Lorentz invariance of the 4-momentum

[redtree]

TL;DR Summary

Lorentz invariance of the 4-momentum is modeled utilizing the energy momentum relation. Instead of this approach, is the Lorentz invariance of the 4-momentum ever modeled as the group action of the Lorentz group acting on the 4-momentum (similar to what is done with 4-position)?

I apologize for the simple question, but here it is.

The energy momentum relation models changes in energy $E$ and 3-momentum $\textbf{p}_3 = \{ p_1,p_2,p_3 \}$ with changes in velocity given $(\textbf{p}_4)^2 = M^2$, where $M$ denotes mass and $\textbf{p}_4$ denotes 4-momentum, where $$M^2 = (\gamma M)^2 - (\gamma M \textbf{v})^2 $$
where $\textbf{v}$ denotes relative velocity, $E= \gamma M$ and $\textbf{p}_3 = \gamma M \textbf{v}$.

However, for the 4-position $\textbf{x}_4$
$$\textbf{x}_4' = \{ x_0',\textbf{x}_3'\} = \Lambda \textbf{x}_4$$
where $\textbf{x}_3 = \{ x_1,x_2,x_3 \}$ and
$$x_0' = \gamma x_0 + \gamma \textbf{v} \textbf{x}_3$$
$$\textbf{x}_3'= \gamma \textbf{x}_3 + \gamma \textbf{v} x_0 $$

Similarly, why not the following for $\textbf{p}_4$?
$$\textbf{p}_4' = \{ E',\textbf{p}_3'\} = \Lambda \textbf{p}_4$$
where $$E' = \gamma E + \gamma \textbf{v} \textbf{p}_3$$
$$\textbf{p}_3'= \gamma \textbf{p}_3 + \gamma \textbf{v} E $$
as this leaves $$(\textbf{p}_4)^2 = (\textbf{p}_4')^2$$

 

[anuttarasammyak]

It seems that you use indexes 0 1 2 3 4. Which are time, space and what is the rest ?

 

[Ibix]

It seems that you use indexes 0 1 2 3 4. Which are time, space and what is the rest ?

He's using 0 as the timelike index and 3 and 4 as subscripts to identify 3- and 4-vectors. It's a slightly confusing notation, particularly as the stated transformation laws seem to be only the (1+1) dimensional form - so effectively 0 is the timelike index, 3 is the spacelike index and 4 isn't an index.

 

[Ibix]

I apologize for the simple question, but here it is.

I'm not sure I understand the question. The point about the transformation law is that it works for all vectors. Four-momentum is a particular four-vector, so needs to be specified by more than its transformation properties. Hence specifying that it is normalised to $M$. We can then use the correspondence principle to equate the components to Newtonian energy and momentum in the low velocity limit. Is that what you were asking?

 

[redtree]

Sorry. I fixed the notation in the original post. (Forgot to bold the vectors)

 

[redtree]

I'm not sure I understand the question. The point about the transformation law is that it works for all vectors. Four-momentum is a particular four-vector, so needs to be specified by more than its transformation properties. Hence specifying that it is normalised to $M$. We can then use the correspondence principle to equate the components to Newtonian energy and momentum in the low velocity limit. Is that what you were asking?

Why does 4-momentum need to be any more specified than 4-position?

 

[Sagittarius A-Star]

Similarly, why not the following for $\textbf{p}_4$?

You can Lorentz-transform the components of each 4-vector, including 4-momentum.

Source:
https://en.wikipedia.org/wiki/Four-vector#Lorentz_transformation

 

[redtree]

$$\textbf{p}_4' = \{ E',\textbf{p}_3'\} = \Lambda \textbf{p}_4$$
where $$E' = \gamma E + \gamma \textbf{v} \textbf{p}_3$$
$$\textbf{p}_3'= \gamma \textbf{p}_3 + \gamma \textbf{v} E $$
as this leaves $$(\textbf{p}_4)^2 = (\textbf{p}_4')^2$$

I agree. So then it's acceptable to transform the 4-momentum as above?

 

[Ibix]

I agree. So then it's acceptable to transform the 4-momentum as above?

Yes. (Edit: as discussed in #12 below, actually the transformation quoted in #8 is not generally correct, but assuming that's just an error, yes.) It's a four vector so can be Lorentz transformed.

The reason you might be interested in a four-vector that is tangent to the path of a particle and has a modulus $M$ is that it is conserved in collisions, which is a property over and above its transformation rules. That's why you need to pick out the tangent vector with that particular modulus.

 

[redtree]

Perfect; thanks!

 

[Sagittarius A-Star]

Instead of this approach, is the Lorentz invariance of the 4-momentum ever modeled as the group action of the Lorentz group acting on the 4-momentum (similar to what is done with 4-position)?

Yes. Another argument: The 4-momentum can be written as
$\mathbf P = {m \over d\tau} (cdt, dx, dy, dz)$
... and the factor before the differential 4-position is an invariant.

 

[Ibix]

Similarly, why not the following for $\textbf{p}_4$?
$$\textbf{p}_4' = \{ E',\textbf{p}_3'\} = \Lambda \textbf{p}_4$$
where $$E' = \gamma E + \gamma \textbf{v} \textbf{p}_3$$
$$\textbf{p}_3'= \gamma \textbf{p}_3 + \gamma \textbf{v} E $$
as this leaves $$(\textbf{p}_4)^2 = (\textbf{p}_4')^2$$

Note that this isn't correct if the bolded items are actually three vectors. The Lorentz transforms in their usual form are $$\Lambda = \left(\begin{array}{cccc}
\gamma&-\gamma v&0&0\\
-\gamma v&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\right)$$so $$p'=\left(\begin{array}{c}
\gamma(E-vp_x)\\
\gamma(p_x-vE)\\
p_y\\
p_z\end{array}\right)$$That means the transformed three-momentum is rather more complicated than just $\textbf{p}_3'= \gamma \textbf{p}_3 + \gamma \textbf{v} E$ unless you are in a one dimensional case where all 3-vector components orthogonal to the frame 3-velocity are zero. (Edit: even ignoring my opposite sign convention.)

 

[Orodruin]

4-position is a special case that is relevant only in an affine spacetime such as Minkowski space. There is no corresponding position vector in GR for example.

The 4-momentum on the other hand has a straightforward generalization to non-affine spacetimes as the tangent of a particle’s world line.

As such, it is even more natural that the 4-momentum follows the vector transformation rules than that the 4-position does.

 

[PeterDonis]

Lorentz invariance of the 4-momentum is modeled utilizing the energy momentum relation. Instead of this approach, is the Lorentz invariance of the 4-momentum ever modeled as the group action of the Lorentz group acting on the 4-momentum (similar to what is done with 4-position)?

These aren't two different ways of modeling Lorentz invariance of 4-momentum. They are two different aspects of the same model. The invariance of the energy-momentum relation $E^2 - p^2 = m^2$ bears the same relationship to the Lorentz transformation of the 4-vector $(E, \vec{p})$ that the invariance of the spacetime interval $t^2 - x^2 = \tau^2$ bears to the Lorentz transformation of the 4-vector $(t, \vec{x})$.

 

[redtree]

Note that this isn't correct if the bolded items are actually three vectors. The Lorentz transforms in their usual form are $$\Lambda = \left(\begin{array}{cccc}
\gamma&-\gamma v&0&0\\
-\gamma v&\gamma&0&0\\
0&0&1&0\\
0&0&0&1
\end{array}\right)$$so $$p'=\left(\begin{array}{c}
\gamma(E-vp_x)\\
\gamma(p_x-vE)\\
p_y\\
p_z\end{array}\right)$$That means the transformed three-momentum is rather more complicated than just $\textbf{p}_3'= \gamma \textbf{p}_3 + \gamma \textbf{v} E$ unless you are in a one dimensional case where all 3-vector components orthogonal to the frame 3-velocity are zero. (Edit: even ignoring my opposite sign convention.)

Could you explain? The bolded subscript "3" are 3-vectors and the bolded subscript "4" are 4-vectors.

 

[DrGreg]

Could you explain? The bolded subscript "3" are 3-vectors and the bolded subscript "4" are 4-vectors.

$$
\left(\begin{array}{c}
p'_x\\
p'_y\\
p'_z
\end{array}\right)
=
\left(\begin{array}{c}
\gamma(p_x-vE)\\
p_y\\
p_z
\end{array}\right)
$$is not compatible with $\textbf{p}'_3 = \gamma \textbf{p}_3 - \gamma \textbf{v} E$, except in the special case where $\textbf{p}_3$ is parallel to $\textbf{v}$ (when $p_y = p_z = 0$).

The general equations for non-parallel $\textbf{p}_3$ and $\textbf{v}$ are $$\begin{align*}
E' & = \gamma E - \gamma \textbf{v} \cdot \textbf{p}_3 \\ \\
\textbf{p}'_3 & = \textbf{p}_3 + (\gamma-1) \frac{\textbf{v} \cdot \textbf{p}_3}{v^2} \textbf{v} - \gamma \textbf{v} E
\end{align*}
$$See https://en.wikipedia.org/wiki/Lorentz_transformation#Proper_transformations

 

[redtree]

I'll take a look at the derivation of this. Thanks

 

[JimWhoKnew]

The reason you might be interested in a four-vector that is tangent to the path of a particle and has a modulus $M$ is that it is conserved in collisions, which is a property over and above its transformation rules. That's why you need to pick out the tangent vector with that particular modulus.

There is some more to it. Imposing the modulus to be $M$ relates it to the concepts of energy and momentum. But these are described by the energy-momentum tensor which transforms as a second-rank tensor. How do the two kinds of transformations fit? To get some idea, we can consider a local Lorentz frame where the massive object is "at rest" (a free fall local frame). In this frame$$T^{\mu\nu}\approx \rho~\delta_0^\mu~\delta_0^\nu \quad ,$$$$~E\approx M\approx \int \rho~dV$$(since $~\sqrt{-g}=1~$ ). $V$ is the volume of the object (assumed "sufficiently" small).
After the second-rank local Lorentz boost $$\rho'\approx T'^{~00}\approx \gamma^2\rho \quad ,$$but$$dV'=\frac{dV}\gamma$$because of Lorentz contraction, so $$E'\approx \int \rho'~dV'\approx \gamma M\approx \gamma E$$ $$p'^{~i}\approx \int T'^{~0i}~dV'\approx -\gamma \beta^i E$$behave like a 4-vector.

 

[pervect]

The way I recall it is this, (t, $\vec{x}$) is Lorentz covariant. Proper time $\tau$ is a world scalar. Therefore (dt/d$\tau$, d$\vec{x}/d \tau$) is Lorentz covariant. The invariant mass m is also a world scalar, so m times the derivative with time, the four-momentum, is also Lorent covariant.