# Deriving Law of Conservation of Energy from E=MC2

1. Feb 19, 2008

### superdirt

Hello,

First post. Thank you for letting me be a part of the discussion.

My question is, can the Law of Conservation of Energy, http://upload.wikimedia.org/math/2/7/3/273abc16486177bf6cc4c8ec4a4a5fc9.png, be derived using E=MC2.

I understand that the E=MC2 can be derived using the LoCE and other formulas, but my question is, can the LoCE be derived from E=MC2 and other necessary formulas. My calculus is not so good these days, but my intuition says that the answer is yes and this is a relatively easy proof.

This question is a part of a larger thought experiment of mine.

Thanks,

Scott

2. Feb 20, 2008

### pam

If "E=MC2 can be derived using the LoCE and other formulas", then writing the steps backwards you can derive the converse. This is true for any derivation.

3. Feb 20, 2008

### George Jones

Staff Emeritus
This depends on what one means by "derivation", and I guess we use different definitions.

I would say that if a derivation is a tautology, then the derivation can be reversed.

For example, I would say that

$$A \Leftrightarrow B \Leftrightarrow C$$

is a derivation of $C$ from $A$ that can be reversed, while

$$A \Rightarrow B \Leftrightarrow C$$

is a derivation of $C$ from $A$ that can't, necessarily, be reversed.

4. Feb 20, 2008

### Lojzek

As I see it both derivations are the same:

1. We find a new experiment where LoCE seems to fail
2. We define a new energy so that LoCE is respected again

Consequences:

If the energy has the defined form, then LoCE is respected.
If LoCE is respected, then energy must have the defined form.

5. Feb 20, 2008

### bernhard.rothenstein

E=mcc

As far as I know E=mcc can be derived without usin conservation of energy

6. Feb 20, 2008

### Lojzek

I have thought of an experiment that (for one simple geometry) derives LoCE from E=mc^2
(or inversely):

You connect two weights with a light stick and fix it into a box so that they can rotate in one plane. Then you close the box and try to accelerate it.
If you calculate the required force of acceleration, it turns out that it is more difficult to accelerate the box if the weights inside are rotating. If the mass of the all box parts is m and kinetic energy of the rotating weights is T, then the required force is:

F=(m+T/c^2)*a

So anyone who would try to determine the mass of the closed box would measure:

M=m+T/c^2

If the box would be taken to pieces, then the total mass would decrease by:

dM=-T/c^2

While the kinetic energy T would emerge.

Conclusion: the conservation of energy is respected only if we set:

dE=dM*c^2

Last edited: Feb 20, 2008
7. Feb 20, 2008

### clem

I think this implicitly assumes E=mc^2 to prove E=mc^2.

8. Feb 20, 2008

### Lojzek

No, because I calculated force as the time derivative of relativistic momentum (and relativistic momentum can be derived from Lorentz transformation (and Newton's law), no need for E=mc^2).
I also used the formula T=m*(gama-1)*c^2. This formula seems derived from E=m*gama*c^2, but it is in fact the other way round: T can be derived by integrating accelerating force over acceleration path, then it can be used to prove E=mc^2 and finaly E=m*gama*c^2 is derived from both formulas.

I imagined the weights inside moving perpendiculary to the direction of box movement. The relativistic momentum is m*gama(v)*v, so the x component of momentum does not depend only on x component of speed.

Last edited: Feb 20, 2008
9. Feb 20, 2008

### superdirt

Thank you for your thoughts, ppl.

I believe this is incorrect. I am with George and his explanation. Example: Being a man implies you are a human, but being a human does not imply you are a man.

That is correct. It can be derived in many ways, at least one of those ways uses the LoCE.

Scott

10. Feb 22, 2008

### rahuldandekar

A related question: In David Morin's book, he says that we cannot "prove" E = gamma*m*c^2 is conserved, or that p = gamma*m*v is conserved, using ONLY the Lorentz Transformations, only that they are good expressions for the conservation of energy in simple collisions, and they reduce to the Newtonian expressions when velocities are small.

He basically says that experiments have shown that these two are conserved, and then goes on to derive the results of relativistic dynamics from these two results. Of course, if they are conserved in one frame, that they are conserved in another follows from the LTs.

But I am not very confident of this approach, surely the expressions for p and E in relativity can be "proven" only from the LTs, in the general case, since they can be proven in the special (and commonly taught) case of grazing collisions.

11. Feb 23, 2008

### 1effect

You are right, here is a very nice proof based on grazing collisions. I remember that TW "Spacetime Physics" has some similar proofs (maybe not as good).

12. Mar 18, 2008

### superdirt

13. Mar 19, 2008

### pmb_phy

It depends. The expression E = Mc^2 was derived based on the assumption that energy is conserved. If we simply postulate E = Mc^2 and assume no other quantity is conserved then we cannot derive conservation of energy because (1) energy is that quantity that is conserved, by definition and thus cannot be proved and (2) without conservation of energy another postulate is required such as the conservation of inertial (aka relativistic) mass or the conservation of momentum.

Therefore if we assume that momentum is conserved (simply because M is defined such that momentum is conserved) then yes, it can be derived. Since M is defined such that momentum is conserned then it can be shown that M is also conserved and as such E = Mc^2 is conserved.

Pete

14. Mar 19, 2008

### yuiop

The proof given in the link posted http://en.wikibooks.org/wiki/Specia...vation_using_the_concept_of_relativistic_mass clearly shows that there is an initial assumption of conservation of energy in the derivation of the momentum equations, so it does not (by itself) prove that energy is conserved in special relativity.

[EDIT] Perhaps it would be more accurate to state their is an assumption of conservation of momentum. When the conservation of momentum is "fixed" so is the conservation of energy.

Last edited: Mar 19, 2008
15. Mar 19, 2008

### 1effect

You are looking at the right wiki page, wrong caption. Look here. User "rahuldandekar" was talking about the derivation of momentum conservation based on grazing collisions. There is a similar proof in Taylor-Wheeler "Spacetime Physics".
I do not have an opinion on the energy conservation but I have proven recently to my satisfaction that the energy-momentum 4-vector is conserved for arbitrary systems of particles using only the LT transforms. I am quite sure you remember that thread.

Last edited: Mar 19, 2008
16. Mar 19, 2008

### 1effect

I don't think so, look at this caption, in the same page.

17. Mar 19, 2008

### yuiop

That section uses an assumption of relativistic 3 force and when you look at the section above it for the derivation of relativistic force transformation you see there is an assumption of the relativistic momentum transformation (p=myv).

18. Mar 19, 2008

### 1effect

You are mixing up the momentum formula ($$p=\gamma m v$$ ) with momentum "transformation". The wiki page had just established the momentum formula from the grazing collision, from base principles. The wiki article goes on to establish the energy formula from base principles by using the momentum formula that was just established from grazing collisions. So, in this sense, the author has established both momentum and energy relativistic formulas from grazing collisions. No use of LT , exactly as user "raduldandekar" conjectured.

19. Mar 19, 2008

### yuiop

You are right that I used the term transformation incorrectly. In the grazing collision they show that without assuming a change in relativistic mass that classical momentum is not conservered and then change the definition of momentum to ensure that momentum is conserved at relativistic speeds.

Some quotes from the text:

"To preserve the principle of momentum conservation in all inertial reference frames, the definition of momentum has to be changed."

and "This means that, if the principle of relativity is to apply then the mass must change by the amount shown in the equation above for the conservation of momentum law to be true."

20. Mar 19, 2008

### 1effect

Correct. This is one of the better wiki pages, the authors put a lot of thought in it.
They show how you can derive the formula for relativistic momentum $$p=\gamma mv$$ from the momentum conservation in an elastic grazing collision. R.C. Tolman was the first to show it (not as well) , Taylor-Wheeler do it in their book.
Based on the derivation for the momentum formula, the authors do a very nice job of deriving $$E=\gamma mc^2$$ . This is exactly what user "raduldandenkar" was asking for.