# When is the momenergy vector invariant?

• Cluemore
In summary, the conversation discusses the concept of "momenergy" and its relation to elastic collisions and the invariance of 4-momentum in different frames. The example of a photon colliding with a stationary electron is used to illustrate the application of this concept. The conversation also mentions the importance of standard naming conventions in understanding complex topics.
Cluemore
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.

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?

Cluemore
"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.

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Cluemore
Ibix said:
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?

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 occurring), 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?

Clueless said:
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 occurring), 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?

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.

Cluemore
Mentz114 said:
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.

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!

Clueless said:
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!

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.

Mentz114 said:
Definately not. The 4-vector is the original tensor. The magnitude is invariant under Lorentz transformations and it obeys tensor algebra, eg uμuνu^\mu u^\nu is a rank-2 tensor.
But surely you know this because the expression P¯⋅P¯−−−−−√=m0c2\sqrt{\bar{P}\cdot \bar{P}}=m_0c^2 says just uμuμ=m20c4u^\mu u_\mu=m_0^2c^4 and m0m_0 and cc 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.

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

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.

vanhees71 said:
The expression ##p_{\mu} p^{\mu}## is invariant under Lorentz transformations, because it's a Minkowski product.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.

What is an open composite system?

vanhees71 said:
It's defined as ##m^2 c^2=p_{\mu} p^{\mu}##,

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##.

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
$$p^{\mu}=\begin{pmatrix} E \\ c \vec{p} \end{pmatrix},$$
$$p_{\mu} p^{\mu} = m^2 c^4.$$

Clueless said:
What is an open composite system?
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##".

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

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.

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?

vanhees71

## 1. What is the momenergy vector invariant?

The momenergy vector invariant is a quantity that remains constant in a closed system, regardless of any internal changes. It is a measure of the total energy and momentum of a system, taking into account both the energy and direction of motion of all particles within the system.

## 2. Why is the momenergy vector invariant important?

The momenergy vector invariant is important because it allows us to understand and predict the behavior of physical systems. It is a fundamental concept in physics and is used in various fields, such as mechanics and thermodynamics, to analyze and solve problems.

## 3. When is the momenergy vector invariant conserved?

The momenergy vector invariant is conserved in a system when there are no external forces acting on it. This means that the total energy and momentum of the system will remain constant over time, even if there are internal interactions or transformations occurring within the system.

## 4. How is the momenergy vector invariant calculated?

The momenergy vector invariant is calculated by adding together the energy and momentum of all particles within the system. This can be done using mathematical equations and formulas, depending on the specific system being analyzed.

## 5. Can the momenergy vector invariant ever change?

In a closed system, the momenergy vector invariant will always remain constant. However, in an open system, where external forces can act on the system, the momenergy vector invariant may change over time. This is because external forces can add or remove energy and momentum from the system, altering the overall invariant quantity.

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