GR and conservation of energy

1. Apr 22, 2008

redtree

Question:
1. A gravitational interaction causes two energy densities to accelerate towards each other.

2. This increases their relative velocity.

3. By special relativity, increasing relative velocity increases (relativistic) mass. Even if the energy involved is small, the energy still must be accounted for.

4. By general relativity, energy densities travel along geodesics. Why would traveling along a geodesic increase the energy of an energy density? In other words, assuming energy is conserved, where does the energy come from to increase the relativistic mass of the accelerating energy density?

2. Apr 22, 2008

Antenna Guy

Consider that if your two "energies" meet, their mutual "increases" cancel.

Regards,

Bill

3. Apr 22, 2008

redtree

If they cancel, then how is energy conserved? Doesn't that "canceling" depend on what happens when they meet?

4. Apr 22, 2008

Antenna Guy

Energy is conserved because their increases in momentum cancel. And no - I don't think that cancelling depends upon what happens when they meet.

Regards,

Bill

5. Apr 23, 2008

Garth

A good question redtree!

Energy is not generally conserved in GR, GR conserves energy-momentum (measured as a particle's 'rest' mass) and that is different.

The measurement of energy is frame-dependent.

In GR, which Emmy Noether http://en.wikipedia.org/wiki/Noether's_theorem]demonstrated[/PLAIN] [Broken] that, as a 'Improper Energy Theorem', energy is only conserved when there is a time-like Killing vector. Time dilation affects energy measurement.

Such a condition exists in the static field around a spherical mass. In your example redtree, when Newton's apple falls towards the Earth, the increase in the apple's kinetic energy is compensated by the increasing time dilation as the apple falls deeper into the Earth's gravitational field, as measured in the Earth's static frame of reference. As measured in this frame of reference total energy is conserved.

However, as measured in the apple's frame of reference the Earth is now falling towards the apple and the Earth's kinetic energy increases. In this case time dilation operates in the opposite sense, the clock is now carried by the apple, and total energy is not conserved.

In the apple's frame of reference the Earth's gravitational field is not static because its strength steadily increases as the Earth approaches; therefore a time-like Killing vector does not exist in this frame.

Garth

Last edited by a moderator: May 3, 2017
6. Apr 24, 2008

redtree

When I mention energy, I am assuming that energy is a function of both rest mass and momentum. My basis for so doing is the following: Working in a natural unit system (c=1), by the relativistic energy-momentum equation, Energy is a function of both rest mass and momentum by the equation E^2 = m^2 + p^2.

Therefore, if one considers Energy as encompassing both rest mass and momentum, then shouldn't energy-momentum be considered simply as energy? And then, with this understanding of energy, why isn't energy conserved in GR? To me, this seems to be a problem with GR and one reason why it is not compatible with Quantum Mechanics.

Regarding time dilation, my understanding is the time dilates with increasing gravitational field. In your discussion of the apple and the Earth, a clock on either object will experience time dilation as both would experience an increasing gravitational field. The clock on the apple would experience an increasing gravitational field from the approaching earth. The clock on the earth would experience an increasing gravitational field from the approaching apple. Granted the approaching earth would cause a greater increase in gravitational field, but time would dilate for both clocks, though the amount of dilation would be less for the clock on the earth. Looking again at your example, time dilation would seem to be greater for the clock on the apple (apple's frame of reference) than the earth (earth's frame of reference), the opposite of what I understood you to write. Please correct me if I am wrong in my understanding.

7. Apr 25, 2008

Garth

No, you are making a dimensionality mistake; energy-momentum is a four-vector whereas energy is a scalar quantity.

The value of the energy is the magnitude of the time component of the energy-momentum vector, but you also have to include in energy-momentum the classical three-momentum of the body, which is the space component of the energy-momentum vector.

A stationary body has rest mass-energy and no momentum, whereas a moving body has rest mass-energy, kinetic energy and momentum.

The (-,+,+,+) signature of the metric in which the time component has the opposite sign to the space components mean that the three-momentum (space-component) of a moving body is subtracted from the total (rest + kinetic) energy (time-component) in the energy-momentum four-vector.

This results in the final magnitude of the energy-momentum four-vector being equal to the original value of the ‘rest’ mass-energy of the body. No matter how fast the body is moving relative to the observer the observer measures the body’s energy-momentum to be a constant value. Energy-momentum is conserved.
That may indeed be one manifestation of the problem, also there is no absolute time in GR, the foliation of space-time into space and time dimensions in GR is dependent on the observer’s frame of reference.
Time dilation is measured as a comparison of two clock rates.

The amount of time dilation is the same for both clocks, i.e. the clock deeper in the field is slower than the one higher up and both observers would agree on this. This is different to SR time-dilation between moving observers where each observer thinks the other’s clock is ‘going slow’.

The clock on the Earth is slower than a clock at altitude, therefore the correction to the measurement of energy acts in the same sense whether you are on the Earth measuring the total energy of the falling apple, or on the apple measuring the total energy of the 'falling' Earth.

As I said in my earlier post, in the case of the observer being on the Earth the increasing kinetic energy of the falling apple (which has a negligible mass and gravitational field) is compensated by the time dilation relative to the clock on the Earth. The Earth bound observer concludes energy is being conserved.

In the case of the Earth falling towards the observer on the apple the energy-correction for time-dilation is in the opposite sense and total energy is not conserved according to this observer.

I hope this helps,
Garth

Last edited: Apr 25, 2008
8. Apr 25, 2008

snoopies622

So the time-component of the momentum four-vector is affected by both velocity-dependent time dilation (the Lorentz factor) and gravitational-time dilation? Is this a case where the Lorentz factor -- part of the momentum four-vector in flat space -- must be replaced with something else in GR that reflects the curvature? In other words, is the standard definition of the momentum four-vector (rest mass times gamma times <c,v(x),v(y),v(z)> inappropriate/inadequate for an object falling in a gravitational field?

9. Apr 25, 2008

pmb_phy

Kinetic energy changes into potential energy. Its the total that is conserved. An increase in relativistic mass comes from an increase in kinetic energy.

I'm curious as to why you refer to it as "energy densities" rather than "mass"? In general energy density is not proportional to mass density.

Pete

10. Apr 25, 2008

pmb_phy

The change in the time-component depends on the gravitational potential and a non-vanishing gravitational potential does not guarentee that the spacetime is curved.

Pete

11. Apr 25, 2008

Garth

You are trying to cobble together SR and GR! Just use GR.

The Lorentz 'gamma' factor is modified by the components of the metric describing curved space-time. Although you can transform the metric into the inertial free-falling coordinate system at any event, i.e. into the SR metric, that only applies in the immediate neighbourhood of the event.

When you consider another body freely falling towards the origin then away from the origin the metric is inescapably non-Minkowskian and curvature modifications of the Minkowski metric become important.

The metric is given by

$$d\tau^2 = -g_{\mu\nu}dx^{\mu}dx^{\nu}$$
where the normal Einstein summation notation is being used and the signature convention is (-,+,+,+).

In the flat space-time case this just gives the Minkowski metric from which the 'gamma' factor is derived. In curved space-time the equivalent to the gamma factor is in general much more complicated and takes into account both velocity and curvature time dilations.

Garth

Last edited: Apr 25, 2008
12. Apr 26, 2008

redtree

Calculating Energy

How might one calculate energy without first knowing momentum and invariant (rest) mass?

13. Apr 26, 2008

pmb_phy

Actually the metric is given by $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$. ds is used rather than $$d\tau$$ because ds represents an arbitrary sapcetime interval whereas $$d\tau$$ represents a proper time interval. The presence of "-" sign is not only unnecessary but is never used in the relativity/math literature. The sign of the dt term is determined by the value of g00. If you leave the minus sign in as you did above the the sign in front of dt will be "+" when written out rather than the "-" that you're thinking of.

Pete

Last edited: Apr 26, 2008
14. Apr 27, 2008

Garth

This is simply a matter of sign convention - I follow the "Landau-Lifshi*z Spacelike (notice the automatic censoring of the website, the * is a 't'!!!) Convention (LLSC) used by Weinberg in "Gravitation and Cosmology" and Misner, Thorne and Wheeler in "Gravitation".

That is why I was careful to specify the signature.

Proper Time Interval is a perfectly good measure of the interval, especially as it encourages the fundamental measurement to be made using a clock.

Garth

15. Apr 27, 2008

redtree

I apologize for the less-than perfect notation, but I'm new to the application.

The energy-momentum 4-vector (in the natural unit system with c=1):

$$\vec{P}$$ = [ E $$\vec{p}$$ ]

The length of the energy-momentum 4-vector

$$\sqrt{P \bullet P}$$ = $$\sqrt{E^{2} - p^{2}}$$ = m

To restate my question:

Given that energy is presumably an independent variable in the 4-vector, how can one determine energy for a location in space-time without knowing both mass and energy?

Or is the energy simply calculated from the rest mass and momentum?

16. Apr 27, 2008

Garth

To restate: energy is frame dependent quantity, the energy of a particle can be calculated in the observer's frame of reference from the particle's rest mass and its momentum, as you have just demonstrated.

Its momentum is obviously dependent on the frame it is measured in, the particle can be moving in one frame and stationary in another.

Garth

17. Apr 27, 2008

redtree

That's exactly my point. Energy is a dependent variable. In GR, energy is a kind of fudge factor that permits the appearance of a conservation law. A hint that energy in GR is more of a mathematical construct than a real measure of energy is seen in the phenomena of dark matter, dark energy and the need for a cosmological constant in cosmology. This entire problem comes about because GR considers gravity as curvature instead of an energy driven reaction, like electromagnetism. This problem is also a primary reason that GR is not compatible with Quantum Physics.

18. Apr 27, 2008

redtree

Frankly, momentum itself may be restated as energy, or even (relativistic) mass for that matter.

A better four vector would be the following:

Energy = [ mass, $$\vec{v}$$ ]

where mass is just mass and doesn't differentiate "invariant" from "relativistic" mass, and velocity ($$\vec{v}$$) is the combination of v(x), v(y), v(z).

The differences between this four-vector and the energy four-vector are the following:
1. The variables cannot be considered as (partial) restatements of each other.
2. A correct theory will conserve energy not because of mathematical constraints but because of physical relationships. Frankly in GR, energy is simply a balancing factor to insure that mass is invariant regardless of momentum, i.e., frame of reference. Furthermore, by GR, momentum may be considered an arbitrary vector given that it is completely dependent on frame of reference.

Again, the indications that energy in GR has more to do with fudging the math than actual measures of energy are the need for considerations of dark energy and dark matter (96% of the mass/energy of the universe) and the cosmological constant to explain known observations in cosmology.

There are answers to these problems, but they are beyond GR.

19. Apr 28, 2008

redtree

Here's a related question:

At what relative velocity must a frame of reference be traveling relative to an electron before the electron appears as a black hole (with an event horizon)? Or, in other words, how much energy must be added to an electron to turn it into a black hole?

20. Apr 28, 2008

redtree

Or put even more starkly:

Assuming finite energy in the universe, what happens in a frame of reference traveling relative to a very, very small mass at a speed such that the momentum of the mass implies an energy greater than the finite energy of the universe? Does this imply that not all frames of reference are equal? Or, is the energy of the universe potentially infinite?