Violation of energy conservation in relativity

[Sensei]

Hello!

 

Subject: Violation of energy conservation in (special) relativity.

 

Possible? plausible? not possible?

 

What is your opinion, and why?

 

Best Regards!

 

[xyzt]

Hello!

 

Subject: Violation of energy conservation in (special) relativity.

 

Possible? plausible? not possible?

 

What is your opinion, and why?

 

Best Regards!

 

Very, very unlikely. What's on your mind?

[ydoaPs]

Hello!

 

Subject: Violation of energy conservation in (special) relativity.

 

Possible? plausible? not possible?

 

What is your opinion, and why?

 

Best Regards!

Energy is conserved within a frame. What goes in comes out. The total energy is constant. That's not to say, however, that energy is invariant, so you have to be careful. The energy tally can be different for different frames of reference.

 

So, conserved, yes. The same for everyone, no.

 

Is there a specific case you want to discuss?

[studiot]

 

xyzt

Very, very unlikely

 

 

And I always thought it was QM that was probabilistic, not relativity.

 

[Sensei]

Imagine rocket traveling at constant velocity v=0.866c. [math]\gamma=2[/math]

 

It has on board accumulator/battery, which is powering laser with wavelength 532 nm (2.33 eV) (in frame of reference of rocket at least).

 

Laser beam is send to polarization filter. It's non polarized light. Therefor 50% of photons pass through filter, and 50% of photons is reflected.

 

Then photons are escaping rocket.

 

They are detected by detectors A and B. That are at rest v=0c. They have the same frame of reference.

 

[math]E_0=\frac{hc}{532 nm}=2.33 eV[/math]

 

Detector A sees 2 photons with energy (relativistic Doppler Effect blue shift):

[math]E_A=E_0 \sqrt{\frac{1+0.866}{1-0.866}}=2.33 eV*3.732=8.7 eV[/math]

 

Detector B sees 2 photons with energy (relativistic Doppler Effect red shift):

[math]E_B=E_0 \sqrt{\frac{1-0.866}{1+0.866}}=2.33 eV/3.732=0.6245 eV[/math]

 

 

In frame of reference of rocket energy spend on emitting photons is 4*2.33 eV = 9.32 eV.

 

We have 2 photons in detector A, and 2 photons in detector B:

2 * 8.7 eV + 2 * 0.6245 eV = 18.644 eV total energy received.

18.644 eV / 4 = 4.66 eV average per photon (in frame of reference of detectors).

 

Now imagine that we have polarized light emitted by laser. We can adjust angle of polarization filter in rocket and adjust quantity of photons passed through and reflected.

Imagine situation where 75% of photons passed through, and 25% of photons is reflected.

 

We have 3 photons in A, and 1 photon in B:

3 * 8.7 eV + 1 * 0.6245 eV = 26.717 eV total energy received.

26.717 eV / 4 = 6.679 eV average per photon

 

Just by rotating angle of polarization filter person on board of rocket can change energy received by all detectors, while in its own frame of reference energy appears to be the same.

Edited January 31, 2015 by Sensei

[xyzt]

 

Just by rotating angle of polarization filter person on board of rocket can change energy received by all detectors, while in its own frame of reference energy appears to be the same.

 

Total energy (just like kinetic energy) is frame-DEPENDENT , the above has nothing to do with conservation of energy. You are mixing frame-(in)dependence with conservation. Totally different concepts.

Edited January 31, 2015 by xyzt

[ajb]

It has already been said, but energy is conserved in relativistic mechanics when measured from a fix inertial frame. As energy is something to to with time translation invariance it is clear that this needs an inertial frame to be picked to be defined properly. The Lorentz transformations mix time and space and so one would not expect a universally agreed upon notion of energy. But this is the hint to what you should be considering, energy and momentum.

 

In relativistic mechanics the mass-shell condition [math]E - p^2 = m^2 [/math] (setting c = 1). That is, although observers may not agree on the energy of momentum of a particle they will agree on the mass.

[MigL]

I always thought GR doesn't say anything about conservation of energy.

Is this a 'shortcoming' of GR ?

Or is energy conservation not a valid global concept ?

[ydoaPs]

I always thought GR doesn't say anything about conservation of energy.

Is this a 'shortcoming' of GR ?

Or is energy conservation not a valid global concept ?

It depends on whether or not you consider the spacetime itself to have energy. [math]\bigtriangledown{T^{\mu\nu}}=0[/math], so there is conservation of 'stuff'-energy.

[swansont]

I always thought GR doesn't say anything about conservation of energy.

Is this a 'shortcoming' of GR ?

Or is energy conservation not a valid global concept ?

 

Energy is conserved only within a frame, and globally you do not have one frame.

[Sensei]

I gave macroscopic example (rocket) with cosmic scale distances. But it's just matter of scale.
The same must be true at micro and quantum scale.

Replace rocket by proton, put in tunnel, and you have CERN experiment.. (except inability to emit directional polarized photons at will)

If I will send electron and positron to spherical vacuum chamber with array of detectors, and one detector will read gamma photon with f.e. 0.25 MeV, then detector at the opposite end of chamber will have to read 0.772 MeV (+ original kinetic energy corrections), as they must sum up to at least 1.022 MeV energy of electron and positron...
From time delay we can read where annihilation happened.
From red shift, blue shift of photons we can read what was their velocity.
Even though chamber is in completely different frame of reference than electron-positron.

ps. I am surprised that nobody pointed me that rocket should be decelerating in 2nd case.

ps2. Every particle is in its own frame of reference.

Edited February 1, 2015 by Sensei

[MigL]

I see your point swansont, but I would still expect energy to be conserved globally, even without GR being able to confirm it.

The alternative opens up a 'can of worms'.