A couple refs:
Four-vectors in Relativity(italics mine)
From my junior year mechanics book Mechanics Third Edition by Keith R. Symon in chapter 14:
– Ring
So where do you describe the location of the remaining mass?
Generally I have seen described in what I have read, 2 distinctions of energy:
“internal energy” – that energy which can be described as a part of a contained structure, such as heat, nuclear binding, anything inside a black hole
and “free energy” – Free photons and kinetic energy.
In those, I have seen mass associated with “internal energy” but not “free energy”.
“Take two photons with parallel velocities – the system has energy but no mass. Now take two photons with antiparallel velocities – the system now has both mass and energy.” – Ring
I understand the reasoning here, but I just generally haven’t heard it described like that.
For mass to exist, there must be inertia and gravitational influence. That is mass by definition. Would you say that the system with parallel velocities is different from the system with anti-parallel velocities in terms of inertia or gravitational influence?
This is just another name for mass.
E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]
In the zero momentum frame (and setting c = 1) E = m or m = E.
Many, but not all, undergraduate texts say this because people tend to think of the constituent parts of the system rather than the system itself.
Mass is not a thing it’s a property of a system. Your question is similar to asking where the energy resides in an electromagnetic wave.
That’s correct. Not all systems that contain energy have mass, e.g. two photons with parallel velocity. In other words you can’t find a zero momentum frame.
Absolutely, the anti-parallel photons change the intrinsic geometry of space-time whereas the parallel ones do not.
A hot object is more massive than the same object cold.
A perfectly reflecting box with a photon bouncing around inside has more mass and higher inertia than the same box without the photon.
If a nuclear weapon is detonated in a vault the vault weighs the same before and after the detonation.
“This is just another name for mass.
E[sup]2[/sup] = m[sup]2[/sup]c[sup]4[/sup] + p[sup]2[/sup]c[sup]2[/sup]
In the zero momentum frame (and setting c = 1) E = m or m = E.” – Ring
The value may be the same, but the units can’t be the same. It seems pretty loose to simply equate energy and mass like this.
"Mass is not a thing it’s a property of a system. Your question is similar to asking where the energy resides in an electromagnetic wave. " – Ring
Yep, so lets subdivide this system. If you break this system into pieces, will the mass of the pieces add up to the mass of the whole? If not, then why not?
"A hot object is more massive than the same object cold.
A perfectly reflecting box with a photon bouncing around inside has more mass and higher inertia than the same box without the photon.
If a nuclear weapon is detonated in a vault the vault weighs the same before and after the detonation." – Ring
Sure, all of these things are “contained”, fitting with the description I gave above, so yes those examples are certainly true. The containment is real. For the other cases, it seems that you are are defining something as contained simply because you can draw an arbitrary “system” line around it.
"Absolutely, the anti-parallel photons change the intrinsic geometry of space-time whereas the parallel ones do not. " – Ring
Do you have a reference to support this? Perhaps if you could demonstrate how the antiparallel photon system could exibit inertia, this would convince me.
“Many, but not all, undergraduate texts say this because people tend to think of the constituent parts of the system rather than the system itself.” – Ring
Alright then. Please provide a supporting reference.
[QUOTE]
Originally posted by DrMatrix *
**That would violate the first law of thermodynamics.[list=1]Thermodynamics[li]You can’t win.[]You can’t break even[/list=1] **[/li][/QUOTE]
- You can’t even quit the game.
I already did.
E[sup]2[/sup] = m[sup]2[/sup]+ p[sup]2[/sup]
(c = 1)
Anti-parallel photons
If p equals 0 then m = E
(and they have the same units)
Parallel photons
If p equals E then m = 0
That’s not a reference, thats an explanation. I was looking for an external source, superior to my Mechanics text for explaining the conservation of mass as you are describing it.
"If p equals 0 then m = E
(and they have the same units) " – Ring
How is that? When I looked at that I had units that certainly did not match. Please describe what specific units you are describing here and show me how those units can be used in 2 independent, valid equations, first representing energy and second representing mass.
He’s using the standard trick of “c=1, and thus I now write m when I mean mc[sup]2[/sup], and p when I mean pc.” If you implicitly put in the appropriate exponents of c in his equations, then, “E = m” works just fine.
This notation firmly brands Ring as a theorist. 
Seriously, this type of shortcut is used all the time, and it confuses the hell out of me. But, it really simplifies “equations”, it’s just that you have to implicitly add back in the factors of c in order to make the units work out.
OK, about those units: In SI, energy and mass are described in different units. However, in relativity, it makes a lot of sense to regard space and time as two aspects of the same thing, and therefore measureable in the same units. If you do this, then c is not 310[sup]8[/sup] meters per second, but rather exactly 1 (with no units). Put another way, in relativistic units, 310[sup]8[/sup] meters = 1 second, in exactly the same way that 2.54 centimeters = 1 inch, and 2.54 centimeters/inch is equal to exactly one. One side effect of this is that mass and energy can be described in the same units: I can have a mass of 511,000 electron-volts, say, or an energy of 10 kilograms. If this bothers you, then you can just stick all the c’s back into the equation, but the physics will be the same.
Now, mass is that part of the energy of a system that can’t be transformed away: In other words, when you go to the zero-momentum frame, all of the energy that’s left is the mass. Two parallel photons (or a single photon, for that matter) don’t have a zero-momentum frame, and as your frame moves faster, the energy of the system becomes arbitrarily close to zero. If the photons are moving away from each other, however, then there is a zero momentum reference frame, and the energy in that frame is the mass. If you go into any other reference frame, the total energy you observe for the photons can only be greater. If you had a large number of photons radiating outward from some central point (so you have spherical symmetry), then an object outside the photon shell would orbit exactly as though you had a chunk of rock there instead, with the same energy.
Why must containment be by some physical object? I could surround that photon explosion with a sphere of tissue paper, if you’d prefer, but the containment would add an arbitrarily small mass to the system, and wouldn’t matter anyway until the photons reach it.
And DrMatrix, Hawking radiation could, indeed, be used as you describe. Black holes are described completely by four parameters: Mass, angular momentum, electric charge, and magnetic charge (if it exists). They totally ignore lepton number, baryon number, and all the other quantum mechanical properties, usually summed up by saying that “Black holes have no hair”. So, for instance, if you have a black hole and drop a hydrogen atom into it, before, you have a net lepton number of 1 (1 from the electron, 0 from the proton, 0 from the hole) and a net baryon number of 1 (0 from the electron, 1 from the proton, 0 from the hole), and afterwards, you have a net baryon number of 0 and a net lepton number of 0 (0 from the hole). The Hawking radiation will then also not conserve lepton number or baryone number, but will produce approximately equal amounts of each. Hawking radiation and proton decay (if it occurs) are the only known processes for converting “normal” matter to useable energy with high efficiency.
“If you do this, then c is not 3*108 meters per second, but rather exactly 1 (with no units).” – Chronos
No, no, no. You can’t just drop units. c=1 is a mathematical convenice, but is still a velocity, and it still has must have a unit, usually light-sec/sec.
“I can have a mass of 511,000 electron-volts, say, or an energy of 10 kilograms.” – Chronos
511,000 eV is not a mass. It is certainly convenient in high-energy experiments to describe a particle in terms of it’s rest energy (or its total energy for that matter), but that doesn’t mean the 2 are equivalent. They are just closely related.
So in such an equation, with c =1, if the mass is 1 kg, the energy is:
1 kg * light-sec / sec. Again we need to be careful to destiguish between value and units. You can never have a velocity with no units.
Crap, that was me using OpalCat’s computer.
Correction, I didn’t square the units, so that should have been an energy of 1 kg * light-sec[sup]2[/sup]/sec[sup]2[/sup]
“If the photons are moving away from each other, however, then there is a zero momentum reference frame” – Chronos
Just because there is a zero momentum reference frame doesn’t mean that these photons can exibit inertia.
Let me ask this of everyone who agrees with this conservation of mass assertion:
If our two anti-parallel particles have rest mass, do you see the kinetic energy as contributing the mass? You pretty much need to right? After all, that kinetic energy could be converted into another form in a collision that could result in additional rest mass.
If anyone is brave enough to answer yes that they want to ascribe mass to kinetic energy in this system, we can go thru the exercise that shows in a relativistic context, ascribing gravitional influnce to kinetic energy leads to contradiction.
And yes folks I am quite familiar with the c=1 construction. I use it in space-time diagrams just like any reasonable person does. That doesn’t mean ya get ta drop the units! For shame!
Dude. I’m 10 years past my particle physics days, but let me see if I can help a bit.
What do you mean by inertia? They attract all other systems with mass in a manner in accordance with GR. Note that photons do not exhibit mass independently because you can translate their energy away by changing frames. But, you cannot do that to a two (non-parallel) photon system - it has an intrinsic energy, which means other objects “feel” a mass.
Who said anything about rest mass? The ‘m’ in “E = m” is not rest mass of a photon, but the mass of the two photon system. How could a collision with something result in additional rest mass? Do you just mean, for example, an atom absorbs one of the photons, the mass of the atom increases by the amount of energy it absorbed from the photon? What is the conceptual problem this gives you?
I don’t think it does lead to contradiction. If you translate yourself into some other frame, thus giving a larger momentum to the two photon system, you’re giving them additional energy. Where is the contradiction?
Actually, as Chronos explained it, you do get to drop the units. If you define 3 x 10[sup]8[/sup] meters to be equal (exactly equal!) to 1 second, then your units do in fact cancel out. You can argue whether it is appropriate to do this, but once you have made that mathematical statement, your units do in fact cancel out.
The statement I think you’re having trouble with is “space and time are just the same things, and so let’s measure them with the same units, seconds.” Once that statement is accepted, poof, no units but mass (in this equation.) I still think theory is a black art, but that’s just because I like playing with lasers and giant superconducting magnets more than equations. But, within the universe of theory, the units go away. I have to stop now before my brain explodes.
“What do you mean by inertia?” – douglips
I mean one of the 2 required properties of mass by definition. All mass has inertia and exibits gravitational influence.
“They attract all other systems with mass in a manner in accordance with GR.” – douglips
That is the second time this assertion has been made without support. Do you have any reliable reference to back up this claim that this system exibits gravitational influence?
“Who said anything about rest mass?” – douglips
Perhaps I should have been more clear. I was proposing a system that used something other than photons-- electrons perhaps. With such a system it can be demonstrated that either you must ascribe mass to kinetic energy, or allow for mass to change.
“I don’t think it does lead to contradiction. If you translate yourself into some other frame, thus giving a larger momentum to the two photon system, you’re giving them additional energy. Where is the contradiction?” – douglips
The contradiction comes when you account for gravitational influence. I can go thru it, but it may take me a little while to prepare.
“Actually, as Chronos explained it, you do get to drop the units. If you define 3 x 108 meters to be equal (exactly equal!) to 1 second” – douglips
Think about what you just said! You just set a unit of time to be equal to a unit of distance! That doesn’t work.
3 x 10[sup]8[/sup] meters is not equal to one second. It is equal to one light-sec. light-sec is a unit of distance, thus it can be translated from meters. seconds cannot be thus translated.
So if we want to make c = 1, we make c = 1 light-sec/sec
"The statement I think you’re having trouble with is “space and time are just the same things, and so let’s measure them with the same units, seconds.” – douglips
This is incorrect. When you describe a space-time interval, your units are in meters. The time factor doesn’t simply show up as seconds-- it is modified. Gimme a sec and I will dig out the simple equation for the measurement of a space-time interval.
Here is the equation for a space-time interval. Bold values are vectors.
x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] + c[sup]2[/sup]t[sup]2[/sup] = I[sup]2[/sup]
As you can see, seconds units here are not lain side by side with meters. All of the space elements in here are simply meters[sup]2[/sup]. The time element is (meters[sup]2[/sup] / sec[sup]2[/sup]) * sec[sup]2[/sup], which nicely reduces to meters[sup]2[/sup]. You cannot trade seconds freely with meters.
Grr. Correction I put a plus sign in the wrong place. Should be:
x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - c[sup]2[/sup]t[sup]2[/sup] = I[sup]2[/sup]