Does it take more energy to go downstairs than upstairs?

On a completely different note, Konrad, your approach to this problem reminds me of a story I heard a long time ago. (This’ll be another long post, but hey, I feel like it.)

It seems that there was a rich man who liked to bet on horse races. He loved the excitement of winning, but absolutely hated to lose. So he decided to invest some money and learn how to win more often. He sought the foremost experts in statistics, biology, and physics, and offered a $10,000,000 prize to the one who could offer him a foolproof system for choosing which horse to bet on.

Lots were drawn to determine the order in which the experts could make their attempts. Each was to be given one year to work on the problem.

The statistician came first. After a year of intense study and work, he came back to the rich man and said, “Sir, I have gathered and analyzed data on all the winning horses since records have been kept. I used the world’s fastest computers, and the most efficient algorithms. I found no consistent pattern which could predict the winner; I’m sorry, but there is no foolproof system.” The rich man thanked him and sent him on his way.

The biologist was next. After a year of intense study and work, he came back to the rich man and said, “Sir, I have measured and examined over three thousand racehorses. I checked their muscle structure, lung capacity, blood chemistry, bone density, and many other factors. But when I compared the winning horses to the losers, there was no factor that could tell them apart for certain. I’m sorry, but there is no foolproof system.” The rich man thanked him and sent him on his way.

The physicist’s turn was next. Just one day after he was given the task, he returned to the rich man’s house with an envelope in his hand and a huge grin on his face. “Sir, this was hardly a challenge,” he said. “I have your foolproof system.”

The rich man, astonished, paid the physicist his ten million dollars, and thanked him profusely. He tore open the envelope, took out the sheet of paper within, and began reading: “Assume a spherical horse travelling in vacuum…”
(Yes, there is a point to this story. Ideal conditions are fine for use in the classroom, but if you try to solve a Real World problem with them, your solutions will make no sense.)

Auraseer: Yes, I understand your point. In fact that’s a joke I often tell.(Except about mathematicians) But I clearly said I wasn’t talking about the real world. My point was to say that the question was answered wrong. It was to say that it was right for the wrong reasons.

What got me to post was that the person who answered the question (like others on this board) shrugged off the idea that you use energy to go downstairs like it was some crazy lunatic theory. Obviously they had no idea what they were talking about.

The whole argument here was not that I disagreed with them on the answer to the real world application, but that they were acting as if they knew what they were talking about, which they obviously did not.

I specifically laid out the terms, they called me crazy, I proved them wrong under the terms I laid out.

Konrad, I’ve gone back and re-read the Mailbag column, and I agree with you that it wasn’t answered well. The SDSTAFF writer did over-simplify it.

(Hey, can it be true? Is the arguing over?)

Is anyone still here? This thread has been inactive for a while now, apparently because everyone’s head is about to explode. But I just got here, so I hope it isn’t too late to try to she some light on the subject.

First of all, as many of you have perceived, there are two issues bound up in the question of the energy required to climb stairs versus the energy required to descend: a basic physics issue (of particular interest to Konrad), and a physiological issue that ultimately resolves the question. From a pure physical point of view, if I could design the human body any way I might wish, descending could cost more energy, less energy, or the same amount of energy as climbing. I could even get some back! It is the physiological mechanics of muscular contraction that ultimately decide the issue, not considerations of basic physics.

Okay, the physics first. “Work” is defined (leaving out calculus) as the product of force times the displacement produced by that force. When work is done, an equal amount of “energy” is transferred from one part of the system to another, often changing form as well. When you climb stairs, your muscles produce forces that act between your feet and the stair treads to raise your body. The force required is your weight (equal to mass m times gravity g); the displacement is the height h of the staircase; hence the total work, and the energy produced by your muscles is E=mgh. Yes, of course, there is also energy required for basal matabolism, static load bearing, gait irregularities, and so forth; but all these are also required to descend, so the energy required specifically to climb the stairs is just E=mgh.

Where does this energy come from and where does it go? I hope we all agree that it comes from ATP metabolized in muscle, and that it goes into gravitational potential energy. I hope we can also agree that gravity is a conservative field, so gravitational energy is completely available to do work. All you have to do is come back down the stairs and that energy is available. Slide down the bannister and the work of friction converts it into heat. The frictional force of the bannister against your rear, times the length of the bannister will always turn out to be mgh. If you just jump off the landing, then the gravitational force does work accelerating you and thereby converting gravitational potential energy to kinetic energy. By the time you get to the floor, your kinetic energy will be equal to mgh. I think we’re all together on this.

But suppose you lower yourself down by walking. First of all, the forces involved in lowering yourself are exactly the same as those involved in raising. It does not take any extra force to go up, or less force to go down (at constant speed). (This point is counterintuitive for many, but it’s true. We can discuss it separately, if necessary.) The distance is the same, too. So if the muscles are exerting the same forces, and moving through the same distance, then they do the same work, right? Wrong. This suggestion is a common error (and a trap physics professors like to lay for the unwary). The forces are the same, but the sign of the displacement is opposite: the direction of travel is reversed. The muscles do not produce mgh of energy in going down, they absorb mgh of energy in letting you down.

Where does this energy come from and where does it go? Well, it comes from the gravitational field. And it goes into the muscles. The mechanism by which the muscles absorb this energy is a physiological question, not a physics question. If we imagine a sort of ideal, frictionless human, as Konrad suggests, then the energy cannot be dissipated as heat. It must be stored somehow. Capture and storage of this energy is certainly not impractical. Electric cars do it all the time through a process called regenerative braking.

The “muscle” of an electric car is an electric motor. In climbing a hill, the motor(s) must do mgh of work converting electrical energy into gravitational potential. Coming back down, the car does work on the motors which run as generators converting the gravitational energy back into electrical energy which is returned to the batteries. The motors/generators don’t do any work in this process; work is done on them.

There’s no fundamental reason that muscles couldn’t work the same way. In such a case, it would not only take less energy to go down stairs, you would actually get your climbing energy (or part of it) back! In physics, we call such a process “reversible.”

Another possibility is that our idealized human would just use controlled friction inside the muscles to provide the lowering force. Ordinary cars do this all the time. That’s how brakes work. The car burns mgh worth of gas to get up the hill, and it converts mgh worth of energy into heat in the brakes on the way down. The frictional force times the distance the brake pads move adds up to mgh by the time you reach the bottom. In this case it costs you exactly zero energy to descend. All the energy is converted to heat where it is unavailable for use by the engine. In physics we call such a process “irreversible.” It costs you no gas (energy) to descend, even though the car is “lowering” itself down the hill. And you certainly don’t get any gas put back in the tank as a result!

There’s a third possibility. We know that muscles consume energy just bearing static loads. When a muscle is contracting isometrically just to hold you up, it is producing a force, but no displacement, so it is doing no work. But it still needs a supply of energy just to produce this force. An isometrically contracting muscle has zero efficiency. This action is analagous to holding your car on a hill by riding the clutch. The engine is doing work to produce a frictional force in the clutch to support the car on the hill, but you aren’t going anywhere so mgh=0. You could use this same technique to descend a hill in a car: point downhill, put 'er in reverse, juice the accelerator and ride the clutch to lower the car down the hill. (Don’t try this at home, folks…) If you use slow engine speed and lots of clutch, you won’t use much gas (energy); if you use fast engine speed and light clutch, you could use tons of gas–all the way up to the maximum power output of the car! Very inefficient! (If I have observed correctly, many carnival rides use clutches and engines in this way.)

So, depending on the physiology of muscles, descending a staircase could require a negative amount of energy (regeneration), it could require none at all (pure braking), or it could require any amount up to the maximum available (“clutch-riding”). But there’s no pure physics argument that can decide the issue.

So the question is, what’s the physiology of muscles? Well, we’ve already been treated to a lucid discussion of the contractile process. Fantastic as it sounds, muscle contraction really does involve a mechanical ratcheting process as myosin filaments crawl along adjacent actin filaments, thereby shortening the muscle. The mechanical bending of the myosins is powered by oxidation of ATP; that is, the energy used by the muscle is equivalent to the amount of ATP oxidized. I hope we’re in agreement that what we’re looking for is this energy, and not gravitational energy dissipated as heat due to irreversible processes.

Evidently, the process involved in extending a contracted muscle (that is, a muscle under load) is much like riding the clutch as you coast downhill, motor revving. There is friction as the actin filaments slide over each other and the myosin filaments between claw away trying to prevent slippage. (By the way, this process inevitably tears up myosin and actin filaments, resulting in soreness.) The only reason any more force is needed (and thus more ATP metabolized) to walk than to stand is to deal with accelerations of the limbs and body. So walking takes more energy than standing still–despite the fact that you are doing no work against gravity and a trivial amount against friction

Geezer: The sign means nothing. No human gets energy by walking down stairs. You can’t get simulate eating a meal by taking the elevator up a tall building and walking down. I know I said ideal human but that’s ridiculous. Ideal means it feels no friction, not that it’s an electric car.

Mommy! Mommy! Make the bad thread stop!

First, I owe everyone an apology. I’m a rookie, err… newbie…, and I made a rookie blunder in not realizing that there was a second page to this thread. Consequently I waxed a bit lengthy without having read all that came before. I sincerely apologize.

But still… I think the essential issue for Konrad and DSYoungEsq, is a pure physics matter, not specifically related to the (rather messy) stair-climbing example. Konrad, I brought out the electric car to illustrate precisely the point you make: human muscle tissue is an irreversible machine. You do get (abssorb) energy walking down stairs, but not in any useful form. It’s all plain old heat. But we can’t even talk about stair-climbing until we settle some basic physics. DSYoungEsq’s comment that it takes energy to stop a falling ball, and your comment that “the sign means nothing” describe the essence of a very interesting pure physics question. I’m not taking sides in a debate; I’d just like to see if we can understand the physics.

First, this significance of the sign. Consider:

A cube of steel sits on a smooth, horizontal surface with a non-zero, finite coefficient of friction. An agent (your finger, for instance) applies a horizontal force F in the +x direction and moves the block through a displacement s in the +x direction. We say that the finger did work on the block. The amount of work is U=Fs. Energy from your body was transferred to the block and finally to the molecules of the block and the table as heat.

Now, Newton’s third law says that if the finger exerts a force F on the block, the block exerts a reaction force F on the finger. This force acts through a displacement s. Does the block do work V=Fs on the finger? No. It doesn’t have any energy available to it to do any work. The block does work V=-Fs on the finger because the reaction force is in the -x direction, that is, opposite the direction of the displacement. The sign means that the (negative) work the block does on the finger is actually the work the finger does on the block. A positive value of work means the object to which the calculations are applied is doing work; a negative value means it is having work done on it. Work is the dot product of the force vector and the displacement vector. The sign counts. Otherwise, inert blocks of steel with no source of energy could do work on fingers.

So do you do work to catch a falling baseball? No; the baseball does work on you. Yes, your hand supplies a decelerating force; and yes, that force acts through a distance; and yes, the force times the distance is good ol’ mgh. But the force your hand exerts is the reaction force to the force exerted by the baseball on your hand. The baseball’s force is the one acting in the same direction as the displacement. Your hand supplies no energy in stopping the baseball; it must absorb mgh to stop the ball.

If you raise a barbell of mass m through a height h, your muscles must metabolize mgh worth of energy to do the required work. To lower the barbell, the problem is not to supply more energy, it is to get rid of the energy the barbell already has–that is, to absorb the energy. If you raise a barbell and then set it on a dashpot so that it gets lowered just as if you lowered it by hand, would the dashpot be supplying energy?

Konrad, are you and DSYoungEsq (and others) suggesting that it is physically necessary to supply energy to a machine that decelerates a moving mass or that lowers a mass through gravity, or am I barking up the wrong tree?

I’m really tired of this one, but:

-After you have walked upstairs, do you not have more potential energy (from being higher up) than when you started?

  • After you have walked downstairs, do you not have less potential energy (from being lower down) than when you started?

Don’t get tired Dex! There’s a nifty little physics paradox here that the original respondent’s spouse, and now Konrad and a few others have put their fingers on. It’s been around for a while, but it’s as interesting as ever. We just have to dig down to the basic question, separated from the details of the stair-climbing example and any semantic and communications issues.

Geezer: As far as the block and the finger goes, A doing work on B means to me that B is absorbing energy.

Yes, it does take energy to lower something down. How? Well look at it this way. Suppose instead of doing it slowly we let it speed up. Now we have something moving with a speed relative to us. What’s one way to stop it? Well we could launch another object at it of the same mass going the same speed. The two will collide and stop. (Without it doing any work on your arm. In fact your arm does work on the second object.)

So we have used energy to stop an object, and we have not gained any ATP from this. Of course all the energy goes into heat. (The amount of heat energy will be twice what you would absorb if you could somehow extract all the energy from the first object.)

If you are decreasing the speed of something relative to you, you can theoretically absorb the energy instead of using more energy to slow it down. That is what is meant by the object doing work on whatever is slowing it down. But that doesn’t mean you have to absorb it, a human doesn’t. When you are walking down stairs, gravity isn’t doing any work on you unless you are speeding up and therefore absorbing its energy. You are using equal but opposite energy instead.

I think Geezer’s point is that, if you were redesigning a human from the ground up, the muscles could be redesigned to work differently.

I also think that he has spent far too much time in Physics 001, and not nearly enough in the Real World. But that’s just IMAO.

Yeah, Aura, but if you were redesigning a human from the ground *down, would it be different?

Ah, AuraSeer, you have seen my aura truly–but, ummmm… you have the sign reversed. Long, long ago in a galaxy far, far away, I was a physics professor and I did spend quite a bit of time in Physics 101. I liked it, too, and was good at it. But, alas, I was incompetent at academic politics and had to go and make my way in the Real World ™. Nowadays I spend way too much time in the Real World building machines that raise and lower things, and not enough engaged in the elegance of pure physics discussing tricky concepts with bright young physics majors like our friend Konrad. You read me right, and I thank you for it.

Interestingly, the main point in my first post was that a pure basic physics argument could not answer the question at hand. You have to know how real muscles work to know if they need to expend energy (ie, burn fuel) in order to lower a mass through gravity. Konrad’s point is that, with suitable idealization, pure physics can give an answer–and that the answer is that it takes equal energy to raise and lower a mass. He’s now given us a lucid and elegant explanation of that assertion.

I’m very interested in this question because I keep seeing it pop up, usually in the form, “Did you know that, as strange as it sounds, it actually takes more energy to descend a staircase (mountain, hill, ladder, etc) than it does to climb it? Isn’t that cool?” Usually, however, the respondent cannot justify the assertion because he or she lacks the facility with physics. Now here’s Konrad, clearly bright and articulate, a physics major, making a somewhat more nuanced claim: it takes exactly the same energy to descend as to climb. I want to know how he draws this conclusion.

So… Konrad. If you’ll permit me, let’s parse your theory.

  1. | |As far as the block and the finger goes, A doing work on B means to me that B is absorbing energy.| |

Yes, it means the same to me, too. Your earliest few posts indicated that we agreed on this, but later material gave me pause. I hope you see my point about the sign. Note that at constant velocity, B is not speeding up, but it is still absorbing energy (and dissipating it as heat). Hold that thought.

  1. | |Yes, it does take energy to lower something down.| |

This is the nub of the discussion. My claim is that it can take energy, but it does not necessarily. It depends on the mechanism employed. These fall into three classes: regenerative (reversible processes required), friction (simplest irreversible process), or “clutch slipping” (energy required to regulate the amount of friction).

  1. | |How? Well look at it this way. Suppose instead of doing it slowly we let it speed up. Now we have something moving with a speed relative to us. What’s one way to stop it? Well we could launch another object at it of the same mass going the same speed. The two will collide and stop. (Without it doing any work on your arm. In fact your arm does work on the second object.)| |

Very nice! This description seems to me to be perfectly rigorous. It’s just that it isn’t general. It describes one method of the class of methods I’ve called “clutch-slipping.” Suppose instead of launching an interceptor of equal mass and speed, you launch one of much less mass–a bullet to stop a brick, as it were. With no bouncing, the interceptor must have the same momentum as the object you are trying to stop, so as mass goes down, speed goes up proportionally. But energy goes up as the square of the speed, so the lightweight interceptor needs to be supplied with much more energy to get up enough momentum to do its job. At the other extreme, suppose you use an interceptor that is much more massive than the object you are trying to stop–a brick to stop a spitball, as it were. Now to match momenta, the brick hardly needs any speed at all, and has near zero energy to bring to the party. This is “clutch slipping” at the limit of pure friction.

  1. | |So we have used energy to stop an object, and we have not gained any ATP from this. Of course all the energy goes into heat. (The amount of heat energy will be twice what you would absorb if you could somehow extract all the energy from the first object.)| |

Right. But perhaps a more rigorous statement would be that we have used momentum to stop an object. The energy used depends on the mass of the interceptor object chosen. By the way, nobody ever claimed that real muscles were reversible and you could gain ATP from descending under muscle power. If I gave this impression, I apologize for being unclear.

  1. | |If you are decreasing the speed of something relative to you, you can theoretically absorb the energy instead of using more energy to slow it down. That is what is meant by the object doing work on whatever is slowing it down.| |

Yes. Except that in your example of “using more energy to slow it down” you still end up absorbing the kinetic energy of the object being stopped. The object being stopped does work on the interceptor. Since you specified an inelastic collision, all that work goes straight to heat, which is absorbed by the matter of the two objects now in contact. I’d say that the energy of the object being stopped was absorbed by the object doing the stopping at the incidental expense of the original energy of the object doing the stopping.

  1. | |But that doesn’t mean you have to absorb it, a human doesn’t.| |

This I don’t understand. In your early posts you indicated that you do absorb it, but that it costs you energy to do so. In your discussion of paragraph 3 you tell what you mean. If something in the problem doesn’t absorb the energy of the object being stopped (or slowed, or lowered), where does the energy go? You say it goes into heat. I agree. Does that not count as being “absorbed” by the body being heated? Do we have a definition-of-terms problem here?

Or is the issue that the energy is absorbed (as heat) by the interceptor particles, but not by the arm that launched them. If so, I’d have to point out that in a muscle, the interceptor particles and the agency that launches them are in thermal contact.

Finally, it’s a non-sequitur to go from “doesn’t mean you have to” to “doesn’t.” To assert that the human doesn’t absorb energy because it doesn’t have to does not follow. You have to look at the mechanism of real muscle contraction to decide the case.

  1. | |When you are walking down stairs, gravity isn’t doing any work on you unless you are speeding up and therefore absorbing its energy.| |

I don’t see how this statement follows. If gravity makes you move, it is doing work on you no matter what your speed profile. Work represents the conversion of gravitational potential energy into kinetic energy if you just fall, or heat if you lower yourself gradually. Either way, you are absorbing energy under the terms of paragraph 1. (There’s also the third possibility of absorbing energy and storing it in springs or batteries or whatever, but we’ve agreed that that is an irrelevancy for muscles since they are irreversible machines.)

Furthermore, you very astutely used the abstraction that gradual lowering consists of an infinite number of infinitesimal drops and catches. So even in your example, gravitational potential is first converted to KE, and then reduced to heat by collision with your interceptor particle. Gravity (or more precisely, the gravitating bodies) are most certainly doing work on each other in raising and lowering processes.

  1. | |You are using equal but opposite energy instead.| |

Fundamentally, the energy used is more or less irrelevant. It is momentum that is doing the heavy lifting, so to speak. If the mass of the interceptor is arbitrary, the energy of the interceptors could be anything–and whatever it is, it all gets absorbed as heat during the collision.

Konrad, I think I understand

Geezer: I think I was a little unclear about the term “absorb”. In some cases I was using it to mean negate and in others to “absorb the energy back”.

But either way, since we know that the human body isn’t reabsorbing any of that energy as ATP and since we know how much energy has to be negated/absorbed and since we’re assuming there’s not much friction we can conclude that you have to do an equal amount of work. You can consider the work negative or whatever but we know it either has to be absorbed or negated and it ain’t being absorbed, so the person is really doing positive work.

Actually I just refuse to accept negative numbers in physics in general…

As far as the momentum vs. energy thing goes it doesn’t make a difference if there is no friction.

I’ve got another interesting problem for you: A chain of total length a is resting on a table with length b hanging off the end of the table. Neglecting friction find the speed of the chain just as its end falls of the table. Integration is verboten.

Auraseer: If you do know anyone who is redesigning the human ask them to make it perfectly spherical and of uniform density. It will makes physics problems much simpler. Then people can just roll down the stairs and it won’t take them any energy to do it.

Well, Konrad, my friend, I think this thread has run itself out. If you refuse to accept negative numbers in physics, then, well, you just aren’t practicing physics, you are practicing religion. That’s fine, but it can’t answer the question at hand, which is a physics question.

It’s too bad, because we’re so close… Consider your statement that

In physics (as opposed to religion or magic) energy doesn’t “negate” energy. (Heck, you’d have to have negative energy to do that.) Even in your example, you assert correctly that all the energy is turned to heat. Where does that heat (which is energy) go? It gets absorbed. It raises the temperature of muscle tissue. How does the energy get converted to heat? Friction–inelestic collisions specifically. The collisions you describe constitute a frictional mechanism internal to the muscles.

When a muscle lowers a weight, if the gravitational energy does not go into storage (e.g. making ATP), and if it doesn’t go into KE (which is what we mean by “lowering”), then it must go into heat. There’s no other choice. Sorry. Muscles absorb energy through friction, not magic. The friction arises between actin and myosin filaments mechanically rubbing against each other.

As far as the energy versus momentum thing goes, it does matter, just as I’ve described it. You’ve got enough physics to do the calculation. If you make the collisions elastic (ie, no friction as you now suggest) rather than inelastic (ie, with friction as you originally described), then the interceptors bounce off with varying speeds depending on their masses. In order to stop the moving object you still have to pick the interceptors so that their momentum change in the collision matches the momentum of the object you are trying to stop. Now you have the problem of stopping all those interceptors ricocheting back at you. Friction will work.

For your chain question, I get v=sqrt(g*(a^2-b^2)/a). (No calc, but I did use those pesky negative numbers.) If you want to discuss how I got this answer we should start a new thread or communicate privately, since the topic here is supposed to be muscle soreness.

Konrad, tell me you were kidding about the negative numbers. Otherwise I feel it my moral duty to warn you that if you refuse to accept negative numbers in physics, you’re just going to implode when you are asked to accept imaginary numbers in physics.

Geezer: Ok for the chain question email me at konrad@axess.com.

I have no problems working with imaginary numbers, I just refuse to accept them. What I mean is that there is no negative number you can’t make positive by shifting the origin. Using negatives is just a convenience. The only case where negatives do what they’re supposed to is things like matter/antimatter.

Ok now back to the stair problem. I agree that all the work goes into heat. As I see it, what’s happening is like 2 balls bouncing off each other in a perfectly elastic collision but being stuck together with a rubber band, they are still going as fast as they were before but now instead of being free the kinetic energy is heat. The stair atoms hit your foot atoms and heats them up while at the same time the average momentum of your body decreases. (Pretend for a moment that instead of walkng down the stairs normally you are stamping your feet so your legs go fast enough downwards that your torso stops moving down and then they hit the stair and stop themselves, it ends up being the same thing, it just makes it easier to put the motion into chunks.)

Does that make sense to you?

Well, you could always just lean over the staircase and go limp. Then you’ll practically not use any energy at all going down… But then again, it will be harder to stand up afterwards…

Amazing! There’s actually someone still following this thread. (Or maybe just laughing at us.) With that encouragement, ummm… what you say makes sense, Konrad, it just isn’t germane. A smooth descent does no work on the stair tread–the tread doesn’t move (remember your idealizing assumptions). Sure, you can stamp your feet going down and use up a lot of energy doing it, but you don’t have to. If you descend smoothly, all the energy goes into heat in the muscle–and the muscle need not supply any additional energy to allow that to happen.

I take your point about looking at this on the molecular level–but it doesn’t help your case. The process you describe takes place at the interface between the actin and myosin filaments in your muscle. As the filaments slide over each other their molecules bang into each other, exchanging energy and momentum. The energy goes into the oscillating chemical bonds and random molecular motions we call heat. The momentum transfer is force (F=dp/dt). That’s what supports your weight as you descend. No energy input required.

In order to prevail, you are going to have to explain why it is that a dashpot can lower a weight gradually with no energy input, but a muscle can’t. There’s no essential difference.

As for the chain… Energy is conserved so the initial potential energy is equal to the final PE+KE.

U1 = -b^2g/2 (Density is arbitrary and cancels.)
T1 = 0
U2 = -a^2
g/2
T2 = a*v^2/2

U1 + T1 = U2 + T2
Solve for v.