denote the acceleration of the ball therefore, the following scalar relation must be satisfied:

 

MA=ma

 

Suppose the ball weighs one kilogram, and accelerates at 10 meters per second squared. Therefore:

 

MA=10

 

The mass of the earth is:

 

6 x 10^24 kilograms

 

Hence:

 

6 x 10^24 A = 10

 

hence

 

A =1.666... x 10^-24 meters per second squared.

 

Now, suppose that initially, you release the ball from a height of say around seven feet above the surface of the earth.

 

In fact, lets make it exactly two meters.

 

So you raise your arm, until the center of mass of the ball is 2 meters above the surface of the earth, and then you let it go.

 

You can use the kinematic equations for constant acceleration to figure out how long it will take the ball to hit the ground.

 

The computation is simple. Here is the formula for the general case of constant acceleration a:

 

D = v0 t + 1/2 a t^2

 

In the formula above, D represents distance traveled, v0 denotes the initial speed, t denotes the amount of time of travel, and a denotes the acceleration.

 

Now in this case the acceleration is ten meters per second squared.

 

The initial speed of the ball, before it is released, in the center of mass reference frame is zero.

 

So here is the formula for D:

 

D = 5 t^2

 

The distance the ball falls is 2, hence:

 

2=5 t^2

 

hence

 

2/5 = t^2

hence

 

4/10 = t^2

 

hence

 

.4 = t^2

 

Now take the square root of both sides of the equation above, and we have:

 

.63245553... = t

 

 

Since we used SI units in the formula, t came out in seconds.

 

So the time of freefall is a little more than half a second.

 

Now, recall the acceleration of the earth in the center of mass reference frame:

 

A = 1.666... x 10^-24 meters per second squared.

 

It's initial speed in the CM frame is zero. Hence the distance it moves whilst falling towards the apple is given by:

 

D = 1/2 a t^2

 

D = 1/2 ( 1.666... x 10^-24 )t^2

 

D = 8.3 x 10^-25 t^2

 

And we know the time till collision is .63 seconds hence:

 

D = 8.3 x 10^-25 (.6324555...)^2

 

Thus:

 

D = 3.3333... x 10^-25 meters.

 

This is a fantastically small distance, in comparison to one meter.

 

The diameter of a hydrogen atom is around

 

1 x 10^-10 meter, which is called an Angstrom.

 

1 x 10^-25 is (1000000000000000) times smaller than that.

 

So thats the distance the earth moves, in the time it takes the ball to fall to earth, when the relative motions of the centers of mass of the earth and ball are viewed in the center of mass frame.

 

Regards

[geistkiesel]

The truthful and correct answer is yes' date=' when what happens is viewed from the center of mass frame.

 

Takes only a second to explain.[/quote']

 

 

Regards

You realize your response is unprovable, do you not? The earth doesn't meet the ball unless in the size of the moon, or another stellar object of comparable size and mass.

[J.C.MacSwell]

Regards

You realize your response is unprovable, do you not? The earth doesn't meet the ball unless in the size of the moon, or another stellar object of comparable size and mass.

 

Conservation of momentum. You cannot change the velocity of the centre of mass of a closed system. If you pick a reference frame where the centre of mass is at rest, it must remain at rest unless an outside force acts upon it.

 

If you mean the Earth/ball example is unmeasurable, you are no doubt correct, but if you sit down where you are standing you will have "raised" the Earth relative to where it would have been otherwise and all other things being equal.

[el bastardo]

Regards

The earth doesn't meet the ball unless in the size of the moon, or another stellar object of comparable size and mass.

Man, I need to brush up on my English... I don't understand a word... well... I understand the words... just not the order... sorry... ima dumas...

[el bastardo]

You realize your response is unprovable, do you not?

I wonder how it could be any other way. Common sense tells me that either something is pushing on my feet or something is pushing on my head... many would say, "Some dam thang's pushin' on 'is hed"... ... But I swear I feel it on my feet... The Earth is dam well growin'... dadblastit... an' so is everything else.

[el bastardo]

not only that...

It's bound to be growing @ 32 ft/sec/sec, right?... or what ever that comes out to in wherever... constant acceleration... for a relative eternity... not that I don't appreciate your explanation, Johnny5... but anybody ever tell you .... well... your wordy little bugger.... but I'm sure we all love you...

[Johnny5]

(Johnny5)... anybody ever tell you .... well... your wordy little bugger

 

Not exactly in quite those words no.

[geistkiesel]

You realize your response is unprovable, do you not? The earth doesn't meet the ball unless in the size of the moon, or another stellar object of comparable size and mass.

 

Conservation of momentum. You cannot change the velocity of the centre of mass of a closed system. If you pick a reference frame where the centre of mass is at rest, it must remain at rest unless an outside force acts upon it.

 

If you mean the Earth/ball example is unmeasurable, you are no doubt correct, but if you sit down where you are standing you will have "raised" the Earth relative to where it would have been otherwise and all other things being equal.

Come on JC MAcSwell, we are talking senssibly here. HJave you had your spot of tea yet? There is no analogue with any physical significance that says I raised the earth to meet me when sitting down. I necessarily closed the distance from my standring position to one closer to the center of the earth, and eveybody knows it did not happen equivalently with me raising the earth, even you know it.

 

One can inject philosphical and theoretical models attempting to equate various symmetrical arangements, but when train stations accelerate and move uniformly while the trains are at rest wrt the train stations one has described a physically impossible conditon to ever produce. Ever. From, the sake of argument I will even grant you any theoretical justification you choose to impose here in the discussion, but what ever science you come up with the fact is "I sat diown to earth", " I did not sit up to earth". trust me on this. JC McSwell.

Hey, JC, you have heard of the phrase, "please sit down" have you? You cannot tell me when some tells you to "please sit down" that you actually wait there until the earth 'raises up' to you do you? hey, I may be wrong in all this and if so then I take in the professional neck of my scientific reputation of having made an incorrect statement on such a fundamenmtal issue regarding a matter of physical law. Did Einstein raise the earth when he sat down.

 

Mybe the earth raisers are actually the causal entity of earthquakes and tunamis??

[geistkiesel]

The earth doesn't meet the ball unless in the size of the moon, or another stellar object of comparable size and mass.

Man, I need to brush up on my English... I don't understand a word... well... I understand the words... just not the order... sorry... ima dumas...

Little balls have no effect on moving big balls next to them, but big balls now move heaven and earth when someone's squeezing them.

.

[geistkiesel]

It will take me awhile to analyze this' date=' but I will.

 

In the meantime, what does kiesel mean?

 

Geist means ghost?

 

Regards[/quote']

kiesel is tranlated literally as "pebbles". Now everybody knows!.

 

Hint I simply broke the motion into convenient segments. The first ct the distance the photon move initially. . vt the distance the frame moves in time t. How far away from the right clock is the right photon after moving ct *the same distance the left photon just moved? Look at the left photon : if d is the distance of midpoint to clock then the left photon has reached vt short of the clock,'s initial position., so then put the right photon also vt short of the right clock, plus the vt the frame has moved when the photons moved a distance ct.

[Johnny5]

kiesel is tranlated literally as "pebbles". Now everybody knows!.

 

Hint I simply broke the motion into convenient segments. The first ct the distance the photon move initially. . vt the distance the frame moves in time t. How far away from the right clock is the right photon after moving ct *the same distance the left photon just moved? Look at the left photon : if d is the distance of midpoint to clock then the left photon has reached vt short of the clock' date=''s initial position., so then put the right photon also vt short of the right clock, plus the vt the frame has moved when the photons moved a distance ct.[/indent']

 

Ok I'll check it out. I was going to do the circular version first, I feel that is a bit more difficult to understand. I myself like challenges, when things are too easy i tend to pay them no mind.

 

I read your post about "squeeze," should i infer that is what you are doing with your attack on SRT? Doing that, in a sense, to the SRTists?

 

If that is what you meant with that remark, it was very funny.

 

Regards

[swansont]

...when train stations accelerate and move uniformly while the trains are at rest wrt the train stations one has described a physically impossible conditon to ever produce.

 

 

Or one has grossly misunderstood/misrepresented special relativity.

 

I wonder which one could have happened.

[el bastardo]

SRT=special relativity theory?... I assume... Ghostpebbles... haha... I know your not trying to say that someone is saying that really little balls "move" really big balls... All I'm trying to say is: ALL the balls are growing proportionally (to each other). There is no gravitational "pull"... anywhere. There is only a push caused by expansion... unless you want to think of a vacuum as being a (gravitational) pull... which is probably one reason why the universe is expanding in the first place... Gravity doesn't exist by the common definition.

[J.C.MacSwell]

I necessarily closed the distance from my standring position to one closer to the center of the earth' date=' and eveybody knows it did not happen [b']equivalently [/b] with me raising the earth, even you know it.

 

[/indent]

 

Equivalently? Who said equivalently? I said you raised the Earth.

 

You close the distance (sit down) 99.9999999999???????? % and the Earth will close (raise up) the remainder...small as it is.

 

If you want a sense of it being equivalent, then the total mass (mass of the Earth) times distance the Earth moves will be equal and opposite to the mass (your mass) times distance that you move. (all other influences notwithstanding).

 

This is basic conservation of momentum.