# When is the momenergy vector invariant?

1. Jun 4, 2015

### Clueless

This has been confusing me for an entire hour or so.

The momenergy vector is defined as mass*((spacetime displacement)/(proper time)). I understand this as much, but I don't know how to apply it to situations. The following snapshot is from a question about a photon colliding with a stationary electron and being backscattered.

I don't understand how the conclusion about the magnitudes of the momenergy being the same can be drawn from knowing that an elastic collision has occured. Kinetic energy is when total kinetic energy is conserved in the collision. But how does that translate to the invariance of momenergy vectors? There is something I am missing, but I don't know what.

2. Jun 4, 2015

### Ibix

This vector is usually called the 4-momentum. It's a good idea to stick to standard naming conventions, especially in a complex topic like this, so you don't have to translate, as well as understand, our replies.

Anyway, the 4-momentum is a 4-vector. Do you know what that means for its modulus when measured in different frames? Do you know its modulus in a particle's rest frame?

3. Jun 4, 2015

### robphy

"Momenergy" is a term introduced in the 2nd edition of Taylor-Wheeler's Spacetime Physics.

Here's a glimpse of the relationship between "elastic collisions" as learned in intro physics and as used in special relativity.
Consider an elastic collision of two massive particles as follows...
Write down the expression for the "conservation of total relativistic-kinetic energy".
Write down the expression for the time-component [i.e. energy-part] of "conservation of total 4-momentum".
Subtract.

Last edited: Jun 4, 2015
4. Jun 4, 2015

5. Jun 4, 2015

### Clueless

I did come across that modulus, but didn't see how it was important. I got the following from https://teachers.web.cern.ch/teachers/archiv/HST2002/Bubblech/mbitu/energymomentum_4vector.htm

Apparently the "length" or modulus of the 4-momentum is invariant for all inertial reference frames, because rest mass is invariant (though you can still have rest mass change in particle annhilation/creation, right?)

https://teachers.web.cern.ch/teachers/archiv/HST2002/Bubblech/mbitu/energy37.gif
Anyway, even with this interesting property, how do I understand the example I gave?

So because rest mass energy is invariant (and no particle annhilation/creation is occuring), we can equate the above? How does kinetic energy conservation (elastic collision) contribute to this conclusion? Or is it unrelated? I know both total momentum and energy is conserved (that seems to form the crux of most of the questions I have encountered concerning collisions), but I fail to see the significance of the elastic collision?

6. Jun 4, 2015

### Mentz114

Working in the rest frame of the matter and supposing it is struck by a photon coming along the x axis so they both recoil in the x direction, we can say for the matter

momentum before $P_0=(m,0,0,0)$, momentum after $P_1= m(\gamma,\gamma\beta,0,0)$.

The photon has momentum $k$ before the collision so the 4-vector is $p_0=(\nu,k,0,0)$ where $k=1/\lambda =\nu$, and after the collision the momentum becomes $k'$. For energy conservation the energy before must be the same as after. The energy is the first term of the four vectors ( with $c=h=1$) and the momentum is in the other terms. I don't have time for more right now.

7. Jun 6, 2015

### Clueless

Hmm...not sure I get it or not. This is a process I have applied in my working, but like I have reiterated before - I don't understand the rationale behind equating the 4-vector for momentum. In fact, the way it is written, the 4 vectors are distinctly different! Is is just a rule of thumb that the magnitudes are invariant? Or does making that choice require some initial conditions?

I just need to know why we are equating what I am seeing. I am terribly sorry if I have been vague at expressing my question.

But thanks Mentz114 so much for responding!

8. Jun 6, 2015

### Mentz114

Definately not. The 4-vector is the original tensor. The magnitude is invariant under Lorentz transformations and it obeys tensor algebra, eg $u^\mu u^\nu$ is a rank-2 tensor.
But surely you know this because the expression $\sqrt{\bar{P}\cdot \bar{P}}=m_0c^2$ says just $u^\mu u_\mu=m_0^2c^4$ and $m_0$ and $c$ are both scalars.

The important thing about tensors is that tensor contractions to scalars are invariant under a big class of transformations. The definition of a tensor is that the contravariant indexes transform with the inverse of the covariant transformation so when a contraction is made the two transforms always cancel to unity.
Anyhow I don't know if this is relevant or helpful.

9. Jun 6, 2015

### Clueless

Wow....I definitely cannot follow what is going on here. I am going to go and study this...ty again!

10. Jun 6, 2015

### vanhees71

The expression $p_{\mu} p^{\mu}$ is invariant under Lorentz transformations, because it's a Minkowski product.

A different but unrelated question is that about conservation laws, and indeed while energy and momentum are conserved for any closed system, because Minkowski space admits space-time translations as symmetries, mass is not conserved in relativity. This is a pretty subtle difference to Newtonian physics, where you can explain why mass is also conserved (due to superselection rules).

An example is the rest mass of a solid body. It's defined as $m^2 c^2=p_{\mu} p^{\mu}$, where $p^{\mu}$ is the total four-momentum (I refuse to use a word like "momenergy" for this well-established quantity which has the proper name four-momentum in the literature) of the body. Now you can heat up the body, which means you add energy to the body (heat $Q$), and this energy increases its invariant mass by the amount $Q/c^2$. Thus if you have an open composite system, which can be somehow "excited", its invariant mass does not need to be a constant.

11. Jun 7, 2015

### Clueless

What is an open composite system?

12. Jun 7, 2015

### Staff: Mentor

Pedantic note: this form of the formula means the units of the LHS and RHS are momentum squared. It's more usual for the units to be energy squared (one reason is that experiments are more usually specified in terms of particle energies), which would make the formula $m^2 c^4 = p_{\mu} p^{\mu}$ (and the units of $p^{\mu}$ would be energy units rather than momentum units).

Of course, many relativists routinely use units where $c = 1$, which makes energy, momentum, and mass all essentially have the same units. In these units the LHS would just be $m^2$.

13. Jun 8, 2015

### vanhees71

Yep, I shouldn't envoke $c$ ;-). I thought the usual convention is to use the units of the spatial part as for the electric current,
$$j^{\mu}=\begin{pmatrix} c \rho \\ \vec{j} \end{pmatrix},$$
i.e.,
$$p^{\mu}=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}.$$
Of course, if you define
$$p^{\mu}=\begin{pmatrix} E \\ c \vec{p} \end{pmatrix},$$
$$p_{\mu} p^{\mu} = m^2 c^4.$$

14. Jun 8, 2015

### vanhees71

It's a system consisting of two or more particles (bound states like atoms, atomic nuclei, or condensed matter) which interact with something else and can get intrinsically excited. Then you define the intrinsic quantities like mass, charge(s) in the restframe of this system, i.e., where the total three-momentum of this system is 0, but with the energy of the composed system, also its invariant mass changes. This is the right interpretation of Einstein's most famous formula "$E=m c^2$".

15. Jun 8, 2015

### pervect

Staff Emeritus
The magnitude of the momenergy 4-vector is the invariant mass of a particle or system. You should find this in your textbook. Can you understand why undergoing an elastic collision doesn't change the invariant mass of either particle?

For extra credit, do you understand why an inelastic collision (assume this is between systems of particles rather than particles) could change the invariant mass?