How does acceleration change length of traveler for motionless observers?

[DimaMazin]

Observers are at distance each from other on axis 'x'. Accelerating spaceship travels lengthways of axis 'x'. Is the spaceship length contracting to center of the spaceship relative to the observers or somehow differently?

[pavelcherepan]

I think that if the velocity vector is along the x-axis and observers are also on the x-axis then they wouldn't observe any contraction.

[DimaMazin]

Length contraction: Three blue rods are at rest in S, and three red rods in S'. At the instant when the left ends of A and D attain the same position on the axis of x, the lengths of the rods shall be compared. In S the simultaneous positions of the left side of A and the right side of C are more distant than those of D and F. While in S' the simultaneous positions of the left side of D and the right side of F are more distant than those of A and C.

 

Let's consider when acceleration has finished its work. All objects and gaps between them are simultaneously accelerated. Are they contracted to tail,head or center?

[Strange]

Length contraction: Three blue rods are at rest in S, and three red rods in S'. At the instant when the left ends of A and D attain the same position on the axis of x, the lengths of the rods shall be compared. In S the simultaneous positions of the left side of A and the right side of C are more distant than those of D and F. While in S' the simultaneous positions of the left side of D and the right side of F are more distant than those of A and C.

 

As far as I can see, this description is correct. With the following caveats:

 

1. The length contraction is due to (relative) velocity, not acceleration.

2. When you say "simultaneous", that can only be measured in a single frame of reference. But I think that is OK, in this case.

 

 

Let's consider when acceleration has finished its work. All objects and gaps between them are simultaneously accelerated. Are they contracted to tail,head or center?

 

I'm not really sure what you are asking here. You have described the situation accurately for a constant velocity. What happens under acceleration is slightly more complicated. For one thing, things are no longer simultaneous between the red blocks in S (or between the blue blocks in S'), and even from one end of a block to the other.

[DimaMazin]

 

You have described the situation accurately for a constant velocity. What happens under acceleration is slightly more complicated. For one thing, things are no longer simultaneous between the red blocks in S (or between the blue blocks in S'), and even from one end of a block to the other.

I want to know the simple. We should consider only in S. They all are equal before motion. And red objects with their gaps are contracted after acceleration . Are they contracted (by the speed) to tail, head or center?

[Strange]

Are they contracted (by the speed) to tail, head or center?

 

I think it just depends what you use as a point of reference. If you measure them from their ends, then the will contract relative to that. If you measure them relative to their centres, then they will contract relative to that.

[DimaMazin]

 

I think it just depends what you use as a point of reference. If you measure them from their ends, then the will contract relative to that. If you measure them relative to their centres, then they will contract relative to that.

I am afraid of that we have relativity of simultaneity in one frame.

[Strange]

I am afraid of that we have relativity of simultaneity in one frame.

 

I'm not sure I understand your point.

[DimaMazin]

 

I'm not sure I understand your point.

Can ends simultaneously be in ends and near to the center of the contraction for all co-located observers?

Edited July 11, 2015 by DimaMazin

[Strange]

Can ends simultaneously be in ends and near to the center of the contraction for all co-located observers?

 

Uh, sorry. Still don't understand.

[DimaMazin]

Let's consider the math : first object starts acceleration on x₁, second object simultaneously starts the same acceleration on x₂, third object simultaneously starts the same acceleration on x₃. First object ends the acceleration on point x₁+dx, second object ends the acceleration on point x₂+dx, third object ends the acceleration on point x₃+dx. They all have the same speed v. The distance between centers of first and second objects was=/x₂-x₁/

At v the distance between centers of first and second objects =/x₂+dx-x₁-dx/

The distances are equal before acceleration and after acceleration in the frame.Tail of train can be the first object, center of train can be the second object, head of train can be the third object.How do observers see the contraction?

Edited July 11, 2015 by DimaMazin

[Strange]

How do observers see the contraction?

 

As you have drawn it, I think: all objects and the distances between their centres will be reduced by the same amount.

 

(Actually, I'm not certain that is completely accurate because you have introduced acceleration. So I am not sure that they do all end up with the same speed. But I should let someone more familiar with the subject comment...)

[DimaMazin]

 

As you have drawn it, I think: all objects and the distances between their centres will be reduced by the same amount.

 

(Actually, I'm not certain that is completely accurate because you have introduced acceleration. So I am not sure that they do all end up with the same speed. But I should let someone more familiar with the subject comment...)

Is the math wrong? Are you not sure that the same acceleration creates the same speed in one frame?

[swansont]

Is the math wrong? Are you not sure that the same acceleration creates the same speed in one frame?

 

You haven't shown any math, so arguing whether it's right or wrong is moot. In post 11 you wrote some variables, but they did not come from the Lorentz transformation, so that's worthless.

 

You speak of simultaneous positions, but simultaneity is relative. Simultaneous in which frame?

 

As far as having the same speed, that's also a matter of which frame you are measuring in. v = at for constant acceleration, but in what frame is that being determined? Because the elapsed time in the other frame depends on not just the time in it's frame, but also the speed and the position. If the elapsed time is not the same, then the speed won't be.

t' = gamma (t-vx/c^2)

[DimaMazin]

 

You haven't shown any math, so arguing whether it's right or wrong is moot. In post 11 you wrote some variables, but they did not come from the Lorentz transformation, so that's worthless.

 

You speak of simultaneous positions, but simultaneity is relative. Simultaneous in which frame?

 

As far as having the same speed, that's also a matter of which frame you are measuring in. v = at for constant acceleration, but in what frame is that being determined? Because the elapsed time in the other frame depends on not just the time in it's frame, but also the speed and the position. If the elapsed time is not the same, then the speed won't be.

t' = gamma (t-vx/c^2)

I consider all these events in one frame of motionless observers.Therefore all the events are simultaneous.Why the Lorentz transformation doesn't happen in one frame of observation for the moving objects?The elapsed time is the same in our frame therefore speed is the same.The observers don't use t' for observation of the events.

Edited July 12, 2015 by DimaMazin

[swansont]

I consider all these events in one frame of motionless observers.Therefore all the events are simultaneous.Why the Lorentz transformation doesn't happen in one frame of observation for the moving objects?The elapsed time is the same in our frame therefore speed is the same.The observers don't use t' for observation of the events.

 

Is that what the math shows (which you haven't shown) or is that just a claim?

[DimaMazin]

 

Is that what the math shows (which you haven't shown) or is that just a claim?

That is my ununderstanding of relativity in one frame.

[J.C.MacSwell]

Length contraction: Three blue rods are at rest in S, and three red rods in S'. At the instant when the left ends of A and D attain the same position on the axis of x, the lengths of the rods shall be compared. In S the simultaneous positions of the left side of A and the right side of C are more distant than those of D and F. While in S' the simultaneous positions of the left side of D and the right side of F are more distant than those of A and C.

 

Let's consider when acceleration has finished its work. All objects and gaps between them are simultaneously accelerated. Are they contracted to tail,head or center?

You cannot maintain simultaneity (agreement) along the axis/direction of acceleration, while accelerating all points equally and for the same duration, (which can only be achieved from the viewpoint of one cross section of the spaceship/rods/spaces).

 

If you mean to do this from the perspective of a constant inertial frame, you cannot do it without stretching or compressing the craft/rods along it's length.

Edited July 12, 2015 by J.C.MacSwell