Massive objects moving too fast

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  • #51
tiny-tim
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I don't think you understood:
I understood fine - your [tex]\vec{V}[/tex] means one thing, and my [tex]\vec{V}[/tex] 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 (= [tex]E_{tot}[/tex] for [tex]\vec{V} = \vec{0}[/tex]). :smile:
 
  • #52
DrGreg
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True, [tex]E_1 E_2 [/tex] is also frame-variant, so how do you prove that [tex]E_1 E_2 - \textbf{p}_1 \cdot \textbf{p}_2 [/tex] is invariant?
Well, first of all, do you accept that the 4-momentum [tex]\left( E/c, \textbf{p} \right) [/tex] is a genuine 4-vector, i.e. it obeys the Lorentz transform, e.g. simplified to one dimension of space

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

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

[tex]\left(mc \frac{dt}{d\tau}, m \frac{d\textbf{x}}{d\tau} \right) [/tex]

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

(Note that [tex]\frac{dt}{d\tau} = \gamma_v[/tex], and [tex]\frac{d\textbf{x}}{d\tau} = \frac{d\textbf{x}}{dt} \frac{dt}{d\tau} = \gamma_v \textbf{v} [/tex].)

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

Now just substitute (1) and (2) into [tex]E_1' E_2' - p_1' p_2' c^4 [/tex] 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 [tex]||\textbf{p}||[/tex] instead of [tex]\textbf{p}[/tex] 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.
 
  • #53
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Well, first of all, do you accept that the 4-momentum [tex]\left( E/c, \textbf{p} \right) [/tex] is a genuine 4-vector, i.e. it obeys the Lorentz transform, e.g. simplified to one dimension of space

[tex] E' = \gamma_u \left(E - u p \right) [/tex] .....(1)
Yes, I am very familiar with this transform

[tex] p' = \gamma_u \left(p - \frac{u E}{c^2} \right) [/tex]...(2) ?
This is the transform for the norm of the momentum, the expression that you need to calculate involves vectors, as in [tex]E_1E_2 -c^2 \vec{p_1} \vec{p_2}[/tex]
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:
[tex]E_1E_2 -c^2 ||\vec{p_1}|| ||\vec{p_2}||[/tex] 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 [tex]\vec{v_1} \vec{v_2}[/tex]



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 [tex]||\textbf{p}||[/tex] instead of [tex]\textbf{p}[/tex] 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.
 
Last edited:
  • #54
Ich
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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?
 
  • #55
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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.
 
  • #56
Ich
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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.
 
  • #57
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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
 
  • #58
pervect
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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.
 

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