Altering perpendicular unit vectors

By Michael Doré (P904) on Sunday, December 26, 1999 - 03:50 pm :

For this question I would like to make the following assumptions:

1) Space and time are continuous.
2) Newton's laws of motion hold.
3) Matter is made of a very large but finite number of point-like particles.

Let the position vector of an arbitrary particle be xi + yj + zk from a fixed origin where i j k are perpendicular unit vectors. Imagine that spontaneously the position of all such particles is changed to (x + y)i + yj + zk . In other words shift all particles a distance d horizontally where d is the vertical distance of the particle from the origin.

What difference, if any, would people in this universe observe? Would it be possible for such people to perform an experiment to determine that the shift had occurred?

Thank you,

Michael

By Dan Goodman (Dfmg2) on Sunday, December 26, 1999 - 05:13 pm :

This might not be what you want to hear, but I think that you have an implicit assumption, that a fixed origin can exist. I think that the best you can do is to look at things relative to some frame of reference which will have to be defined in terms of something to do with the positions of particles if we are actually to observe it. If everything was translated, then so would our frame of reference, and so we wouldn't notice anything, as all the relative positions are exactly the same. Someone with some knowledge of general relativity might be able to correct me on these points.

By Michael Doré (P904) on Sunday, December 26, 1999 - 06:30 pm :

Many thanks for your prompt reply.

I agree that it is meaningless to talk of a fixed origin if all particles are translated by the same amount. However here, I am translating particles by different amounts - it depends on their vertical displacement. In effect I am swapping the vector j for i + j . I am not certain but I think that under the assumptions I made there it still wouldn't be possible to detect a change.

Thanks again,

Michael

By Michael Doré (P904) on Sunday, December 26, 1999 - 09:10 pm :

Sorry, I forgot to mention that the velocities of all the particles are changed in exactly the same way, at the same time that the positions are altered. (So that a particle moving with velocity ui + vj + wk changes to velocity (u + v)i + vj + wk .) If the velocities were unaltered it would clearly be possible to detect the change in the position. If a particle was at (0,0,0) and was heading for a stationary particle at (0,1,0) then when the swap occurred the second particle would now be at (1,1,0) and the first particle would be heading for (0,1,0) still. When the particles didn't collide you would be able to detect that something strange had happened.

If you do alter the velocities appropriately I suspect that it would be impossible to detect any difference but I'm not really sure at all. Is it possible to generalise the set of transformations for which no change can be detected? Clearly transformations such as xi + yj + zk going to:

(x + A)i + (y + B)j + (z + C)k

don't make a difference. I believe the same is true for it going to

A(xi + yj + zk )

or even

Axi + Byj + Czk . Also swapping around the i s j s and k s will not be detectable but I'm not too sure when you actually mix the perpendicular vectors together.

Thanks,

Michael

By Dan Goodman (Dfmg2) on Monday, December 27, 1999 - 03:44 am :

Doh! I didn't notice it wasn't a straight translation, sorry about that. The first thing to say is that there is no absolute frame of reference, so you would have to make this transformation from some particular frame of reference, but you have already said this. I'm pretty sure you can't do these transformations under general relativity, as the maximum speed is

c

, the speed of light, but your question was about Newtonian mechanics, let's see if we can get anywhere with that.

First off, you can make a fixed rotation, fixed translation, or translate over time without it being observable. This is because they are inertial frames of reference.

Any transformation you apply to your position vectors will also need to be applied to your velocity vectors, otherwise there will obviously be a way to detect the change.

Onwards, your transformation will be detectable I think, here's why. Imagine you have a very heavy particle at 0, and another particle in orbit about it at a distance $r$ in the plane $z = 0$ , wait until the particle is at a point $(0, r, 0)$ travelling with velocity $(r ω, 0, 0)$ where $ω$ is the angular velocity. Afterwards, the position will be $(r, r, 0)$ and the velocity will be $(r ω, 0, 0)$ . However, a particle in orbit always has a velocity vector perpendicular to its position vector, which is not true after the transformation, so it will no longer be in a circular orbit, this will be detectable.

I think that $(x, y, z) \to (A x, A y, A z)$ will be detectable, although I can't think of a way to do so at the moment. I think that the transformations above are the only ones that will be work.

By Michael Doré (P904) on Monday, December 27, 1999 - 07:34 pm :

Well I am not completely sure whether the particle moving on the ellipse would look any different from the circle considering that the equipment used to measure it will have also become lopsided.

I think that things may well look instantaneously the same but you could actually perform experiments to tell that the transformation had occurred. For example you could have a massive particle at (0,0,0) and two light identical particles at (1,0,0) and (0,1,0). All 3 particles are intially stationary. If there were no transformation then the light particles would collide with the massive particle at the same time as one another due to gravitation, however if there is a transformation then the one at (1,0,0) would collide first as it is nearer than the one at (1,1,0).

The same goes for (x,y,z) to (Ax,By,Cz) ? I thought this would not make a difference at first because in special relativity if a rocket is travelling close to the speed of light, its contents are squashed in one dimension only relative to Earth (the dimension the motion occurs in). Yet anybody inside won't notice any difference. This transformation seemed to be like (x,y,z) to (Ax,y,z).

However there is an actual squashing of space in special relativity and this is rather different from simply decreasing the separation of the particles. So perhaps I should change the question so that actual space is changed by the vector transformations rather than just the positions of the particles. This means that when you calculate the forces between particles using the inverse square law you would have to take account of the fact that the space was stretched/squashed. I think that now the transformation (x,y,z) to (Ax,By,Cz) is undetectable.

Thanks,

Michael

By Dan Goodman (Dfmg2) on Tuesday, December 28, 1999 - 01:09 am :

If the particle was moving on an ellipse, then the focus would be where the particle is, but the centre wouldn't be, this change would be detectable.

As to the squashing of space, I'm not really competent enough to be able to answer it, but I suspect that any such transformation would reduce to "If you suitably changed all physical laws so that the change was undetectable, then the change would be undetectable", which isn't that impressive.

Nice idea though, it would be nice to think that the universe is constantly changing in bizarre ways which we didn't even know about.

By Michael Doré (P904) on Tuesday, December 28, 1999 - 12:31 pm :

Yes, but as you say, if no-one can detect/observe the change it is fairly meaningless.

For the (x,y,z) to (Ax,Ay,Az) case I think that maybe if you reduced all the velocities of the particles by a factor of sqr(A) and didn't alter any laws of physics then the change would be undetectable. For example with A = 2, everything I can think of would happen 2sqr(2) times more slowly - for example the period of a pendulum, the period of a planet on a circular orbit, the time a ball takes to hit the ground. This doesn't sound detectable. For an elliptical orbit, the rate at which the planet swept out area would go up by sqr(2) - as this is equal to its instantaneous tangential speed multiplied by its instantaneous distance. Now as the overall area for a period goes up by 4, the orbital time goes up by 4/sqr(2) which is 2sqr(2) as we were hoping.

So is it possible to prove that this transformation is undetectable? Well the force on every particle would go down by 4 (as all macroscopic forces are caused by inverse square forces so if you scale the system up the forces quarter).

So imagine that these two universes were running alongside each other. (I only do this for convenience.) Let f(t) be the position vector of a particle in the first universe. In the second universe, if all events are indeed just scaled up by 2 and slowed down by 2sqr(2), the position vector of the same particle is given by 2f(t/2sqr(2)). Differentiating once gives f'(t) and 1/sqr(2) f'(t/2sqr(2)) so the velocity at each event is reduced by sqr(2) as we were hoping. Differentiating again gives f''(t) and 1/4 f''(t/2sqr(2)) meaning that at each event the force is quartered. Therefore non-detectability is consistent with the laws of physics.

Applying A = -1 (reflection) suggests that every velocity should be multiplied by -i and events should happen 1/sqr(2)i-1/sqr(2) times quicker! Which is interesting although it may just mean rotate the velocity by 90 degrees.

So if we allow allow ourselves to change the velocities but not the laws of physics - so far the following transformations are undetectable:

reflection
rotation
translation
enlargement (by uniform amount)

Is this as far as we can go?

Many thanks,

Michael

By Dan Goodman (Dfmg2) on Tuesday, December 28, 1999 - 01:37 pm :

Under the enlargement transformation x -> Ax things will indeed take longer, won't this mean that the change will be detectable? Atomic clocks work on the rate of radioactive decay (or maybe they don't, but it is something subatomic), I don't think that this would be affected by increasing the separation of the particles, so I think that we could observe a slowdown in time (balls falling to the ground for instance), and hence this transformation. Another possibility is transformations applied to 4D Minkowski space-time, i.e. where all "events" are given coordinates (x,y,z,t) where t is the time of the event. If t was suitably scaled up along with x,y,z might you be able to make it undetectable? For instance (x,y,z,t) to (Ax,Ay,Az,At)?

By Michael Doré (P904) on Tuesday, December 28, 1999 - 04:20 pm :

Well, I haven't really got a clue as to whether radioactive decay would be affected by increasing the separation, because it is governed by the strong and weak nuclear forces rather than electrostatics. I think that scaling things up in the quantum universe would cause you all sorts of problems because there is a minimum length called the Planck length. If you scaled up atoms then the ratio of the atomic radius to the Planck length would change, and this would result in instability (for reasons I don't understand). Which is why I am only considering this transformation in the Newtonian universe where space and time are continuous and all fundamental forces are inverse square.

As for scaling up Minkowski 4-D space-time... If you have two identical planets in a circular orbit around their centre of mass, the period is proportional to r^3/2 . So multiplying the lengths and periods by 2 doesn't appear to work. So I think you would have to scale it up like (x,y,z,t) to (Ax,Ay,Az,Asqr(A)t).

Thanks for your help,

Michael