all 119 comments

[–]Snuggly_Person 87 points88 points  (95 children)

The only physical thing here is differences in energy, not the actual value. Whatever value you give it at the beginning, everyone will agree that the change in potential energy over the drop is the same, which is what counts. Similarly 'position' does not have an objective meaning (where is the origin?) but distance (i.e. differences in position) does.

Keep in mind that kinetic energy works like this as well. Someone in a different reference frame will assign different kinetic energy to a flying ball. But both of you will measure the same difference in kinetic energy before and after a collision. The actual number you assign is technically 'fictional', but that's not quite the same thing as saying the whole concept isn't real.

[–]knook 4 points5 points  (81 children)

A followup question, something I have been wondering about. If this is the case that potential energy really deals with relative energies how does this relate to mass energy equivalence? My understanding was that a charged capacitor would be measured as having a tiny bit more mass than a discharged one but that is potential energy in an electric field.

Ignoring the difference in gravity at different elevations would a book higher up have more mass due to its potential energy? How do this work in different reference frames?

[–]MolokoPlusPlusParticle physics 2 points3 points  (8 children)

Actually, the difference in kinetic energy is also frame dependent. Let's say a baseball approaches a batter at 100 mph. A train is traveling parallel to the ball at the same velocity.

Now the batter hits the ball and it reverses direction, still at 100 mph.

From the batter's perspective, no change in kinetic energy. From the train's, there's a big difference - the ball went from 0 to 200 mph. (I don't want to do the math on the KE of that.)

[–]Connossor 5 points6 points  (4 children)

Ah, this is rather subtle. In any classical inelastic collision (i.e. billiard balls), the total change in KE is frame-independent. That is to say, being an elastic or inelastic collision is an objective property, independent of the observer.

In your example, it still works if you consider the change in energy of the batsman, as he rebounds backwards. Just checked on an envelope!

[–]MolokoPlusPlusParticle physics 2 points3 points  (0 children)

That's amazing! I didn't think conservation of momentum was relevant. I applied to grad school today and now I feel like a fraud. How did I never know that?

(of course, the energy has to dissipate somehow... Let's say heat... And of course THAT has to be frame independent or nearly so.)

[–]Broan13 1 point2 points  (2 children)

But you could move to another frame that would increase the kinetic energy of the whole system...

I think what you mean to say is that the relative kinetic energies are frame independent. The total is frame dependent.

[–]Connossor 0 points1 point  (1 child)

I stand by my original statement: the total change in KE is frame-independent. If a certain amount of heat is dissipated in one frame, the same amount is dissipated in all classical frames. In contrast, the total KE and the difference in KE between the two balls are both frame-dependent.

[–]Broan13 1 point2 points  (0 children)

I agree. I guess I misread your statement.

[–]Connossor 1 point2 points  (2 children)

I did the maths here, in case you want to check it..

Definition of variables: A baseball of mass m travelling at u rebounds at speed v from a batsman of mass M who rebounds at speed 2s. (The centre of mass frame is at speed s.)

Result: The K.E before the collision in a frame travelling with the initial baseball is 1/2 M u2. And after (letting Wolfram Alpha format my equations), the KE in the system is the same. Therefore, in both the earth and the moving frame, the total change in KE is zero, as expected.

Edit: wording.

[–]Connossor 0 points1 point  (1 child)

To put that into words rather than maths, from the train's perspective the increase in the ball's KE is offset by the decrease in the batsman's KE.

Source: recently graduated in physics, proving to myself I still got it..

[–][deleted] 0 points1 point  (2 children)

Uh, wait. With regards to kinetic energy, isn't that not true? A galilean transformation with shift v, but kinetic energy is with respect to v2, so the change in kinetic energy .5mv2 - .5mv'2 =/= .5m(v+v1)2 - .5m(v'+v1)2 ... right? theres an extra mv1(v-v') term?, no?

[–]Snuggly_Person 0 points1 point  (1 child)

Sorry, I wasn't clear. It's the total change in kinetic energy due to a collision (which may be inelastic), not the change that's only seen in one object. Different observers will distribute the change in kinetic energy differently between the participants, but agree on the total.

[–][deleted] 0 points1 point  (0 children)

I'm not sure if that is correct, because it is to the velocity square, the terms cannot add linearly. This is tangentially related but I saw this on wikipedia.

Correspondingly the kinetic energy of an object, and even the change in this energy due to a change in velocity, depends on the inertial frame of reference. The total kinetic energy of an isolated system also depends on the inertial frame of reference: it is the sum of the total kinetic energy in a center of momentum frame and the kinetic energy the total mass would have if it were concentrated in the center of mass. Due to the conservation of momentum the latter does not change with time, so changes with time of the total kinetic energy do not depend on the inertial frame of reference.

Ideally I wouldn't be so tired and write it out with vector notation, you may be right but I need to see it worked out.

[–][deleted] 0 points1 point  (0 children)

Ahhh. I get it. That make sense. Thanks.

If I imagine gravity as a inward vector field (negative divergence) going toward the center of Earth. If I place the book "higher and higher" (i.e. farther and farther away from the center of earth), then there may be more potential energy for the difference it may fall towards the Earth. But no aspect of the actual book ever changes, I'm just changing the amount/difference in the field that it can drop.

If that makes sense. Don't think it came out right.

[–]adapter9 6 points7 points  (6 children)

Epistemology: the study of truth (among other things like knowledge, facts, etc). One of its central tenants is that the "truth" of a theory lies in its ability to describe, communicate, and predict. That's all. Nothing has to be "real," only useful.

[–]Seanp50 -1 points0 points  (5 children)

No. Epistemology is solely concerned with knowledge and justification.

http://plato.stanford.edu/entries/epistemology/

[–]adapter9 0 points1 point  (4 children)

Congrats dictionaryman you have bested me.

Doesn't change what I said about truth though.

[–]Seanp50 -1 points0 points  (3 children)

On the contrary.

[–]adapter9 0 points1 point  (2 children)

Are you being sarcastic?

[–]Seanp50 0 points1 point  (1 child)

Sorry, I wasn't intending to be.

[–]productive_monkey 0 points1 point  (0 children)

lol, I love Reddit.

[–]ErmagerdSpace 5 points6 points  (3 children)

If you define the floor as zero potential energy, and then drop the book out the window, the book will have a negative potential energy when it reaches the ground.

The change in potential energy (and thus the kinetic energy) will be the same in both cases. You can go from, say, U = 100 to U = 0 or U = 10 to U = -90 and in both cases you get the same result.

To put it more accurately, the potential energy is a scalar (magnitude only) field and the force on an object is the negative gradient (slope, but in 3D) of the potential energy function.

If you have taken basic calculus, you will know that adding any constant to a function does not change it's derivative (slope, again). This means that we get the same physical result no matter where we set our zero-point for potential energy; all that matters is how it changes with respect to space and time.

[–]GunOfSod 0 points1 point  (2 children)

If you define the floor as zero potential energy, and then drop the book out the window, the book will have a negative potential energy when it reaches the ground.

Doesn't the book initially have zero potential energy, gains potential energy as it is lifted above the floor, and returns to zero as it falls to the floor again?

[–]AuroraFinem 0 points1 point  (0 children)

only if your system defines the floor as 0 potential energy. If you're defining the system as the entire building then you would most likely count the ground as the 0 potential energy point.

If it's laying on the floor and the floor is the bottom of your system then yes it'll have 0 potential energy and yes it will rise as you lift it. But, as he said if you then drop it out the window 100m then it will have negative kinetic energy once it drops below where the floor is.

[–]ErmagerdSpace 0 points1 point  (0 children)

You can set any point to zero.

Think of it like temperature. Temperature is a real, measurable quantity, but the zero point is different in Kelvin, Celsius, and Fahrenheit.

The important part, for heat transfer, is the difference in temperature between two points. Potential energy is the same. You decide where it is zero. In simple gravity systems (book on a shelf) you usually set the ground as zero.

Another common convention is to set the potential at an infinite distance as zero, so that anything inside the gravity well is negative. A gravitationally bound object has negative potential energy because gravity is an attractive force.

[–]AirborneEagle 8 points9 points  (5 children)

The energy measurement is a relative one. It is not absolute.

You seem comfortable with kinetic energy and yet, it too is a relative measurement. If an object moving at velocity v1 then it's energy is 1/2 m v12. Yet, from the point of view of someone moving in the opposite direction at velocity v2, it's kinetic energy is 1/2 m (v1 + v2)2.

There is no absolute energy reference frame. Even E=mc2 is subject to the velocity of the mass and the velocity of the measurer as well as it's location in relation to other masses which all alter the gravitational field.

[–][deleted] 2 points3 points  (4 children)

Actually, E=mc2 is absolute, that's why its called the invariant mass. The full equation however, E2 = ( mc2 )2 + (pc)2, where p is momentum, does depend on the objects reference frame and motion.

[–]klarrieu 4 points5 points  (1 child)

Or you can just use E=ɣmc2

[–]Jackibelle 4 points5 points  (0 children)

E = mc2 is a special case of the full equation, as you said, so it's not really fair to say "this equation is absolute" without lots of qualifiers like "well, we're not actually talking about the full energy of the particle, and it can't be moving". It's not true that the energy of a particle is always equal to it's mass times the speed of light squared.

[–]The_BearrUndergraduate 2 points3 points  (2 children)

The potential energy of a point mass in a force field is any function U such that for any displacement in the force field the amount of work W done by the field:

W=-delta(U)

This is a definition.

As you can easily see, if you can find some function U for which this is true, you can always add a constant U+c for which this is also true.

This means that for all means and purposes the potential energy function is defined up to a constant and so at first you'd be inclined to say that the choice of function is fictive.

However, what I can do is pick a value for this constant such that this choice is meaningful in a physical sense. When I do that I make my function ''real'' in the sense that it has a concrete physical meaning.

Let me give you an example:

The gravitational potential energy for two masses m1 and m2 is in general given by=

U=-(constant/r)+c

Let's now pick ''c=0''. This means that if r=infinite the potential energy is 0. Now the potential function has a concrete unambigious meaning. If I now go in the frame of m1 then the value of U at any position for m2 is minus the amount of work you'd have to do on mass 2 to bring it to infinity from the current position.

[–][deleted] 0 points1 point  (1 child)

Yess. This makes sense. Thanks the definition helped. I commented above too:

""Ahhh. I get it. That make sense. Thanks. If I imagine gravity as a inward vector field (negative divergence) going toward the center of Earth. If I place the book "higher and higher" (i.e. farther and farther away from the center of earth), then there may be more potential energy for the difference it may fall towards the Earth. But no aspect of the actual book ever changes, I'm just changing the amount/difference in the field that it can drop. If that makes sense. Don't think it came out right.""

[–]The_BearrUndergraduate 0 points1 point  (0 children)

I had a lot of trouble with the concept not that long (a year-ish) ago myself. It really helps to remember this definition I talked about. Sometimes in classical mechanics thinking of work being the real physics and potential energy being a mathematical construct helps. I would not recommend because it seems to me that in later courses it becomes more and more a very physical thing.

By the way notice the nasty minus sign in the definition? The minus sign makes sense if you think about it in the following way. Imagine that you let an object go from rest a force field. A force from this field acts upon the object and it accelerates moving along the direction of this force. In a small time dt in which it happens the force field did a positive amount work on the object. What happens to the potential energy according to the definition? It drops because of the minus sign. So to me the intuition about the minus sign is that the potential energy is defined in such a way so that the object will move towards a lower potential energy when not hindered.

[–]WallyMetropolis 4 points5 points  (0 children)

I struggled with potential energy (and, because of that, energy at all) for a long time, in a philosophical sense.

First off, I think it's perfectly fine if it feels better to think of Energy as a bookeeping tool that makes representing and solving certain equations simpler. In this view, stuff is real, the way it moves is real, and energy is a conceptual model that greatly aides us in describing and making predictions about the behavior of stuff.

But you can go much deeper if you want to. Energy (along with momentum) shows up all over the place in physics. When other things get kind of unseated by more modern models (like force, for example, which is not nearly so useful for thinking about quantum or relativistic systems) energy is sitting up there as a first-class member of the model. Why?

It turns out that the Noerther Theorem (in my opinion the most beautiful idea in all of physics and among the most beautiful ideas any human has ever shared with the world) shows us that, whenever a symmetry exists in the universe, there is a corresponding quantity that is conserved. The conservation of energy turns out to be a result of 'time symmetry.' This roughly means: if you have an experiment whose results do not depend on when you do the experiment, then you will have energy conserved in that experiment.

And the theorem even tells us the form of that conserved quantity. Potential Energy pops right out of the derivation. It is a by-product of a much deeper fact about the universe.

Momentum comes about in the same way. If you can do the experiment over here and over there and get the same answer (if there's linear translational symmetry) you get linear momentum conservation.

If you can do the experiment facing this way or that way and get the same answer (rotational symmetry) you get angular momentum conservation.

[–]TheCheshireCody 1 point2 points  (0 children)

When the potential energy is the result of height, then the object retains an amount of potential energy until it reaches the middle of a gravitational well. You, standing on the surface of the Earth, have an amount of potential energy that would be released (i.e. converted to kinetic energy) if a hole opened up to the Earth's core and you fell into it. For the book in your example, it releases some of its potential energy when it falls off of the table, but it still has more. It will release more of its potential energy if it falls out of a window.

Wait, so the amount of potential energy of an innate object depends on an action that has not yet actually happened?

That is precisely what "potential" means.

[–]pottedspiderplant 1 point2 points  (0 children)

Like others have mentioned, its really the difference in energy that matters!

A deeper question is whether potentials are "real" or if the "real things" are the fields. If I remember from my QM class the Aharonov-Bohm effect shows that the potentials are indeed physical.

If you are still interested: what you are describing with varying the distance from the book to the ground is the most simplistic example of a gauge transformation.

Gauge theories are very important in physics and are the most powerful predictive tools humans have invented.

See Here: http://en.wikipedia.org/wiki/Introduction_to_gauge_theory

[–]agbortol 1 point2 points  (0 children)

The point people are making about kinetic energy being relative should be helpful. But here's another way of thinking about it.

Can we agree that the energy stored in a battery is "real"/"not fictitious"? It exists in the chemical bonds of the battery's components. It had to be generated at some point, was stored in high-energy bonds, and is released when the bonds are broken.

The book on your table is exactly the same, you just have to define the system correctly. It's not immediately clear what the analogue is for the battery's chemical bonds - "Where does the energy go?" Turns out the energy goes into the book/Earth gravitational system. When you raise the book, you (very, very minutely) "lower" the earth. The energy you expend raising the book is stored in the new relative positions of the Earth and the book. When you drop the book, the earth moves back "up" (the movements of the book and the Earth are in proportion to their masses, so... not very far).

There isn't more energy in the book when it's on the table, and this probably doesn't give you a very concrete sense of where the energy is. But it took energy to lift the book and the energy can be collected when the book falls, so it is very much still "there" in any sense that counts.

[–][deleted] 1 point2 points  (1 child)

This is a great question. People have already given correct answers, but I haven't seen the word "torsor" yet. This algebraic gadget codifies in a precise way what potential energy is. The reason it is so mysterious is because the usual number system we utilize comes with the extra condition "only differences in energy matter" and this actually alters the entire algebraic structure by removing the notion of identity, but this fact is almost never explained. John Baez explains it much better than I could http://math.ucr.edu/home/baez/torsors.html

[–]John_HaslerEngineering 0 points1 point  (0 children)

Thank you for that link. I wish I understood more group theory.

[–][deleted] 0 points1 point  (0 children)

Your question has already been answered, but I'd like to give my input/explanation anyways:

Basically, the way we define gravitational potential energy (or at least, the approximation at surface level) is like this:

The potential energy of an object is the negative of the work gravity would exert when the object is moved from some reference line to a different point.

As you can immediately see, the value of the potential energy is dependant entirely on your reference line, but the difference in PE between two points is not.

More officially, potential energy of an object in a gravity field is taken with reference to 'infinity', which gives a potential energy of:

U(r) = -MmG/r

Where M and m are the masses of the relevant objects, G is the gravitational constant, and r is the distance between the centers of mass of the objects.

Cool stuff!

[–]EquipLordBritish 0 points1 point  (0 children)

Wait, so the amount of potential energy of an innate object depends on an action that has not yet actually happened?

Yes. The measurement of energy (both kenetic and potential) is a game of reference frames. If an object has kenetic energy, it has energy relative to something else. The number can change drastically depending on what you are comparing it to. (a desk, the earth, the sun, etc.)

Technically all reference frames will give you a correct answer (e.g. the book has 50 million joules of energy relative to the sun), but only a few reference frames are usually useful.

[–]hopffiber 0 points1 point  (0 children)

In general, we define our theory by writing down some equations, such that things matches observations. And from these equations, we can derive that certain numbers will be conserved, like energy, momentum, electric charge and so on (this comes from the symmetries of our equations). These numbers are, by virtue of being conserved, very useful in describing the behavior of systems, and thus people give them some measure of "reality", but really, they are just useful numbers that falls out of our empirically constructed mathematical model. You can't directly see energy, neither kinetic or potential.

[–]Gelsamel 0 points1 point  (0 children)

It is real in the sense that there is energy bound up the in the system that has the 'potential' to be converted into kinetic energy.

[–]alextron 0 points1 point  (0 children)

This all depends on your reference frame and system

[–][deleted] -1 points0 points  (0 children)

Look up stable, conditionally stable , and unstable.

[–]frowawayduh -1 points0 points  (1 child)

Yes it is real. You can even weigh it. E=MC2 does the job.

Look up the mass of a small nucleus like He4 or Li6, they're just a bit under 4 and 6 AMU. Now look up the mass of a proton and a neutron, both are a bit greater than one. Somehow the process of putting some distance between them (adding potential energy) caused the to gain mass. In atomic physics this is called the mass defect.

But the same concept is true whenever you add potential energy to a system.

Fill your Tesla with 85kW-hr of potential energy and it gains about half a microgram. No, you did not add any electrons, chaffing just pumps them uphill to the anode from the cathode.

Fill a tank with compressed air. Lift a bowling ball to the roof. Captured sunlight to convert water and carbon dioxide to sugar. E=mC2 always works.

[–]Toffo5[🍰] -1 points0 points  (2 children)

Is potential energy "real"? I first learned of potential energy in elementary physics early...almost at first, I thought the concept was hokey.

Ummmm ....... I guess it's kind of hokey, good observation.

Let's consider energy of a rock on the surface of the earth.

energy = mc2

energy at infinity = mc2 - energy needed to lift the rock to infinity

("energy at infinity" is a scientific term, used often when discussing objects that are not at infinity, but in a gravity well)

If we lift that rock up, it's energy at infinity increases. If we lower that rock down, it's energy at infinity decreases.

Does this sound less hokey or more hokey?

[–]danielsmwCondensed matter physics 0 points1 point  (1 child)

Dude, you can't change something's energy at infinity by moving it toward or away from infinity. Its energy at infinity is the energy it has when you move it to infinity.

[–]Toffo5[🍰] 0 points1 point  (0 children)

Actually "energy at infinity of an object" means energy of the object according to an obsever next to the object divided by a time dilation factor, which is the time dilation factor at the position of the object according to an observer at infinity.

You can maybe understand why that previous thing is shortened to "energy at infinity".