Is holding something in a gravitational field doing work ?

[bunburryist]

Is holding something in a gravitational field "doing work"?

My son and I are on opposite sides of this question - if I am holding something in my hand in a gravitational field, am I doing work? My position is this - if I constantly accelerate a ball in space I am doing work. Since acceleration is equivalent to being in a gravitational field, and since holding a ball on Earth is in a gravitational field, I am doing work when I hold it. My son's position is that since the thing I am holding is not moving (there is a net acceleration of zero) I am doing no work. Is it simply that we are disagreeing about what is the relevant frame of reference - mine being the Earth's gravitational field, his being my body?

 

[Razzor7]

You're not doing any work on it.

 

[uart]

Bunburryist, what about if you place the object on a table and allow it to resist the gravity instead of your hand? Is the table doing work on the object? If so then where is it's energy coming from.

 

[hdunham]

Have your son hold a 5 lb weight straight out at arms length for 5 minutes, then ask him if he felt like he did any work.

Yes, it's a frame of reference difference. As your son holds the mass, have him imagine he is on the moon. The displacement is obvious.

The NET work is zero if you are just holding the ball still. Gravity is doing work to accelerate the ball toward Earth and you are doing equal and opposite work in holding it still, balancing out gravity. That can be seen by taking your son's frame of reference at which the displacement of the ball is zero, thus zero work.

One must not be confused between the work one does against gravity, the work gravity does, and the net work.

 

[Hootenanny]

Have your son hold a 5 lb weight straight out at arms length for 5 minutes, then ask him if he felt like he did any work.

Yes, it's a frame of reference difference. As your son holds the mass, have him imagine he is on the moon. The displacement is obvious.

The NET work is zero if you are just holding the ball still. Gravity is doing work to accelerate the ball toward Earth and you are doing equal and opposite work in holding it still, balancing out gravity. That can be seen by taking your son's frame of reference at which the displacement of the ball is zero, thus zero work.

One must not be confused between the work one does against gravity, the work gravity does, and the net work.

Could you please quantify the work done by gravity on the 5 lb weight, assuming that the weight is held stationary by the boy?

 

[Topher925]

hdunham, I disagree. No work is being done. Work is a force applied over a displacement. In equation form:

W = F x d

You are providing the force, but where is the displacement? There isn't any. If you dropped the weight, or picked it up, then work would be done. I understand your moon analogy, but the force is not being applied in the same direction as the displacement.

 

[atyy]

When you accelerate the ball in empty space, you are acting like a gravitational field for the ball - you are doing the work that the gravitational field would be doing if the ball was free falling in a gravitational field. So the analogy should be you in empty space equals gravitational field, but you the way you told it is you in empty space equals you in a gravitational field. If you pushed the ball hard in empty space, and some one pushed equally hard the other way, no one would be doing any work. So when you oppose the pull of gravity on the ball, neither you nor the field is doing any work either (that's my guess).

 

[hdunham]

yeah, ok, I'm going to have to eat crow. it's still early. what i meant to say was forces, not work. net FORCE is zero, so as with displacement its zero.

my bad.

any displacement due to rotation is tangential to the force so the dot product is zero.

sorry!

 

[atyy]

But in general, work and energy are frame dependent.

 

[Renge Ishyo]

As many have said you are doing no work *on the ball* and your son is right. What confuses people is that they "feel" as if they are doing work. Indeed, there IS work being done in this example, but it is work done *inside your body* to contract your muscles so that you can hold the ball (the muscles are constantly being tugged into position with the expense of ATP so that you can support the ball). But you are not transferring any of that work to the ball, because you have not moved the ball against gravity through a distance. In effect, all that work done by your muscles is just wasting away as heat, and your body "feels" this expense whether you move the ball against gravity (doing work on the ball) or keep the ball at the same height above the ground (doing no work on the ball).

 

[bunburryist]

A balls natural state in a gravitational field would be to fall (follow a geodesic). That is, when it is in freefall relative to the Earth it is "standing still" (following a geodesic) in spacetime. So anything we do to force the ball to deviate from it's "standing still" in spacetime would be an acceleration. So when we hold the ball from falling we are constantly accelerating it against it's natural state of falling. Does this make sense?

We can, I think, agree that work is being done as rockets on my feet accelerate me and the ball in my hand. Let's have me holding a scale with the ball on the scale, so that, as I accelerate, it registers "weight." Imagine that there is another person next to me in space who is accelerating along with me. This person might, mistakenly, believe that since the ball is "standing still" in his frame of reference, that there is no work being done. He might even conclude that we were merely in a gravitational field, that that is why the ball is "pressing" on the scale, and that no work was being done. (We could replace my body on Earth with a rocket that pushes just hard enough to keep the ball suspended.) Of course, he is mistaken. I am doing work (work is being done) as the ball is accelerating constantly. So whether it is my legs or rockets on my feet, if the ball registers weight, it is undergoing an acceleration. If it is undergoing an acceleration, then there must be work being done.

Someone next to me on the Earth is analogous to my fellow rocketeer. He is, along with the ball, me, etc. constantly accelerating relative to the Earth's gravitational field.

Does this make sense?

 

[Renge Ishyo]

So anything we do to force the ball to deviate from it's "standing still" in spacetime would be an acceleration. So when we hold the ball from falling we are constantly accelerating it against it's natural state of falling. Does this make sense?

That's simply not the correct way to apply the concept of work. Ignoring complicated relativistic considerations you can look at the issue this way, doing work on a something raises the energy of that something regardless of what frame of reference you look at it in. It is clear that by holding the ball static at a certain height for 20 seconds and then dropping it a certain distance and by holding the ball at that same certain height for 5 minutes and then dropping it a certain distance that in both cases the ball would have the same kinetic energy as it hits the ground (even if some measured values differ based on the frame of reference, the *energy change* between start point and end point would be measured the same regardless).

If you believed that you were doing work on the ball by holding the ball in place you would predict that the ball held for 5 minutes would have much more kinetic energy when it was dropped through the same height as the ball held for 20 seconds. Experiment shows this not to be the case. In fact, the ball has the same final kinetic energy in both examples (and this is why, by definition, no "work" is being done on the ball while you hold it in place).

 

[Dale]

Is it simply that we are disagreeing about what is the relevant frame of reference?

Yes, it's a frame of reference difference.

But in general, work and energy are frame dependent.

I just wanted to briefly expand on the idea of energy and reference frames. Work is defined by f.d, but since d is frame-dependent f.d is naturally also frame dependent. Let's consider three cases:

a) Ball on a table in a uniform gravity field. As the ball sits on the table the real upwards normal force exerted by the table is counteracted by the real downward gravity force exerted by the earth. The ball does not accelerate, the displacement is 0 in this reference frame, so the work is zero and the KE is also 0.

b) Ball on a table in a uniformly accelerating rocket as considered from an inertial reference frame. As the ball sits on the table the real upwards normal force exerted by the table is not counteracted by any force. The ball does accelerate, the displacement is non-zero in this reference frame, so the work is non-zero and the KE increases.

c) Ball on a table in a uniformly accelerating rocket as considered from the rocket's accelerated reference frame. As the ball sits on the table the real upwards normal force exerted by the table is counteracted by the ficticious inertial force. The ball does not accelerate, the displacement is 0 in this reference frame, so the work done is 0 and the KE is also 0.

The usual convention on the surface of the Earth is to use the reference frame as shown in a), but it is just a convention. There are no absolute answers to typical energy questions, it is all relative to the reference frame chosen.

 

[russ_watters]

If it is undergoing an acceleration, then there must be work being done.

Does this make sense?

No. That's not how work is defined. Work is force times distance. Where's the distance here?

What you are describing is analagous to a circular orbit. There is a constant acceleration in an orbit, yet no work being done because the force is perpendicular to the direction of motion.

In your example of equivalence between gravitational force and acceleration, you are right that you can look at the force from different frames. But in no frame is there motion, so in no frame is any work being done.

 

[Renge Ishyo]

There are no absolute answers to typical energy questions, it is all relative to the reference frame chosen.

The *change* in energy is independent of the reference frame chosen (you just have to be careful that you are measuring the initial and final parameters from the same reference frame). For example, neglecting friction if I push a box measured initially to be at rest on Earth with 5N through a distance of 1M, the kinetic energy it would gain would be 5J. I would say the initial kinetic energy here is 0J and the final is 5J. The change in kinetic energy is 5J-0J = 5J.

Now say you watch the same situation while you look down at me in space. Due to the spinning Earth you see the box as moving with an initial kinetic energy of 250J in some direction. If I then on Earth exert a force on the box of 5N through a distance of 1M along this same direction, you would measure the final kinetic energy of the box in space as 255J, not 5J. Yet the *difference* in energy that we calculate would be the same in both frames of reference. In space you would calculate, 255J-250J=5J.

But wait wait wait Renge you say (as Dalespam said above in his example c), let us suppose that I start measuring the box at rest at 0J in a reference frame and then my reference frame accelerates along with the ball so that as you add your 5J to the ball it always appears that the ball is at rest. I would calculate the change in energy to be 0J-0J=0J! Energy changes are relative!

No, they are not. The flaw in this case is that the observer didn't measure the final parameter and the initial parameter in the *same* chosen reference frame (the laws of physics require that you chose one reference frame and make both the final and initial measurements relative to that). The number calculated was off, because the observer didn't take into account that his reference frame had changed between the time he made his initial measurement and final measurement. His result was a calculation error brought about by an error in measurement, the box still gained the same amount of energy in all three examples. The laws of conservation of energy survived relativity in this way, and so did the son's explanation of work.

 

[Andy Resnick]

As Renge points out, there is no mechanical work being performed on the ball. However, your body (or whoever is holding the ball) is performing chemical work to maintain their arm in a nonequilibrium configuration. If you like, this is simply non-PV work.

 

[cesiumfrog]

In your example of equivalence between gravitational force and acceleration, you are right that you can look at the force from different frames. But in no frame is there motion, so in no frame is any work being done.

No.. in every *inertial* frame (that is, relative to any free-fall observer) work is being done to lift the ball's trajectory. (The energy is supplied from the work of the floor on the person, not from the person's metabolism. The effect originates from the curved space-time of GR: no single flat inertial frame can be applied to the entire system, and as has already been noted, energy is always frame-dependent.)

 

[russ_watters]

I understand that, but in this example, there is no free-fall observer, nor any easy/convincing way to create one.

 

[LukeJD]

I would agree that you are exerting a force, because you are accelerating the ball at 9.8m/s^2 in the opposite direction from the center of the gravitational field. However mathematically, Work is defined by Force x Displacement, there is no displacement. So no work is being done.

 

[atyy]

Someone next to me on the Earth is analogous to my fellow rocketeer. He is, along with the ball, me, etc. constantly accelerating relative to the Earth's gravitational field.

Does this make sense?

Well, perhaps your fellow accelerating rocketeer should be equivalent to a person free falling in a gravitational field. So he is not analogous to someone standing still next to you on earth. While you are standing on earth, a free falling observer will see 'mistakenly' think that you are doing work on the ball. (I'm not sure the 'mistakenly' is really meaningful, just using it to parallel your description of your fellow accelerating rocketeer 'mistakenly' thinking no work is done.)

 

[Gear300]

You're not doing work on the object, but your body is doing work on itself. Your muscles are constantly contracting, which is why you'll start to eventually feel strain.

 

[Dale]

The *change* in energy is independent of the reference frame chosen

Actually, even the change in energy is frame dependent. For example, consider an object of mass 1 at rest. The four-momentum p0 = (1,0,0,0) (in units where c=1), if you exert a force on it in the x direction accelerating it to .6 c the four-momentum is p1 = (1.25,.75,0,0) for a change in energy of .25. Now, if you boost p0 and p1 by .5 c you get p0' = (1.15,0.58,0,0) and p1' = (1.87,1.59,0,0) for a change in energy of .72. Just like time is relative so is energy.

I believe that the only time the change in energy is independent of the reference frame is when you are considering only inertial frames in Newtonian physics.

 

[bunburryist]

Imagine that a rocket takes off, and inside the rocket is a ball on a scale. As the rocket accelerates, the ball “pushes down” on the scale – it has “weight.” The whole idea of equivalence, in this context, is that people in the rocket wouldn’t know (just from experiments they could do in the rocket) whether they were undergoing an acceleration or whether they were feeling the effects of a gravitational field.

So the rocket accelerates for a certain length of time, all of which the ball exhibits weight on the scale – evidence of the acceleration of the rocket. Suppose the rocket turns off its engine. The ball would no longer have “weight” – it would not push on the scale any more. No more work is being performed by the rocket.

Relative to the rocket, the ball has no more energy than before it took off. Someone on the rocket would conclude (mistakenly, as would someone holding a ball and believing he didn’t do any work) that “since the ball has no more energy now than when we started, we must not have done any work.” That is mistaken because it is using the wrong frame of reference relative to which it must be determined if the ball has more energy. It is the frame of reference from whence the rocket/ball started that the ball has more energy relative to. To people in that frame of reference, the ball and the rocket both have more energy than before it took off.

When we hold a ball, we are like the person on the rocket. We can hold it all day long, and even though it has weight (evidence of our doing work), it doesn’t seem to have any more energy than it did an hour or a day ago – within our reference frame. So we conclude that we have done no work. So who/what in our ball holding example is equivalent to the person where the rocket took off from, relative to which the ball does have more energy? It would be someone in free fall.

The longer you hold a ball in a gravitational field, the more energy it has relative to someone in freefall in the gravitational field. That is, the longer you hold the ball, the faster it is accelerating away from the free-falling person. But who is doing the accelerating – you, the earth, and the ball, or the free-faller? Although we, on the earth, learn to think of ourselves as being at rest, and falling things as being those which are accelerating, it isn’t the person in free-fall who is undergoing an acceleration. It is the ball – and you and the Earth - that is accelerating. Remember, feeling the force of a gravitational field is equivalent to the acceleration. As you hold the ball, it is moving away from the free-faller faster and faster.

Imagine someone standing on a trap door next to you as you hold the ball. You can hold the ball all the live-long-day, and it will not have any more energy from your friends’ perspective, just as it wouldn’t from yours. Suppose now that the trap door opened and your friend was to start a free fall. The ball you are holding, you, and the Earth would move away from/past him faster and faster – it would be accelerating. We tend to think that it is the falling person who is accelerating, but that is mistaken. It is we, the balls we hold, and the ground we are standing on that is accelerating. Remember – if the person falling had a ball on a scale it would not have weight – he would not be undergoing an acceleration. That is, no experiment he could do would show that he was either accelerating or was under the effect of a gravitational field. He might as well be very far removed from any massive body – he wouldn’t know the difference.

If the ball you are holding is undergoing an acceleration, then it should have more and more energy as time goes by, right? How can this energy be used by the free-falling person, showing that work has been done and that the ball (and you and the earth) have more energy? How can he measure the fact that you, the ball, and the Earth have more energy with each passing moment? Suppose that your falling friend was connected to a very long comb shaped device with paddles sticking out, so that one passes near the ball every ten seconds. If someone on the comb device wants to measure the kinetic energy of the ball as it passes by, he adjusts the next paddle slightly so it hits the ball. As the comb device falls faster and faster, the ball is moving faster and faster in the other direction, so that with each passing paddle (if it was to be made to hit the ball) there would be a greater ability to “do work” (impose a force on the paddle).

So who is undergoing an acceleration? Is it the ball you are holding (along with you, the earth,etc.), or the comb-device with your free-falling friend? If there was a ball on a scale on the comb device there would be no evidence of acceleration or of feeling the force of a gravitational field (the ball would not “weigh” anything on their scale, but would, like the rest of the comb device, be in free-fall). But the ball you are holding on your scale continuously has weight – evidence that it is undergoing an acceleration.

There is a tendency to think that the gravitation field is imparting more and more energy to your falling friend (doing work) as he accelerates away from you. But it is you who is doing work. Actually, it is you, the ground you’re standing on – the whole Earth – everything that stops you from following a geodesic – doing work – gaining more and more energy in your free-falling friends frame of reference. He is “standing still” in space-time – following a geodesic. So it’s not merely the ball that gets more and more energy – it is the whole structure of you, earth, etc. that is moving faster and faster relative to the free-falling friend, as all of it has more energy relative to the free-falling friend, and so is capable of doing more and more work in his frame of reference.

You and your friend were originally in the same frame of reference. When the trap door opened it wasn’t him that accelerated. Rather, it was him who stopped accelerating as he went into free-fall – his scale would stop registering weight of his ball on his scale showing that he was neither accelerating nor under the effect of a gravitational field. It would be equivalent to the rocket stopping it's acceleration. Imagine two rockets side by side with balls on scales. If one rocket stopped accelerating, it would appear to the one who was accelerating that he (the one who stopped accelerating) was falling. But it is really the one on the rocket that is still firing that is accelerating.

I know I’m going on and on about this, but I really am trying to make sense of it. I understand that within the context of the people holding the ball that the ball does not gain energy as it is being held. On the other hand, feeling the effects of a gravitational field is equivalent to acceleration, and it takes work to perform an acceleration. I think there is some subtle aspect of GR that I either don’t understand completely or I understand incorrectly.

Maybe I should have posted this in the general relativity section!

 

[Landru]

Do the chemical reactions which occur in the arms muscles which work to resist the effect of gravity count as 'work'? Do the chemical reactions themselves constitute force * distance?

 

[Dale]

Do the chemical reactions which occur in the arms muscles which work to resist the effect of gravity count as 'work'? Do the chemical reactions themselves constitute force * distance?

Yes, the myosin fibers burn ATP which pulls the thin filament. Therefore on a molecular level f.d>0 so there is work being done internally. The external work done may be 0, implying 0% efficiency. In my examples I used a table precisely to avoid this confusion.

 

[Dale]

Relative to the rocket, the ball has no more energy than before it took off. Someone on the rocket would conclude (mistakenly, as would someone holding a ball and believing he didn’t do any work) that “since the ball has no more energy now than when we started, we must not have done any work.” That is mistaken because it is using the wrong frame of reference relative to which it must be determined if the ball has more energy. It is the frame of reference from whence the rocket/ball started that the ball has more energy relative to. To people in that frame of reference, the ball and the rocket both have more energy than before it took off.

It is not mistaken, there is no "wrong frame of reference". The people on the rocket are free to use their rest frame, the pepole on the surface of the Earth are free to use theirs. The free-falling observer is free to use theirs. None of them are wrong even if they disagree.

 

[bunburryist]

It is not mistaken, there is no "wrong frame of reference". The people on the rocket are free to use their rest frame, the pepole on the surface of the Earth are free to use theirs. The free-falling observer is free to use theirs. None of them are wrong even if they disagree.

I know there is no "wrong" frame of reference - thanks for correcting me. But this makes my point exactly - that in some frames of reference I am doing work when I "hold a ball still in my hand." What I meant by "mistaken" was their assumption that their frame was the right frame, and so there was no work being done when in some frames there are work being done. For every frame "in which no work is being done" when holding an object in a gravitational field it seems to me that there are an infinite number within whose frames of reference work is being done.

 

[russ_watters]

That's all well and good, but your example in the OP is about a house, not a rocket. You look out your window and see it sitting on the ground. There is no free-falling frame of reference to compare to and conclude you are doing work on the object.

You could take another object in your other hand and drop it and say that relative to that object, you are doing work on the object you are holding, but there is still a problem with that: in less than a second, that ball hits the floor and stops moving. There is no accelerating frame that can be created where you are always doing work on the object you are holding.

 

[Dale]

You don't have to attach a reference frame to a physical object, so there is nothing mathematically wrong with a free-fall reference frame near the surface of the earth. But I agree with your fundamental point that it is a rather weird and cumbersome reference frame to use for this problem.

 

[bunburryist]

That's all well and good, but your example in the OP is about a house, not a rocket. You look out your window and see it sitting on the ground. There is no free-falling frame of reference to compare to and conclude you are doing work on the object.

You could take another object in your other hand and drop it and say that relative to that object, you are doing work on the object you are holding, but there is still a problem with that: in less than a second, that ball hits the floor and stops moving. There is no accelerating frame that can be created where you are always doing work on the object you are holding.

If someone was falling into a black hole I could hold the ball as long as I like and, from his frame of reference, the ball would be accelerating.

The main idea I'm trying to get straight is the relationship between acceleration and work, and to what extent acceleration and gravitation are really equivalent. While my friend and I a standing on the Earth we are in a gravitational field. This is supposed to be equivalent to an acceleration. So if his trap door opens and he goes into free fall, I am still in the gravitation/acceleration state. Relative to him, I, the ball, and the Earth are accelerating. As time passes, we have more and more energy which he can use. If in his frame of reference we are getting more and more energy, mustn't there be some work being done in some sense? How can we be acquiring more and more energy if no work is being done?

I suspect (perhaps incorrectly) that this "work" is in fact illusory, and must in some way tie into the way we conceive of gravitation, curved space-time, etc. I don't know. This is what I'm trying to figure out.

 

[Renge Ishyo]

I believe that the only time the change in energy is independent of the reference frame is when you are considering only inertial frames in Newtonian physics.

Work is a Newtonian concept. Of what value is it to try to describe what it might be in situations where we cannot even test our conclusions? The point of the example I gave earlier was to show how "honest" calculation results made in accelerated frames cannot be trusted because the laws of physics weren't created (and therefore, don't necessarily have to hold) in such frames to begin with.

I suspect (perhaps incorrectly) that this "work" is in fact illusory, and must in some way tie into the way we conceive of gravitation, curved space-time, etc. I don't know. This is what I'm trying to figure out.

Work is a term used to describe an input or output of energy to a system. We can describe what it is only so far as we have experiments that can verify that it behaves the way we think it behaves. Part of the problem with modern physics is that most of the new arguments put forth these days are completely untestable (both by logic and by experiment).

Therefore, when such explanations are offered up as "evidence" all that one can do is just shrug and say "o.k., maybe so" or better yet "I'll believe that when its proven...which more than likely will be never. Until then I will stick with testable conclusions." If you adopt the latter stance then the question is answerable and your son wins because the theories you have put forth that require the term work to be redefined have not been tested. If you adopt the former stance then you both lose and neither side can win because the "true nature of work" is something that might be beyond the reach of science. The only difference I suppose is that in the former case he doesn't win, and maybe that's what you're going for?

 

[Dale]

Work is a Newtonian concept. Of what value is it to try to describe what it might be in situations where we cannot even test our conclusions? The point of the example I gave earlier was to show how "honest" calculation results made in accelerated frames cannot be trusted because the laws of physics weren't created (and therefore, don't necessarily have to hold) in such frames to begin with.

I honestly don't know what you are talking about. Relativity is one of the most well-tested theories ever. And there is nothing dishonest about using accelerated reference frames, the principles are well-understood.

 

[Renge Ishyo]

I honestly don't know what you are talking about. Relativity is one of the most well-tested theories ever. And there is nothing dishonest about using accelerated reference frames, the principles are well-understood.

I suppose in a less roundabout way what I am asking is do we need relativity to answer the original question of this thread? The problem is that the solution is no good if the answer is so complicated that nobody can understand it; isn't it better to remove unnecessary complications? There is nothing dishonest about using accelerated reference frames (so long as you know what you are doing and are accounting for it in your measurements...it is very easy to screw these up and obtain faulty conclusions as I tried to show above), but such an approach does not serve a purpose here except to possibly confuse the issue. Not that I blame the dad much for pushing the discussion along those lines (hey, when I notice I have made a mistake in chess I usually give up playing for the win and play for a tie instead...call it competitive nature or whatever...).

 

[Dale]

OK, to make things clear:

In the usual context no work is done holding something stationary in a gravitational field. So Dad loses to Son.

However, work and energy are fundamentally frame-variant concepts (even in Newtonian physics) and there do exist reference frames in which work is done holding something stationary in a gravitational field. So Dad can claim a draw if he wants.

 

[Renge Ishyo]

OK, to make things clear:

In the usual context no work is done holding something stationary in a gravitational field. So Dad loses to Son.

However, work and energy are fundamentally frame-variant concepts (even in Newtonian physics) and there do exist reference frames in which work is done holding something stationary in a gravitational field. So Dad can claim a draw if he wants.

Fair enough I suppose. Although if I was the son I would counter this with "could the situation of a man holding a ball physically exist in the conditions needed for the latter?" And if I was the father I would counter this by grounding him (because I could).

 

[Dale]

Fair enough I suppose. Although if I was the son I would counter this with "could the situation of a man holding a ball physically exist in the conditions needed for the latter?"

Yes, of course it could physically exist. Reference frames are just coordinate systems. Changing reference frames is just a way of looking at the same problem differently. So, although it is silly, there is no physical reason that would prevent you from using a free-falling reference frame for this problem.

And if I was the father I would counter this by grounding him (because I could).

 

[russ_watters]

Yes, of course it could physically exist.

So, although it is silly, there is no physical reason that would prevent you from using a free-falling reference frame for this problem.

Could it? How? Describe this reference frame to me.

 

[bunburryist]

Believe it or not, I've never had to ground my son! But seriously folks . . .

What I'm going to take away from this is what my son and I were going back and forth about in the first place - that he was talking about a local frame "guy holding a ball in his hand" and I was talking about something relativistic (accelerating rockets, etc.).

Thanks for all of your responses.

 

[russ_watters]

Were you really? Or are you changing the scenario now because you don't want to be beaten by your son?

I always hated it when people would do that when I was a kid. No one ever wants to be wrong in front of a kid.

I once won a bet with my 7th grade math teacher. He wouldn't actually admit he was wrong, but he bought me the soda anyway.

 

[Renge Ishyo]

Were you really? Or are you changing the scenario now because you don't want to be beaten by your son?

This thread is another great example of how to apply the relativity defense:

Person 1: "Here, measure the length of this bookshelf for me. It should be 14 inches."

Person 2: "It's 14 and 1/2 inches."

Person 1: "No it's not! Let me see that."

(Person 1 measures it out..."crap, it's 14 and 1/2 inches. What to do? I need relativity!")

Person 1: "Well I can see how you may have been led to believe that this bookshelf is 14 and 1/2 inches long, but in actual fact if you account for relativistic effects it is just as true that this bookshelf is 14 inches long. You are just failing to account for its length as it approaches the speed of light."

("Phew, good ole relativity to the rescue again!)

 

[Phrak]

My son and I are on opposite sides of this question - if I am holding something in my hand in a gravitational field, am I doing work? My position is this - if I constantly accelerate a ball in space I am doing work. Since acceleration is equivalent to being in a gravitational field, and since holding a ball on Earth is in a gravitational field, I am doing work when I hold it. My son's position is that since the thing I am holding is not moving (there is a net acceleration of zero) I am doing no work. Is it simply that we are disagreeing about what is the relevant frame of reference - mine being the Earth's gravitational field, his being my body?

There are two different contexts at work here. One is Newtonian physics, which I presume your son is speaking with, the other is General Relativity.

I bring this up because you say "acceleration is equivalent to being [stationary] in a gravitational field." This is true in Relativity, but not generally true in Newtonian physics.

As you may have gathered from previous posts, work is defined as force acting over a distance.

Your son is operating under the assumption (I presume) that the force of gravity is balanced by the opposite force applied by your hand. So with no net force there can be no work done.

On the other hand, in your case, without a gravtiational force, forces are not balanced, so there's at least some chance of work being done.

 

[Renge Ishyo]

Whether or not you use Relativity or Newtonian physics you still end up with the problem that experimentally the ball releases the same amount of energy when it is dropped to the ground whether you hold it for 2 seconds before dropping it or 2 minutes. If this is true it implies that no work is done on the ball by simply holding it for a period of time regardless of which model you use to study it. But of course, this is the beauty of taking the relativistic approach. Even faced with such a situation as coming dangerously close to violating the law of conservation of energy, you can still use the theory to "distort" time and make 2 seconds the same thing as 2 minutes and dodge the issue entirely...

 

[atyy]

OK, since this has been going on and on, let me chip in with another scenario for you guys to play with. There is nothing special or general relativistic about all this. It all makes sense within Newtonian theory. Let's pretend the Earth is flat, and the gravitational field of the Earth uniform. There is a man standing on the surface holding a ball. In Newton's theory, the free falling observer (she) is not an inertial observer - she is an accelerated observer. She will therefore feel an "inertial" force due to her acceleration that exactly cancels the "real" gravitational force on her due to the attraction of the earth. She will see the ball being accelerated towards her, and conclude that there is a net force on the ball. The downward force on the ball is the attraction of real gravity. The first upward force on the ball is the reaction (3rd law) provided by the man's hand against the ball's weight on the man's hand. The second upward force on the ball is the "inertial force" due to her acceleration. Since the reaction provided by the man's hand on the ball is equal and opposite to the attraction of gravity on the ball, she will conclude that the ball is accelerating towards her due to the "inertial" force. Therefore, she will conclude that neither gravity nor the man is doing any work, but that the "inertial" force is doing work. So maybe although work is being done, it is not necessarily being done by the man holding the ball.

 

[Dale]

Could it? How? Describe this reference frame to me.

Reference frame 1: Standard reference frame at rest wrt the surface of the earth. Origin at the ball at t=0, x north, y west, z up.

Position of the ball: r(t) = (0,0,0)
Displacement of the ball: d(t) = r(t)-r(0) = (0,0,0)
Velocity of the ball: v(t) = dr/dt = (0,0,0)
Acceleration of the ball: a(t) = dv/dt = (0,0,0)
Net force: f(t) = ma = (0,0,0)
KE of the ball: KE(t) = mv²/2 = 0
Work done on ball: f.d = (0,0,0).(0,0,0) = 0
Son wins

Reference frame 2: Free falling frame. Axes coincident with unprimed frame at t=0.

x' = x
y' = y
z' = gt²/2 + z

Position of the ball: r'(t) = (0,0,gt²/2)
Displacement of ball: d'(t) = r'(t)-r'(0) = (0,0,gt²/2)
Velocity of the ball: v'(t) = dr'/dt = (0,0,gt)
Acceleration of the ball: a'(t) = dv'/dt = (0,0,g)
Force on ball: f'(t) = ma' = (0,0,mg)
KE of the ball: KE'(t) = mv'²/2 = mg²t²/2
Work done on ball: f'.d' = (0,0,mg).(0,0,gt²/2) = mg²t²/2
Dad weasels out a draw

But of course, this is the beauty of taking the relativistic approach. Even faced with such a situation as coming dangerously close to violating the law of conservation of energy, you can still use the theory to "distort" time and make 2 seconds the same thing as 2 minutes and dodge the issue entirely...

Please note that I did not use special relativity above. Even with classical Newtonian physics energy is clearly frame variant and energy is clearly conserved in both cases. I think you misunderstand the idea of reference frames even in Galilean relativity.

 

[atyy]

Reference frame 2: Free falling frame. Axes coincident with unprimed frame at t=0.

x' = x
y' = y
z' = gt²/2 + z

Position of the ball: r'(t) = (0,0,gt²/2)
Velocity of the ball: v'(t) = dr'/dt = (0,0,gt)
Acceleration of the ball: a'(t) = dv'/dt = (0,0,g) -> net force, normal force not canceled out
KE of the ball: KE'(t) = mv'²/2 = mg²t²/2
Work done on ball: KE'(t) - KE'(0) = mg²t²/2 - 0 = mg²t²/2
Dad weasels out a draw

But the work is done not by dad (mum?) holding the ball. It is done by the "inertial" force of the accelerated frame.

 

[Renge Ishyo]

The greatness of relativity as a theory lies not in its power to contradict objective reality. Einstein was very careful when he formulated it to stipulate that it must agree completely with observed classical phenomena for it to be of any use. It is indeed useful when it is (correctly) applied in that it can account for extreme phenomena that lie beyond the boundries of classical mechanics (and indeed our common experiences) while still reducing down to the exact classical rules mathematically so that our observable facts which we can rely on are not betrayed. Using relativity to describe classical phenomena doesn't make a person's interpretation "more true" here on Earth anymore than using a lot of extra digits for pi makes a mathematicians calculation "more true." The approximations should agree to a very close degree in either case if the mathematician has done his job right. So should the classical physicist and the relativist.

That is not what is happening here. Here we have a clear cut question with a simple answer as far as classical physics is concerned. No, the man is not doing any work by holding up the ball. What is trying to be argued here is that in certain reference frames or at certain speeds, relativity says that classical physics is wrong and the man is doing work in such frames. So "O.K., maybe so." But at least as far as Einstein is concerned the two answers should have agreed. Furthermore, nobody has addressed my query that if work is being done on the ball in *any* reference frame, why it releases the same amount of energy classically when it is dropped regardless of how long you hold it there. But I suppose it doesn't matter, because I cannot say whether the relativistic concepts are incorrect because I can't test them or experience such frames of reference.

 

[Phrak]

...
Reference frame 2: Free falling frame. Axes coincident with unprimed frame at t=0.

x' = x
y' = y
z' = gt²/2 + z

Position of the ball: r'(t) = (0,0,gt²/2)
Displacement of ball: d'(t) = r'(t)-r'(0) = (0,0,gt²/2)
Velocity of the ball: v'(t) = dr'/dt = (0,0,gt)
Acceleration of the ball: a'(t) = dv'/dt = (0,0,g)
Force on ball: f'(t) = ma' = (0,0,mg)
KE of the ball: KE'(t) = mv'²/2 = mg²t²/2
Work done on ball: f'.d' = (0,0,mg).(0,0,gt²/2) = mg²t²/2
Dad weasels out a draw
...

Dale,

I think what you have done is to demonstrate that Newtonian mechanics is generally valid only in Galilean reference frame, haven't you?

 

[Renge Ishyo]

Actually, I think I have found a non-relativistic way for the father to weasel his way into a tie on this bet.

O.k., suppose a man is holding a ball static in a gravitational field. Further suppose that the ball is at a lower temperature than the man's hand. In such a situation net work *would* be performed by the man's hand on the ball at the molecular level even if both objects were held "perfectly still". At the molecular level, the hot vibrating molecules in the man's hand would do work on the colder molecules in the ball increasing their vibration. The work done on the ball can be verified experimentally as the temperature increase of the surface of the ball. At equilibrium no net work would be done, but so long as the temperatures were different you can have net work in this system. This isn't quite what the original bet inferred, but it wasn't specific enough to remove this interpretation either...

 

[Dale]

But the work is done not by dad (mum?) holding the ball. It is done by the "inertial" force of the accelerated frame.

Technically work is always done by the net force, but you could make a reasonable case for that view since the difference between the primed and unprimed frames (in terms of forces) is the inertial force. In general, inertial forces can do work in their reference frame and can even be associated with a potential.

The greatness of relativity as a theory lies not in its power to contradict objective reality. Einstein was very careful when he formulated it to stipulate that it must agree completely with observed classical phenomena for it to be of any use. It is indeed useful when it is (correctly) applied in that it can account for extreme phenomena that lie beyond the boundries of classical mechanics (and indeed our common experiences) while still reducing down to the exact classical rules mathematically so that our observable facts which we can rely on are not betrayed. Using relativity to describe classical phenomena doesn't make a person's interpretation "more true" here on Earth anymore than using a lot of extra digits for pi makes a mathematicians calculation "more true." The approximations should agree to a very close degree in either case if the mathematician has done his job right. So should the classical physicist and the relativist.

That is not what is happening here. Here we have a clear cut question with a simple answer as far as classical physics is concerned. No, the man is not doing any work by holding up the ball. What is trying to be argued here is that in certain reference frames or at certain speeds, relativity says that classical physics is wrong and the man is doing work in such frames. So "O.K., maybe so." But at least as far as Einstein is concerned the two answers should have agreed. Furthermore, nobody has addressed my query that if work is being done on the ball in *any* reference frame, why it releases the same amount of energy classically when it is dropped regardless of how long you hold it there. But I suppose it doesn't matter, because I cannot say whether the relativistic concepts are incorrect because I can't test them or experience such frames of reference.

I don't see the relevance of any of these comments since I didn't use any special relativity in my recent calculations. I even went out of my way to explicitly point out the fact that I didn't use any SR. I used strictly classical mechanics and implicitly assumed v<<c, so as you say, any relativistic corrections would be negligible.

What I showed is that even in classical physics energy and work are frame variant. This really has nothing to do with SR. This is purely classical mechanics (aka Galilean relativity) and you "experience such frames of reference" every day.

Btw, regarding your query, in the other reference frames the remainder of the energy goes into changing the energy of the earth.

I think what you have done is to demonstrate that Newtonian mechanics is generally valid only in Galilean reference frame, haven't you?

It is not demonstrated by what I did above, but yes you are correct. Newton's third law is violated in non-inertial reference frames since the inertial forces do not form 3rd-law pairs. But otherwise classical mechanics works fine in non-inertial reference frames and these analysis techniques are well understood.

 

[ValenceE]

Hello to all,

Earth’s reference frame is accelerating as it rotates around itself and the sun.

Anything that is on Earth’s ground or attached to it is accelerated accordingly.

Dad is holding a ball, sitting in a comfortable chair, itself sitting on the living room floor which, through the house’s construction, sits on Earth’s ground.

The ball that dad’s holding, all the way down to Earth’s surface, is accelerated accordingly.

If the Earth is doing work, keeping anything inert on it’s surface, then so is dad.


We can certainly use Earth’s reference frame as a free fall frame. I mean it’s big enough so some of us can get their thrills doing free fall jumps. So, no need to go too far in the explanations…


Regards,

VE