# Relativity questions

1. Nov 5, 2004

### kakarukeys

Can anybody help me?

(1) Lorentz Group preserves the scalar product of Minkowski space, what about Poincare Group?

(2) Lorentz Transformations include spatial rotations, why only conservation of Energy-Momentum is associated with Lorentz symmetry (but not Angular Momentum)?

2. Nov 5, 2004

### Fredrik

Staff Emeritus
1. A Poincaré group transformation is a map

$$x\mapsto\Lambda x+a$$

Such that $$\Lambda^t\eta\Lambda=1$$. The Minkowski space "scalar product" is not preserved when $$a\neq 0$$

2. Conservation of energy is associated with translations in time. Conservation of momentum is associated with translations in space. Conservation of angular momentum is associated with rotations.

3. Nov 5, 2004

### Garth

To conserve either energy, momentum or angular momentum you therefore have to have a preferred frame of reference. Energy-momentum is a frame independent geometric object and is invariant in any frame of reference and therefore the conservation of energy-momentum is the one associated with Lorentz symmetry.

Garth

4. Nov 5, 2004

### Fredrik

Staff Emeritus
Huh?! What makes you say that?

No it isn't. The energy-momentum tensor is of course a tensor, so it's components are different in different frames. The momentum four-vector is also a tensor (by isomorphism) so it's components also depend on the frame.

The "momentum squared"

$$p^2\equiv-(p^0)^2+(p^1)^2+(p^2)^2+(p^3)^2$$

is frame-independent, but so is the square of any other four-vector.

For those who don't know this already (I think Garth does), the invariant value of p² is denoted by -m² (since it is always negative, unless the particle is massless or a tachyon). The quantity m defined in this way is just the mass. (Some people prefer to call it "rest mass").

(I'm using relativistic units (c=1) and a -+++ metric).

5. Nov 5, 2004

### jcsd

Fredik, most would say that tensors are frame independent ('cos the motivtaion behind tensors are frame indepednt expressions of physical laws), after all choosing a frame is just choosing a particualr basis (when we talk about a vector in different bases we still think of it as the same 'object').

6. Nov 5, 2004

### Fredrik

Staff Emeritus
I know that of course, but Garth said that energy-momentum "is an invariant quantity in all reference frames" (my emphasis) and that doesn't even make sense unless it refers to the components of the tensors. Nothing can be more frame dependent than the components of a tensor.

Last edited: Nov 5, 2004
7. Nov 5, 2004

### jcsd

I took it to mean something along the lines of: if we change the basis a four vector will still be the same mathaematical object, though with differnt compoents, but a three vector will change into a different 3 vector.

8. Nov 5, 2004

### Fredrik

Staff Emeritus
Maybe that's what he was thinking, but it's not what he said. You're probably right though.

9. Nov 5, 2004

Staff Emeritus
The components of a four vector change with frame, but its magnitude remains invariant. So the magnitude of the energy-momentum four vector is invariant, though its composition as energy and momentum change with frame.

10. Nov 6, 2004

### Garth

Thank you SelfAdjoint & jcsd. Perhaps I should have made myself more clear, I thought the word geometric object, which exists in 4-space independent of any 3-space foliation of space-time, would have been self explanatory. [It is basic see MTW page 48]
In the general case in GR the energy or momentum of an particle in an inertial frame of reference is not conserved. You need a time-like Killing vector which does not in general exist (although special static cases in which the metric is time-independent can be considered).
The preferred frame of reference I referred to is that one in which the energy or momentum are conserved.

Garth

Last edited: Nov 6, 2004
11. Nov 6, 2004

### pervect

Staff Emeritus
Although we've been over this before, Fredrick hasn't, so there's probably a few things I should say.

Let's say we have a small body orbiting the sun. If we use Newtonian mechanics, it's reasonably obvious that the momentum of the small body is not conserved. The momentum of the system as a whole is conserved, but not just the momentum of just the small body.

The energy of the small orbiting body is conserved, as long as one takes into account the gravitational potential energy. But the energy of the small orbiting body is not conserved if one does not add the gravitational potential energy to the small bodies kinetic energy.

So this is how it works in Newtonian mechanics - let's jump to relativity.

If we let the energy-momentum 4 vector of a small object orbiting a large object be P, then the quantity

Pa ka

is a constant of motion. Here ka is the "Killing Vector" that represents the timelike symmetry of the system. There are several ways of determining what the Killing vectors are, one of them is that they must satisfy "Killings equation"

$$\nabla_a k^b + \nabla_b k^a = 0$$

This is equivalent to saying that the covariant derivative $$\nabla_a k^b$$ must be a completely antisymmetric tensor. Note that a unit vector in the time direction in the exterior vacuum region of the Schwarzschild metric is a Killing vector. This makes it a timelike Killing vector, which yields a conserved energy. There are space-like Killing vectors in the Schwarzschild metric as well - these yield conserved momenta (i.e. the angular momentum of the orbiting body). This is an example of Noether's theorem in action - time symmetries give conserved energies, space symmetries give conserved momenta. Noether's theorem applies to this situation because geodesics are an extremum of an action.

Because Pa ka is a conserved quantity for an object following a geodesic, and because ka is a unit vector in the time direction for the Schwarzschild metric in Schwarzschild coordinates, P0 is a constant of motion for a particle orbiting a massive body in the Schwarzschild metric in Schwarzschild coordinates.

Opinions vary on what name to give this quantity, some people here like to call it the energy of the particle, but I'm not totally convinced that this is correct. This has been the topic of a reasonably endless debate. Note that the simplest form of the expression as written occurs only when Schwarzschild coordinates are used (it's coordinate dependent, not coordinate independent).

Names aside, it's very useful to know that this quantity is conserved.

However, this quantity is not the same as the total energy of the system. If the "small" orbiting body is the Earth, this does not give the total energy of the Earth-sun system, for instance.

The problem of finding the energy of a system in General Relativity is not the same as the problem of finding conserved quantites for orbits (particles following geodesics). It's a different problem entirely, though there are some similarites. It turns out that the energy of a system in GR can be defined if either one of two conditions is met. One condtion which permits a conserved energy to be defined is if the system has a time-like Killing vector, as the Schwarzschild metric does. When this condition is satisfied, one has both a definition of energy, and an observer-independent notion of "where" the energy is. In this special case one can separate the total energy of the system out in terms of kinetic energy and gravitational binding energy, as one does for the Newtonian problem.

The second, more general condition that conserves energy in GR is is that the system exists in an asymptotically flat space-time. If this condition is met, and there is no time-like Killing vector, the total energy of the system can be defined, but the energy can no longer be localized as it was in the previous case. To be really precise, there are at least two notions of energy in asymptotically flat space-time, the Bondi energy and the ADM energy. These two notions differ in how the "bookeeping" of the energy stored in gravitational waves is handled. These two different notions of energy can otherwise be shown to be equivalent, though it's a difficult enough job that Wald,s textbook "General Relativity" for instance, only gives references to where this is proved rather than proving it in the textbook itself.

The sci.physics.faq "Is energy conserved in General Relativity" is also a useful reference (and less technical than what I wrote), it is located here

There's another notion of conservation of energy in GR that's worth mentioning. This is a differential conservation law which applies to the stress energy tensor. That makes a total of 4 different notions of energy and it's conservation that have been discussed so far!. The basic equation is

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

This is yields 4 equations (for b=0,1,2,3, the sum over the repeated indiex "a" is implied). These equations can be interpreted as a conservation of energy in an infinitesimal region of space-time, and as a conservation of momentum in an infinitesimal region of space-time. Unfortunately, these equations do NOT have anything useful to say about a finite region of space-time. A useful web page discussing the stress-energy tensor and how these equations correspond to the conservation of energy in an infenitesimal cube is located http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/Sec12.html [Broken]

Garth is unhappy with the fact that energy (total system energy in a non-infinitesimal volume of space-time) is not conserved for arbitrary space-times in general relativity, and has created an alternate theory of gravity called SCC in which energy is conserved for arbitrary space-times. I'm sure he'll be happy to post the link to his papers.

Last edited by a moderator: May 1, 2017
12. Nov 6, 2004

### Garth

Thank you perfect for that excellent and informative post.

The link to a post on Self Creation Cosmology (SCC) can be found

here.

Garth

13. Nov 6, 2004

### rv

i am new here, but i do have a question about proper time what do
the following mean-
1."the proper time is the time as seen by an observer in the rest frame of the particle"
2. "light signals in four dimensions also move along straight lines but the proper time along the line is zero"

14. Nov 6, 2004

Staff Emeritus
1. A rest frame of a particle means the state of an observer who sees the particle at rest, i.e. with zero velocity. As long as the particle is massive, there will always be such a rest frame; you're moving alongside it at the same speed as it is going relative to other observers. Then you experience the same time that it does, and your clock shows it. This time, accumulated at the particle as if it were standing still, is proper time. Proper time provides a parameter for the world line of the particle.

2. Light signals have no rest frames; it is impossible in special relativity to see them standing still, so they experience no proper time, and their world lines, which in special relativity are straight lines, have "accumulated length" equal to zero. Because they are strainght they are geodesics (straightest lines), because the have length zero they are null geodesics.

15. Nov 6, 2004

### Garth

Hi rv welcome to the Physics Forums!

The question is, "Why should the geodesic of a light ray be a null geodesic?"

So how do we combine space and time?

The formula Einstein came up with, with a little help from his former lecturer Minkowski, was that using Pythagoras, you had to subtract the square of the time dimension:-

ds2 = dx2 + dy2 + dz2 - c2dt2.

Now a light year is the distance light travels in one year; so if you think about a super nova exploding 1000 light years away, and being seen on Earth today, you would be seeing that supernova as it exploded 1000 years ago, it has taken that length of time for the light to reach us. But what is the interval or space-time separation between the event of that explosion and the event of it being seen on Earth?

According to the formula I gave, which is called the metric of space-time, the interval is (1000) squared minus (1000) squared, which is zero! The space-time separation along a light beam is always zero. That is why it is called a null geodesic.

Garth

Last edited: Nov 6, 2004
16. Nov 7, 2004

### kakarukeys

noted.

Fredrik,
I think it should be $$\Lambda^t\eta\Lambda=\eta$$

the conservation law should look like: $$\frac{d}{d\tau}p^a = 0$$

17. Nov 7, 2004

### Fredrik

Staff Emeritus
Oops. You're right of course.

18. Nov 7, 2004

### Fredrik

Staff Emeritus
It didn't even occur to me that you might be talking about general relativity. Since the question was about SR I just assumed the answer was too.

19. Nov 7, 2004

### Fredrik

Staff Emeritus
Pervect: Thanks for that excellent explanation about energy in GR. I understand SR very well, but some of the things you said (about GR) were new to me. I've read the first few chapters of Wald's book, but not the chapter that explains this.

Last edited: Nov 7, 2004
20. Nov 8, 2004

### rv

Thank you Garth and SA for defining proper time, and it may be an obsession but here is yet another question about it;
consider black holes and schwarzchild solution: the solution becomes singular at r=2m called critical radius and after some calculations we find that , the proper time for a particle falling to the criticle radius is still some finite qty .
for values of r<2m we switch coordinates and b/c the gravitational field is changing fast we take g(mu,nu) as time dependent, ok we arrive at a now non singular solution at r=2m . but the particle cannot communicate with the outside world as again the signal to an outside observer will take forever to reach. now this is fine but can we calculate proper time inside the b'hole if so then in what coordinate system. and once we know the proper time then we do know every thing dont we about inside?
incidently i am reading general theory of relativity by dirac.