# Relative time question?

Suppose mass A is at rest and soon to be accelerated ( pushed ).

Mass A can not be accelerated all at once because it will take some time for the energy field which is to accelerate mass A to pass through mass A.

As the energy field passes through mass A the edge of mass A closest to the source of the energy field will start to move first.

As the energy field continues to passes through mass A the middle of A will start to move and then finally the edge of mass A which is farthest from the energy field source will start to move.

If one edge of mass A is moving before the other would mass A be compressed or does mass A perceive that all of its mass starts moving at the same time?

Duane

It depends on the field. If the field applies the same acceleration everywhere ( ie it is homogenous) then all of the body begins to move at once. I'm thinking of gravitation here. If the field varies in strength across space, then the body will experience tidal compression and/or decompression. The earths field is not homogenous so a falling bodies experiences some compression.

If you just push something on one side, the impulse travels through the material at the speed of sound, which varies from one material to another.

But that's a different situation from the gravitational field.

Suppose mass A is at rest and soon to be accelerated ( pushed ).

Mass A can not be accelerated all at once because it will take some time for the energy field which is to accelerate mass A to pass through mass A.

As the energy field passes through mass A the edge of mass A closest to the source of the energy field will start to move first.

As the energy field continues to passes through mass A the middle of A will start to move and then finally the edge of mass A which is farthest from the energy field source will start to move.

If one edge of mass A is moving before the other would mass A be compressed or does mass A perceive that all of its mass starts moving at the same time?
The mass is compressed and the whole mass does not start to accelerate at the same time.

Consider a mass of length l with location A being the location of the applied force and B being the location that is a length l removed from A.

Code:
             +----------------------+
Force ---> A |         MASS         | B
+----------------------+
<---------- l --------->
Then there are two factors to keep in mind:

1. The propagation speed of the force from A to B

Forces in materials do not propagate at the speed of light but at far less speeds, usually at speeds near the speed of sound.

We could measure this by placing two, Einstein synchronized clocks, one at location A and another one at location B before the acceleration starts. Then during the acceleration we would be able to measure that B accelerates later than A.

2. Acceleration in flat space-time

Place an, Einstein synchronized, clock and an accelerometer at both location A and B and have it record resp. the start times and the amount of acceleration during the acceleration phase. Then we will measure that, as mentioned above, the acceleration at B happens not only later than at A, but is also less than at A.

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Lets define it better.

The energy wave is light that was started instantaneously.

Mass A is partially transparent

Mass A has minimum mass to be acceleration easily and to prevent the slowing of the light passage as much as possible.

Just to make it easier I will define Mass A is a cloud of particles.

The light wave front is partially absorbed as it passes through Mass A.

Particles in Mass A are accelerated as the light wave front passes through them and the momentum of the light is absorbed by the particles.

Point (1) is where the light wave front enters Mass A.

Point (2) is where the light wave front leaves Mass A.

From the point of view of an observer at the light entry point (1) the entry point of A is moving before the leaving point (2) of A.

From the point of view of an observer at the light leaving point (2) of A the entire mass A starts moving at the same time because the information that the entry point (1) is moving willl arrive at the same time as the light wave front.

Does this mean that from the perspective of point (1) of A, A is compressed
But from the perspective of point (2) of A, A is not compressed?

Is this a temproary condition or not?

Duane

Sorry duordi but your second post does not make much sense to me.
Let me know if I can help you any further with the explanation I gave regarding your first posting.

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Duordi, your second post has a number of misunderstandings about the way light interacts with matter. Ordinary light does not 'push' a cloud of particles ( smoke ?) in the way you suggest. We would have noticed by now.

I apologize if I am incorrect but it is my understanding that light carries momentum.
If a photon is absorbed by a particle the particle has a momentum change, which requires a velocity change in the particle.

Only a few of the photons are absorbed as the light passes through the cloud of particles to cause the effect.

The general intent was not to discuss particle physics but to discuss the differing observation of observers at point (1) and point (2) in Mass A.

I guess I do not understand your problem with this.

Duane

russ_watters
Mentor
From the point of view of an observer at the light leaving point (2) of A the entire mass A starts moving at the same time because the information that the entry point (1) is moving willl arrive at the same time as the light wave front.
Knowing the distance between point 1 and 2 and knowing the speed of light, someone at point 2 will not conclude that the entire mass starts moving at the same time, but will be able to calculate when the wave front hit point 1 relative to point 2.
Does this mean that from the perspective of point (1) of A, A is compressed
But from the perspective of point (2) of A, A is not compressed?
The cloud/object is compressed and both will observe it to be so.

If the people at (1) and (2) are surfers and the wave is a wave on the ocean, the principle is the same - it is just a relativity of simultenaety issue.

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I thought that all co-ordinate systems are equally valid?

Knowing the distance between point 1 and 2 and knowing the speed of light, someone at point 2 will not conclude that the entire mass starts moving at the same time, but will be able to calculate when the wave front hit point 1 relative to point 2. The cloud/object is compressed and both will observe it to be so.

If the people at (1) and (2) are surfers and the wave is a wave on the ocean, the principle is the same - it is just a relativity of simultenaety issue.

What you are saying is that both (1) and (2) will "calculate" the same condition but observe a different condition and that reality is at point (3) which is your co-ordinate system with zero velocity with respect to mass A with an un-accelerated condition as A was before the encounter.

I guess I am confused on this.

I thought that all co-ordinate systems (1) , (2) and (3) are equally valid?

Duane

russ_watters
Mentor
What you are saying is that both (1) and (2) will "calculate" the same condition but observe a different condition and that reality is at point (3) which is your co-ordinate system with zero velocity with respect to mass A with an un-accelerated condition as A was before the encounter.

I guess I am confused on this.

I thought that all co-ordinate systems (1) , (2) and (3) are equally valid?

Duane
They are all equally valid and observers in each can even calculate what others will see in other reference frames.

So is mass A compressed as viewed by (1) or not compressed as viewed by (2) or is one of the "relative time events" where there is a disagreement when an event occurs?

I think the third option is the correct one, am I right?

The part I really need an answer to is will mass A stay compressed or uncompressed as the case may be to each observer?

If there was a spring between (1) and (2) would it transfer a force or not assuming it was at zero force before the velocity change of mass A.

(1) would expect a force of compression and (2) would not.

These can not both be correct can they?

Duane

pervect
Staff Emeritus
If events A and B are related by a light beam passing from A to B, everyone will agree that A happens first and that B happens after A.

Your example appears to me to be of this form - a light beam causes both A and B to accelerate, so the beam can be considered to pass from A to B. This implies that everyone will agree on the order of events A and B.

The technical name for this sort of separation is "null separation". It occurs when B is "on the light cone" of A.

If A and B were far enough apart that a signal from A could not reach B, they would be space-like separated. In this case B is "outside the lightcone of A". In this case, there is no definite ordering for A and B.

If A and B are close enough together that a signal from A can reach B and travel further, the separation is "timelike". In this case B is "inside the lightcone of A". In this case as well, all observers agree on the order of A and B.

If I consider A as the entry point of the photons and B the leaving point instead of (1) and (2).

You are right I put B ( the second event ) on the light cone of A ( the first event).
Because B is on the light cone of A it is possible for B to consider the events at A and B to happen at the same time but certainly not in reverse order.

My question has to do with an apparent compression of the mass from the perspective of point A which considers that A started to move before B compressing the object.
While B considers the two events to happen at the same time causing no compression.

If there is really a compression of the mass from point A’s view and no compression from B’s view would they experience equal and opposite gravitational forces between A and B?

After the acceleration A and B are not moving or accelerating with respect to one another.

There is no time rate of change variation between them after the event is there?

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