# Einstein energy and thermodynamic energy

1. Oct 18, 2004

### Iraides Belandria

Thermodynamics textbooks indicates that the total energy of a system E is given by the following equation
E = U + Ek +Ep
where U is the internal energy due to atomic or molecular interactions, Ek is the kinetic energy due to the velocity of the system ,and Ep is the potential energy relative to a reference system.
Now, according to relativity theory
e = mc2
Then, my question is ¿ how is related the einstein energy term ,e, with the the thermodynamic energy, is e=E, or e = Ep, or e= Ek or e = U ?

2. Oct 19, 2004

### pervect

Staff Emeritus

"Is energy conserved in General Relativity?"l

Then you can try the following threads here on physics forum

Energy not defined in GR?

Mass of a body

I think there was at least one other thread on this topic here recently, but its name escapes me.

In spite of some of the scary thread names, GR does have a concept of energy. But it doesn't have a concept that applies to arbitrary space-times. See the FAQ and the discussions for more info. Some of the discussions here meander a bit (well, a lot).

3. Oct 19, 2004

### franznietzsche

E = mc^2 is not according to relativity theory. Its is derivation is unrelated.

E=mc^2 is the equivalent rest energy, it is possessed by any mass without other forms of energy, its is unrelated to thermodynamics (I believe...)

4. Oct 20, 2004

### Iraides Belandria

e=U

Franznietzsche:
I agree that e is not equal to E, and I think, that e=U ¿ What do you think about it?. I read the articles advised by Pervect, but I did not find any hint to my question. Finally, ¿How can we relate thermodynamics and relativity?. ¿Are they separated ? ¿ Why people talks about thermodynamics relativistic?.

5. Oct 20, 2004

### pervect

Staff Emeritus
Energy is conserved in general relativity only in particular circumstances, the most important of which is asymptotically flat space-times.

The first reference I gave, the sci.physics.faq reference, talks about the conditions in which energy is conserved, which is vital to understand "up front".

The simplest case to analyze is a space-time which is static AND asymptotically flat, in which case the energy of the system can be expressed in a variety of ways. One of the ways the energy of a system with these properties can be computed is as a volume intergal of the stress-energy tensor

$$M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV$$

here

This is actually probably not the most convenient description of energy in General Relativity, though it's the closest to the thermodynamic example you cite, in that the above intergal can be broken into parts that look like "rest energy" plus kinetic energy (that's the first term in the intergal the T_ab term), plus another part that looks like binding energy (that's the second term, the T g_ab term). [See the previous thread for more detail on this interpretation].

The total mass (aka energy) of a system can also be expressed also as a surface intergal, similar to the way that electrostatic defines a conserved charge.

One formula for the surface intergal from (from Wald) , also for a static asymptotically flat space-time is:

$$M = -\frac{1}{8 \pi} \int_S \epsilon_{abcd} \nabla^c k^d$$

which I discuss a little bit over here

over here

The easiest way to think of this intergal, though, rather than to wade through unfamilar tensor notation and Killing vectors / Killing's equation is to note that in a static asymptotically flat space-time, gravity is nearly Newtonian "far away" from the main body. You can then define the mass of a system as the equivalent Newtonian mass that would cause distant bodies to follow the same orbits, to experience the same "gravitational forces". The surface intergal is very similar to Gauss's law (the electrostatic definition of charge enclosed by a surface) - the intergal of the normal force over the surface of a sphere.

Note that gravitational energy CAN NOT be localized to a specific point in space-time. This makes it very different from the energy in an electrostatic field, which can be localized in this manner.

Non-static space-times are a bit trickier to analyze - it's a lot more convenient to go to the center of mass of a system, and compute the energy in a coordinate system where the c.of. m is not moving. However, in standard relativity, this condition is not essential to defining energy. Some posters feel strongly that the center of mass should play a more pivotal role in defining energy - note that this is not "textbook standard" relativity theory, though.

6. Oct 20, 2004

### pervect

Staff Emeritus
A few more usefull comments occured to me. The approach to energy that "breaks down" the energy into various types (binding, rest mass, kinetic) can really be justified only when you have a true timelike Killing vector which exists in the strong field region . Which implies a static spacetime.

You can get the total energy in a non-static space-time with asymptotically timelike killing vectors, which exists only in the weak field regions. This, however, requires the surface intergal approach to take advantage of the asymptotic Killing vectors in the only region where they exist. You can't categorize the energy into subtypes with this approach, because the analysis occurs entirely in the weak field region, far separated from the source.

For a spherical star, MTW also breaks down the total energy into categories. (In the section on spherical stars, in the region of pages near 600 as I recall). I haven't been able to compare his breakdown to mine to be sure they are identical at this point, because the formulas I found for the timelike Killing vectors inside the star for the case I analyzed (the simplest possible one, one with a constant density) were rather messy.

Last edited: Oct 20, 2004
7. Oct 20, 2004

### Mike2

I have my doubts that there is an energy conservation law even in special relativity. For it takes time for the forces to travel between objects. So what would seem to be an energy conservation law at the local level would not apply when particles will have travelled extra distant by the time they feel each others force fields.

8. Oct 20, 2004

### pervect

Staff Emeritus
Energy is well defined in special relativity (except for gravitational energy, of course), via the mechanism of fields.

The total energy of two interacting charged particles has components both in the field and in the particles themselves. There is no period of delay during which the energy vanishes - the energy flows smoothly from particle to field, from field to particle.

The energy in the electromagnetic field is proportional to E^2 + B^2 (with some extra unit conversion factors thrown in here & there depending on the system of units used).

One other thing I should point out is that the electric field of a charge moving at a constant velocity always points directly away from or towards the charge. It does not point towards a "delayed position" of the charge, it points directly at the current position of the charge.

see for instance

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

Last edited by a moderator: Apr 21, 2017
9. Oct 22, 2004

### Iraides Belandria

About this comment, I consider that the equivalent rest energy refers to any mass without kinetic or gravitational potential energy, but with atoms or molecules with translational, rotational and vibrational movements. In other words with a mass dotated with internal energy, U. Therefore, the equivalent rest energy, e, should be equal to U, the thermodynamic internal energy term.

10. Oct 23, 2004

### pervect

Staff Emeritus
This is a little awkward, because of course translational, rotational and vibrational movements are kinetic energy. I think what you mean is that you are measuring the total energy (excluding gravitational energy but including kinetic energy) of matter, considered here as a swarm of particles, in the rest frame of the swarm, so that the total momentum of the swarm is zero.

If you add into this the energy stored in any non-gravitational fields (the binding energy if you will), you'll be very close to having the sort of energy that's used in the stress-energy tensor. All you need to do is to take the density of this sort of energy per unit volume, and you'll have the main component of the stress-energy tensor, the energy density per unit volume. This is called T00

Because your swarm of particles is in a coordinate system where the total momentum is zero, all of the off diagonal terms of the stress energy tensor will be zero.

The stress-energy tensor describes not only the density of energy in space, but of momentum. By choosing the rest frame of the swarm, we've eliminated the momentum terms.

But there are still three other terms on the diagonal of the stress-energy tensor left, though - there are 4 terms on the diagonal of this 4x4 tensor, and we've defined one of them, called T00.

These off-diagonal terms represent pressure. In GR, pressure generates gravity. In an ideal fluid, the pressure will be isotropic (the same in all direction), Thus pressure = T11 = T22 = T33. (Remember that energy was T00).

You can artificially generate situations where you have a swarm-of-particles where the pressure is not isotropic, but this is already getting long so I won't go into that unless someone wants to know.

If you skim through

Baez's GR tutorial

Or his fuller paper

here

and if you get past the math you'll see that T00 + T11 + T22+T33 is an important quantity. It's what causes the volume of a swarm of stationary particles to decrease, i.e. gravity.

11. Oct 29, 2004

### Iraides Belandria

Given this explanation, ¿ How can we write the first law of thermodynamics including relativistic effects?

12. Oct 30, 2004

### pervect

Staff Emeritus
I believe the usual approach is ultimately based on the equation

$$\nabla^a T_{ab} = 0$$

This statement can be interpreted as saying that the energy is conserved in a zero-volume piece of space-time. (Unfortunately it doesn't generalize to saying that energy is conserved in a finite volume.)

There are also some equations on MTW (pg 560 if you hapen to have the book) but I dont quite follow all the steps in their derivation of the formula.

They wind up with

$$d\rho = \frac{\rho + P}{n}dn + nTds$$

and since the derivative works in any direction, this becomes

$$\nabla_u \rho = \frac{\rho + P}{n} \nabla_u{n}+ n T \nabla_u{S}$$

here $$\rho$$ is T00, the energy density per unit volume, P is the isotropic pressure,

 in an ideal fluid P = $$T_{11} = T_{22} = T_{33}$$
but I'm not sure that actually works for this case where the fluid has energy components other than translational motion
edit]

n is the density of baryons per unit volume, T (unsubscripted) is the temperature, and S is the entropy per baryon

Last edited: Oct 30, 2004
13. Nov 1, 2004

### Iraides Belandria

Pervect

I think that this equation is not correct because the units of energy density and pressure in the first term are not consistent.

14. Nov 1, 2004

### pervect

Staff Emeritus
I should mention that the equation uses relativistic units as does the whole text.

This means that G=1 (the gravitational constant), and c=1, and so is G/c^2.

The usual unit used is the cm. We measure distance in cm. We measure time in cm/c. We measure mass in cm * c^2 / G. We measure energy in units of cm as well (since E=mc^2 and c =1).

So energy per unit volume has units of cm / cm^3 = cm^-2

Force / unit area also has units of cm^-2

Force = mass * acceleration = cm * cm / cm^2 = dimensionless
and of course area = cm^2

15. Nov 2, 2004

### Iraides Belandria

Dear pervect, I appreciate your dedication to my original question, but I still have an almost phylosophical doubt ¿ if thermodynamic properties are independent of time and trajectory, why the first law changes in a relativistic frame? ? Why the internal energy of matter changes ?.

16. Nov 3, 2004

### pervect

Staff Emeritus
I'm not sure I understand the question ??

I see that I've omitted mentioning, unfortunately, that P and $$\rho$$ are measured in the rest frame of the fluid with the approach where I quoted some equations from MTW.

The stress-energy approach is a little cleaner, in my opinion, but it's probably not as related to traditional thermodynamic quantities so it may be unfamiliar.

There's a discussion of the classical and relativistic stress-energy tensor's http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec12.html [Broken]
which may be helpful in explaining the stress-energy form of the differential conservation law, but beware of minor notational differences, e.g. their T44 is my T00).

Basically if you take a very small cube of space-time, the energy flowing into the cube must equal the rate of change of the energy density stored in the cube. Momentum must also be conserved in the same way.

The equation in terms of pressures and energy densities basically makes the same statement, it's just in terms of different variables.

Note also that the work done by gravitational fields will be zero in the limit as the size of the cube face goes to zero. This explains why the differential form of the first law of thermodynamics doesn't get tangled in the problem that energy in the gravitational field must exist but cannot be localized.

Last edited by a moderator: May 1, 2017
17. Nov 16, 2004

### Iraides Belandria

Pervect,
I have dedicated some time to understand your suggestions and to read the articles, but after this reading, I still think, that the internal energy of matter , U, is independent of trajectory and time and should coincide with the relativistic energy at rest, in other words,
e=u
therefore, for a pure sustance in a rest reference frame at zero height.
du= de= d(mc2)= dQ-dw
where Q and w are heat and work , respectively. u and e are defined at the beginnig of the thread.