# Where did the energy go?

Suppose you are running with speed of 5m/s (with respect to ground, mass=20kg)

and there is a mechanism by which all of your kinetic energy is converted to electrical energy which is used to light a bulb of constant power

Case 1: 'L' who is at rest on ground reports the duration for which the bulb has glowed

Case 2: 'M' who is running opposite to you with speed 5m/s also reports the duration for which the bulb has glowed

Do both the cases have same answer???

How much amount of energy will be converted to electrical energy in both the frame?? 5 years, 9 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

I agree with @Shashwat Shukla , if you look into the mechanism of the process, suppose you press a spring and the energy stored in spring is converted to electrical energy,

Then the spring is only compressed by the relative motion of your hand and spring. if the spring's tip was moving as fast as you, Then the energy is not stored in spring and instead simply remains as the kinetic energy of the particles as seen from the frame forever.

But ofcourse, this simply means that our assumption that a device can exist which converts the whole kinetic energy in all frames into electric energy is impossible.

Suppose it is his pedalling that does the job, then only the circular motion of the pedals affects the energy stored and any free translation of all particles causes no change.

Another perhaps more rigorious statement would be to state that energy is stored or removed when the body is deformed and deformity requires relative motion between its various particles.

Now you might ask, when we throw a stone, energy is stored as kinetic energy but no deformity occurs? Well consider the whole system, We are throwing the stone, it is moving forward, so we are actually pushing it and hence our hand is actually moving slightly faster than the stone at each instant to provide the stone its extra energy and hence there is relative motion here as well.

- 5 years, 9 months ago

Thank you @shashwat shukla and @Mvs Saketh

- 5 years, 9 months ago

I agree. However the term $deform$ for non-contact forces must be extended to mean relative motion (and not just change in the shape of the body).

- 5 years, 9 months ago

Yes, both cases have same answer. And in both the ground frame and in the frame of 'M', the amount of energy converted is identical and equal to $1/2*20*25=250J$

- 5 years, 9 months ago

Then what about the other 250 joules in frame of M?? Do the efficiency of mechanism changes just by changing the reference frame???

- 5 years, 9 months ago

Hmm. I thought you might have done that too. check out these links 1

wiki

- 5 years, 9 months ago

Sorry my bad.... Other 750 joules.... If M calculates the kinetic energy then it will come out to be 1000 joules

- 5 years, 9 months ago

So in the frame of M if only 250 J is bein converted then for M the efficiency of mechanism changes.... My doubt is should it really happen? ?? Does efficiency changes by change of reference frame? ?

- 5 years, 9 months ago

And accepted that kinetic energy is frame dependent.....but is total energy of the system also frame dependent..???

- 5 years, 9 months ago

A) Please explain what you mean by efficiency, seeing as all the energy is being converted anyways.

B)The ambiguity arises because you haven't really defined this 'mechanism'.

As I understand it, what you are saying is this: Look at the man in a particular reference frame, calculate his kinetic energy and convert it all into electrical energy. And your question is, why it isn't the same in all frames.

That's precisely because the mechanism is $frame \quad dependent$ (meaning that you have one machine corresponding to each frame, and not one unique machine) and this 'mechanism' cannot actually exist(because if it did exist, then as you said, the amount of light emitted will be the same in every frame).

So why can't it exist? We have to analyse this 'mechanism'. Well, while this may not be perfectly rigorous reasoning, look at it this way:

One possible machine is as follows: Make the man push against a big spring which is coupled to a dynamo. It is immediately obvious as to why the duration of the bulb glowing will be the same in all frames: the $relative$ velocity between the spring and the man at all times and in all frames will be the same.

Yet a more general way to say the same is to note that the way in which energy is converted/transferred from one form to another is via an interaction of bodies through forces. The important point is that these $forces \quad are \quad frame \quad independent$(if we are working with inertial frames of reference) as this is just Einstein's equivalence principle. And thus, as the forces remain the same, the work done by the man on the device is also the same(this work appears as electrical energy).

C)The total energy is also not conserved under a change of reference frame simply because the K.E is not. If you're wondering if some 'other' form of energy like P.E can change so that the net effect is that total energy is a constant, my explanation remains the same: Any form of potential energy exists only by virtue of the body having done work against a force (like gravity or electrostatic force),which as mentioned before, is frame invariant. But is there any other form of energy apart from K.E and P.E?

Not that I know of; as vibrational, rotational energy etc. are also just other forms of K.E

- 5 years, 9 months ago

Hey thank you so much...... I did figure out this ambiguity....actually wot I really wanted to ask if the total energy of the system is going to be the same in different inertial frames or not. .. Kindly consider the other problem which I posted n point out my mistake

- 5 years, 9 months ago

You're welcome. But, why is part C of my answer unsatisfactory(it pertains to total energy)?

- 5 years, 9 months ago

I think to include several forces would make it complicated so I thought to start with a basic isolated system consisting of the mass point on which no forces act whatsoever . . . . . For such a system the total energy should be kinetic plus the energy in form of mass. .

- 5 years, 9 months ago

To tell you the truth, I don't know how to explain it to you. That means that i don't understand it well either. @Azhaghu Roopesh M @Mvs Saketh @Shashwat Shukla , help us out here?

- 5 years, 9 months ago

I don't know either... But I thought that the total energy of the isolated system should be constant. .. That must also include the energy in form of mass... So here is my argument kindly consider it: If you have a mass point of mass M which is moving in space with velocity V and no force acts on it at all. . . We can not interact with this isolated system . . .what we can do is observe it from different INERTIAL frame . . So I tried to calculate the total energy of the system (KINETIC +IN FORM OF MASS) I was careful enough to use the relativistic equations instead of newtonian equation for energy and I also used the Lorentz transformation for relative velocity rather than galliean transformation And I am getting different total energy in different frame ..

So does it really come out to be so? Does total energy of the isolated system depends upon the frame?

Or I have done some mistake ?

- 5 years, 9 months ago

Please take a look at my arguments.

- 5 years, 9 months ago

OK.... So we can conclude that.... The total energy of the isolated system will depend upon the frame and will be conserved in evry frame... Right?

- 5 years, 9 months ago

Refer Shashwat's answer above. And yes, it does depend on the frame.

- 5 years, 9 months ago

Ah , I guess I didn't notice my name being mentioned here ! Sorry about that .

How about anyone of you make a question on this just to check the conceptual clarity of others , what say ?

- 5 years, 9 months ago

Thanks Raghav , I didn't know that a site named Physics Stack Exchange existed ! I was too engrossed in the Maths version of it !

- 5 years, 9 months ago

We can also say that energy or energy transformations require the presence of matter(Einstein)(vaguely). Since some properties of matter(like kinetic energy) depend upon the frame of reference,,we can say that energy transformations (in this case) will depend upon the relative motion of the runner and the device(which has to be some form of matter).

- 5 years, 8 months ago

The question itself is incomplete ... When you say " there is a mechanism by which all of your kinetic energy is converted to electrical energy" Then kinetic energy in which frame you are talking about..??

- 5 years, 6 months ago