# Mass-energy of a system?

1. Jun 26, 2006

### tim_lou

hi, this is my first post here. i've always loved physics... here's just some thoughts about a problem:

let's say i have a particle at rest relative to me (inertial frame) and has a rest mass of $m_0$.

now lets say in another frame of reference moving in the positive x direction with velocity v relative to me. what would be the apparant mass of that particle? simply $\gamma{m_0}$?

if so then,
lets say i take a closer look at that particle, and it is actually consisted of 2 particles. these 2 particles are moving relative to each other and they are interacting with each other via some forces. now, if i sum up the apparant masses of the two particles and the mass equilvalent of the potential energy between the two, would i get $m_0$? what is the determinig factor of whatever not one is calculating the "rest mass" of a system? what determines if a system is at "rest" relative to me? the center of mass (does this idea still apply in relativity)? or is the concept of "rest" mass irrelevant? if it is irrelevant, why (as a whole the system is still a "particle")?

regardless of what the above answers are. Let's say i get an answer of m ("rest mass" of the system). Then when the mass of the system is measured in the moving frame i said above, would the answer be $\gamma{m}$?

if all these work, is it possible to calculate the electrical potential of moving charges and gravitional potential of moving masses using this idea?

*btw, sry for my bad english, if anything needs to be clarified, please tell me. any help is appreciated on this problem.

Last edited: Jun 26, 2006
2. Jun 26, 2006

### actionintegral

I wanted to answer your questions, but upon careful reading, I counted 9 questions. Perhaps you should make 9 separate posts, one at a time. I suspect that after the first few posts, you will be able to answer the questions yourself.

3. Jun 26, 2006

### tim_lou

well... i wouldnt wanna flood the forum..
i have been thinking about these questions for some time. it is mainly for the derivation of potential(gravitional and electrical) of moving objects, all i want to know is if this approach to the problem might be viable (if my logic is correct). you don't have to answer them all, just one or two is good enough...those questions are all related anyway.

well, the more important ones are:
can i treat a system as on object and calculate mass by suming all energy and divided by c^2?
can i relate the mass in one frame to another simply using gamma*m0 and how do i know if a system is at "rest" (to find rest mass)?

Last edited: Jun 26, 2006
4. Jun 26, 2006

### actionintegral

Well, the notion of "rest mass" is by definition. You start by saying "I am at rest with an object" as a given.

Now imagine you are looking at 100 objects, and they are all at rest.
You can easily find the center of mass of the objects!
Now imagine you are looking at a snapshot of the 100 objects, but now each object has its own velocity. By giving an object enough velocity, you can make it as heavy as you like! You can move the center of mass
as close as you like to one of the objects by that object enough velocity.

5. Jun 26, 2006

### tim_lou

so center of mass doesnt really mean anything in relativity correct?

then how would I know if a system is at "rest" relative to me? it is easy to say an atom is at "rest" (neglect any quantum effect) but if we look closer, there are electrons, protons moving all around the place. so "what" "makes" the atom at rest? mathematically?

Last edited: Jun 26, 2006
6. Jun 26, 2006

### actionintegral

Go back to the 100 objects at rest. You can calculate their center of mass.

Next consider a person moving by. He will see the mass of each object
increase by some amount because he is moving. Each object will seem heavier to him. But by the same amount - so he will calculate the same
center of mass as you. (of course, it will be moving past him, just like you are)

7. Jun 26, 2006

### tim_lou

but what if the 100 objects are moving, let's say circling around? let's say in this system, one object is moving to the positive x direction at v. however, according to a frame also moving to the positive x direction at v, the object will be at rest and the apparant mass will be LESS. it will become some form of vector addition according to lorenz transformation.

if the center of mass idea still works in relativity (i doubt it since the derivation of it from torque and equilbrium must change according to relativity) then, the net increase factor will be $Delta\gamma$ of the center of mass according to a moving frames, am i right? i guess the tricky part is the "center of mass" of relativity.....and what exactly is the definition of "rest" when looking at a system

when an object is at rest, and the object is made up of orbiting moving particles, what exactly is at "rest" mathematically(in relativity)?

Last edited: Jun 26, 2006
8. Jun 26, 2006

### actionintegral

Well, let's proceed in a step - by - step fashion. Imagine that each object
has a fixed velocity in a random direction. Then the relativistic mass of each object is a fixed amount! And so you can still calculate the center of mass, which may be stationary or moving in a straight line.

9. Jun 26, 2006

### actionintegral

Now give each object a random, time dependent velocity like a swarm of bees. At each instant, we can take a snapshot of the system and calculate the center of mass.

In the next instant, maybe an object has moved a bit. And maybe it's relativistic mass has changed a little. So we might expect the center of mass to have moved a little bit. But we can always calculate a center of
mass at each instant.

10. Jun 27, 2006

### pmb_phy

Yes.
No. If the particles are interacting you have to take into account the speed of each particle in the system (relative to the observer's frame of reference), the (potential) energy associated with the interaction of the particles and the stress in the field surrounding the particles. This requires the use of the so-called stess-energy-momentum tensor.
Whether or not the total momentum of ther system is zero or not.
Yes. The idea still applies in relativity. The center of mass has a very well defined meaning. However the value is not observer independant. But a quantity can be well defined and still be observer dependant. See

http://www.geocities.com/physics_world/sr/center_of_mass.htm
Yes.
I'm not 100% sure but I think so.

Pete

Last edited: Jun 27, 2006
11. Jun 27, 2006

### pervect

Staff Emeritus
Gravitational potential energy cannot be defined in general in GR. So you are probably about to get into deep water and make some mistakes.

Take a look at

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

before starting.
First of all, even in SR there are at least two defintions of mass. You seem to be using a rather old-fashioned defintion, the "relativistic" mass. Nowadays, the invariant mass is much more popular.

The invariant mass of a single particle is given by

M = sqrt(E^2/C^2 - P^2/C^4)

where E is the energy of a particle, and P is its momentum.

This formula works for an isolated system in SR as well as a single particle.

For either a single particle or an isolated system in special relativity, the invariant mass is constant for all observers. This is what makes it handy. It is thus a property of the system itself, and independent of the obsever. (Note that the "relativistic" mass depends on both the system and the observer!).

See for instance the sci.physics.faq

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html" [Broken]

For a non-isolated system in SR, the invariant mass of the system is not necessarily constant under a Lorentz boost. This rather interesting, I'm not sure how widely known it is, but it's been discussed a lot on the board. A specific example, if you have a system that's under external pressure, E^2/c^2 - P^2/c^4 is not constant for observers moving at different velocities.

Going back to isolated systems:

You can see that the invariant mass of a particle is just E/C^2, and when the total momentum of a system is zero, it's invariant mass is also just E/C^2.

You can also see that the invariant mass of a system is equal to its relativistic mass when the momentum is zero.

Things get much more complicated in GR. There are at least three defintions of mass in use (probably more), and while energy is conserved locally (the differential form of enregy consevation is built into the equations of GR itself), there is no single conserved number that can represent the "energy" of the universe. (See the FAQ I mentioned previously about energy conservation in GR).

[/quote]

A system is at rest if its momentum is zero - that's probably the easiest criterion.

Last edited by a moderator: May 2, 2017
12. Jun 28, 2006

### tim_lou

thank you for the detailed answers, things make a lot more sense now. so a system having zero total momentum is at "rest"

doesn't
M = sqrt(E^2/C^2 - P^2/C^4)
come from
$E^2=(mc^2)^2+(pc)^2=(\gamma{m}c^2)^2$?
oh i see, this definition can be used for a whole system, since net momentum can be easily found and it takes care of potential energy brillantly.

now pervect said that this definition of rest mass is invarient under lorenz transformation(as long as there is no external force and the sytem is isolated).... which will be perfect to derive potential of a moving charged particle! thx guys for the help.

Last edited: Jun 28, 2006
13. Jun 28, 2006

### pervect

Staff Emeritus
Take a look at

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

Note that the curl of E is not zero.

Now if the E field were the gradient of some potential function, i.e. if
$E = \nabla V$

then it's curl would be zero, for the curl of a divergence is always zero.

Therfore the electric field of a moving charge cannot be described by a scalar potential, it is not the divergence of any potential function.

You might also be interested in http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf

This gives the E-field of a moving charged particle.

The good news is that this gives you the solution for the electromagnetic case. The bad news is that this result cannot be applied directly to gravity, though it does give some general insight as to roughly what happens.

Note that Gauss's law says that the intergal of an electric field over the normal surface is equal to the enclosed charge, and independent of the velocity of the charge.

This will NOT be true for gravity.

Even measuring the "gravitational field" is quite problematical. An electric field can be measured at a point by comparing the acceleration of a neutral particle to a charged test particle.

A gravitational field cannot be measured in this manner, because gravity affects everything.

Gravitational fields can be defined in the case of a static geometry, but there is unfortunately no generalization of this to a non-static geometry. The case of a moving, isolated particle is a non-static geometry.

If you want to get correct answers for the gravitational case, work the problem in the rest frame of the dominant mass.

The equations of motion for a particle of small mass in a Schwarzschild metric are well known, see for instance

http://www.fourmilab.ch/gravitation/orbits/

If you don't have a dominant mass, you've got a serious problem. You basically have to solve the 2-body problem in GR. This requires numerical approaches.

You can tell when you have a dominant mass by the fact that the frame in which the total momentum of the entire system is zero is basically at rest with respect to one of the masses in the system.

Also note that while the gravitational field of a single moving particle has severe difficulties in being well defined because the system is not static, the aggregate field of a bunch of moving particles does not necessarily have this problem.

See for instance

in which I talk about the gravitational field of a sphere of hot gas. A hot gas can be thought of as an ensemble of moving particles with zero average mopmentum. It can be seen from this analysis that the problem is not trivial. Bascially, the gravitatational field due to the hot gas greater than the gravitational field due to the energy density (E/c^2) of the gas alone. The sphere enclosing the hot gas, though, is actually lighter when it is under tension. The total mass of the system turns out to be equal to E/c^2.

The techncial way of putting it is this: both pressure AND energy cause gravity. For an isolated static system, the intergal of the pressure terms will always be zero, and thus you'll find the mass of an isolated system is always E/c^2.

In the sphere of hot gas, the pressure in the center of the sphere causes its contribution to gravity to be greater than E/c^2 alone. The tension in the shell surrounding the sphere causes its gravitational contribution to be lower the E/c^2.

The analysis technique I described used above works only for static systems, BTW. You'll need something like the ADM mass (and not the Komar mass) to handle the mass of the sphere of hot gas if you let it explode or escape.

The Komar mass has a relatively simple analogy to Gauss's law, but applies only to static geometries. The ADM mass does not have any simple analogies, but applies to any asymptotically flat space-time.

If your space-time is neither asymptotically flat nor static, there isn't any general defintion of energy of a system or mass of a system in GR. (I think I mentioned the appropriate FAQ).

This is probably too much information to give all at once, but I'm not sure what parts of it can be cut.

Last edited by a moderator: Apr 22, 2017
14. Jun 29, 2006

### pmb_phy

For that relationship to be valid for a system consisting of particles alone the particles must not interact in anyway whatsoever, with the exception of contact forces. Otherwise one must take into account the the stress-energy- momentum of the field which exists between particles in order for them to interact when the particles are seperated by a finite distance.

Pete

15. Jun 29, 2006

### actionintegral

I would like to keep this thread alive, if possible. I am interested in seeing where the original poster is going. But you guys jumped way out of my league! I am still sitting there at the center of mass of a swarm of moving objects! So going back to the original post...

Now that we have established that it makes sense to move to the center of mass of a set of relativistic objects, is the objective calculate the potential due to these objects, given that each object is a) massive and b) charged?

16. Jun 29, 2006

### tim_lou

wot!!!!:surprised

thank you the all these information....

i wont pretend to knows things that i dont... all these stuffs are really way out of my league as actionintegral said....

my tensor math is basiclly no math at all. i can do some calc 3 vectors stuff but...

the way i understand gauss law is when a charge isn't moving. gauss law can be easily derived from the inverse square coulumb's law using a imaginary sphere.

i know how electromagnetic wave equation is derived from maxwell's equation. since c is constant in lorentz's transformation, the wave equation must be invarient in lorenz's transformation. should maxwell's equations be then invarient under lorentz's transformation as well? i dont know if working backward from the wave equation would give me maxwell's equations without any assumption that it is true.... i still need to work on that...

i hate the way how they derive electric field in moving charge. they simply say$\norm E_y=\gamma{E_y'}$ without any justification and $\norm E_x=E'_x$
so the fields are basically like... contracted?

perhaps my questions are silly... i'm sorry, but still...i'm not a fan of believing because of not-knowing-anything. i found a whole bunch of articles about retarded potentials and the derivation of electromagnetic field based on 'em....thought i found some easier ways but...i'll just go through those and try to get as much as i can get based on my math....

as for the "stress-energy-momentum"... i really appreciate your input... but sorry, i honestly have no idea what they are (need to learn more physics )... i guess i'll just try some other ways.
And gravity, oh man gravity... i'm already scared when i think of GR and the 5 pages einstein field equations. i guess i'll stay away from it until i learn more physics.

also, pervect is right... i totally forgot about changes in magnetic field which results in a non-static field in a moving charge. so even if i find the potential function, i cannot simply take the gradient of it...darn!:uhh: so, non of my methods that use energy could work now...

Last edited: Jun 29, 2006
17. Jun 29, 2006

### actionintegral

Well - you said potential. If the objects are charges, you can get the coulomb potential. If the objects are masses, you can get the gravity potential. You can then just add those up. The magnetic part is a little trickier because that's a vector but the idea is the same.

Now, for someone moving past you, they will see a little different picture.
The velocities will be a little different, some extra magnetic fields will be around, and the masses will be larger.

But nothing really mind-blowing.

18. Jun 29, 2006

### tim_lou

but the thing is. when something's moving, the electric field changes, producing magnetic fields. at the sam time, magnetic field changes, producing a component of electric fields. these fields relate in a way according to maxwell's equation but are not explicitedly formulated... (i still needs to proof to myself how maxwell's equation (specially gauss's law) is invarient in lorentz transformation) the mathematics gets a whole lot complicated...
now i gotta go step by step.... from the basics to the complicated....

19. Jun 29, 2006

### pervect

Staff Emeritus
Yes, definitely, Maxwell's equations are invariant under Lorentz transformations.

You can see more details if you look at some of the earlier web pages. pages

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_11.pdf
starts off with a disucssion of Gauss's law and how charge is invariant under Lorentz transforms.

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_12.pdf
explains how surface charges vary as a result of Lorentz transform. Charge is invariant, but area is not invariant, thus charge / area changes as is described on this webpage.

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_13.pdf
builds on the previous slide to discuss how electric fields vary with velocity. The results are used in

http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_14.pdf
and
http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

present the results, I think you've seen them.

If you've got the time, you might want to look at earlier web pages. The eariler pages discuss background, notational details, etc, including some of the material I presented on invariant mass.

Page 9 talks about how forces transform. Page 16 builds on the result of page 9 and page 15 to illustrate that force is always
electric field * charge.

The complete list of figures for this website is at
http://www.phys.ufl.edu/~rfield/PHY2061/

under "lecture notes" (it's frame driven, so until you click on "lecture notes" you won't see the list of figures).

So far your questions appear to be right on the mark.

As far as retarded potentials go, you can derive a set of vector potentials for the electromagnetic field, known as the Lienard-Wiechert potentials. (A scalar potential doesn't exist as I remarked earlier). You should get the same result with this approach, but it will tend to be more abstract and less physically motivated - though quite useful.

One can also get into tensors to treat the same material, but I won't go into a great deal of detail since you haven't studied them yet. I will mention that the "Faraday tensor" is the single tensor that represents both electrostatic and magnetic fields, though. So if you ever get interested in the tensor approach, the Faraday tensor is the one you want to read about for electromagnetism.

Last edited by a moderator: Apr 22, 2017
20. Jun 29, 2006

### pmb_phy

If you put a little effort into this you'll be able to grasp what this tensor is. I made a web page to describe this tensor. Please see

http://www.geocities.com/physics_world/sr/mass_tensor.htm

The potential that is of use here is not the coulomb potential, its the potential energy of a multi-particle system. The total potential energy of a static system of charged particles equals the work required to bring all the particles in from infinity to their final position. The mass of this system as measured from the rest frame of these particles will be the sum of the proper masses of all the particles and the total mass-energy associated with the electric field. However if you transform to a frame of referance moving relative to the rest frame then this is no longer true. The total inertial mass of the system will equal the ratio of the magnitude of the total 3-momentum of the field divided by the speed of the system. This sum will include the relativistic masses of all the particles in addition to the mass associated with the stress in the field. Neglecting the stress in the field will lead to an apparent paradox. An example was given by Rindler and Denur in the Am. J. Phys. See

This demonstrates the importance of field stress in calculating the relativistic mass and the total 3-momentum of the system.

Pete

Last edited: Jun 29, 2006
21. Jun 29, 2006

### pervect

Staff Emeritus
Consider the electric field E of just a single moving charge. There is no potential function V such that the field is given by

Ex = $\partial V / \partial x$ Ey = $\partial V / \partial y$ $Ez = \partial V / \partial z$

as I mentioned earlier. I get the impression from your post that you are falesly assuming that there is such a scalar potential function and asking what it is.

Thus it makes no sense to ask what "the potential is" - it doesn't exist - unless one is talking about a 4-potential and not a vector potential. I don't think you are talking about a 4-potential here, though.

22. Jun 29, 2006

### pervect

Staff Emeritus
A quick note on magnetic vector potentials and Lienard-Wiechert potentials, for readers with vector calculus.

By introducing the familar concept of "voltage" (electric potential) and adding to it the concept of the "magnetic vector potential", one can automatically satisfy two of maxwells equations.

The electric potential V is just a scalar function of time and position.
The magnetic vector potential $\vec{A}$ is a vector (3-vector) which is also a function of time and position, just like the electric potential.

The approrpriate defintions are
$$E = \nabla V - \frac{\partial \vec{A}}{\partial t}$$
$$B = \nabla \times A$$

The last equation can be taken as the defintion of the magnetic vector potential $\vec{A}$ (which is not unique, but has a gauge degree of freedom,.

Specifically one can add the divergence of any scalar function to the magnetic vector potential, for the curl of a divergence will be zero, without affecting any fields.

This is very similar to the way that one can add any scalar function of time to the electric potential V.

The non-uniquness of V and A are known as the "gauge degree of freedom" of electromagnetism.

It can be seen that two of Maxwell's equations are automatically satisfied with these defintions - the divergence of B is guaranteed to be equal to zero, the curl of E is also guaranteed to be the right value.

The remaining two of Maxwell's equations yield a D'Alembertian equation for V and A.

The solution of V and A in terms of sources (charges & currents) is given by the Linenard-Wiechert potentials

See for instance

http://en.wikipedia.org/wiki/4-potential

(V,$c\vec{A}$) are often taken together as a single 4-vector - which, like other 4-vectors, is Lorentz invariant.

23. Jun 29, 2006

### tim_lou

thank you for all your input pervect... I'm sort of a slow and stubbborn person... i do not easily get in the ways of assumptions and hypothesis... and i have to get a very VERY firm grips of fundamental ideas before applying them. so i'll take a lot of time to learn all these new stuffs. and yes, uh... Lienard-Wiechert potentials, i've been trying to understand that too. actually, i was trying to derive it without any help or hint...(thats why this post was created)

so, i see, when things are moving. charge density is "shrinked". so, if i use the relationship of current density vector and drift velocity, and transform the drife velocity in a moving frame... then the current density can be "redefined" this way,right?

also... now i see why the L-W potential equation has 2 functions!! i was wondering what the heck the vector function was!

a somewhat offtopic question: i see all your great resources come from university of florida. are you a professor over there? sorry for this personal question, you dont have to answer it if you dont want to.

24. Jun 30, 2006

### pervect

Staff Emeritus
No relation here to the U of Fla. They do have a good website on E&M though, I link to it a lot.

You look like you're about at the stage where you could easily but usefully spend about 300+ in textbooks.

For E&M I'd suggest Purcell for the undergraduate level

Purcell: Berkeley Physics Series Vol 2.

(140 bucks new from amazon)

Later on, or if you are interested in tensors, I'd suggest Jackson for an E&M textbook (graduate level) in addition to Purcell (better as an undergraduate introduction).

Jackson: Classical Electrodynamics

(97 new from amazon)

Learning tensors in E&M will be a big help in learning GR if you want to continue on.

And of course, for the Lagrangian and Hamiltonian approach to physics, Goldstein, "Classical Mechanics" (that's 118 new).

You can get by lower if you buy used, of course.

25. Jun 30, 2006

### pmb_phy

Note: One cannot speak of the energy/invariant mass of a system of pointlike particles. The energy of such a configuration diverges in such cases.

I don't believe that this is a correct usage of the term "invariant." If you know of a textual source of such usage can you please post the reference? Thanks.

An invariant quantity is a quantity which remains unchanged upon changes in coordinates (e.g. a tensor of rank 0). Another usage of invariant is, within the domain of special relativity, the statement of the principle of relativity which states that
. I believe the correct term is Lorentz covariant (e.g. a vector is a "covariant object"). The meaning of the term "covariant" depends on the context in which the term is used. E.g. within the domain of GR. I.e. the Principle of general covariance is stated as
The definition of invariant equations is different than the definition of covariant equations. Invariant equations are those equations whose form is unchanged by coordinate transformations and quantities such as the speed of light etc. are also invariant.

Pete

Last edited: Jun 30, 2006