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### Author Topic: The weights of moving objects  (Read 7126 times)

#### Atomic-S

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##### The weights of moving objects
« on: 09/03/2007 06:19:02 »
Let us suppose, upon a very large gravitating planet whose size is much larger than the apparatus to be discussed but whose g value at the surface is "reasonable", (we would be talking about a spherical shell light-years in diameter, well removed from other objects in the universe), the following experiment is conducted:

A large "flat" table exists, that is, of negligible curvature over the dimensions of the experiment. Set into this table, aligned in the x direction but offset and side by side in the y direction, are the platforms of two very fast acting scales (balances).

Two objects move frictionlessly upon the table, one from the -x direction, aimed across one of the scale platforms, and the other from the +x direction, aimed across the other platform. They are moving at equal but opposite speeds, rapidly as a significant fraction of the speed of light. They are so moving that they each will cross the platforms of their respective scales at the same time, as seen by an observer standing on the planet.

Relativistically, each will have a greater-than-rest mass, and the two masses will be equal because the speeds are equal in magnitude. Thus, when each object crosses its scale platform, each scale will record its weight, and the weights will be equal.

Now let us view this experiment from within a vehicle moving at the same speed as one of the objects.  In this reference frame, the one object is at rest, but the other is moving at a high speed. Thus, their relativisitic masses will not be equal. Both objects will still seem to cross their respective platforms at the same time. And the scales will still seem to record equal weights for each, even though the masses now are not the same.

Explanation?

#### another_someone

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##### The weights of moving objects
« Reply #1 on: 09/03/2007 06:55:00 »
You cannot use the balances to measure the mass of the objects in motion - have you ever tried to measure a moving mass on a balance?

A balance scale merely measures force, not mass; and there will be many consituants to force that have nothing to do with mass (more technically, weight, not even mass) when an object is in motion.  You have to remove the dynamic component of force by causing the object to stop moving before you can use a balance to measure its weight.

That aside, you also have to take into account that the observer is observing the measurements, which means the whole notion of simultaneousity does not necessarily work, as he may see two events as simultaneous in one frame of reference, but not simultaneous in another frame of reference.

#### Atomic-S

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##### The weights of moving objects
« Reply #2 on: 18/03/2007 04:08:11 »
You cannot use the balances to measure the mass of the objects in motion - have you ever tried to measure a moving mass on a balance?
Suire you can. The balances could be so constructed that each is preloaded by internal spring or whatever, with the anticipated weight. The internal preload would have been programmed to release at just the time when  the object hits the pan. The correctness of the weight will be verified by the complete non-reaction of the balance to the object crossing it. (Were the weight different than anticipated, some kind of kinetic impulse would be generated, which will not happen if the pre-load is calculated in advance correctly). Keep in mind that the object is moving along the surface of the pan face, so that its motion is of no dynamical significance as relates to what the pan records unless the object creates some kind of a vertical deflection, which will not happen in this situation

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A balance scale merely measures force, not mass; and there will be many consituants to force that have nothing to do with mass (more technically, weight, not even mass) when an object is in motion.
(see previous argument.)
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You have to remove the dynamic component of force by causing the object to stop moving before you can use a balance to measure its weight.
Done.

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That aside, you also have to take into account that the observer is observing the measurements, which means the whole notion of simultaneousity does not necessarily work, as he may see two events as simultaneous in one frame of reference, but not simultaneous in another frame of reference.
assuming that the events are differently located in some way in each frame. However, special relativity tells us that two events that, in one frame, occur at the same time and at the same place, will be simultaneous as well as colocated in all frames. That is the situation here, effectively, by reason of the close proximity of the balances and the fact that the objects cross them at the same time in the ground frame. Therefore they cross them at the same time in all frames.
[/quote]

#### another_someone

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##### The weights of moving objects
« Reply #3 on: 18/03/2007 13:00:19 »
You cannot use the balances to measure the mass of the objects in motion - have you ever tried to measure a moving mass on a balance?
Suire you can. The balances could be so constructed that each is preloaded by internal spring or whatever, with the anticipated weight. The internal preload would have been programmed to release at just the time when  the object hits the pan. The correctness of the weight will be verified by the complete non-reaction of the balance to the object crossing it. (Were the weight different than anticipated, some kind of kinetic impulse would be generated, which will not happen if the pre-load is calculated in advance correctly). Keep in mind that the object is moving along the surface of the pan face, so that its motion is of no dynamical significance as relates to what the pan records unless the object creates some kind of a vertical deflection, which will not happen in this situation

Lets ignore the relativistic issue right now, and just look at the simply Newtonian mechanics at (relatively) low speed.

A balance will measure the weight of an object, in other words, it measures the downward force of an object, in otherwords it will measure the propensity of an object to fall down.

If you take a small lump of lead in your hand, and drop it, it will fall down immediately on the spot.  If you put a balance under that lump of lead, you can measure the force that is causing the lead to fall down.

If you fire that lead horizontally out of the barrel of a gun (say, about twice the speed of sound - still very slow by relativistic standards), that same bullet travel over a mile before it falls down - the effective force causing it to fall down is overwhelmed by the inertia causing it to move horizontally, so the nett force of the bullet is far more horizontal than vertical, and placing a balance under the path of the bullet (ignore aerodynamics and frictional effects - just say you coated the balance with Teflon, and performed the experiment in a vacuum) will just have the bullet skit across the surface of the balance rather than pressing down on the balance.  In other words, the apparent weight of the bullet in flight will appear to be almost zero (and their is nothing realistic about this).
« Last Edit: 18/03/2007 13:01:55 by another_someone »

#### lightarrow

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##### The weights of moving objects
« Reply #4 on: 20/03/2007 17:30:26 »
If you fire that lead horizontally out of the barrel of a gun (say, about twice the speed of sound - still very slow by relativistic standards), that same bullet travel over a mile before it falls down - the effective force causing it to fall down is overwhelmed by the inertia causing it to move horizontally, so the nett force of the bullet is far more horizontal than vertical, and placing a balance under the path of the bullet (ignore aerodynamics and frictional effects - just say you coated the balance with Teflon, and performed the experiment in a vacuum) will just have the bullet skit across the surface of the balance rather than pressing down on the balance.  In other words, the apparent weight of the bullet in flight will appear to be almost zero (and their is nothing realistic about this).

No, that's not correct. The weight force doesn't change at all. If you release a stationary piece of lead, it'll travel 9.81 metres down in one second; if you fire it horizontally at 100,000 m/s, in one second it will travel 100,000 metres horizontally and 9.81 metres down. Nothing changes.

Edit 1.
Sorry, I was not clear. I intended that "weight force doesn't change at all" for the reasons another_someone seemed to say.
Infact, weight force DOES change, it increases, because of the object's speed.

Edit 2.
In one second an object falling down doesn't travel 9.81 metres but 9.81/2 = 4.9 metres.

« Last Edit: 22/03/2007 19:15:57 by lightarrow »

#### lightarrow

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##### The weights of moving objects
« Reply #5 on: 20/03/2007 17:36:20 »

but I'm not sure to have understood all the details.

#### another_someone

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##### The weights of moving objects
« Reply #6 on: 20/03/2007 18:58:29 »
If you fire that lead horizontally out of the barrel of a gun (say, about twice the speed of sound - still very slow by relativistic standards), that same bullet travel over a mile before it falls down - the effective force causing it to fall down is overwhelmed by the inertia causing it to move horizontally, so the nett force of the bullet is far more horizontal than vertical, and placing a balance under the path of the bullet (ignore aerodynamics and frictional effects - just say you coated the balance with Teflon, and performed the experiment in a vacuum) will just have the bullet skit across the surface of the balance rather than pressing down on the balance.  In other words, the apparent weight of the bullet in flight will appear to be almost zero (and their is nothing realistic about this).

No, that's not correct. The weight force doesn't change at all. If you release a stationary piece of lead, it'll travel 9.81 metres down in one second; if you fire it horizontally at 100,000 m/s, in one second it will travel 100,000 metres horizontally and 9.81 metres down. Nothing changes.

I did not say the downward force changed, I said the direction of the nett force changed because of the inertia of the bullet, and the addition of orthoganal forces.

The bullet will have a downward acceleration of 9.81 ms-2, but by one second later, it will be a long long way away (incidentaly, the 9.81 ms-2 is acceleration, not velocity).

Yes, if a bullet was travelling at 100,000 m/s, and you scales remaid in step with it, they would feel the full downward acceleration of 9.81ms-2, but that is not the scenario we have here, so what the stationary scales will feel is a composite of the horizontal inertia, and the vertical acceleration.

I know there is a problem that I am mixing acceleration and speed, but since the actual motion of the bullet is tangental to the balance, thus it is in effect decelerating towards the balance horizontally, before passing the balance, and then accelerating away again.
« Last Edit: 20/03/2007 21:03:50 by another_someone »

#### lyner

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##### The weights of moving objects
« Reply #7 on: 21/03/2007 14:29:46 »
Quote
I did not say the downward force changed, I said the direction of the nett force changed because of the inertia of the bullet, and the addition of orthoganal forces.

'Inertia' is not a meaningful term in physics. You can choose from 'Mass', 'Momentum' or 'Force' for a serious argument. The nearest, allowable, equivalent of 'Inertia' must be 'Mass'- i.e. how hard it is to get  the object moving or change its direction ( as in Newton's Second law). Mass is a scalar quantity - no direction involved..

Newton (and you have to try very hard to ignore him) tells us that there is no 'orthogonal force' on the object, only the downward weight force. If there were such a force, the object would slow down or speed up and we said there was no friction in the original model.
All the moving object has is momentum (mass times velocity).
If you're going to have a cat's chance in hell of understanding SR problems you must, first, get your classical mechanics right.

There's another problem with this model. The objects will be, effectively, in orbit , or free fall.  If  the table is 'horizontal' both objects will skim past and leave the surface on a, more or less, tangential  (straight line / very low curvature) path and there would be no vertical force. Why bother with the idea of weight, under these circumstances, anyway? The masses will be perfectly well behaved, according to  relativity. GR will come into play because of the  Earth's near presence, which makes it more complicated.

I guess, if the planet were HUGE enough to regard the ground as flat enough and the attraction regarded as not 'central' but in one direction, then the Mass of the attractor would be huge. Is it a fruitful thought experiment?

« Last Edit: 21/03/2007 14:48:40 by sophiecentaur »

#### another_someone

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##### The weights of moving objects
« Reply #8 on: 21/03/2007 16:09:02 »
Quote
I did not say the downward force changed, I said the direction of the nett force changed because of the inertia of the bullet, and the addition of orthoganal forces.

'Inertia' is not a meaningful term in physics. You can choose from 'Mass', 'Momentum' or 'Force' for a serious argument. The nearest, allowable, equivalent of 'Inertia' must be 'Mass'- i.e. how hard it is to get  the object moving or change its direction ( as in Newton's Second law). Mass is a scalar quantity - no direction involved..

Sorry about the ambiguity - the term I probably wanted was momentum rather than mass or inertia (as you say, mass has no vector, only a scalar; but momentum has a vector).

Newton (and you have to try very hard to ignore him) tells us that there is no 'orthogonal force' on the object, only the downward weight force. If there were such a force, the object would slow down or speed up and we said there was no friction in the original model.

But, in the view of the balance, the object is speeding up and slowing down in one direction, and slowing down and speeding up in another.

Let me try and explain with a crude drawing:

The bullet (in orange) will skim over the top of the balance (in blue), since we do not intend that the bullet will hit slam bang into the balance.

As the bullet is a long distance away (the upper diagram), its total velocity vector of the bullet is almost parrallel to the line of sight that the balance has of the bullet; but as the bullet gets closer to the balance, its velocity vector shifts away from the line of sight of the balance; until as the bullet passes over the balance, its velocity vector is perpendicular to the line of sight of the balance; and thus the bullet must be appearing to accelerate and decelerate from the perspective of the balance (aside from the acceleration of gravity).

If the bullet was simply resting upon the balance, the only acceleration it would see is the downward acceleration caused by gravity, and this would remain constant over time (not effected by the position of the bullet, since the position of the bullet does not change).

I hope this makes more sense.

#### BillJx

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##### The weights of moving objects
« Reply #9 on: 21/03/2007 20:50:11 »

A balance scale merely measures force, not mass;

I'm not up to a relativistic discussion, but my understanding was always that a balance scale measures mass, a spring scale measures force.  The balance scale compares two masses regardless of the gravitational field.

Am I missing something in your argument, or when you wrote "balance scale" were you thinking of a spring scale?

#### ukmicky

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##### The weights of moving objects
« Reply #10 on: 21/03/2007 22:03:29 »
Quote
Now let us view this experiment from within a vehicle moving at the same speed as one of the objects.  In this reference frame, the one object is at rest, but the other is moving at a high speed. Thus, their relativistic masses will not be equal. Both objects will still seem to cross their respective platforms at the same time. And the scales will still seem to record equal weights for each, even though the masses now are not the same.
But their masses are still the same.  A moving objects rest mass is the same as if it were standing still. a bullet travelling at near the speed of light has the same rest mass and so still weighs the same as it does when its standing still. I believe.
OR HAVE I UNDERSTOOD IT INCORRECTLY
« Last Edit: 22/03/2007 00:09:03 by ukmicky »

#### lyner

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##### The weights of moving objects
« Reply #11 on: 22/03/2007 19:10:44 »
First, let's assume it's all happening slowly, wrt the speed of light.
Each bullet will be , effectively, in orbit or at least there will be some ' centrifugal force ' on it, as seen by the observer on the table. It will, therefore, either have (experience) no force on it from the table  or else a  reduced force.
That is the only meaning for the term 'weight'.
A ball, rolling across the pan of a balance will, also, 'weigh' a bit less, for the same reason.
Secondly, if you want it to go fast wrt the speed of light then you are into all sorts of complications: the motion is either not linear, but along a circumference, (so not an inertial frame) or else it is in a straight line , so the direction of the  (radial) gravitational field will be changing : also not an inertial frame.   The  geometry (symmetry ) of the situation would be taking the two bullets to the perigee (nearest point) of what is effectively their orbits around the planet so, still no 'weight' at all, for either of them because they are in free fall, whatever happens to their masses because of relativity.
Furthermore, I really don't think it is a valid model to suggest an isotropic gravitational field.   Did anyone ever come across one of THEM anywhere?
This has been bending my mind far more than it should. Let's do the angels on a pinhead instead.

#### lightarrow

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##### The weights of moving objects
« Reply #12 on: 22/03/2007 20:00:29 »
Quote
I did not say the downward force changed, I said the direction of the nett force changed because of the inertia of the bullet, and the addition of orthoganal forces.
'Inertia' is not a meaningful term in physics.
However, in physics "inertial field" is a meaningful concept. For example, centrifugal and Coriolis forces are inertial forces.
Quote
Newton (and you have to try very hard to ignore him) tells us that there is no 'orthogonal force' on the object, only the downward weight force. If there were such a force, the object would slow down or speed up and we said there was no friction in the original model.
All the moving object has is momentum (mass times velocity).
Here you are referring to "rest mass" or "relativistic mass"?
It's better to avoid talking about "relativistic mass" in physics, to avoid problems in some subjects.
If you meant "rest mass", then your statement is not correct relativistically:
p ≠ mv;    p = γ*mv;  where γ = 1/sqrt[1-(v/c)^2].
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There's another problem with this model. The objects will be, effectively, in orbit , or free fall.  If  the table is 'horizontal' both objects will skim past and leave the surface on a, more or less, tangential  (straight line / very low curvature) path and there would be no vertical force. Why bother with the idea of weight, under these circumstances, anyway? The masses will be perfectly well behaved, according to  relativity. GR will come into play because of the  Earth's near presence, which makes it more complicated.
I guess, if the planet were HUGE enough to regard the ground as flat enough and the attraction regarded as not 'central' but in one direction, then the Mass of the attractor would be huge. Is it a fruitful thought experiment?
If I have understood well your point of view, you are saying that, in the model, he should account for centrifugal force too and that it's not possible to avoid using GR because of the huge mass of the planet. About centrifugal force, since it's equal to mv2/R, it can become negligible if R is very large.
About GR, if R is very large then the gravitational field is approximately uniform so you can replace the large planet with an inertial field, that is, making the experiment inside an accelerating rocket.

However, where is the problem in considering a huge object? Our galaxy, for example, could replace the huge planet. Consider that, to maintain gravity on its surface to an acceptable level, the planet's density should decrease as 1/R; if R becomes ≈ 5000 times the earth's radius and you want the same g of the earth's surface, then the density have to become comparable to that of the air.
« Last Edit: 22/03/2007 20:02:09 by lightarrow »

#### lightarrow

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##### The weights of moving objects
« Reply #13 on: 22/03/2007 20:12:41 »
Quote
Now let us view this experiment from within a vehicle moving at the same speed as one of the objects.  In this reference frame, the one object is at rest, but the other is moving at a high speed. Thus, their relativistic masses will not be equal. Both objects will still seem to cross their respective platforms at the same time. And the scales will still seem to record equal weights for each, even though the masses now are not the same.
But their masses are still the same.  A moving objects rest mass is the same as if it were standing still. a bullet travelling at near the speed of light has the same rest mass and so still weighs the same as it does when its standing still. I believe.
OR HAVE I UNDERSTOOD IT INCORRECTLY
Yes, rest mass doesn't change, it is relativistically invariant: it's exactly for this reason that the concept of rest mass, which now all physicists call simply "mass", is so useful in physics.
However, an object's weight is m*g only if the object is at rest with respect the scale; if it's not, then the weight is m*g*γ;  γ = 1/√(1-β2).

So the paradox is resolved saying that what counts is the speed of the object relative to the scale, not its absolute speed.

#### lyner

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##### The weights of moving objects
« Reply #14 on: 24/03/2007 00:03:52 »
However, where is the problem in considering a huge object? Our galaxy, for example, could replace the huge planet. Consider that, to maintain gravity on its surface to an acceptable level, the planet's density should decrease as 1/R; if R becomes ≈ 5000 times the earth's radius and you want the same g of the earth's surface, then the density have to become comparable to that of the air.
Not unreasonable as a thought experiment. I do agree with your sums here but have you considered what sort of material, with such a low density, could be envisaged to be strong enough to support a whole planet of that density? Any equivalent 'gas' giant would have non uniform density- very dense and compressed at the centre and low on the outside. I guess, as an alternative, you could have a more dense and massive planet with a very high tower, on which you could carry out your measurements.

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So the paradox is resolved saying that what counts is the speed of the object relative to the scale, not its absolute speed.
I think this sums up the answer to the relativity bit very well.
We're really saying that two masses travelling side by side will experience a force  of
G m1  m2 / d squared, however fast they are going - fair enough.
AND
If they're travelling past each other then the force will be modified by  the γ factor, caused by their relative velocity.
This could, presumably, be seen by very careful observation of the paths of projectiles.
Can I go home now?

« Last Edit: 24/03/2007 00:21:23 by another_someone »

#### thebrain13

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##### The weights of moving objects
« Reply #15 on: 01/04/2007 21:32:58 »
im sorray if I skipped someones valid answer, but this is what I believe is correct. The speed of the bullet would make zero difference, a bullet at rest would push the direction of the gravitational source the same as a moving one would.

The strength of gravity is not equal to all frames of reference. A moving projectile (hence moving slower in time) views the strength of gravity as being greater than if it were at rest. If an object were traveling near earth so that time was advancing twice as slow, he would view the strength of gravity as 19.62 ms^2 or 9.81x2.

#### lyner

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##### The weights of moving objects
« Reply #16 on: 02/04/2007 10:30:59 »
Quote
The speed of the bullet would make zero difference, a bullet at rest would push the direction of the gravitational source the same as a moving one would.

What do you mean by this statement? Are you referring to the force of attraction between the two objects?

Quote
A moving projectile (hence moving slower in time) views the strength of gravity as being greater than if it were at rest.
If both objects have zero acceleration then they are both observing from inertial frames. Each will experience the same force. Of course, this force would cause an acceleration which, in turn, would mean that neither was in an inertial frame any more!
As these object are moving past each other, they can only deduce the force between them by the paths they follow. The bullet will 'see' a more massive earth  hurtling past  and acting on it for an unchanged time (his clocks all work the same as far as he can see). Likewise, the Earth will see a more massive bullet going past. They will observe (I think) the same relative motion (the amount of deflection from a straight line) and will conclude the same thing about the product of their masses.
OR, will the bullet, being much less massive, accelerate more than the Earth and, hence, have more of a relativistic change in what it does or sees?
« Last Edit: 02/04/2007 10:42:12 by sophiecentaur »

#### Atomic-S

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##### The weights of moving objects
« Reply #17 on: 08/04/2007 04:40:15 »
This subject has become confusing. I have posted a related but somewhat different topic at http://www.thenakedscientists.com/forum/index.php?topic=7163.new#new that may avoid some of the difficulties.

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##### The weights of moving objects
« Reply #17 on: 08/04/2007 04:40:15 »