# Does change in Lorenz contraction w/ respect to time = Lorenz Velocity

1. Apr 25, 2014

### duordi134

Length contraction velocity question.

Suppose there are two sets of binary neutron stars in mirrored synchronous orbits one light year apart with zero differential velocity between the orbits center of mass.

Both sets of binary stars are orbiting very fast causing velocities with respect to one another close to the speed of light.

An observer on one of the neutron stars records that one of the neutron stars from each orbit set are traveling directly towards each other and one of the neutron stars from each orbit set are traveling directly away from each other.

For an observer on one of the neutron stars a large Lorenz contraction of the distance between two of the the neutron stars make the neutron stars temporarily appear to be 1/2 light year apart.

A short time later the the same neutron stars are traveling parallel due to their orbit progression and the Lorenz contraction vanishes causing the neutron stars now to appear to be one light year apart.

This pattern continues with each orbit.
Because the differential velocity is gradually changing with the orbits the Lorenz contraction must also gradually change.

There is an apparent "Lorenz Velocity" and "Lorenz Acceleration" as the neutron star moves from
1/2 light year distance to 1 light year distance and back.

I understand the interaction of the neutron stars gravitational acceleration toward one another must depend on the Lorenz contracted distance and so the Lorenz contracted distance should be considered real from a physics standpoint.

Should I consider the apparent Lorenz induced velocity and the Lorenz induce acceleration as real
or fictitious?

The Lorenz contraction depends on differential velocity.

Should the change in distance (Lorenz Velocity) caused by change in the Lorenz contraction with respect to time be included in the velocity term of the Lorenz contraction formula?

In the example above the "Lorenz velocity" exceeds the speed of light so please explain how it is to be included in the Lorenz contraction equation if it is to be included.

If I put a faster than light speed into the Lorenz contraction equation I will get imaginary numbers.

Duordi

2. Apr 26, 2014

### Staff: Mentor

You understand incorrectly. Gravity in relativity (meaning general relativity) does not work like it does in Newtonian mechanics, and it certainly doesn't work like Newtonian gravity with relativistic corrections to distances. The interaction between the neutron stars in each individual pair, in the simplest case where the mutual orbits are circular in the center of mass frame, is unchanging. The interaction between the two binary pairs will cause the centers of mass of the two binaries to slowly approach each other (assuming no other objects are present), but assuming that the size of each binary pair is much smaller than one light-year (which it should be if the orbital velocities of the stars in each binary are relativistic in the center of mass frame), the interaction between the pairs will be basically unaffected by the orbital motions of the individual stars in each binary.

3. Apr 26, 2014

### duordi134

PeterDonis response

If what you just "said" is true than the Lorenz contraction is truly an illusion which
has no real affect in the physical world.

If I have two stars traveling toward each other at one light years distance and two
stars traveling parallel to each other at one light years distance and there is no difference
in gravitational acceleration toward one another Lorenz contraction has no real significance.

For distant objects Lorenz contraction has no affect on gravity and if you are on one star or the other you will be moving with the star and see no Lorenz contraction.

Does Lorenz contraction have any consequences in physical reality?

Duordi

4. Apr 27, 2014

### WannabeNewton

This is in some sense or another a bit of a thorny issue. Consider for example the Ehrenfest paradox: http://en.wikipedia.org/wiki/Ehrenfest's_paradox. We know that in a frame-invariant sense the disk experiences stresses due to the torque applied to the disk upon setting it into rotation-the stresses experienced by the disk are measurable (physical observables) and as such the associated quantities characterizing the stresses will be present in all frames. However each frame will (potentially) have a different explanation as to why the stresses come about in the disk upon application of the torque. In a suitable frame (e.g. the global inertial frame in which the disk was initially at rest) and under a suitable tangential acceleration program, one can attribute the inception of the stresses in the disk to the obstruction* of tangential Lorentz contraction from taking place upon being set into rotation-this is clearly a frame-dependent explanation (for example, the explanation for the stresses in the instantaneous rest frame of an infinitesimal segment of the rim of the disk might depend on the relativity of simultaneity instead depending on the setup) but in this sense one can use Lorentz contraction (or lack thereof in this case) as the reason for why the disk experiences stresses upon being set into rotation from rest.

*The tangential Lorentz contraction of infinitesimal rulers placed tangential to the rim of the disk will only arise if the disk is undergoing Born rigid motion-it can be shown (and in fact the Ehrenfest paradox indirectly shows) that one cannot Born rigidly set a disk into rotation.

Last edited: Apr 27, 2014
5. Apr 27, 2014

### Staff: Mentor

Yes. Two examples that have been extensively discussed here are the cosmic ray muon lifetime measurements (analyze using coordinates in which the muon is at rest while a length-contracted earth approaches it) and the way in which magnetic fields arise around a current-carrying wire (see https://www.physicsforums.com/showthread.php?t=229034,there's a short section on length contraction there)

Last edited: Apr 27, 2014
6. Apr 27, 2014

### Staff: Mentor

The Lorentz-contracted distance between two points is the distance between those two points - there's nothing else that we could use as the "distance". It's as real as any other frame-dependent measurable thing.

You should treat them as you would any other coordinate velocity or coordinate acceleration. They appear in your calculations as a result of your choice of coordinates (in this case, momentarily comoving inertial frames of one of the stars) so you either include them in your calculations or you transform to coordinates in which they don't appear in your calculations, whichever is easier. This is similar to classical centrifugal force and acceleration, which are also fictitious. But please do remember that in this context "fictitious" does not mean "not real", it means "can be made to disappear by choosing different coordinates".

No. The $v$ that appears in the Lorentz transforms is the velocity of the origin of an inertial coordinate system, not the velocity of any object being described by those coordinates. In this problem, none of the stars are at the origin of such a coordinate system.

7. Apr 27, 2014

### duordi134

For Nugatory

Nugatory.

There is a lot here! that I need to understand by reading on my own.
I will be back when I think I have a handle on it.

I think my key misunderstanding it this statement:

"No. The v that appears in the Lorentz transforms is the velocity of the origin of an inertial coordinate system, not the velocity of any object being described by those coordinates. In this problem, none of the stars are at the origin of such a coordinate system."

If I have a star moving toward the earth and a star at rest with respect to the earth at one light years distance are they in an inertial coordinate systems which will cause different Lorenz contractions or am I just
restating the same thing I did before.

Thanks much

Duordi

8. Apr 27, 2014

### Staff: Mentor

The two stars are at rest in two different inertial coordinate systems.

In one system (call it system #1), star #1 is at rest and the earth is moving.

In another system (call it system #2), both star #2 and the earth are at rest.

9. Apr 27, 2014

### Staff: Mentor

[The comments below are assuming the global inertial coordinate systems that are possible in special relativity - general relativity brings in a whole bunch of other complications that we don't need to mess with here]

No, and you may be being confused by the way that we say "in" a coordinate system or "in" a frame of reference instead of the more precise "using coordinates". (If you're going to insist on using the word "in" - it's convenient, I do it myself when I'm in a hurry - then you could say that everything is always "in all coordinate systems" or "in all frames").

If I have a star that is at rest with respect to the earth, and another star moving towards the earth at speed $u$, I can choose any coordinate system and then use it to describe the motion of all three objects. For example:

Using coordinates in which the earth is at rest (that is, has velocity zero), star A is moving with velocity $u$, and star B is moving at velocity zero (that is, is also at rest).

Using coordinates in which star A is at rest, the earth and star B are both moving with velocity $-u$.

Now consider a spaceship flying from the earth towards star A at speed $w$ relative to the earth. Using coordinates in which that spaceship is at rest, the speed of the earth and star B is $-w$. (The speed of star A using these coordinates is not the $u-w$ that you would expect from classical mechanics but we don't need to worry about that now - just note that it will be very close to $u-w$ if $w$ is small compared with the speed of light so there's no huge contradiction of everyday experience and common sense here). Note that I don't actually need the spaceship to use this coordinate system; what I'm really doing is choosing the origin of the system such that the earth is moving with speed $-w$ relative to the origin and the spaceship is just a convenience to help me visualize the path of the origin.

All three of these coordinate systems (and as many other as I please... I can pick any value of $w$ less than $c$) can be used to calculate and describe the trajectory of all three bodies. There is nothing magic or special about the coordinate systems in which some of the objects are at rest, although some calculations may be easier using one coordinate system or another (you wouldn't want to land an airplane on the surface of the earth using instruments that report positions using coordinates in which the earth is not at rest!).

The Lorentz transformations relate the values of the coordinates in one coordinate system to the values of the coordinates in another coordinate system. The only thing that matters is the relative velocity of the origins of the two coordinate systems - whether there are objects like stars and planets and spaceships and airplanes moving around is irrelevant to the transforms.

10. Apr 27, 2014

### duordi134

Let me give it a try.

Ok, so I said one star was moving and the other was stationary with respect to earth..

So I chose a frame of reference with the earth and one star at rest and the other star moving toward Earth.

The Earth should observe the star approaching it as being closer than the star which is defined as being stationary with respect to earth.

Can the stationary star think that the moving other star is right next to it when we think the stars are much different distances from us. Did I just change the frame of reference going from earth to the stationary star or are they both the same frame of reference? I think jtbell said these two are in the same frame of reference.

Acceleration is not allowed so the distant stars can not be orbiting one another because that would cause them to be accelerating toward one another.

Duordi

Last edited: Apr 27, 2014
11. Apr 27, 2014

### Staff: Mentor

The distance between the earth and the star that is not at rest relative to the earth (I carefully do not say "the moving star" as I might be using coordinates in which that star is at rest and the earth is moving) is of course constantly changing, so whenever we speak of that distance we have to specify a time.

But with that said... If star A is moving relative to the earth with speed $u$ and star B is at rest relative to the earth at a distance of one light-year when using coordinates in which the earth is at rest, all in a straight line, then there is a moment when A passes very close to B (if they were perfect point objects I'd say that they were in the same place at that moment, but colliding our two stars messes up the story). At that moment....

Anyone anywhere in the universe using any coordinates whatsoever will get the same results for the distance from A to B at that moment: it is zero, because they're both in the same place at that moment. The distance has to come out zero in all frames and using all coordinate systems; otherwise we'd have the absurd result that whether the stars collide or not depends on what coordinates we use to describe their motion.

By the same logic, the distance between earth and star A must be the same as the distance between the earth and star B at that moment when A and B are colocated.

But what is that distance? If we use coordinates in which the earth and star B are at rest, that distance is one light year because that's how we set up the initial conditions. These are, of course, natural coordinates for an observer on earth or star B to use. However, if we use coordinates in which star A is at rest while star B and the earth are moving towards it, we'll get a different length-contracted distance. But there's never a contradiction or inconsistency - no matter which coordinates you use, at the moment that A and B are colocated, the distance between A and earth is equal to the distance between B and earth, and the distance between A and B is zero.

He did and they are.

Earth and star B are both in the same frame of reference, which is to say that there exists a coordinate system in which they are both at rest, and they agree to use that coordinate system to describe the universe.

12. Apr 27, 2014

### duordi134

When and Where.

Given that we choose a frame of reference stationary with the Earth and stationary with star A and hold star B to be traveling toward Earth at close to light speeds.

The distance between the Earth and star B which is moving with respect to the Earth is 1/2 light year.
The distance between star A and star B is zero.
The distance between Earth and star A which is not moving with respect to Earth is one light year.

Now suppose both stars send a light show to Earth because of the close encounter between star A and star B.

Will we receive the signal from both stars at the same time one light year later?

I am expecting somewhere here to have a disagreement as to when all these events happen.
Am I right?

Duordi.

13. Apr 27, 2014

### Staff: Mentor

Ok, but then this...

...is false, because you said we were using star A's and Earth's rest frame, not star B's rest frame. You need to be careful to keep your frames straight.

Ok. (Note that this must be true in any frame.)

Ok here, since you already stipulated this was the case in the star A/Earth rest frame.

Yes (assuming that "we" are on Earth).

You mean, one year later in the frame we are using (the star A/Earth rest frame), in which the light has covered a distance of one light-year? Yes.

Now, if you were to do this calculation in star B's rest frame instead, you would find that the light was emitted at the same time from both stars (since star A and star B passing each other is an invariant, it's a single event in any frame) and received on earth at the same time from both stars, but the reception event would only be half a year after the emission event, and the light would only cover half a light-year.

What you can't do is mix two different frames in a single calculation; you can't calculate the time it takes star A's light to travel to Earth in one frame, and the time it takes star B's light to travel to Earth in a different frame, and expect to get consistent results. Similarly for the distances traveled. See below.

Why? Light is traveling from one event (star A and star B passing each other) to another event (Earth receiving the light). The path of the light through spacetime is independent of which reference frame we choose. As soon as you specify that the light is emitted when star A and star B pass each other, i.e., from a single event in spacetime, you have guaranteed that the light from both stars will be received at the same event, i.e., at the same time on Earth.

The mistake you appear to have made is to try to calculate the time/distance traveled for star A's light in one frame, and the time/distance traveled for star B's light in a different frame. As I noted above, this will not give consistent results. You need to pick a single frame in which to calculate.

14. Apr 27, 2014

### Staff: Mentor

[boldface above is added by me]

The key point here is that the distance between two objects at any given moment is not determined by the speed of the objects, it is determined by the speed of the observer who is measuring the distance (more precisely, by the frame whose coordinates we use to describe the distance).

Using coordinates in which none of earth, star A and star B are at rest, which is to say the frame of some other observer who is moving relative to all three, we would still find that when the distance between A and B is zero, the distance from earth to A is the same as the distance from earth to B - although it would be neither 1/2 lightyear nor one lightyear. We would also find that A is at rest relative to the earth, which is to say that A and the earth are moving at the same speed.

Last edited: Apr 27, 2014
15. Apr 30, 2014

### duordi134

Nugatory, sorry for the delay.

I am having a hard time both understanding and accepting how things really work.

I think I understand how this works now.
I have to be moving to see Lorenz contraction and time delay so I must choose a frame of reference which is moving with me and the distant object must remain stationary.

It seems odd that I can change the distance to star B simply by choosing a reference frame.
So...

I get out my telescope and measure the distance to star B with triangulation and luminosity.
The earth is moving with respect to star B.
How far will star B be away? That is to say which frame of reference will the measurement
give me the result for 1/2 light year or 1 light year?

I drop a candy rapper out my zero mass spaceship and time how
long it takes for the candy rapper to fall to star B.
Does the candy rapper fall as if star B is 1/2 light year away or 1 light year away?

In other words does the laws of physics select a frame of reference for me?

Duordi

16. Apr 30, 2014

### Staff: Mentor

To be precise, you can change the "coordinate distance" to star B by choosing a reference frame. But no direct physical observable corresponds to this "coordinate distance"; it's a calculated number that depends on the frame. You can't change any direct physical observables by choosing a reference frame.

And the distance you measure will quite possibly *not* be the same as the "coordinate distance" you picked by choosing a reference frame. For it to be the same, you must choose your frame carefully. Frames are arbitrary choices, but direct physical measurements are not.

The key thing is not whether the earth is moving with respect to star B, but whether your telescope is. I'll assume your telescope is on earth in what follows.

It depends on how fast your telescope is moving relative to star B; but the effect will *not* be what you think. See below.

It's not a matter of frames of reference. It's a matter of how your telescope is moving, relative to star B, as it makes the measurement. If your telescope is moving relative to star B, the Doppler effect will change the triangulation (I assume you actually mean parallax here, i.e., you are looking at the star from different angles and seeing how its position relative to more distant stars changes) and luminosity results you get: moving toward star B will (to a first approximation) make it seem brighter, and moving away will make it seem dimmer, and both motions will make its apparent parallax decrease.

The combined effect of these things will *not* give a distance result that matches either 1/2 light year or 1 light year; that's because those numbers are coordinate numbers you have to calculate, not numbers you directly observe. (It's straightforward to correct the telescopic observations for the Doppler effect, since you can measure the spectrum of star B and determine the Doppler shift directly from that. If you correct your direct measurements that way, you will get a distance of 1 light year.)

In general, neither, because relativistic gravity is not the simple Newtonian force you are used to. But if your spaceship is at rest relative to star B when the object is dropped, its time to fall will be what you would expect for a distance of 1 light year.

If your spaceship is moving relative to star B when the object is dropped, a simple Newtonian calculation for an object starting at rest but using the length-contracted distance will *not* give the right answer for time to fall. The right answer will be closer to what you would expect from a Newtonian calculation using a distance of 1 light year but an initial velocity that is nonzero.

17. May 1, 2014

### duordi134

So Lorenz contraction does not really exist?

So the physical world always uses a frame of reference which avoids Lorenz contraction,

and we hove no equations of motion that are better than Newton's in their ability
to predict an objects location.

Is it possible that Lorenz contraction does not happen?
There are many cases in which equations will give solutions which are not found
in reality.

Perhaps reality does not allow the frame of reference required for Lorenz contraction.

Duordi.

18. May 1, 2014

### Staff: Mentor

No. Consider that:
1) Magnets work as a result of Lorentz contraction. We have equations that describe how magnetic forces arise from electric fields that are observed from a frame moving relative to the source of the electric field, and the Lorentz contraction is an essential part of their derivation. Try section 7 of the FAQ from https://www.physicsforums.com/showthread.php?t=229034.
2) Particle lifetime measurements (in section 4 of the same FAQ) are usually described as tests of time dilation, but if you transform to a frame in which the particle is at rest, they're tests of length contraction. A muon created by a cosmic ray striking the top of the earth's atmosphere will reach the surface of the earth. An earthbound observer will explain that the muon lives long enough to make the 200km journey from top of the atmosphere to surface of the earth because of time dilation; but an observer comoving with the muon and watching the surface of the earth approach at close to $c$ will (just as correctly) explain that the distance from top of atmosphere to surface of earth is Lorentz-contracted enough that the surface of the approaching earth can reach the muon before it's had enough time to decay.

19. May 1, 2014

### Staff: Mentor

Huh? How did you get that from what anyone has said?

20. May 2, 2014

### duordi134

Preffered frame of reference could be a lowest energy state

No direct indication of Lorenz contraction can be tested because we are not able to measure accurately enough in the conditions required.

The Lorenz contraction can be assumed because time delay can be measured and simply by changing the preferred reference frame Lorenz contraction is required.

Unfortunately this does not address my question.

I agree that the mathematics of any chosen preferred reference frames will give consistent results.

I am suggesting that for whatever reason reality does not use certain frames of reference. Nature only needs one correct equation to get the right answer and we cannot tell which of the infinite frames of reference may have been used because they will all give consistent mathematical results.

Suppose physical reality chooses the frame of reference which holds the universal energy to be as low as possible. Choosing a Reference frame which causes Earthâ€™s atmosphere to Lorenz contract requires the Earth to be traveling close to the speed of light. Because most of the mass of the universe must also travel similar speeds in this frame of reference the universal energy must have an enormous value. If the muon is traveling at close to light speed and the remainder of the universe is close to stationary the universal energy content is close to the lowest possible value.

I am not saying all reference frames are not valid or even that there is a preferred reference frame from a mathematics point of view but that there may be a reference frame reality uses for one reason or another.

Can we really say that because the math works the physical universe must use every reference frame?

Duordi

21. May 2, 2014

### Staff: Mentor

That's like asking whether a Euclidean two-dimensional plane "must use" both polar and Cartesian coordinates.

The physical universe doesn't use any reference frame. Reference frames are conventions that we apply to the physical universe when we want to associate numerical values (also coordinates) to events to facilitate our computations.

22. May 2, 2014

### duordi134

I 'll take that as a no.

Duordi

23. May 2, 2014

### Staff: Mentor

OK, as long as you understand that no one has ever suggested that the physical universe does "use every reference frame".

24. May 5, 2014

### duordi134

I should explain myself.

I should explain myself.

The suns gravity causes an acceleration on the Earths oceans.
The direction of this force is toward where the Sun actually is.
This is as if physical reality has decided that gravity considers the field generator
(the sun) to be held stationary while the interaction mass (the Earth) as considered
to be in motion.

Light from the sun is different.
Light we receive from the sun is not pointed where the sun actually is but is pointed where the sun was 8 minutes ago. This is the expected result if a preferred reference frame was used which held the Earth stationary and attributed all motion to the Sun.

Based on this it seems that the physical universe is selective when choosing a preferred frame of reference.

25. May 5, 2014

### Staff: Mentor

Reality doesn't use frames of reference at all. Frames of reference are abstractions that we construct in order to model reality mathematically.

No, Nature just gets the answer. Nature doesn't use equations; it just is. Equations are abstractions that we construct in order to model reality mathematically.

No, we can say that, because the math works, we can *model* reality using any reference frame we like. But that doesn't mean our model *is* reality. It's still just a model.