Energy, once again weird unclear thing

[Mad_Eye]

hi.. once again my thoughts bumped into weird energy question, yet, very basic...

so say there is car, which mass is 1kg (yes, very realistic indeed), accelerating from 0 to 2 m/s.
to viewer on the ground the energy at start was 0 and in the end 2J, the work done by the engine to the car is 2J.
but now, and I'm quite sure most of you get the problem as soon as i say the next sentence, say I'm walking against the car direction in 1m/s speed. so for me, the initial speed is 1m/s and the final speed is 3m/s, as you can calc yourself, for me the work the engine did is now 4J.

we can get the same results if we use the W=F*S definition.

am i missing something? or does work depends on the viewer?
if so, does it not related to very practical things, such as the amount of fuel was burned and so on?


thank you very much indeed :D

 

[Dale]

does work depends on the viewer?

Yes, energy is frame variant or coordinate dependent.

if so, does it not related to very practical things, such as the amount of fuel was burned and so on?

It does relate to the amount of fuel burned in any given frame. Energy is frame variant, but it is also conserved.

Go ahead and work out the full problem in both frames taking into account the conservation of momentum and the KE of the earth. For this example it is ok to use any arbitrary large mass for the Earth and to consider linear momentum rather than angular momentum.

 

[Mad_Eye]

Yes, energy is frame variant or coordinate dependent.

i did NOT ask about energy, but about work. i know obviously that energy is frame variant.. but work as well?

It does relate to the amount of fuel burned in any given frame. Energy is frame variant, but it is also conserved.

Go ahead and work out the full problem in both frames taking into account the conservation of momentum and the KE of the earth. For this example it is ok to use any arbitrary large mass for the Earth and to consider linear momentum rather than angular momentum.

i really have no idea what the Earth movement, mass, and even momentum, have to do with that...
and what do you mean by "work out the full problem"? what exactly should i do?

 

[Dale]

i did NOT ask about energy, but about work. i know obviously that energy is frame variant.. but work as well?

Yes.

i really have no idea what the Earth movement, mass, and even momentum, have to do with that...
and what do you mean by "work out the full problem"? what exactly should i do?

Assume some large mass of the earth. Start with the Earth and car initially at rest and use conservation of momentum to determine the the initial and final velocites of the Earth and the car. Use those to determine the change in the energy of the system. Then do the same except starting with the Earth and the car initially at 1 m/s.

 

[russ_watters]

A couple of things not noted yet:

we can get the same results if we use the W=F*S definition.

Work isn't force times speed, it is force times distance.

as soon as i say the next sentence, say I'm walking against the car direction in 1m/s speed. so for me, the initial speed is 1m/s and the final speed is 3m/s, as you can calc yourself, for me the work the engine did is now 4J.

am i missing something? or does work depends on the viewer?
if so, does it not related to very practical things, such as the amount of fuel was burned and so on?

Yes, what you are missing is that the car isn't doing any work against you, so that frame of reference doesn't relate to how much energy the engine expended. The car is sitting on the ground, so the work done by the engine can only be calculated in that frame.

 

[cesiumfrog]

Work isn't force times speed, it is force times distance.

"Speed"? Doesn't s conventionally stand for displacement?

Yes, what you are missing is that the car isn't doing any work against you, so that frame of reference doesn't relate to how much energy the engine expended. The car is sitting on the ground, so the work done by the engine can only be calculated in that frame.

What? Only?
"That frame" isn't even inertial..

 

[Dale]

The car is sitting on the ground, so the work done by the engine can only be calculated in that frame.

The work done can be calculated in any frame. You will get different amounts of work done, but as long as you consider the entire system you will get conservation of energy. Here is a quick example of the kind I was encouraging the OP to work out:

Let's say that we start with the 1 kg car and a 1000 kg Earth both initially at rest and the car accelerates to 2 m/s. By conservation of momentum the Earth will accelerate to -0.002 m/s. The change in the KE of the car is 2 J and the change in the KE of the Earth is 0.002 J, so the engine did 2.002 J work which is the change in the total KE of the system.

Now suppose both the car and the Earth are initially traveling at 1 m/s and the car accelerates to 3 m/s. By conservation of momentum the Earth will decelerate to 0.998 m/s. The change in the KE of the car is 4 J and the change in the KE of the Earth is -1.998 J, so the engine again did 2.002 J work which is the change in the total KE of the system.

So you can calculate the work done by the engine in any frame, you just have to be careful and not neglect important parts of the system.

 

[russ_watters]

"Speed"? Doesn't s conventionally stand for displacement?

Ehh, that's the engineer in me... didn't realize that displacement is s. But from the OP it looks like he's multiplying force times speed: he doesn't say what his force is, but when he talks about the first case having a speed of 2 m/s and 2J of KE and the second case being 4m/s and 4J, that's what it looks like to me. This can't be since KE is a square function of speed.

What? Only?
"That frame" isn't even inertial..

The work done can be calculated in any frame. You will get different amounts of work done, but as long as you consider the entire system you will get conservation of energy. Here is a quick example of the kind I was encouraging the OP to work out... Now suppose both the car and the Earth are initially traveling at 1 m/s and the car accelerates to 3 m/s. By conservation of momentum the Earth will decelerate to 1.998 m/s. The change in the KE of the car is 4 J and the change in the KE of the Earth is -1.998 J,

Before I get to a full response, isn't the Earth's new speed 0.998 m/s?

[edit]That little issue aside, *I don't think it is important to consider the change in Earth's speed since it
earth is so much larger that the Earth's change in energy is near zero. In either case, yes, you can calculate work in any frame, but at the end of the problem, you're still subtacting one from the other to find the work done by the car against the ground. You're just converting frames at the end of the the problem instead of selecting the relevant frame at the start of the problem.

[edit2] *Nevermind, I figured out what you did. I didn't realize the Earth's energy change would be significant for Earth (that's me not being a physicist...). Nevertheless, at the end you are subtracting the work done on Earth as seen by the person from the Earth done on the car as seen by the person, which gives you the answer in a car vs Earth frame. You're still just converting frames at the end of the problem instead of selecting that frame from the beginning, aren't you?

 

[Dale]

isn't the Earth's new speed 0.998 m/s?

Oops, quite right. I have it correct on my notes and copied it down wrong.

[edit]That little issue aside, I don't think it is important to consider the change in Earth's speed since it Earth is so much larger that the Earth's change in energy is near zero.

Only in the frame where the Earth is at rest. In other frames the Earth's change in energy is significant even if the change in velocity is not. In this example the change in the Earth's energy is approximately equal to the amount of work done by the engine, hardly negligible.

 

[russ_watters]

See late second edit...I had to do the problem again with the Earth's mass at a million kg to prove that second part to myself.

 

[Mad_Eye]

very interesting!
but didn't you tell me work is frame variant?
my common sense says it is not, and so you are showing yourself in this example..

so did i not understand you correctly, or is there still cases in which the work done by a certain force would be different in different frames?

thanks again :)

btw, i think it is better to show it with parameters..

 

[russ_watters]

I thin the disagreement DaleSpam and I were having was largely semantec in nature. The end result of the exercise is that Dale calculated the work differently, injecting different reference frames into one of the calculations, and came out with the same answer as if he only used the one frame. So no, work done does not change when viewed from a different reference frame.

 

[Dale]

Work done does change when viewed from a different reference frame. My point above is that energy is still conserved, even though the work is different in different reference frames. Energy is a frame-variant but conserved quantity, you just need to pay attention to the whole system.

Although my above example does not show it since energy is the timelike component of the four-momentum it "dilates" the same way that time does when you start talking about relativistic velocities.

 

[DaTario]

i did NOT ask about energy, but about work. i know obviously that energy is frame variant.. but work as well?



i really have no idea what the Earth movement, mass, and even momentum, have to do with that...
and what do you mean by "work out the full problem"? what exactly should i do?

very interesting!
but didn't you tell me work is frame variant?
my common sense says it is not, and so you are showing yourself in this example..

so did i not understand you correctly, or is there still cases in which the work done by a certain force would be different in different frames?

thanks again :)

btw, i think it is better to show it with parameters..

You should approach this problem thinking acceleration is not frame invariant (in changing between inertial frames) ,mass, displacement and angles also not, so the product of force, displacement and cosine of the angle also won't be.

Relativity put far from this discussion, obviously...

Best wishes,

DaTario

 

[Dale]

but didn't you tell me work is frame variant?
my common sense says it is not, and so you are showing yourself in this example..
...
btw, i think it is better to show it with parameters..

Me too, I think that it is better showing general relationships. Since you prefer parameters then it is much easier:

The four-momentum is given by:
(E/c,p)
Which transforms as a four-vector.
(E'/c,p')=L.(E/c,p)

Therefore, like all four-vectors, the components are frame variant.

 

[Mad_Eye]

so can you give an example in which the work is different between one frame and another?

and i have no idea what four momentum is :P

 

[Dale]

so can you give an example in which the work is different between one frame and another?

I already did, and so did you in your OP.

The change in the KE of the car is 2 J ...
The change in the KE of the car is 4 J

The work done on the car is different in each frame.

Another simple example:
A stewardess on a jet traveling at 300 m/s ground speed holds a cart in place next to row 14 using 10 N of force for 1 s, how much work did she do on the cart in the plane's frame and in the ground's frame?
in the plane's frame: w = f.d = 10 N 0 m = 0 J
in the ground frame: w = f.d = 10 N 300 m = 3 kJ

Again, work is different in each frame, but energy would still be conserved.

 

[DrZoidberg]

Of course the amount of fuel burnt will be the same in each reference frame. Fuel consists of atoms. Even relativistic effects can not change the number of atoms. They can change their size. They can change the speed with which the electrons move around the nucleus etc. but not their number.
In your original example the amount of work done by the car engine will be the same in both reference frames. When you move towards the car with 1m/s it's kinetic energy is higher. However in that reference frame the ground is also moving with 1m/s. Therefore the ground will do part of the job of accelerating the car. The engine doesn't need to deliver the full 4J but only 2J.
If you look instead at a rocket the situation is similar. The amount of energy delivered by it's engine is the same in each reference frame. However the energy is distributed differently. Part of the energy goes into the rocket and part goes into the burnt fuel that is shot out the end. So if you change the reference frame and the energy put into the rocket e.g. increases then the energy put into the burnt fuel will decrease or vice versa.

 

[Dale]

In your original example the amount of work done by the car engine will be the same in both reference frames. When you move towards the car with 1m/s it's kinetic energy is higher. However in that reference frame the ground is also moving with 1m/s. Therefore the ground will do part of the job of accelerating the car. The engine doesn't need to deliver the full 4J but only 2J.

Exactly. Well said.

 

[Mad_Eye]

I already did, and so did you in your OP.
The work done on the car is different in each frame.

Another simple example:
A stewardess on a jet traveling at 300 m/s ground speed holds a cart in place next to row 14 using 10 N of force for 1 s, how much work did she do on the cart in the plane's frame and in the ground's frame?
in the plane's frame: w = f.d = 10 N 0 m = 0 J
in the ground frame: w = f.d = 10 N 300 m = 3 kJ

Again, work is different in each frame, but energy would still be conserved.

hmm, I'm not talking about work done on a body, but work a force does..

is the work force does, frame-invariant?

 

[DaTario]

Work is integral of F dot ds.

Non relativistically speaking:
mass is frame invariant
displacement is frame invariant
acceleration is frame invariant
angles are frame invariant

So work of a force is frame invariant


Best wishes

DaTario

 

[Dale]

hmm, I'm not talking about work done on a body, but work a force does..

is the work force does, frame-invariant?

No. The distance is frame variant so the work done is also.

 

[Dale]

Work is integral of F dot ds.

Non relativistically speaking:
mass is frame invariant
displacement is frame invariant
acceleration is frame invariant
angles are frame invariant

So work of a force is frame invariant

No, displacement is frame variant. See the airplane example above.

 

[DaTario]

No, displacement is frame variant. See the airplane example above.

Sorry. Let me insist on this. In classical physics displacement is frame invariant.


Best wishes

DaTario

 

[Dale]

In classical physics displacement is frame invariant.

No, it is not.

Given the Galilean transform from classical physics:
t' = t
x' = v t + x
y' = y
z' = z

Then:
Δx' = x1' - x0' = (v t1 + x1) - (v t0 + x0) = v (t1-t0) + (x1-x0) = v Δt + Δx ≠ Δx

Therefore, displacement is not frame invariant.

 

[DaTario]

Dale Span, You are correct, I was thinking of spatial intervals (distances), which are different from displacements.

Best wishes

DaTario

 

[Dale]

I was thinking of spatial intervals (distances), which are different from displacements.

This is correct, spatial intervals have $\delta t = 0$ (unlike displacements in the formula for work) so they are frame invariant under the Galilean transform as shown above.

 

[Mad_Eye]

sorry, but I'm still not quite understand it...
in classical mechanics ONLY,
is work frame-invariant, or not? what did you agree then?
if the work a force does, is frame-variant, can someone please give me an example?
because as far as i understand, in my example the engine is doing the same amount of work in both frames (2.002J)

and why is the work 2.002J if you are calculating it via F*S?
(in the second frame, the displacement is bigger, and the force should be the same)

thanks.

p.s.
it's now came to me that the force moving the car, is the friction the Earth is exerts, and the force moving the Earth back, is the friction the car ,exerts it's actually not the same force. is it a mistake to sum both work up and call it "the work the engine does"?

p.s.2
i know it has nothing to do with this post, but does anybody know any good physics book, that explains things deeply, in such level (or more of course); giving examples like this and solve them etc.?

 

[Dale]

I have already given you several examples worked out in detail and I have already stated in the simplest and clearest possible language that work is frame variant. What more can I possibly do?

I believe that you need to take my advice from post #2 and simply sit down and do the work yourself. My doing the work for you is obviously not helping. Come up with some arbitrary situation, calculate the change in momentum of each object in the system using Newton's 3rd law, calculate the work done by each force using the definition of work, and calculate the change in kinetic energy of each object individually and the system as a whole. Then perform a Galilean transform to a different frame, and re-do the calculations using the same definitions in the new frame. Look at the work done by each force, look at the change in energy of each body, look at the change in momentum. Is the work the same or different? Is energy conserved or not? This is something that you have to do for yourself if you are really interested in learning.

If you get stuck or if you just want confirmation, then I will be glad to check your work for you. If you are having trouble getting started then try this as an introductory problem:

Two masses, 1 kg and 10 kg, on a frictionless plane interact via a 3 N repulsive force which acts for 2 s. Calculate the work the force does on each object, the initial and final velocities and kinetic energies if (a) they are initially at rest (b) they are initially moving at 4 m/s with the larger mass directly in front of the smaller and (c) they are initially moving at 4 m/s with the smaller mass directly in front of the larger and (d) they are initially moving at 4 m/s with the two masses side-by-side.

 

[Mad_Eye]

fine, thank you.
i've checked up your example, and in all cases, the work is the same (19.8J), if you consider both forces of action and reaction as one.
so it is just confusing you are keep telling me work is frame variant, and giving me example in which the work is the same in all frames.

i guess i miss something, and this is what I'm trying to understand.
for example, is it alright to consider the work of both forces of action and reaction, as one work?

and, as i see it, there are two different issues, the work done on a body, and the work done by a body. are BOTH of them, frame variant? (just asking).
(for example, in my original example, the work done by the engine, is the same in all frames, but the work done on the car is different from one frame to another).

another question,
is it possible that the change of the KE of the whole system would be different, in two different frames?

thanks.

 

[Dale]

"in the ground frame: w = f.d = 10 N 300 m = 3 kJ"
isn't the work is still 0, because the angle between the force and the distance is 90

Thanks for putting some effort in. Sorry, my description was a bit unclear. The stewardess is holding the cart stationary by pushing horizontally so that it doesn't roll backwards towards her, so the force is approximately parallel to the displacement.

 

[Mad_Eye]

Thanks for putting some effort in. Sorry, my description was a bit unclear. The stewardess is holding the cart stationary by pushing horizontally so that it doesn't roll backwards towards her, so the force is approximately parallel to the displacement.

i see.
and I've edited my post (quite a lot :D)

 

[Dale]

i've checked up your example, and in all cases, the work is the same (19.8J), if you consider both forces of action and reaction as one.
so it is just confusing you are keep telling me work is frame variant, and giving me example in which the work is the same in all frames.

i guess i miss something, and this is what I'm trying to understand.
for example, is it alright to consider the work of both forces of action and reaction, as one work?

As you did the problem I am sure that you noticed that each force individually did a different amount of work in each frame. Thus work is frame variant. (Look at the definition of work. Work is defined for an individual force, not for a 3rd law pair of forces).

You also noticed that once you included the conservation of momentum (Netwon's 3rd law) for the system then you found that energy was also conserved for the system. In other words, if you want to exert this much force on these objects using something like a spring then you need to compress it with 19.8 J. Thus since all frames see the spring lose 19.8 J of PE and all frames see the system gain 19.8 J of KE then all frames agree that energy was conserved despite the fact that the work done by each force was frame variant.

 

[DaTario]

Let me please insert a question which I repute as pertaining to this subject. When passing from one inertial frame to another is it correct to say that we are introducing that famous additive constant to work (energy)?

Trying to be clearer: Energy concept works well even if we sum a constant to all energy levels. Does the frame variance imply sum of a constant to all possible works ?


Best Regards

DaTario

 

[Mad_Eye]

You also noticed that once you included the conservation of momentum (Newton's 3rd law) for the system then you found that energy was also conserved for the system. In other words, if you want to exert this much force on these objects using something like a spring then you need to compress it with 19.8 J. Thus since all frames see the spring lose 19.8 J of PE and all frames see the system gain 19.8 J of KE then all frames agree that energy was conserved despite the fact that the work done by each force was frame variant.

What exactly do you mean by "the energy is conserved"?
I didn't found the energy conserved, I've just seen the total amount of work is the same in all frames, but I have no idea where that energy came from.

(And if already in that, where indeed this KE (19.8J) came from?)