[JP]

Hi Cheryl J; having worked through this thread I find myself wondering if your original question was answered.  I think it may have been, but that could be because I had my own pre-conceived idea as to what it should be.

I would be fascinated to know your thoughts.

I think I'm more confused than ever.

At the risk of confusing matters more...

If you understand mass as a measure of the "resistance" of something to being pushed faster, then it does increase as the speed increases.  In brief, since the speed of light is a limit, it takes more and more work to push something and speed it up as you get closer to the speed of light.  The number of particles stays the same, so the amount of matter doesn't increase in that sense. 

Any time this gets brought up, it'll start an argument over the correct definition of mass.  The problem is there are other definitions which don't agree with the above.  For slowly moving things all definitions DO agree, but at high speeds they differ.  The standard, textbook definition that's taught these days does not increase as speed increases, but I'll leave that to others to debate. 

[JP]

okay, what about that weird meter stick thought experiment where different observers pass it going different relative speeds. Does the meter stick really become shorter as the observers approach the speed of light. Is there "less" of the meter stick? Because in this experiment it doesn't sound like the meter stick's mass, matter, or inertia has changed at all, the observers are different.

No more matter is created or destroyed in the meter stick.  The number of particles in it stays the same.  The shape of the particles will change as they'll now be measured to be short.

[lightarrow]

Let's take a example that gives us a totally new, and well earned, headache

Not to me  []

Consider yourself heating up a gram of some, very, temperature resistant metal. You've weighted it before but after it gets heated you weight it again, finding it to weight more.

Correct.

One way to describe it might be to consider the particles making the material accelerating inside the metal as they gain 'energy' from heat, moving agitatedly.

Correct.

Can we then discuss those particles as gaining a relativistic mass, or not?

You can discuss it, but there is no need of relativistic mass to describe that fact.
In physics is Very important to give attention to which is the physical system considered. The system of particles inside the gram of metal *it's not* the simple "sum" of the particles. Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?

There is a way of saying that "the whole is more than the sum of its parts". Something similar in physics (or, at least, in this case).

[yor_on]

Heh

I can see you and Pete gearing up to a argument Lightarrow
But I think you can use the idea to argue a 'relativistic mass', if you want?
You can translate the particles agitations into accelerations, and a added '(mass)energy' confined inside the metal.

It depends on how you define 'gravity' though. Using the original equivalence it must be a constant uniform acceleration, but to me all accelerations should present you with a inertia/gravity. The squid those guys used to create a dynamical Casimir effect should have weighted a 'little' more if I'm thinking right there.

[Bill S]

[quote = Cheryl J]okay, what about that weird meter stick thought experiment where different observers pass it going different relative speeds. Does the meter stick really become shorter as the observers approach the speed of light. Is there "less" of the meter stick? Because in this experiment it doesn't sound like the meter stick's mass, matter, or inertia has changed at all, the observers are different.[/quote]

I think it all depends on the frame of reference.  In trying to get my head round this I resorted to a thought experiment that can easily be turned into a real one.  All you need is a blank wall, a directional light and a metre (OK, the spell checker doesn't like UK spelling, but I'm too old to change  ) stick.  Hold the stick parallel to the wall and stick and shadow are the same length.  Rotate the stick and the shadow becomes shorter.

In its own F of R, the stick is the same length, but in the F of R of the wall (or the observer) it is measurably shorter.

The scientists on the forum will probably throw up their hands in horror, but it helped me.

[Bill S]

[quote = JP] If you understand mass as a measure of the "resistance" of something to being pushed faster, then it does increase as the speed increases. [/quote]

Assuming that the Higgs particle has really been found, this would seem to herald a new era in physics.  How will it influence our understanding of mass?

[yor_on]

It all depends on how you think of it Chery. Physicists like to find the smallest common nominator for things, and when we (they) talk about mass then that should be 'mass-energy'. Just exchange matter for mass-energy and it will make more sense. Physically at least. So it's not 'more particles' in the matter weighting more after being heated, but there is definitely more 'energy' inside it, as heat.

[bizerl]

At the risk of departing on an entirely new tangent, does the theoretical "Higgs Boson" come in to consideration? I was led to believe that this is what actually gives something "mass" (whatever that is, now there appears to be different types  []). Does more energy "create" more Higgs Bosons?

Is the increase in mass phenomenon the same as the length contraction phenomenon, and dependent on how it is measured (or indeed, defined)?

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

[Phractality]

Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?

If the two photons are somehow bound together, the pair would act like a particle. Since we're in the mainstream forum I'll refrain from speculating on what force or field could conceivable bind a pair of photons together. A change in relative velocity, dv, of the center of the pair (whether they're bound or not) is equivalent to looking at the pair from a different reference frame, having velocity dv relative to the center of the pair. SR gives the ratio of the pair's energy and momentum in the two reference frames. At non-relativistic speeds, the momentum ratio for a given velocity difference is the mass of the pair. That's what inertial mass is ...  M = dp/dv.

I don't accept the claim that a photon has no mass. A bound pair of photons (if there is such a thing) would have a rest mass. The radiant energy of the photons would become the rest mass of the pair (or particle).

In the reference frame centered on the pair, the mass would be M = E/c². In a different reference fame it would be greater by the SR factor, gamma. Mathematically, you would get the same result if the photons are not bound together, but considered as the sum of the two free photons.

If you consider only one photon, it also has different momenta in different reference frames. A change of reference frame gives a change of momentum. If dv is in the direction of v, then dp/dv is the inertial mass of the photon. So a photon has mass. If dv is not parallel to the path of the photon, you have to apply the relativistic form of M = dp/dv. At relativistic speeds, M varies; dp = Mdv +vdM, which, I believe, turns M into a hyperbolic function of v. (Above my pay grade.) Perhaps the mass of an individual photon is different in different directions; I lack the math skill to settle that question.

[yor_on]

m_o= rest mass (SR)
m= can be relativistic mass, or rest mass depending, as far as I've seen.

Einstein is said to not have used that formula. It's a abbreviation made later, by others.
Take a look here for one side of the discussion Rest Mass Versus Relativistic Mass.

I find the idea of a rest mass simpler, as I don't have to consider accelerations relative different uniform motions, because even though you could, if considering a Higgs field, assume that a acceleration then results in more 'energy' locally expressed, what about the changed uniform motion after the acceleration? Gaining a different (although still 'relative') speed, but not a different energy, locally? All as I interpret relativity naturally.

[Spacetectonics]

Good question !
Particle gets mass when interacting with higgs field(H.F),higgs field is through out the universe and it is an energy field.
particles considered as wave"behavior" in QM >
wave/Pare. interacting with H.F creating mass >
Nature always wants to be in its lowest state and that is why H.F born>
In QM Wave Interacting with the lowest known state of energy in universe creating mass>
mass is a quantitative measure of an object's resistance to acceleration>
If acceleration reaches C mass will be Infinite.
!!Based on this apparently, argument will be "Resistance"!!
Quantitative measurements are those which involve the collection of numbers.I put this in a post
And yet I  don't know the answer ,as everyone comes up with different solutions!

Cheers

[JP]

[quote = JP] If you understand mass as a measure of the "resistance" of something to being pushed faster, then it does increase as the speed increases.

Assuming that the Higgs particle has really been found, this would seem to herald a new era in physics.  How will it influence our understanding of mass?
[/quote]

I'm not a Higgs expert, but from what I understand it won't be too revolutionary to have found the standard Higgs.  This is because it's been part of the Standard Model for a while (since the 1960s, I think) and people have spent a lot of time thinking about its implications.  If we find the standard Higgs particle, it validates the model, but doesn't introduce new physics. 

Now, if we find a Higgs that isn't exactly as the model predicts, find more than one Higgs-like particle or don't find it at all, we'd need to retool our theories. It would, however, save people time in looking for alternatives to the Higgs mechanism.

There are other reasons why the Higgs won't completely revolutionize our idea of mass.  As Yor_on mentioned earlier it explains the inertial mass, or resistance to pushing, of simple particles.  But most matter is due to particles bound together with forces, and the energy of these bonds also creates mass which the Higgs doesn't explain (as far as I understand it, at least). 

Another reason is that the Higgs provides a mechanism for inertial mass--the Higgs field makes particles resist changes in velocity.  However the magnitude of a particle's mass is determined by how strongly it interacts with the Higgs field.  There's no theoretical reason why an electron couples to the field with one strength and a muon with another.  We determine the strengths experimentally.  Presumably, if the Higgs is discovered (and even now that there's very strong evidence for it), physicists will start working on theories to explain the interaction strength.

[lightarrow]

Do you remember when I showed that a system of 2 photons with opposite velocities has non zero invariant mass?

If the two photons are somehow bound together, the pair would act like a particle.

There is no need of binding and, anyway, two photons don't bind together, but I see what you want to say.

Since we're in the mainstream forum I'll refrain from speculating on what force or field could conceivable bind a pair of photons together.

Good choice  []

A change in relative velocity, dv, of the center of the pair

How do you define the centre in a system of two photons?

(whether they're bound or not) is equivalent to looking at the pair from a different reference frame, having velocity dv relative to the center of the pair. SR gives the ratio of the pair's energy and momentum in the two reference frames. At non-relativistic speeds, the momentum ratio for a given velocity difference is the mass of the pair. That's what inertial mass is ...  M = dp/dv.
I don't accept the claim that a photon has no mass.

It's not a claim: if it had (invariant) mass, it would have infinite energy.

A bound pair of photons (if there is such a thing) would have a rest mass.

But even unbound, the system has invariant mass, what do you want more?  []  If two photons escape one from the other, or if staied close each other as in a sort of atomic system, it wouldn't make difference: the system has/would have invariant mass for the same reason.

[lightarrow]

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).

[Phractality]

How do you define the centre in a system of two photons?

Tough question! [] I'll have to ponder this for a while. Let's not even think about the effect of the expansion of space; just confine the discussion to short distances and times where space does not expand appreciably.

For openers, let's consider only inertial reference frames whose relative motion is restricted to the x direction, which is the direction of relative motion of the two photons, which are moving in opposite directions parallel to the x-axis. (Later we may try to generalize to other reference frames.) Let's define the origins of all these reference frames as the point in space-time where the two photons pass closest to one another, at x = 0, t = 0. In all such reference frames, the center of the two-photon system is the origin, but the origins of different reference frames only coincide at the instant when the photons pass one another.

I guess you have to start with a reference frame in which both photons have equal energy and equal and opposite momenta. In that reference frame, the energy of each photon is E, and  the momentum of each photon is p = E/c. Since the momenta are equal and opposite, the momentum of the system is zero.

Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'₁= E/2, and the other photon has energy E'₂ = 2E. Do I have that correct? My brain is about to trip a circuit breaker, here. Someone, please let me know if I got the photons' energy right before I proceed to dig myself into a deeper hole.

My postulate is that, at low velocities, the two photon system has inertial mass M = dp/dv = E/c². At higher velocities, I suspect that formula turns into a hyperbolic function. Rats! I hate hyperbolic functions!   [xx(]

Thinking ahead: In any one reference frame, the center of the two-photon system is fixed, so a two-photon system can't have a non-zero velocity in an inertial reference frame. However, when you change to another reference frame, you move that center. (This gets into a gray area between SR and GR. I'm not sure I'll be able to handle the math.) A gradual change from one reference frame to another by small increments, dv, will gradually move the center of the system. So the rate of change of reference frame's velocity relative to the first reference frame (dv/dt) imparts motion to the system's center. I don't know yet whether the systems center has acceleration or uniform velocity while the reference frame's velocity is changing at a constant rate. If the momentum of the system changes at the rate dp/dv, that is the inertial mass of the system.

[lightarrow]

How do you define the centre in a system of two photons?

Tough question! []
Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'₁= E/2, and the other photon has energy E'₂ = 2E. Do I have that correct?

No. In the first case:

E' = E*sqrt[(1-beta)/(1+beta)] = E*sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)] = [2 - sqrt(3)]E ~ 0.268E;

in the second case:

E' = E*sqrt[(1+beta)/(1-beta)] = E*sqrt[(1+sqrt(3)/2)/(1-sqrt(3)/2)] = [2 + sqrt(3)]E ~ 3.73E.

[Phractality]

How do you define the centre in a system of two photons?

Tough question!
Next, consider a reference frame moving in the +x direction at .866 c relative to the first reference frame. Gamma = 2; so in this reference frame, the photon moving in the +x direction has energy E'₁= E/2, and the other photon has energy E'₂ = 2E. Do I have that correct?

No. In the first case:

E' = sqrt[(1-beta)/(1+beta)] = sqrt[(1-sqrt(3)/2)/(1+sqrt(3)/2)] = 2 - sqrt(3) ~ 0.268E;

in the second case:

E' = sqrt[(1+beta)/(1-beta)] = sqrt[(1+sqrt(3)/2)/(1-sqrt(3)/2)] = 2 + sqrt(3) ~ 3.73E.

Thanks for the correction. I had a feeling I got it wrong. I know how to do the math, but the math corner of my brain was in full revolt.

I'm wondering, now, if I should consider the center of the two-photons to be the center of energy; equivalent to center of mass. Applying inverse square law to the energy of each photon to get a ratio of each photon's distance from the center. My brain hurts; maybe I'll just play solitaire, instead. 

[lightarrow]

Thanks for the correction. I had a feeling I got it wrong. I know how to do the math, but the math corner of my brain was in full revolt.

I'm wondering, now, if I should consider the center of the two-photons to be the center of energy; equivalent to center of mass. Applying inverse square law to the energy of each photon to get a ratio of each photon's distance from the center. My brain hurts; maybe I'll just play solitaire, instead. 

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

[Phractality]

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

Suppose you know that a pair of equal photons were emitted in opposite directions from a certain point in space-time (as determined by detected particles left behind). Define the origin of an inertial frame at that point, with the photons on the x-axis. Then you can assume they exist at points x = ct and x = -ct.

It gets more interesting and challenging when you try to describe those same photons in a different inertial frame at relativistic speed relative to the first frame. If you reflect the two photons off of perfect mirrors to make their paths parallel but not collinear, it gets a bit more challenging. I don't even want to think about accelerating reference frames, but I'm afraid they are necessary to describe the inertia of the two-photon system, which is given by dp/dv (for small values of dv). Accelerating an observer relative to the two photon system is equivalent to accelerating the center (of perhaps center of mass/energy) of the system in the observer's frame. But acceleration takes us out of the realm of SR, and the math gets way to hairy for my puny brain to handle.   []

[Bill S]

I'm not a Higgs expert, but from what I understand it won't be too revolutionary to have found the standard Higgs.  This is because it's been part of the Standard Model for a while (since the 1960s, I think) and people have spent a lot of time thinking about its implications.  If we find the standard Higgs particle, it validates the model, but doesn't introduce new physics. 

Point taken, but if we have found the standard Higgs particle, then, presumably, we have found the Higgs field.

If we have found the Higgs field, must we not have co-ordinates for an "absolute space". 

I could be wide of the mark, but the thinking goes something like this:  If we are saying that a particle's mass arises from its motion through the Higgs field, the Higgs field must be stationary, in an absolute sense, or the masses of particles would vary, depending on their direction of movement through the Higgs field.