Components of 4-momentum

  • Thread starter Rudipoo
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  • #1
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Firstly, apologies for the notation.

The 4-momentum of a massive particle (rest mass m) is defined by

p=mu

where u is the 4-velocity. Thus in a frame S in which a particle has 3-velocity u the components of p are

[p]=gamma*(mc,mu)

How can we then identify the zeroth component of this with the energy/c and the spatial component as the 3-momentum measured in S? In my lecture notes it just makes this assertion and I don't know where that comes from.

Thanks
 

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  • #2
PAllen
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Firstly, apologies for the notation.

The 4-momentum of a massive particle (rest mass m) is defined by

p=mu

where u is the 4-velocity. Thus in a frame S in which a particle has 3-velocity u the components of p are

[p]=gamma*(mc,mu)

How can we then identify the zeroth component of this with the energy/c and the spatial component as the 3-momentum measured in S? In my lecture notes it just makes this assertion and I don't know where that comes from.

Thanks

One argument is from conservation. If you accept that 4 momentum is conserved, then note that for Lorentz coordinates, the first component is always positive and is conserved separately from the other 3 components. This suggests it is total energy.

Partly, this is just a matter of definitions of mathematical objects that lead to physically meaningful predictions. In other words, it is defined this way because it works.

Note that the first component being energy is very much tied to 'common conventions' : use of locally Lorentz coordinates, with a particular order of listing components. There are coordinate independent definitions of these things that can be more useful in general relativity.
 
  • #3
K^2
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[p]=gamma*(mc,mu)

How can we then identify the zeroth component of this with the energy/c and the spatial component as the 3-momentum measured in S?

[tex](\gamma mc^2)^2 = \gamma^2(mc^2)^2 = \frac{c^2}{c^2 - v^2}(mc^2)^2 = \left(1 + \frac{v^2}{c^2 - v^2}\right)(mc^2)^2 = m^2c^4 + \gamma^2 m^2 v^2 c^2 = m^2c^4 + p^2c^2 = E^2[/tex]

Or are you having trouble with p = γmv?
 
  • #4
bcrowell
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How can we then identify the zeroth component of this with the energy/c and the spatial component as the 3-momentum measured in S?

I would start with the paper by Einstein "Does the inertia of a body depend upon its energy content?"
http://fourmilab.ch/etexts/einstein/E_mc2/www/ . Einstein describes a thought experiment that shows that mass and energy are equivalent.

Next you can expand [itex]p_0[/itex] in a Taylor series, giving [itex]p_0=m+mv^2/2+\ldots[/itex] (in units with c=1). That shows that in the Newtonian limit, [itex]p_0[/itex] corresponds to the rest mass plus the Newtonian kinetic energy.

Now we step back and ask ourselves what we expect the laws of physics to be in special relativity. We expect there to be conservation laws, and we know by the correspondence principle that these conservation laws must somehow reduce to the conservation laws of nonrelativistic mechanics at low velocities. We also expect that these conservation laws should be valid in any frame of reference. But in general when you write an equation involving quantities like m and v in one frame of reference, that equation will *not* be valid when you change to another frame of reference in SR. The only way it will be frame-independent is if the quantities on the two sides of the equation are both Lorentz scalars, or both four-vectors (or higher-order tensors).

Based on these considerations, if there is going to be a conservation law in SR that plays the appropriate role, the only possible conservation law is conservation of the momentum four-vector, and its timelike component should be interpreted as mass-energy. That doesn't mean that there *will* be such a conservation law, only that if there is, it must have these properties. But experiments do confirm it.
 
  • #5
jambaugh
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Another aspect of this unification. When you look at canonical formulations of mechanics you find a duality between (potential) symmetries and (potentially) conserved quantities, for example between rotation and angular momentum.

The dual of translations in space (parametrized by x,y,z) are the momenta P_x, P_y, P_z. The dual of translation in time is energy. So if you unify space+time you get a dual vector momentum+energy.

A related way to see it is that action units are action = momentum * distance = energy*time.
 
  • #6
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Firstly, apologies for the notation.

The 4-momentum of a massive particle (rest mass m) is defined by

p=mu

where u is the 4-velocity. Thus in a frame S in which a particle has 3-velocity u the components of p are

[p]=gamma*(mc,mu)

How can we then identify the zeroth component of this with the energy/c and the spatial component as the 3-momentum measured in S? In my lecture notes it just makes this assertion and I don't know where that comes from.

Thanks

if this wasn't apparent, note:

[Tex] E=mc^2\Rightarrow mc=\frac{mc^2}{c}=\frac{E}{c} [/Tex]

Then Energy is the first component of the momentum... divided by some super common number in physics...
 

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