[JP]

Bill, the inertial mass of a particle does depend on its motion.  The important bit is that the way the interaction is measured is observer dependent, so that what matters is the relative velocities between observer and observee, which keeps the whole thing in line with special relativity.

[jopie64]

Yor_on, here's the same question, but in a slightly more extreme form:  If you have a box made of perfect mirrors and you inject some light into it, the box's energy has now increased.  If it's sitting still next to you, its mass increases (by E=mc², which holds for stationary objects).  So clearly its mass, measured at rest, went up.  Since invariant mass is supposed to not change with reference frame, and the rest frame is a reference frame, its invariant mass also went up.  Additionally, if you try to push it, you'll find its inertial mass went up. 

But photons individually have no mass?  How did it gain mass?

To what I've learned from physics, photons dont have mass, but they do have impulse. Photons can 'push' things when something absorbs or reflects it. The mirrors reflect the photons. So when you push the box and accelerate it a bit, photons push harder against the side you push. So you feel resistance. Even so photons are following the spacetime curve of gravity, so they tend to move more down then up in the box. So more photons reflect to the bottom then to the top of the box. This way the box feels heavier.

My question is, does this increased mass you feel, because of the 'impulse pressure' (or whatever you call it), also generate gravity?

[bizerl]

To what I've learned from physics, photons dont have mass, but they do have impulse. Photons can 'push' things when something absorbs or reflects it. The mirrors reflect the photons. So when you push the box and accelerate it a bit, photons push harder against the side you push. So you feel resistance. Even so photons are following the spacetime curve of gravity, so they tend to move more down then up in the box. So more photons reflect to the bottom then to the top of the box. This way the box feels heavier.

My question is, does this increased mass you feel, because of the 'impulse pressure' (or whatever you call it), also generate gravity?

I like this idea, however it would only increase the weight of the box if it was already in a gravitational field, but not necessarily the mass.

It seems like one of those ideas that depend on how "mass" is defined and measured.

[JP]

Yes!  The box is a source of gravity, whether you call it "mass" or not, it certainly contains energy and energy is a source of gravity.

[Pmb]

But mass creates a gravitational field. Whereas inertia doesn't, in my opinion.

By definition, the quantity which generates a gravitational field is an objects active gravitational mass. Since the active gravitational field of a body increases with the speed of a body it follows that an objects active gravitational mass also increases. Since the active gravitational mass of a body equals the body’s inertial mass it follows that the body’s inertial mass also increases with speed.

That's the distinction I'm referring to.

You weren’t the person that post was responding to.

Mass also implies the amount of "stuff" …

The term “matter” does is not well defined and is only be used in a vague sense. Inertia really refers to the idea that Newton referred to when he spoke of “quantity of motion” which refers to the quantity m in the relation p = mv where p is defined as in F = force = dp/dt.

So does that mean that things moving near the speed of light have a larger gravitational field than they would otherwise?

The faster a body moves the stronger its gravitational field

[yor_on]

But that doesn't count for uniformly moving bodies Pete, right?

Or do you mean that it is strictly observer dependent, and so needs two bodies in relative motion versus each other? But that way 'gravity' would 'fluctuate' with what observer we have in relative motion, relative what body's gravitational field he measures. My thought has been, and still is, that uniform motion no matter its speed, as measured relative something else, has no effect on the gravitational field surrounding it?

Assuming that we have a buildup of gravity depending on uniform motion wreaks havoc to relativity as I think, because to me it implies a 'global speed definition' in where you locally do have a 'absolute definition' of what a speed is, not relative.

[Pmb]

But that doesn't count for uniformly moving bodies Pete, right?

Sure it does. Why wouldn't it?

Or do you mean that it is strictly observer dependent, and so needs two bodies in relative motion versus each other?

That question is not clear. Please rephrase. Why d you need two bodies? What function do these bodies serve?

Do you do you think that, for some reason, something that is observer dependant requies two bodies in relative motion?

But that way 'gravity' would 'fluctuate' with what observer we have in relative motion, relative what body's gravitational field he measures.

I don't understand what you're talking about wqen you speak of "gravity would fluctuate". Do mean that the gravitational field changes with the speed of the source? Its an odd thing but the fact that the strength of a gravitational source depends on the velocity of the source is a fundamental fact in GR but its such a little known fact on these internet forums. I suppose its because so many people who use the concept of mass being independant of its motion in SR also assume that they should also imploy that same definition in the mass of a gravitational source in GR. Just look at how much confuses it causes???

My thought has been, and still is, that uniform motion no matter its speed, as measured relative something else, has no effect on the gravitational field surrounding it?

Why? What led you to that conclusion? It certainly isn't one that arrives at through calculation, that's for sure.

Consider quantities which define the strength of the gravitational field such as the Christofell symbols or the components of the metric tensor. Since these quantities become velocity dependant when one invokes a coordinate transformation from one inertial frame to another then the new field strengths will become velocity dependant.

Here is a few examples of gravitational fields for which the gravitational field in the "rest frame" is given as well as in a frame moving relative ti the object


 object whose gravitational field is velocity dependant.
http://home.comcast.net/~peter.m.brown/gr/grav_moving_rod.htm
http://home.comcast.net/~peter.m.brown/gr/grav_moving_sheet.htm
http://home.comcast.net/~peter.m.brown/gr/grav_moving_sheet.htm

[Pmb]

Physicists like to find the smallest common nominator for things, and when we (they) talk about mass then that should be 'mass-energy'.

This phrase makes no sense to me at all!  In what sense are you saying that “Physicists like to find the smallest common nominator for things” and what does that have to do with mass-energy? By the way, when one is speaking of mass-energy one is speaking about the kinetic energy of a body that is moving. E.g. If T is the stress-energy-momentum tensor of, say, a gas then T^00 (the energy density) is also referred to as the mass-energy density. T^00 includes kinetic energy and not just rest energy.

[Pmb]

But photons individually have no mass?  How did it gain mass?

If you keep thinking like this you’re going to keep confusing yourself. I fail to understand why you folks like to make life hard for yourself with this bizarre definition of mass?

Two ways of looking at this; 1) photons have mass m according to E = mc^2. This is the relativistic mass of the photon. The mass of a system of photons is then the sum of the relativistic masses of all the photons.

(2) mass = invariant mass of photons – I.e. for a box of photons the box itself allows there to be a frame of reference in which all the photons are confined. The total energy in the box is the sum of all the energies of all the photons in the box. In the rest frame of the box the total momentum of all the photons is zero. That means that the energy in the box is the rest energy. The mass of the photons in the box is therefore defined through E = mc^2

[Pmb]

What type of mass does the "m" represent in the good old E=mc²?

Invariant mass (the one sometimes also called "rest" mass or "proper" mass).

The term “invariant mass” typically refers to a system of particles and not to a single particle. In anycase the concept of invariant mass can only be used in flat spacetime

[Pmb]

I don't accept the claim that a photon has no mass.

Smart man!

If that is your opinion then you might enjoy reading my article on the subject. It’s at
http://arxiv.org/abs/0709.0687

[Pmb]

You can't localize a photon, so you can't do that.

Everything I’ve seen in this thread speaks mostly about classical physics, e.g. relativity. In relativity one uses classical photons, which is basically a point particle having a classical trajectory but zero proper mass. Such a thing can be localized.

[yor_on]

Maybe I am unclear?

Are you telling me that gravity is observer dependent?

[yor_on]

As for why I doubt it is that all experiments in a uniformly moving body is said to act out the same, meaning that whatever relative motion you achieve does nothing to change it. If I now assume that the relative motion indeed is non-relative, meaning that there is a definite change in the gravitational field locally measured then that should invalidate that statement. If we on the other hand treat gravity as something locally unchanging, but when involving two bodies measuring one, observer dependent? I need to think about that one, a damn lot. My original thought is that gravity is constantly dynamically updated in a universe, obeying 'c' as information between relatively moving bodies, but not that they also will measure a different gravity, depending on from where you do the measurement? That would hurt my head terribly to assume

[yor_on]

Or maybe not? But I still need to think about it One could assume that the 'energy' of a universe is a constant one, the same no matter what observer dependencies exist, and then include gravity into that. But it still makes my head ache a little.

[yor_on]

Actually been thinking a little and I think it must be correct. Gravity is observer dependent, meaning that when you measure the gravity of other uniformly moving objects that should change with your relative speed. But locally it won't change for you, meaning that different uniform motions, locally measured, won't change your weight, as you measuring it on a weight scale. Then there is this other type of description in where some solutions to a non-rotating black hole present you with a 'infinite space' and so a weaker gravity, as observed inside the event horizon (locally). Can that also be called a observer dependency? I guess it can thinking of it.

If now that was what you meant?

[Pmb]

Maybe I am unclear?

Are you telling me that gravity is observer dependent?

Depending on what youi mean by "gravity" yes.

[lightarrow]

You can't localize a photon, so you can't do that.

Everything I’ve seen in this thread speaks mostly about classical physics, e.g. relativity. In relativity one uses classical photons, which is basically a point particle having a classical trajectory but zero proper mass. Such a thing can be localized.

Classical photons? Which movie is it?  []

[Pmb]

Classical photons? Which movie is it?  []

I don't understand what you mean, "Which movie is it?"

lightarrow - Have you ever heard of the terms "classical photon" and "classical electron"? Perhaps some use the term “scalar photon” or “scalar electron” instead. I read an article where similar such term(s) were used, rather than forcing someone to explain that what it means. While you may be using it as some sort as slang, I have no idea what it means which means that other people don’t either. It you mean photon then please say photon and he same with electrons. We then won’t have to waste space by trying to explain terms or explain what was a joke and then razz the person who didn’t get the joke.

[lightarrow]

Classical photons? Which movie is it?  []

I don't understand what you mean, "Which movie is it?"

lightarrow - Have you ever heard of the terms "classical photon" and "classical electron"?

"classical electron": yes
"classical photon": no. The reason is because of qm history: a classical electron was a starting point for Bohr and Sommerfeld when they described the atom. But a classical photon couldn't have any meaning, because m = 0 in this case.
But that was history. Now we know that a precise trajectory of particles is impossible, they don't have at all. Wavefunctions are wat replaced them.

Phractality wrote about defining the centre of a system of two photons: you don't even know where is a photon, and you want to find such a thing?