Why is the 3 momentum represented with a negative sign in the 4-vector?

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In summary: In other words, in a reference frame where the magnitude of the 4-vector is positive, the magnitude of the 4-vector would be positive if the reference frame were moved to another inertial frame, but would be negative in a frame where the magnitude of the 4-vector is "-1". So the convention is to use the "-" sign to indicate that the magnitude of the 4-vector is invariant.The reason that the "magnitude" of a 4-vector is invariant between different reference frames is because the 4-vector is a combination of the 3-momentum and the 4-velocity. The "magnitude" of the 4-vector is simply the sum of the "magn
  • #1
Davio
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Hey guys, what exactly is a 3 momentum? I can't any references to it anywhere on the net, which actually tells me what is it, I know a energy 4 momentum is, px , py, pz e/c, but that's not much help!
 
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  • #2
Three momentum is simply a vector containing all three momenta,

[tex]\boldmath P \unboldmath = \left(\begin{array}{c}p_x \\ p_y \\ p_z\end{array}\right)[/tex]
 
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  • #3
3-momentum is the "spatial-part" of the 4-momentum.

Here is one place to consult:
http://www2.maths.ox.ac.uk/~nwoodh/sr/ [Broken]
 
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  • #4
Davio said:
Hey guys, what exactly is a 3 momentum? I can't any references to it anywhere on the net, which actually tells me what is it, I know a energy 4 momentum is, px , py, pz e/c, but that's not much help!

In a 3D reference frame a line of arbitary length and direction from the origin of the frame can be described in terms of x,y and z component. Using pythagorous theorem we can find the length of that arbitary line from [tex] L =\sqrt{x^2+y^2+z^2}[/tex] You could call that length the 3 length sometimes abreviated to ||L||. When we include an additional dimension of time to the 3 spatial dimensions then we have the 4 length [tex] \sqrt{x^2+y^2+z^2-(ct)^2}[/tex]

or [tex]\sqrt{||L||^2-(ct)^2}[/tex]

Similarly 3 velocity ||v|| = [tex] \sqrt{v_x^2+v_y^2+v_z^2}[/tex]

and 3 momentum ||p|| = [tex] \sqrt{p_x^2+p_y^2+p_z^2}[/tex]

(Edited to fix the typo pointed out by ehj)
 
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  • #5
Shouldn't it be?
[tex]\sqrt{||L||^2-(ct)^2}[/tex]
 
  • #6
ehj said:
Shouldn't it be?
[tex]\sqrt{||L||^2-(ct)^2}[/tex]

Yes, sorry about the typo. :(
 
  • #7
robphy said:
3-momentum is the "spatial-part" of the 4-momentum.

Here is one place to consult:
http://www2.maths.ox.ac.uk/~nwoodh/sr/ [Broken]

I should clarify that, given a 4-momentum vector,
the 3-momentum is essentially the "spatial-part" according to a given observer.
That is, the 3-momentum is [obtained from] the vector-component of the 4-momentum that
is [Minkowski-]perpendicular to an observer's 4-velocity.
From a given 4-momentum vector, different observers will determine different 3-momentum vectors.

Given a 4-momentum [tex]\tilde p[/tex] and an observer's 4-velocity [tex]\tilde u[/tex] (with [tex]\tilde u \cdot \tilde u=1[/tex] in the [tex]+---[/tex] convention),
Write out this identity [a decomposition of [tex]\tilde p[/tex] into a part parallel to [tex]\tilde u[/tex], and the rest perpendicular to [tex]\tilde u[/tex]]:
[tex]\tilde p = (\tilde p \cdot \tilde u)\tilde u + (\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex].

The 4-vector [tex](\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex] is "purely spatial" according to the [tex] \tilde u [/tex] observer [check it by dotting with u], and can be thought of as a three-component vector in [tex] \tilde u [/tex]'s "space" by projection. That projected vector is the 3-momentum of the object according to [tex]\tilde u [/tex].

(Note, however, that the 4-vector [tex](\tilde p - (\tilde p \cdot \tilde u)\tilde u) [/tex] is generally NOT "purely spatial" according to another observer [tex] \tilde w [/tex]. To [tex] \tilde w [/tex], that 4-vector has both nonzero spatial- and temporal-parts.)
 
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  • #8
Hey guys, feel a bit silly, of course its just the partial part! Whilst we're on that topic though, why is it -(ct)^2 and not positive? Is it because CT=Distance, L^2+D^2 =p^2 ? Surely not though because, L ^2 = x^2 +y^2 +z^2
 
  • #9
Davio said:
why is it -(ct)^2 and not positive?

With the "-" sign in that position, the "magnitude" of a 4-vector is invariant between different inertial reference frames. With a "+" sign instead, the "magnitude" is not invariant.
 

1. What is momentum?

Momentum is a physical quantity that describes the amount of motion an object has. It is calculated by multiplying an object's mass by its velocity.

2. How is momentum measured?

Momentum is measured in units of kilogram-meters per second (kg•m/s).

3. What is the equation for momentum?

The equation for momentum is p = mv, where p is momentum, m is mass, and v is velocity.

4. What is the principle of conservation of momentum?

The principle of conservation of momentum states that in a closed system, the total momentum before an event must be equal to the total momentum after the event. This means that momentum cannot be created or destroyed, only transferred between objects.

5. How does momentum affect collisions?

Momentum plays a crucial role in collisions. In an elastic collision, the total momentum of the objects before and after the collision remains the same. In an inelastic collision, some of the momentum is lost as heat or sound, but the total momentum still remains constant.

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