# Massive objects moving too fast

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.

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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.

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:

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

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tiny-tim
Homework Helper
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! 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." ?

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tiny-tim
Homework Helper

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? 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.

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
Homework Helper
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 $$\vec{V}$$.)

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
Homework Helper
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 mtot 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, 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
Gold Member
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.

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...

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tiny-tim
Homework Helper
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 $$\vec{V} = \vec{0}$$). 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.

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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?

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tiny-tim
Homework Helper
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!!! 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)

tiny-tim
Homework Helper
I don't think you understood:

I understood fine - your $$\vec{V}$$ means one thing, and my $$\vec{V}$$ means another - they're not supposed to be the same. I pointed that out!

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

No, you didn't!! If you had, you'd have got:

where M is a constant (= $$E_{tot}$$ for $$\vec{V} = \vec{0}$$). DrGreg
Gold Member
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?
Well, first of all, do you accept that the 4-momentum $$\left( E/c, \textbf{p} \right)$$ is a genuine 4-vector, i.e. it obeys the Lorentz transform, e.g. simplified to one dimension of space

$$E' = \gamma_u \left(E - u p \right)$$ .....(1)
$$p' = \gamma_u \left(p - \frac{u E}{c^2} \right)$$...(2) ?

If you don't know why 4-momentum is a 4-vector, it can also be defined as

$$\left(mc \frac{dt}{d\tau}, m \frac{d\textbf{x}}{d\tau} \right)$$

where m is invariant mass and $$\tau$$ is the proper time of the particle being measured (also an invariant).

(Note that $$\frac{dt}{d\tau} = \gamma_v$$, and $$\frac{d\textbf{x}}{d\tau} = \frac{d\textbf{x}}{dt} \frac{dt}{d\tau} = \gamma_v \textbf{v}$$.)

So just apply the operator $$m \frac{d}{d\tau}$$ to the standard time-space Lorentz transform to get the energy-momentum Lorentz transform (1) and (2).

Now just substitute (1) and (2) into $$E_1' E_2' - p_1' p_2' c^4$$ and see what you get.

Your "different approach" to finding the energy & momentum ought to work too, and, yes, it is "just the good ol' centre of mass and its velocity" as tiny-tim says (but expressed in a different way to the Wikipedia "Invariant mass" article -- as is often the case, there's more than one way to define a concept). I haven't done the calculation myself but I feel confident both are equivalent.

And the reason I used $$||\textbf{p}||$$ instead of $$\textbf{p}$$ in the relevant formula is because that's how you calculate the norm of a 4-vector -- it's the metric "ds" but applied to the 4-momentum instead of to (cdt, dx). The Wikipedia "Invariant mass" article was a bit lax about including ||.|| norm symbols but I've corrected that now.

The Wikipedia article on "Center of mass frame" only calculates the Newtonian version. However the description in words in the first two paragraphs is still valid in the relativistic case.

Well, first of all, do you accept that the 4-momentum $$\left( E/c, \textbf{p} \right)$$ is a genuine 4-vector, i.e. it obeys the Lorentz transform, e.g. simplified to one dimension of space

$$E' = \gamma_u \left(E - u p \right)$$ .....(1)

Yes, I am very familiar with this transform

$$p' = \gamma_u \left(p - \frac{u E}{c^2} \right)$$...(2) ?

This is the transform for the norm of the momentum, the expression that you need to calculate involves vectors, as in $$E_1E_2 -c^2 \vec{p_1} \vec{p_2}$$
So, you cannot use eq (2). The transform for vectors is much more complicated.
This was the point of all the previous posts.
Unless , you decide to calculate the much easier quantity:
$$E_1E_2 -c^2 ||\vec{p_1}|| ||\vec{p_2}||$$ but this is not the dot product of the vectors. As you can see even from this example what needs to be computd is the dot product, not the product of the norms. When you do that you get stuck with the term in $$\vec{v_1} \vec{v_2}$$

Your "different approach" to finding the energy & momentum ought to work too, and, yes, it is "just the good ol' centre of mass and its velocity" as tiny-tim says (but expressed in a different way to the Wikipedia "Invariant mass" article -- as is often the case, there's more than one way to define a concept). I haven't done the calculation myself but I feel confident both are equivalent.

The wiki page deals with classical mechanics only, I think that the correct formula for relativity is the one I calculated earlier.

And the reason I used $$||\textbf{p}||$$ instead of $$\textbf{p}$$ in the relevant formula is because that's how you calculate the norm of a 4-vector -- it's the metric "ds" but applied to the 4-momentum instead of to (cdt, dx). The Wikipedia "Invariant mass" article was a bit lax about including ||.|| norm symbols but I've corrected that now.

The wiki page is very sloppy. Thank you for correcting it.

The Wikipedia article on "Center of mass frame" only calculates the Newtonian version. However the description in words in the first two paragraphs is still valid in the relativistic case.

Again, the wiki page deals with classical mechanics only, I think that the correct formula for relativity is the one I calculated earlier.

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Ich
I'm jumpin up for a last time, there is too much delay.
1effect said:
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.
Because
Ich said:
Sums of four vectors are four vectors
What is not clear about that?

I'm jumpin up for a last time, there is too much delay.

Because

What is not clear about that?

Because it is not sufficient for the sum of the two four-vectors to be a four vector. When you do the math you realize the problem. Look at my exchange with DrGreg.

Ich
Because it is not sufficient for the sum of the two four-vectors to be a four vector.
Of course it is sufficient, because all four vectors have invariant norm. That's pretty much the definition of a four vector.

Of course it is sufficient, because all four vectors have invariant norm. That's pretty much the definition of a four vector.

If you looked at the computations, thugh the norm of the energy-momentum 4-vector is invariant, it appears that the property does not transfer to the sum of energy-momentum 4-vectors

pervect
Staff Emeritus
I'm afraid I've lost track of this thread, which seems to have wandered around a lot. I will make a few remarks in the hope that people will listen.

The energy-momentum of a point particle (which has zero volume) or a closed system, can be described by a 4-vector.

The energy-momentum of a non-closed system with a non-zero volume is not, however, in general a 4-vector.

See for instance http://arxiv.org/abs/physics/0505004 which mentions this point, though it is primarly concerned with thermodynamics.

The energy-momentum of an object with finite volume is not a covariant physical entity because of the relativity of simultaneity

Another important point:

While the total-energy momentum of a general system is not necessarily a 4-vector, the distribution of energy-momentum can in general be covariantly described by a rank 2 tensor, the stress-energy tensor, however.