A matter of weight

What do you weigh? More to the point, what does that mean?

What is weight?

What it isn't: mass

Like many people, I find out what I weigh by standing on the bathroom scale. Usually, the answer is approximately 80 kilograms. Or so I try to convince myself.

Well, there's a problem with that. Kilograms is a measure of mass, not weight. But what's mass?

To answer this, we rely on Newton's laws. Mass is resistance to force. If you act on a mass, \(m\), with a force, \(\pmb{F}\), then it has an acceleration, \(\pmb{a}\), and the three are related by \[ \pmb{F} = m \pmb{a}. \]

Actually, things are a bit more complicated than that, but as long as the mass isn't changing, it's OK.

Then if I have a reliable way of exerting a known force, I can compare the masses of two different object by comparing the accelerations imparted to both. And if I have a reference mass (and there is one for the kilogram) then I can work out the mass of anything else I act on with this force in terms of the reference mass.

OK, so that's mass (well, the quick version) sorted out. But what is weight, and what is my bathroom scale really telling me?

What weight is absolutely

The first definition that you are likely to meet in a physics text is that it is the force exerted on a body by gravity: sometimes the actual force (as a vector), but more usually the magnitude of the force.

It is a remarkable fact that the force exerted by gravity on an object is exactly proportional to its mass, so that all objects, whatever their mass, fall with the same acceleration (in the absence of confounding factors like air resistance).

In SI units (i.e. those used by the civilized world), the unit of force is the Newton, and the unit of mass is the kilogram. At the surface of the earth, the force of gravity is approximately \(9.8 \text{N}/\text{kg}\); a mass of 1 kilogram weighs about \(9.8\) Newtons.

We can work with this. If I have access to materials which can be compressed or extended by acting on them with a force, then I can build a machine which measures the (magnitude of the) force I exert on it. And if I know that what people really want to know is their mass, I can calibrate it so that if the force acting on it is the force that gravity exerts on their body, then the scale shows the mass that gravity must be acting on, rather than the actual force.

This is typically how a bathroom scale works.

And is fine and excellent, as long as I know what the force of gravity is where I use the bathroom scale. There is a minor issue caused by the fact that the force of gravity isn't quite the same all over the surface of the earth; but the variation is fairly small (from \(9.78 \text{N}/\text{kg}\) to \(9.83 \text{N}/\text{kg}\), and for everyday purposes we can pretty much ignore this variation.

But now we bump up against a bit of a problem. For the scale to work properly, we have to stand still. You'll have noticed that if you jump up and down on it, the displayed weight oscillates along with you. It's less likely that you've taken it into a lift, but if you have you'll have noticed that when the lift has an upward acceleration you show a higher weight, and if it has downward acceleration you show a lower weight.

The snag is, of course, that to find your weight, the scale has to hold you stationary in the earth's gravitational field, and measure the force it takes to do that.

This raises a slightly awkward question: how do you know when the scale is holding you stationary?

This could well seem like a rather stupid question. All you have to do is look around.

But what if you are in an enclosed room? How could you tell the difference between being (say) on the moon, with a weight of approximately one sixth of your weight on earth, and being in a lift which has a downward acceleration of about five sixths the acceleration due to earth's gravity, or indeed being in a rocket a long way away from any body of mass, accelerating at about one sixth of the acceleration due to earth's gravity?

It turns out, that's really hard to do. You can't do it by making local measurements, which is a heavily coded way of saying that no matter how good your measuring apparatus is, if you work in a sufficiently small laboratory, over a sufficiently small time span, you can't detect the difference. (If you allow non-local measurements, which effectively means that you work in a region big enough that the force of gravity varies appreciably, then you can measure the resulting tidal effects: here is a nice look at tidal effects from the point of view of an orbiting astronaut, from @ColinThe Mathmo.

This is a bit of a blow. We have what looks like a perfectly sensible definition, but it turns out to be hard to use. So we do that standard trick. We replace it by a somewhat different definition, which is easier to use, but doesn't look quite so sensible.

At least, it doesn't look quite to sensible to start with. Once you get used to it, you can convince yourself that it is in fact the sensible definition, and it's the original definition that's misled.

What weight is relatively

So what do we do instead?

We accept that since what we can measure (reasonably) depends on the state of motion of the equipment and body, we build that into the definition.

We now define the weight of an object in a given frame of reference to be the (magnitude of the) force required to hold it stationary in that frame of reference.

So, if your scale says you weigh a particular amount, that's what you weigh: you don't have to know whether you're on the moon, or whether you have to compensate because the scale is accelerating.

This does have it advantages. Principally, it means that you weight what you feel as if you weigh. If you're in a rocket taking off, or visiting a heavy-gravity planet (we can dream) you weigh more if you're standing still on the earth. If you're in a lift plummeting you towards an untimely end, or on the moon, you weigh less than if you're standing still on the earth. And that's all there is to it.

So, to the question I've been avoiding up to this point: is an astronaut in orbit, say on a space station, weightless?

There are two answers to this.

• No. The force on gravity on him is slightly less than it is on the surface of the earth, so he weighs a little less than he would at home.
• Yes. If he tries to stand on a scale, it reads zero; it takes no force to keep him stationary in his space-station, so in the frame of reference of the space-station, he is weightless.

And both answers are correct. Whether the astronaut is weightless or not depends on the definition of weight that you prefer.

But in either case, the astronaut is definitely not outside the gravitational influence of the earth; gravity is still pulling on him with a very significant force.

I prefer the second definition, because it's the one that doesn't require you somehow to know the situation outside your laboratory; it only requires you to make local measurements. But that doesn't make it right, only preferable from some points of view.

At least, that's the situation if you live in a universe where gravitation acts according to Newtonian theory. But in fact we don't really, and there's a better way of trying to understand gravity, and then we have a better reason for preferring the second definition.

What weight is relativistically

There's a great mystery in Newtonian mechanics: why is it that the mass which interacts via gravity is exactly the same as (or at least, exactly proportional to) the mass which resists acceleration? Very sensitive experiments have been carried out to test this, and they've always come out unable to detect a difference. They agree to no worse that one part in one hundred billion.

So, why should this be?

The current answer is that there is a better way to think about gravitational influence than as a force. In general relativity, we have a way of relating the distribution of matter to a distortion in space and time, which changes the notion of a straight line: we generalize this to a kind of curve called a geodesic. So a particle moving in a gravitational field is one which travels along a geodesic: there is no force acting on it, it's just (just!) that the nature of space and time mean that it doesn't travel along the same trajectory as it would in the absence of gravity.

Then any particle (with no external forces acting on it) travels along a geodesic which is determined purely by a starting point and a starting velocity; mass does not have a part in the trajectory of a the motion of a particle in a gravitational field. Of course, part of the joy of this is that it does give very nearly the same answer as Newtonian gravity.

Very nearly, but not quite the same as Newtonian gravity. And with careful observation we find that the general relativity version of gravity matches measurements better than the Newtonian one.

But now there is no force of gravity.

If I'm standing on the surface of the earth, the only force acting on me is the force of the earth pushing me off the geodesic that the curvature of space and time wants me to travel along. If I'm on the space station, there is no force acting on me at all.

So in this framework, the only definition of my weight that makes any sense is that it is the magnitude of the force required in some frame of reference to keep me stationary in that frame of reference.

This looks superficially the same as the second definition in Newtonian gravity, but it's really quite different in principle. In the Newtonian case, it's avoiding the issue that there's no easy way to access the actual force of gravity, and settling for what is easy to measure. In the general relativistic case, it's saying that weight is an illusion caused by being in an unnatural frame of reference.

I try to take some consolation from that as my weight seems to gradually, but inexorably, increase.