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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

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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

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

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

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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 [tex]\frac{m}{\sqrt{1\,-\,v^2/c^2}}[/tex]?

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

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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

The increased mass, rest-mass times [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], 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

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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!

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You just have to be careful!

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

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.

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You will be much better off applying the accepted standard of only using rest mass.Yes, but the question is whetherB'snormal 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 ...

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.

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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", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], 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!

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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!

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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.

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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.

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I agree.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.

(...and mine is not!)My taste is to say that the mass increases.

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.

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).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!

For the benefit of anyone who doesn't know why, the invariant mass

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

which reduces to

[tex]\Sigma E = m_{tot} c^2 [/tex]

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

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

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

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

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

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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 massmof a collection of particles can be defined by_{tot}

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

which reduces to [tex]\Sigma E = m_{tot} c^2 [/tex] 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:

[tex]\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}[/tex]

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

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

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Lorentz equations do not require they be applied to mass.We apply the "gamma factor", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], 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!

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.

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

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

Yes, but the question is whetherB'snormal 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

- #23

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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:

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

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

The simple fact is that "Most modern practicing science"

- #24

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Hi AB! (btw, I

{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

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

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 the same for all observers (no gamma factor); and

it can be used as [tex]m_0[/tex] in all equations I know of

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!

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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.

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- #26

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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 ("[tex]E_x = E_x\,'[/tex]");

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

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

(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

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I don't disagree there are important interrelationships between Energy Momentum and Mass.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 certainlynotindependent - 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"doesregard 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.

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.

- #28

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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?

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For the benefit of anyone who doesn't know why, the invariant massmof a collection of particles can be defined by_{tot}

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

which reduces to

[tex]\Sigma E = m_{tot} c^2 [/tex]

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

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

I am getting a different result, for a system of particles:

The total energy is:

[tex]E=\Sigma e= c^2 \Sigma \gamma_i m_i[/tex]

The total momentum is :

[tex]\textbf{P}=\Sigma \textbf{p}=\Sigma \gamma_i m_i \vec{v_i} [/tex]

[tex]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} ) [/tex]

On the other hand:

[tex]Q^2= \Sigma ||\textbf{p}||^2 =\Sigma \gamma_i^2 m_i^2 v_i^2 [/tex]

so:

[tex]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[/tex]

Comparing with your definition, I arrived to:

[tex]m_{tot}=\Sigma \gamma_i m_i > \Sigma m_i[/tex]

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- #30

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I'm not following that.

Anyway, both your formula and

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 [tex]m_i[/tex] and velocities [tex]\vec{V}\,+\,\vec{v_i}[/tex] (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:

[tex]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,[/tex]

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

- #31

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1. Why would we bother with [tex]V[/tex], when we have all the measurable quantities [tex]v_i[/tex] ?I'm not following that.

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

2. I think I pointed out that the quantity :

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

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

[tex]E^2-c^2 \Sigma ||\textbf{p}||^2[/tex]

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

[tex]E^2 -c^2 ||\Sigma \textbf{p}||^2 [/tex]

Note that both formulas need to include [tex]|| \textbf{p}||[/tex] instead of [tex] \textbf{p}[/tex] from the wiki page. There is nothing stopping you from using [tex] \textbf{p}[/tex] 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.

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- #32

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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?In this example, [tex]M[/tex] is clearly not an invariant due to the term [tex]p_1p_2[/tex]

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.

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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 [tex]v[/tex] makes the quantity [tex]M[/tex]

The total energy is:

[tex]E=\Sigma e= c^2 \Sigma \gamma_i m_i[/tex]

The total momentum is :

[tex]\textbf{P}=\Sigma \textbf{p}=\Sigma \gamma_i m_i \vec{v_i} [/tex]

[tex]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} ) [/tex]

Same problem as in the wiki example:

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

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- #34

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Why would we bother with [tex]V[/tex], when we have all the measurable quantities [tex]v_i[/tex] ?

Because this is for an object with internal moving parts, and:

(a) we very often don't have the [tex](v_i)[/tex]s;

(b) even when we do, it's difficult to work out each of the the [tex](\vec{V}\,+\,v_i)[/tex]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 [tex]\frac{1}{\sqrt{1\,-\,V^2/c^2}}[/tex], which is both much easier and more intuitive!

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Because this is for an object with internal moving parts, and:

(a) we very often don't have the [tex](v_i)[/tex]s;

(b) even when we do, it's difficult to work out each of the the [tex](\vec{V}\,+\,v_i)[/tex]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 [tex]\frac{1}{\sqrt{1\,-\,V^2/c^2}}[/tex], 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." ?

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