Massive objects moving too fast

[Coughlan]

I've been puzzling over this for quite sometime and have not been able to find it anywhere here which is stunning. What if you had a structure that was moving fast enough to have it's relativistic mass be...1 kg for now. And we had this 1 kg mass zipped by something that was 0.5 kg for sake of argument. What would be the "grativistic(sp?)" forces at work here? Should I try and use Newton universal law to figure it out? or would the speed negate any other forces because it is just simply moving too fast. Here's another thought: Is gravity affected by time or is time affected by gravity?

Thank you for reading and if it has already been said please just post a link or a few keywords I can search for. :)

-Coughlan

 

[HallsofIvy]

No, that gravitational force, at a specific instant, would be calculated, in special relativity, by using Newton's gravitational formula with the distance between the objects and their masses at that instant.

I specified "in special relativity" because, of course, general relativity has a completely different way of computing gravity.

 

[Mentz114]

I'm not clear on your scenario, but if it involves gravitational interaction between masses which are in relative motion, there is a newer theory than Newton's gravity, called general relativity which deals with this.

Is gravity affected by time or is time affected by gravity?

Clocks are affected by gravity. Obviously the gravitational field at a certain place may vary if masses are moving nearby.

 

[tiny-tim]

is this the question?

I think what you're asking is:

Gravitational force depends on mass and distance.

Is that mass the ordinary mass (rest-mass) m, or, for something moving at a very fast speed v, is it the increased mass \frac{m}{\sqrt{1\,-\,v^2/c^2}}?

Erm … I don't actually know the answer to that!

 

[Coughlan]

Well what I was referring to was this. In order to calculate a relitivistic mass you need to take into account its energy. Thus the fast you go the more mass you have. So what if this high speed object with its new mass passes a stationary object half its weight? What happens? Is the second object attracted to the first? I hope that helps

 

[AcidBathSDMF]

I'm going to take a guess here. From my understanding, at relativistic speeds, an object doesn't gain mass, per se. By that I mean it doesn't gain particles or anything. Gaining mass at high speeds is a bad interpretation because it leads to weird conclusions when you talk about a mass traveling at, say, a 30 degree angle between the x and y axis. Does it gain more mass in the x direction than the y since the velocity component is higher in that direction? As you can see, different masses in different directions of the same object doesn't make much sense. But, if you interpret mass as a quantity that tells you the resistance to change in its trajectory, I think it makes more sense. Then, in my analogy about traveling at 30 degrees, its conclusion would be that it's harder to accelerate the mass in the x direction than in the y direction. So the big point is, if something is going at a relativistic speed in a particular direction, it harder to make that object go any faster in that direction than if it were going at a lower speed. In your question, I assume you're talking about some big object passing beside another big object at a high speed. I think you should break things down into vector components. Let's say object B is at the origin in the x-y plane, and object A is traveling in the y-direction and passed the x-axis at some distance "d" from the origin. Since object A is going beside object B, and the gravitational force is one of attraction, object B wouldn't cause A to go any faster in the y-direction--it's pulling it towards the origin. So the effect would be that object A would gain a velocity component in the -x direction, toward object B. Since object A is not traveling at relativistic speeds in this direction, it's normal mass would be used. If you want the situation of the gravitational pull being in the same direction as the relativistic velocity, you need General Relativity. Special Relativity does not apply to non-inertial frames. Hope this helps.

 

[tiny-tim]

increased mass is the same for each direction

I'm going to take a guess here. From my understanding, at relativistic speeds, an object doesn't gain mass, per se {snip} different masses in different directions of the same object doesn't make much sense {snip}

Sorry, but this is just wrong. Momentum in each direction is rest-mass times velocity-in-that-direction times the same factor, \frac{1}{\sqrt{1\,-\,v^2/c^2}}, where v is the total velocity.

The increased mass, rest-mass times \frac{1}{\sqrt{1\,-\,v^2/c^2}}, is the same for all directions.

Since object A is not traveling at relativistic speeds in this direction, it's normal mass would be used.

Yes, but the question is whether B's normal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity, so I'd rather let someone else comment on it.

 

[Oliver981]

You are trying to unify Special Relativity with Newtonian gravitation, I guess.
At the moment I have no idea about how doing that, but on one point I am quite sure. Newton's gravity has as framework the Euclidean space and three-vectors, while Special Relativity resides in the space-time (Minkowski space) and uses four-vectors.
So, Newton's law cannot be used as it stands, but needs to be generalized to the four-dimensional case, as one usually does for Electromagnetism, introducing the four-potential A^{\mu}.
However, the generalization of Newton's law to Minkowski space (if it is possible) has not to be trivial!
If we assume it possible, then one uses four-momenta and four-forces and should be able to deal properly with Newtonian gravity.

I think I have been right!

 

[tiny-tim]

No. All Newtonian laws can be converted into four-dimensional laws.

You just have to be careful!

Usually, it just involves multiplying the mass by \frac{1}{\sqrt{1\,-\,v^2/c^2}}.

In electromagnetism, for example, E and B are parts of a six-component tensor in four-dimensional space, and would be different for an observer with a different velocity, but for any particular observer the usual three-dimensional rules for E and B apply.

 

[RandallB]

Yes, but the question is whether B's normal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity ...

You will be much better off applying the accepted standard of only using rest mass.
Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

Energy is defined by applying gamma to momentum no more than that is needed.
No need to assume that the momentum value change is the result of a change in any actual mass inside the equation for momentum. Physicist today find it makes much better sense to view the momentum itself as being altered relativisticly by time and distance changes defined in the gamma factor. Those effects are what is important not the idea that it “looks as if the mass changed”. That is all that has happened, the relativistic effects are real. Not what only seems like a change in mass from our point of view.

 

[tiny-tim]

You will be much better off applying the accepted standard of only using rest mass.
Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

We apply the "gamma factor", \frac{1}{\sqrt{1\,-\,v^2/c^2}}, because the Lorentz equations tell us to. That then tells us what effects will be apparent, not the other way round!

Energy is defined by applying gamma to momentum no more than that is needed.

What?

Physicist today find it makes much better sense to view the momentum itself as being altered relativisticly by time and distance changes defined in the gamma factor. Those effects are what is important not the idea that it “looks as if the mass changed”. That is all that has happened, the relativistic effects are real. Not what only seems like a change in mass from our point of view.

I don't really understand what you're saying here.

I'm quite happy to accept that there's more than one way of looking at the same mathematics, and which you choose is largely a matter of taste.

My taste is to say that the mass increases. That's perfectly consistent with the mathematics.

If you want to state clearly what your taste is, I'll probably be happy to agree with you that that's fine too.

I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

 

[Oliver981]

Electromagnetism is a very special case! In fact Einstein conceived Special Relativity starting from it (in his pioneer work "Zur Elektrodynamik bewegter Körper", "on the electrodynamics among bodies").
He also said that Special Relativity was already contained in Maxwell equations, for which Galileo transformations give problems.
The same problems are absent for Newtonian gravity, so one has to investigate some sort of "gravitodynamics among bodies"! :-)
Namely, which are the relativistic effects in Newtonian gravity? And how do you build the gravitational four-potential?

Anyway, in my opinion the rest mass, being a scalar, should be used as the source of the gravitational force. The gamma factor would enter through Lorentz transformations!

 

[tiny-tim]

Electromagnetism is a very special case!

So far as I know, all forces are affected in the same way by the Lorentz transformation.

The same problems are absent for Newtonian gravity

No, they're there, but gravity is many many orders of magnitude weaker than electromagnetism (think how fast an electron can be accelerated form rest across a cathode ray tune, and how long gravity would take to pull it the same distance!), so nobody noticed them.

 

[Coughlan]

Wow thanks for the response. I had no idea that this topic was so unclear. I figured that I was just missing something. I still don't know what to think but it has been a great thought provoking experience non the less!

 

[DrGreg]

There's a big problem trying to reconcile Newtonian gravity with special relativity. In Newtonian gravity, two masses exert equal and opposite forces on each other simultaneously. In relativity, there's no such thing as "simultaneous", so if the masses are moving relative to each other, you are in trouble.

I stand to be corrected, but I believe you have no choice but to use general relativity. I'm no expert, but I recall other answers to similar questions saying that, in GR, gravity relates to energy-momentum rather than to (rest) mass. So even photons have some (tiny) gravitational influence.

 

[Coughlan]

...I was afraid someone was going to say that. Unfortunately I couldn't do Tensor Calculus to save my life. As a matter of fact I'm barely hanging on with regular calculus. So I guess I'll just have to take your guys' word for it. Thank you for the prompt responses! :)

 

[DrGreg]

I'm quite happy to accept that there's more than one way of looking at the same mathematics, and which you choose is largely a matter of taste.

I agree.

My taste is to say that the mass increases.

(...and mine is not!)

I would suggest to everyone that when we talk about mass, we should make it clear whether we mean "relativistic mass" (proportional to energy, including kinetic energy) or "rest mass" (="invariant mass"). The modern fashion is for "mass" to mean "invariant mass". In these forums, many readers are struggling to work out what mass is, so we should avoid ambiguity.

I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).

For the benefit of anyone who doesn't know why, the invariant mass m_tot of a collection of particles can be defined by

(\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2.

which reduces to

\Sigma E = m_{tot} c^2

in the box's frame, where the total momentum is zero. For each individual particle of invariant mass m,

E^2 = m^2 c^4 + ||\textbf{p}||^2 c^2, so E > m c^2, and m_{tot} > \Sigma m.

 

[jtbell]

The "gravitational field of a moving object" has been discussed fairly regularly here. Note particularly Pervect's posts in the following threads.

https://www.physicsforums.com/showthread.php?t=205073

https://www.physicsforums.com/showthread.php?t=153173

https://www.physicsforums.com/showthread.php?t=150342

 

[tiny-tim]

invariant mass

Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).

As you say, invariant mass = rest mass.

And you agree something can have "increased invariant mass".

Which is the same as "increased rest mass"!

Which is why my taste is to regard mass as increased in all situations!

(And don't forget, there's nothing unusual about rotational motion contributing to invariant/rest mass: the electrons in you and me have very fast rotational motion which does contribute to increasing our "rest mass"! )

For the benefit of anyone who doesn't know why, the invariant mass m_tot of a collection of particles can be defined by

(\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2.

which reduces to \Sigma E = m_{tot} c^2 in the box's frame, where the total momentum is zero.

No. If a wheel is rotating so that its rim is moving with speed U relative to the centre of the wheel, and if the whole wheel is moving parallel to its axis with speed V relative to us, then the Lorentz-increased mass (energy) of the rim is the rest-mass multiplied by:

\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}​

Even in the box's frame, the Lorentz-increased mass (energy) of the rim is still multiplied by \Large\frac{1}{\sqrt{1 - U^2}}.

 

[Coughlan]

thank you jtbell. I will look when my boss isn't here!

 

[RandallB]

We apply the "gamma factor", \frac{1}{\sqrt{1\,-\,v^2/c^2}}, because the Lorentz equations tell us to.
If you want to state clearly what your taste is, I'll probably be happy to agree with you that that's fine too.

I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass!

Lorentz equations do not require they be applied to mass.

I did state it – I use gamma on momentum and NOT on mass as most modern science does.

In the rotational case you are comparing the effects of inertia.
Inertia is not defined by changing mass but by Energy, Momentum, and invariant Mass. You are simply attempting to justify changes in Energy and Momentum following relativistic rather than Classical rules by attributing the changes in Energy and Momentum to actual changes in the physical mass of an object.

Most modern practicing science does not find it necessary to use that way of thinking as a crutch to understanding what is going on. They are able to think in terms of Energy and Momentum as being independent things that need to be treated relativisticly, that’s all.
True you can treat Energy and Momentum Classically by correctly applying relativistic effects only to mass, most of modern science just does not do that.

I suppose either way might be OK, with the possible exception that applying it to mass might require using a preferred reference frame- but I’m not sure on that point. Probably not that important a distinction since a preferred frame is desirable to some anyway.

If you want to survey how most scientists think, I think you find I’m correct on how the majority view the issue.

 

[AcidBathSDMF]

Sorry, but this is just wrong. Momentum in each direction is rest-mass times velocity-in-that-direction times the same factor, \frac{1}{\sqrt{1\,-\,v^2/c^2}}, where v is the total velocity.

The increased mass, rest-mass times \frac{1}{\sqrt{1\,-\,v^2/c^2}}, is the same for all directions.



Yes, but the question is whether B's normal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity, so I'd rather let someone else comment on it.

I'll have to rethink my reasoning--thanks for pointing out the troubles. I found a different thread on here that explains what I wanted to say in a much better way. Here is the conclusion, and I'll give a link to the mathematics:
"Given this form of the dynamics equation for special relativity it becomes evident that a mass "changing with speed" is not the correct explanation for why a massive object can not be accelerated up to the speed of light. The mass m here does not change with speed. The correct explanation is that given this law of motion an arbitrary amount of ordinary force produces a diminishing coordinate acceleration as the coordinate velocity approaches c due to the time dilations involved in relating coordinate to four-vector expressions. These time dilations are in turn due to the Lorentzian structure of spacetime(the Lorentzian structure being why we defined the momentum four vector in terms of a Lorentz transform in the first place). So the reason really that nothing with mass can be accelerated up to c speeds in special relativity is ultimately that spacetime has a Lorentzian structure."

https://www.physicsforums.com/showthread.php?t=11072

 

[tiny-tim]

Lorentz equations do not require they be applied to mass.

You've quoted me out of context (and without giving the post's reference).
I was simply denying that I was getting the gamma factor from what you called the "apparent effect", as in:

You will be much better off applying the accepted standard of only using rest mass.
Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

We apply the "gamma factor", \frac{1}{\sqrt{1\,-\,v^2/c^2}}, because the Lorentz equations tell us to. That then tells us what effects will be apparent, not the other way round!

I did state it – I use gamma on momentum and NOT on mass as most modern science does.

Sorry … that's too unclear for me to either agree or disagree with.

In the rotational case you are comparing the effects of inertia … {saving space} … preferred frame is desirable to some anyway.

I'm really not following most of this.

For example, you say "They are able to think in terms of Energy and Momentum as being independent things that need to be treated relativisticly" - but Energy and Momentum are most certainly not independent - they're components of the same four-vector, and they get mixed up by the Lorentz transformation in the same way that time and position do.

The simple fact is that "Most modern practicing science" does regard my mass as already taking into account the "increased mass" of the electrons whizzing around inside me. No-one would try to define it or measure it any other way.

 

[tiny-tim]

physicists make measurements, and give those measured things names

Hi AB! (btw, I wish you'd use paragraphs! )

{snip} The mass m here {of something accelerating towards the speed of light} does not change with speed. {snip}

Yes, in that the rest-mass or inertial mass is the same for all observers.

But that doesn't help with the case of something with moving parts, which has an increased inertial mass even when the things as a whole is stationary!

I can't see any useful reason for avoiding incorporating the gamma factor to make what is after all a very straightforward physical measurement!

Ultimately, I think the difference here is between physicists and mathematicians.

For example, mathematicians will say that:

the reason really that nothing with mass can be accelerated up to c speeds in special relativity is ultimately that spacetime has a Lorentzian structure.

But physicists are entitled to say "we want to relate the maths to things which we, as experimental physicists, can actually measure."

Now, no experimental physicists in their right minds are going to measure the rest-mass of the individual bits of me, or of a vehicle containing a rotating flywheel. They'll measure the mass as a whole, either by weighing or by pushing.

It might be better if they called it the "resistance-to-movement" or "inertia" - but:

it is measured in the same units as mass; and
it is the same for all observers (no gamma factor); and
it can be used as m_0 in all equations I know of​

so they prefer to give it a name with the word "mass" in it.

Some call it "rest-mass", but that's slightly misleading because it takes account of some parts which aren't at rest.

Some call it "inertial mass", which is probably best because it is a measure of its resistance to movement (which is, of course, the way it is usually measured).

A mathematician may say "you are missing the fundamental nature of the group structure of space-time."

But a physicist should reply "you are missing the simplicity of your equations, and the necessity of applying them to observable measurements in the actual universe!"

This thing - whatever we call it - is a perfectly valid measurement, and we are entitled to put brackets in the equations round the terms that describe it, and to defend those brackets against all attacks!

 

[pervect]

The "gravitational field of a moving object" has been discussed fairly regularly here. Note particularly Pervect's posts in the following threads.

https://www.physicsforums.com/showthread.php?t=205073

https://www.physicsforums.com/showthread.php?t=153173

https://www.physicsforums.com/showthread.php?t=150342

Thank you for mentioning this. Yes, it is a frequently asked question, and many of the answers given here have been rather speculative.

There are some fine technical points here which make the gravitational field of a moving mass hard to even talk about rigorously.

If you think of the gravitational field of a moving mass as being like the electric field of a moving charge, i.e. concentrated in the transverse direction, however, while you will be wrong, you won't be far wrong. By "not far wrong", I mean that you'll probably get the right order of magnitude when you translate the not-well-defined concept of "gravitational field" into experiment so you can actually measure it.

I have the feeling from skimming the threads that most posters here aren't even familiar with how relativity treats the electric field of a relativistically moving particle (a much simpler question. See for instance

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

and look and numbers above and below 14 in the url to get a fuller treatment.

 

[tiny-tim]

Lorentz equations on electric field

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

Goodness! That page is one of the most confusing treatments I've seen for a long time.

It obscures the simple facts that:

(a) electric force in the direction of movement is unchanged ("E_x = E_x\,'");

(b) electric force perpendicular the direction of movement is multiplied by the "gamma factor" ("E_y = \gamma E_y\,'"). ​

(And, of course, a magnetic field is also produced "from nothing" - but its effect on the observer is zero, since by definition the observer is stationary, and therefore unaffected by magnetic fields!)

If you think of the gravitational field of a moving mass as being like the electric field of a moving charge, i.e. concentrated in the transverse direction, however, while you will be wrong, you won't be far wrong.

So, using special relativity, the gravity from a moving object at closest approach (so the field is at that moment perpendicular to the line of motion) would be multiplied by the gamma factor: in other words, at closest approach the gravitational mass would equal the "Lorentz-increased mass" (but using general relativity, presumably the answer is different).

 

[RandallB]

For example, you say "They are able to think in terms of Energy and Momentum as being independent things that need to be treated relativisticly" - but Energy and Momentum are most certainly not independent - they're components of the same four-vector, and they get mixed up by the Lorentz transformation in the same way that time and position do.

The simple fact is that "Most modern practicing science" does regard my mass as already taking into account the "increased mass" of the electrons whizzing around inside me. No-one would try to define it or measure it any other way.

I don't disagree there are important interrelationships between Energy Momentum and Mass.
All I’m saying you do not at all need to assume the something like momentum must be defined by a changing mass. And that most modern applied science does not. That you think otherwise only tells me you haven’t discussed the issue with real scientists to see how modern science actually does this kind of work.
Arguing with me won’t change the facts. And you your opinion will depend on who you listen to. So with that there is no further advice I can give you.

 

[tiny-tim]

Measuring the mass of something with moving parts

All I’m saying you do not at all need to assume the something like momentum must be defined by a changing mass.

I agree you don't need to, but you can if you want, and it seems both clearer and more consistent with the way physicists actually measure the mass of something with moving parts.

And that most modern applied science does not.

You haven't answered my (implied) question: when "most modern applied science" measure my mass, do they measure the "increased mass" of the electrons whizzing around inside me (as I say they do), or do they somehow regard the mass or momentum of the electrons separately, and do a calculation different from that of the rest of me?

 

[1effect]

For the benefit of anyone who doesn't know why, the invariant mass m_tot of a collection of particles can be defined by

(\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2.

which reduces to

\Sigma E = m_{tot} c^2

in the box's frame, where the total momentum is zero. For each individual particle of invariant mass m,

E^2 = m^2 c^4 + ||\textbf{p}||^2 c^2, so E > m c^2, and m_{tot} > \Sigma m.

I am getting a different result, for a system of particles:
The total energy is:
E=\Sigma e= c^2 \Sigma \gamma_i m_i

The total momentum is :
\textbf{P}=\Sigma \textbf{p}=\Sigma \gamma_i m_i \vec{v_i}

E^2-(\textbf{P}c)^2=c^4 \Sigma m_i^2 +c^2 \Sigma \gamma_i \gamma_j m_i m_j (c^2-\vec{v_i} \vec{ v_j} )

On the other hand:

Q^2= \Sigma ||\textbf{p}||^2 =\Sigma \gamma_i^2 m_i^2 v_i^2

so:

E^2-c^2Q^2=c^4 (\Sigma \gamma_i m_i)^2-c^2\Sigma \gamma_i^2 m_i^2 v_i^2

Comparing with your definition, I arrived to:

m_{tot}=\Sigma \gamma_i m_i > \Sigma m_i

 

[tiny-tim]

Dependence on velocity of the system as a whole

I'm not following that.

Anyway, both your formula and Dr Greg's formula are simply working out energy/momentum for a system of particles without any mention of the velocity, V, of the system as a whole.

The question for the "rest mass" or "invariant mass" or "Lorentz mass" or "increased mass" or … of an object with velocity V but with moving internal parts is:

What are the properties of a system of particles of masses m_i and velocities \vec{V}\,+\,\vec{v_i} (where the "+" denotes the relativistic rule for addition of velocities)?

If you work it out (I can't be bothered right now :zzz:), it's:

E_{tot}\,=\,\frac{M}{\sqrt{1\,-\,V^2/c^2}}\qquad,\qquad P_{tot}\,=\,\frac{M\,\vec{V}}{\sqrt{1\,-\,V^2/c^2}}\qquad,​

where M is a constant (= E_{tot} for \vec{V} = \vec{0}).

(Incidentally, I suspect the original formula came from http://en.wikipedia.org/wiki/Invariant_mass,
where it refers to the total mass of the particles from a decay of one particle as being equal to that of the original particle.)

 

[1effect]

I'm not following that.

Anyway, both your formula and Dr Greg's formula are simply working out energy/momentum for a system of particles without any mention of the velocity, V, of the system as a whole.

1. Why would we bother with V, when we have all the measurable quantities v_i ?

2. I think I pointed out that the quantity :

E^2-(\textbf{P}c)^2 is wrong because it produces terms in v_i v_j. This can be seen as well from the example . In this example, M is clearly not an invariant due to the term p_1p_2

3. I am of the opinion that the correct quantity is:

E^2-c^2 \Sigma ||\textbf{p}||^2

4. DrGreg is of the opinion that the correct quantity is :

E^2 -c^2 ||\Sigma \textbf{p}||^2

Note that both formulas need to include || \textbf{p}|| instead of \textbf{p} from the wiki page. There is nothing stopping you from using \textbf{p} as long as you realize that it will produce a frame-variant value for the total mass of the system. If this is what you want, fine, it is all a matter of definitions.

 

[Ich]

In this example, M is clearly not an invariant due to the term p_1p_2

That's nonsense. Both p1 and p2 are variant, so how could p1²+p2² be invariant? Why don't you calculate the example explicitly if you don't believe that M is invariant?
And, of course, DrGreg's formula is correct, and yes, the invariant mass of a system of particles depends on their relative velocities, but is invariant with respect to the motion of the system as a whole.

 

[1effect]

That's nonsense. Both p1 and p2 are variant, so how could p1²+p2² be invariant? Why don't you calculate the example explicitly if you don't believe that M is invariant?
And, of course, DrGreg's formula is correct, and yes, the invariant mass of a system of particles depends on their relative velocities, but is invariant with respect to the motion of the system as a whole.

Before you jump up, I think you are confused: the dependency on speed v makes the quantity M vary when translating from frame S to frame S' in relative motion with speed u because v transforms into v'.

The total energy is:
E=\Sigma e= c^2 \Sigma \gamma_i m_i

The total momentum is :
\textbf{P}=\Sigma \textbf{p}=\Sigma \gamma_i m_i \vec{v_i}

E^2-(\textbf{P}c)^2=c^4 \Sigma m_i^2 +c^2 \Sigma \gamma_i \gamma_j m_i m_j (c^2-\vec{v_i} \vec{ v_j} )

Same problem as in the wiki example:

M^2= m_1^2 + m_2^2 + 2\left(E_1 E_2 - \vec{p}_1 \cdot \vec{p}_2 \right) \,

 

[tiny-tim]

Why would we bother with V, when we have all the measurable quantities v_i ?

Because this is for an object with internal moving parts, and:
(a) we very often don't have the (v_i)s;
(b) even when we do, it's difficult to work out each of the the (\vec{V}\,+\,v_i)s (with relativistic addition);
(c) it's simpler to do the calculation when V = 0 (giving a number M in units of mass), and then just multiply by \frac{1}{\sqrt{1\,-\,V^2/c^2}}, which is both much easier and more intuitive!

 

[1effect]

Because this is for an object with internal moving parts, and:
(a) we very often don't have the (v_i)s;(b) even when we do, it's difficult to work out each of the the (\vec{V}\,+\,v_i)s (with relativistic addition);

Not at all.

(c) it's simpler to do the calculation when V = 0 (giving a number M in units of mass), and then just multiply by \frac{1}{\sqrt{1\,-\,V^2/c^2}}, which is both much easier and more intuitive!

Anyway, both your formula and Dr Greg's formula are simply working out energy/momentum for a system of particles without any mention of the velocity, V, of the system as a whole.

So, how do you measure "the velocity, V, of the system as a whole." ?

 

[tiny-tim]

radar?

So, how do you measure "the velocity, V, of the system as a whole." ?

Erm … well, if it's a rocket, for example, you could use radar …

Am I missing something?

 

[1effect]

Erm … well, if it's a rocket, for example, you could use radar …

Am I missing something?

Yes, you are missing quite a bit, it is a system of particles each one having its own speed v_i. So, how do you measure V?

 

[Ich]

M²=E²-P² is the square norm of a four vector, a scalar. Scalars are invariant.
(\Sigma E)^2 -( \Sigma P)^2 is the square norm of a sum of four vectors.
Sums of four vectors are four vectors, therefore the square norm is also invariant.
Invariant mass has two important properties:
1. It is not additive - m!=m1+m2 in general
2. It is, well, invariant.

 

[1effect]

M²=E²-P² is the square norm of a four vector, a scalar. Scalars are invariant.
(\Sigma E)^2 -( \Sigma P)^2 is the square norm of a sum of four vectors.
Sums of four vectors are four vectors, therefore the square norm is also invariant.

I know that the energy-momentum four vector is frame - invariant. I am not clear how you are getting that the sum of four vectors is also frame invariant. How do you get rid of the terms dependent of velocities?

 

[tiny-tim]

You just measure it!

it is a system of particles each one having its own velocity \vec{v}_i. So, how do you measure \vec{V}?

You just measure it!

For example, if the system is the system of particles that make up a rocket, or that make up a box full of hot gas, then you just measure the velocity \vec{V} of the outside of the rocket, or of the box!

How would you measure the velocity \vec{V} of a box containing hot gas?

(Measure the velocities of all the individual molecules, and then take an average, perhaps?)

(The relevance is that you may have two identical boxes of gas, one hot and one cold, and want to know what the "inertial mass", ie resistance to acceleration, of each is at various velocities \vec{V}.)

 

[1effect]

You just measure it!

For example, if the system is the system of particles that make up a rocket, or that make up a box full of hot gas, then you just measure the velocity \vec{V} of the outside of the rocket, or of the box!

How would you measure the velocity \vec{V} of a box containing hot gas?

(Measure the velocities of all the individual molecules, and then take an average, perhaps?)

(The relevance is that you may have two identical boxes of gas, one hot and one cold, and want to know what the "inertial mass", ie resistance to acceleration, of each is at various velocities \vec{V}.)

One more time, this is a system of particles of random velocities v_i, there is no "box".

 

[tiny-tim]

Flies are random … but flywheels aren't!

One more time, this is a system of particles of random velocities v_i, there is no "box".

1effect, when you joined this thread at post #29, you were commenting on Dr Greg's post, in which he commented on a rotating flywheel:

Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).

For the benefit of anyone who doesn't know why, the invariant mass m_tot of a collection of particles can be defined by

(\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2.

which reduces to

\Sigma E = m_{tot} c^2

in the box's frame, where the total momentum is zero. For each individual particle of invariant mass m,

E^2 = m^2 c^4 + ||\textbf{p}||^2 c^2, so E > m c^2, and m_{tot} > \Sigma m.

You then disagreed on his calculations (I'm staying neutral on that!), and gave your own calculations.

But Dr Greg and I (and presumably you too) were talking about a highly organised system of particles - a rotating flywheel - not a system of particles of random velocities.

In my terminology, the individual bits of the rotating flywheel have different velocities \vec{v}_i, and the flywheel as a whole has velocity \vec{V}.

Do you agree?

 

[1effect]

1effect, when you joined this thread at post #29, you were commenting on Dr Greg's post, in which he commented on a rotating flywheel:



You then disagreed on his calculations (I'm staying neutral on that!), and gave your own calculations.

But Dr Greg and I (and presumably you too) were talking about a highly organised system of particles - a rotating flywheel - not a system of particles of random velocities.

In my terminology, the individual bits of the rotating flywheel have different velocities \vec{v}_i, and the flywheel as a whole has velocity \vec{V}.

Do you agree?

For the third time, I am talking about a system of particles with a random distribution of velocities v_i and with NO "box".

 

[DrGreg]

Sorry, I haven't been able to respond for several days.

First, the argument I gave in post #17, although it was in response to tiny-tim's example in post #11 of a rotating wheel, works perfectly well for any system of particles with arbitrary velocities. I made no assumption about a "rigid body". So my argument also shows that a hot object has more invariant mass than an "identical" cold object (the thermal energy being, of course, kinetic energy of constituent molecules).

The equations I used are the same equations (rearranged and in different notation) as appear in the Wikipedia article Invariant mass.

What I did do was choose a frame in which the total momentum \textbf{P} = \Sigma \textbf{p} is zero. (As invariant mass is invariant, I can choose to calculate via any frame I wish.) This frame is called the centre of momentum frame (a.k.a. centre of mass frame) (I'm British, that's how we spell "centre").

1effect has asked, how do you measure the overall velocity of a system of particles? The answer is, it is the velocity of the centre of momentum frame (relative to the observer asking the question).

For those readers who understand 4-vectors, all the above can be expressed in terms of 4-momentum \left( E/c, \textbf{p} \right). It can be shown this is indeed a 4-vector (you use the Lorentz transform to change inertial frames), and it's additive (by the laws of conservation of energy and momemtum). Invariant mass is just the norm ("length") of this 4-vector, which is automatically an invariant (frame-independent) quantity. To find the centre of momentum frame, calculate the total 4-momentum (the sum of the individual particle 4-momenta), which will be a timelike 4-vector, and choose a coordinate system in which this vector is straight up the "t" axis.

In post #31, 1effect claimed that the formula (with c = 1)

M^2 = m_1^2 + m_2^2 + 2\left(E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 \right)

for the invariant mass of a two-particle system was "not invariant" due to the presence of the "variant" term \textbf{p}_1 \cdot \textbf{p}_2. However the term E_1 E_2 isn't invariant either and the "variants" cancel out. In fact the expression E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 is (when c = 1) the "inner product" of two 4-momentum vectors and is therefore invariant.

 

[1effect]

In post #31, 1effect claimed that the formula (with c = 1)

M^2 = m_1^2 + m_2^2 + 2\left(E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 \right)

for the invariant mass of a two-particle system was "not invariant" due to the presence of the "variant" term \textbf{p}_1 \cdot \textbf{p}_2. However the term E_1 E_2 isn't invariant either and the "variants" cancel out. In fact the expression E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 is (when c = 1) the "inner product" of two 4-momentum vectors and is therefore invariant.

True, E_1 E_2 is also frame-variant, so how do you prove that E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 is invariant? I am getting

E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2=\gamma_1 \gamma_2 m_1 m_2( c^2-\vec{v_1} \vec{v_2}). This is still not an invariant.

Another question, if E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 is indeed invariant, why do you use ||\vec{p}|| instead of \vec{p} in your formulas?

As always, thank you for your help.

I got a different approach. Given a system of particles of masses m_i and distribution of velocities \vec{v_i} you can always find a mass M and a velocity \vec{V} such that:

E=c^2 \Sigma (\gamma_i m_i)=c^2 \gamma(V) M

and

\vec{P}=\Sigma \gamma_i m_i \vec{v_i}=\gamma(V) M \vec{V}

So:

\vec{V}=\frac {\Sigma \gamma_i m_i \vec{v_i}}{\Sigma \gamma_i m_i}

and

M=\frac{\Sigma \gamma_i m_i}{\gamma(V)}

M is clearly not an invariant, so though E^2-c^2 \vec{p} \vec{p}=M^2c^4 doesn't mean anything, since M is not an invariant.
I think that it can be shown that M>\Sigma m_i ...but I don't know if it has any value...

 

[tiny-tim]

Given a system of particles of masses m_i and distribution of velocities \vec{v_i} you can always find a mass M and a velocity \vec{V} such that:

E=c^2 \Sigma (\gamma_i m_i)=c^2 \gamma(V) M
and
\vec{P}=\Sigma \gamma_i m_i \vec{v_i}=\gamma(V) M \vec{V}

Isn't that just the good ol' centre of mass and its velocity?

So:

\vec{V}=\frac {\Sigma \gamma_i m_i \vec{v_i}}{\Sigma \gamma_i m_i}

and

M=\frac{\Sigma \gamma_i m_i}{\gamma(V)}.

I think these are the same as my E_{tot} and P_{tot} for \vec{V} = 0 (careful!: my \vec{V} is completely different from your \vec{V})

If you work it out (I can't be bothered right now :zzz:), it's:

E_{tot}\,=\,\frac{M}{\sqrt{1\,-\,V^2/c^2}}\qquad,\qquad P_{tot}\,=\,\frac{M\,\vec{V}}{\sqrt{1\,-\,V^2/c^2}}\qquad,​

where M is a constant (= E_{tot} for \vec{V} = \vec{0}).

 

[1effect]

Isn't that just the good ol' centre of mass and its velocity?

While your formulas are identical to the ones I derived, looking at the wiki definition , the answer is "no", the formulas do not appear to have anything with the "center of mass" (see also my post to DrGreg). I did not use any of the assumptions about center of mass in my derivation.

I think these are the same as my E_{tot} and P_{tot} for \vec{V} = 0 (careful!: my \vec{V} is completely different from your \vec{V})

You can't make \vec{V} what you like it to be, \vec{V} has its own formula, see my post.
As an aside, you cannot claim that "M is constant because it is equal to E_{tot} at \vec{V}=0" since E_{tot} is a frame-variant quantity.

 

[1effect]

What I did do was choose a frame in which the total momentum \textbf{P} = \Sigma \textbf{p} is zero. (As invariant mass is invariant, I can choose to calculate via any frame I wish.) This frame is called the centre of momentum frame

The derivation is done for classical mechanics, it is not clear (at least to me) how does this translate for relativity. The equations should contain the terms in \gamma(v_i) . I am not seeing that.

I am guessing that the relativistic formulation would be the stuff that I got earlier:

\vec{V}=\frac {\Sigma \gamma_i m_i \vec{v_i}}{\Sigma \gamma_i m_i}

What do you think?

 

[tiny-tim]

You can't make \vec{V} what you like it to be, \vec{V} has its own formula, see my post.

Oi! You don't have the copyright on \vec{V} (erm … you don't do you? …)

You and I can each use the letter \vec{V} to mean whatever we like!

As an aside, you cannot claim that "M is constant because it is equal to E_{tot} at \vec{V}=0"

I didn't! You've mis-copied that!

 

[1effect]

Oi! You don't have the copyright on \vec{V} (erm … you don't do you? …)

You and I can each use the letter \vec{V} to mean whatever we like!

I don't think you understood:

\vec{V}=\frac {\Sigma \gamma_i m_i \vec{v_i}}{\Sigma \gamma_i m_i}

I didn't! You've mis-copied that!

I cut and paste from your post #46 (see bottom)