The Momentum Four-Vector

8. The Momentum Four-Vector#

You have been introduced to the displacement 4-vector, the Lorentz transformation and to the velocity 4-vector. These quantities can be used to locate objects and determine their motion. Borrowing an idea introduced by Newton to predict the behavior of dynamical systems, we introduce the momentum 4-vector. The 4-velocity is a proper 4-vector in that its components in two different reference frames are related by the Lorentz transformation, and its size is a Lorentz scalar as demanded by the Michelson- Morley experiment.

Knowing that if I multiply a proper 4-vector by a Lorentz scalar, I will get another proper 4-vector, I introduce:

(8.1)#\[[p_4]=m_0[v_4]\]

where \(m_0\) is the mass as measured by an observer in the rest frame of the object. As is true with the time interval as measured in the rest frame, the ‘rest mass’ , the mass measured in the rest frame of the object, is a Lorentz invariant.

8.1. Momentum#

The components of the momentum 4-vector introduced in equation (8.1) are

(8.2)#\[\begin{split}[p_4] = m_0[v_4] = m_0 \gamma \frac{[dx_4]}{dt}= \begin{bmatrix} icm_0\gamma\\ m_0\gamma v_x\\ m_0\gamma v_y\\ m_0\gamma v_z \end{bmatrix}\end{split}\]

The three spatial components are almost the classical components of the momentum of a particle, which is usually defined as \(\vec{p}\equiv m\vec{v}\). What Newton missed was the factor of \(\gamma\). It is not surprising that this happened as the value of \(\gamma\) for the things Newton could observe is different from the value 1 by just a few parts in \(10^{12},\) hardly worth taking into account.

So, let us re-define the physical momentum of a particle as:

(8.3)#\[\vec{p}\equiv m_0 \gamma \vec{v}\]

which means Equation (8.1) becomes

(8.4)#\[\begin{split}\boxed{ [p_4] = \begin{bmatrix} icm_0\gamma\\ p_x\\ p_y\\ p_z \end{bmatrix} }\end{split}\]

To check if this 4-vector does follow the dictates of the Michelson-Morley experiment, we use the form for the 4-momentum found in Equation (8.2) and take the dot product of the 4-vector with itself:

(8.5)#\[[p_4]^2 = m_0^2\gamma^2(-c^2+v_x^2+v_y^2+v_z^2)\]
(8.6)#\[[p_4]^2 = -m_0^2c^2\gamma^2(1-v^2/c^2) = -m_0^2c^2\]

Since \(m_0\) is an invariant and so is \(c\), the size of the momentum 4-vector is a Lorentz invariant. Once again, it’s negative.

If a system is made up of more than one particle, the total 4-momentum for the system is the sum of the momentum 4-vectors of each of the pieces. It is the size of the net (sum of momentum of each of the pieces) 4-momentum that is a Lorentz scalar.

8.2. Momentum and Lorentz Transformations#

The components of the momentum 4-vector transform from those measured by one inertial observer to those measured by a second inertial observer moving with respect to the first with \(\beta_R\) in the \(+x\) direction in the same manner as any other 4-vector:

(8.7)#\[\begin{split}[p_4]' = {\cal L}_x(\beta_R)[p_4] = \begin{bmatrix} \gamma_R & -i\beta_R\gamma_R & 0 & 0\\ i\beta_R\gamma_R & \gamma_R & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} icm_0\gamma\\ m_0\gamma v_x\\ m_0\gamma v_y\\ m_0\gamma v_z \end{bmatrix}\end{split}\]

Carry out the matrix multiplication to get

(8.8)#\[\begin{split}[p_4]' = \gamma_Rm_0\gamma \begin{bmatrix} i(c-\beta_Rv_x)\\ -\beta_Rc+v_x\\ v_y\\ v_z \end{bmatrix}\end{split}\]

factor out a \(c\) and compare with the primed variables:

(8.9)#\[\begin{split}[p_4]' = \gamma_R\gamma m_0c \begin{bmatrix} i(1-\beta_R\beta_x)\\ \beta_x-\beta_R\\ \beta_y\\ \beta_z \end{bmatrix} = \begin{bmatrix} icm_0\gamma'\\ p'_x\\ p'_y\\ p'_z \end{bmatrix}\end{split}\]

From this we can deduce predictions like

(8.10)#\[\gamma' = \gamma_R\gamma(1-\beta_R\beta_x)\]

or

(8.11)#\[p'_x = \gamma_R\gamma m_0c(\beta_x-\beta_R)=\gamma'm_0v'_x\]

Divide Equation (8.11) by Equation (8.10) to get the velocity addition formula again!

8.3. Example of Transforming Momentum#

Example 8.2: Suppose that two observers in different reference frames were measuring the momentum of the particle in Example 8.1 above. Say the primed observer is moving with \(\beta_R=0.866\) in the \(+x\) direction with respect to the unprimed observer. The unprimed observer would measure a momentum 4-vector of:

\[\begin{split}[p_4] = \begin{bmatrix} i6.00\times10^{-19}~{\rm kg~m/s}\\ 5.20\times10^{-19}~{\rm kg~m/s} \end{bmatrix}\end{split}\]

The components of the momentum 4-vector in the primed co-ordinate system are found by using the Lorentz transformation:

\[\begin{split}[p_4]' = \begin{bmatrix} 2 & -i1.73\\ i1.73 & 2 \end{bmatrix} \begin{bmatrix} i6.00\times10^{-19}~{\rm kg~m/s}\\ 5.20\times10^{-19}~{\rm kg~m/s} \end{bmatrix}\end{split}\]
\[\begin{split}[p_4]' = \begin{bmatrix} i3.00\times10^{-19}~{\rm kg~m/s}\\ 0.0 \end{bmatrix}\end{split}\]

The primed frame of reference is moving at the same speed as the particle, so the primed observer is in the rest frame of the particle, hence the physical momentum should be zero. Which it is! The time component in the rest frame should be the rest mass times \(c\), which you can easily check is the case here.

8.4. Momentum and Energy#

The size of the momentum 4-vector can be found by squaring each of the components in Equation (8.2) and adding them up.

(8.12)#\[[p_4]^2 = p_x^2+p_y^2+p_z^2-(\gamma m_0 c)^2 = p^2 - (\gamma m_0 c)^2\]

where we take \(p\) to be the physical momentum (including the \(\gamma\)!). Multiplying both sides by \(c^2\) and doing a bit of algebra gives

(8.13)#\[\boxed{ (cp)^2 +(m_0 c^2)^2 = (\gamma m_0 c^2)^2}\]

The term on the right of Equation (8.13) can be identified by making use of the correspondence principle. Expanding the \(\gamma\) on the right using a Taylor expansion about \(\beta = 0\) gives:

(8.14)#\[\gamma = (1-\beta^2)^{-1/2} \rightarrow 1 +\frac{1}{2} \beta^2 + \frac{3}{8} \beta^4 + \frac{15}{48} \beta^6 + ...\]

If we assume that \(\beta\) is small enough that we can ignore all the higher order powers, then \(\gamma \approx 1+ \beta^2/2\) and \(\gamma m_0 c^2\) becomes approximately

(8.15)#\[\gamma m_0 c^2 \approx \left(1+\frac{1}{2}\beta^2\right) m_0 c^2 = m_0c^2 + \frac{1}{2}m_0 v^2\]

The last term on the right is recognizable as definition of the kinetic energy of the particle, according to Newtonian mechanics. The dimensions of all the terms in Equation (8.15) are energy. As the velocity becomes exactly equal to 0, the term on the right (and all the terms in the expansion, of course) disappear, but there still is a term left in this ‘energy’ equation.

The term \(m_0 c^2\) can therefore be interpreted as an energy associated with the particle when it is at rest. This we therefore dub the rest energy of the particle, giving rise to that most famous of equations,

(8.16)#\[\boxed{ E_0 = m_0 c^2}\]

To be completely accurate, then, the kinetic energy should really include all those higher order terms. Not just \(1/2mv^2\), but a \(v^4\) term and a \(v^6\) term and a \(v^8\) term and so on. This expansion means that the closer \(v\) gets to \(c\), the more the KE must increase for small increases in \(v\). Or to put it another way, the same increase in KE will yield a much smaller increase in \(v\) if \(v\) is already close to \(c\). That energy goes more and more into the higher order terms rather than into increasing \(v\). The closer you get to the speed of light, the more energy it takes to go any faster. This is another way of saying that nothing can move faster than light. It would take an infinite amount of energy.

Note

How did people for centuries not notice the rest energy of a particle? The rest energy of a person with a (rest) mass of 100 kg is \(E =100~{\rm kg}(3\times10^8~{\rm m/s})^2 = 9\times10^{18}\) joules. This is an unbelievably large number. The total yearly electrical demand of the entire US is on the order of \(10^{19}\) joules (according to the EIA). This is about the same order of magnitude!!! Two reasons nobody noticed this massive amount of energy. Most importantly, only a change in energy is connected with anything happening. Energy that doesn’t change has no measurable implications, so you would never know it was there. Secondly, a 100 kg person does not just vanish, in Newton’s time or now, so that \(10^{19}\) joules of energy isn’t available to be associated with anything else you could measure. Until people discovered that the rest energy of particles can change (through processes like fusion or radioactivity), there was no reason to notice an energy associated with rest mass, because it never changed. I even hear people today talk about a “Law of Conservation of Mass,” even though there is no such law. Mass can and does change, freeing up that rest energy to be associate with motion. This energy is the cause of making the Sun shine, nuclear power plants, or nuclear bombs. Mass is most definitely not conserved.

If the total energy \(\gamma m_0 c^2\) reduces to just the rest energy when the particle is not moving, the fact that it increases with the particle’s speed (\(\gamma\) gets bigger than one) suggests that the remaining energy above and beyond the rest energy would be the kinetic energy. This conclusion is supported by the fact that when \(\gamma\) is small but not zero, \(\gamma m_0 c^2\) is approximately the rest energy plus the classical kinetic energy (Eq. (8.15)). The \(1/2 mv^2\) rule for kinetic energy that we all learned in introductory physics is therefore only an approximation. A full equation would be

(8.17)#\[\boxed{ E_{\rm total} = \gamma m_0 c^2 = E_0 + KE }\]

We can use Equation (8.13) to say that

(8.18)#\[\boxed{ E_{\rm total}^2 = (\gamma m_0 c^2)^2 = E_0^2 + (cp)^2 }\]

or we could turn it around and solve for the momentum of a particle given its energies:

(8.19)#\[cp = \sqrt{E_{\rm total}^2 - E_0^2}\]

Expanding the square using Equation (8.17):

(8.20)#\[cp = \sqrt{E_0^2+2E_0KE+KE^2 - E_0^2} = \sqrt{KE^2+2E_0KE}\]

You could pull the KE out of the square root, to get

(8.21)#\[\boxed{ p = \frac{KE}{c}\sqrt{1+2\frac{E_0}{KE}}}\]

In the world of fast moving objects, the momentum and energy of the system are connected as shown in equation (8.18).

Note

One of the subjects where the rules of relativity cannot be ignored is in the realm of high energy particle physics. In that regime, it is often useful to use units of eV for energy (or keV or even MeV). An eV is the amount the energy of an electron changes when moving through a potential difference of one volt. One volt is a joule per coulomb, so because the charge on an electron is \(1.602\times10^{-19}\) C, the value of one eV in SI units is \(1.602\times10^{-19}\) J. Because of Equation (8.18), we know that \(cp\) has dimensions of energy, so it is often more convenient when you know the energy in eV to express the momentum in eV/c. Then you don’t have to worry about converting everything to kg m/s. See the example 8.3 in the margin, below, for how easy eV units can be.

This means that the momentum 4-vector can be written in terms of the physical momentum and total energy of the system as:

(8.22)#\[\begin{split}\boxed{ [p_4] = \begin{bmatrix} iE_{\rm tot}/c\\ p_x\\ p_y\\ p_z \end{bmatrix} }\end{split}\]

which means the time component of the momentum four vector is telling you the total energy of the system. Many people therefore call the momentum four vector “the energy-momentum four vector”.

Warning

Please make a careful note that Equations (8.17) and (8.18) are NOT the same equation! Do NOT think that \(cp\) is the kinetic energy. Each term in Equation (8.18) is squared, and the sum of squares is not the same thing as the square of a sum! The term \(cp\) is clearly related to the kinetic energy – they have the same dimensions and they both go up and down together – but they are not the same thing. You can see that they are related but not equal in Equation (8.21).

Equation (8.17) can be used to find the speed of a particle of known rest and kinetic energy. If the total energy is rest plus kinetic, and the total energy is \(\gamma\) times the rest energy, then

(8.23)#\[E_{\rm tot} = \gamma E_0 = E_0 + KE \rightarrow \boxed{\gamma = 1 + \frac{KE}{E_0}}\]

This suggests an easy test for deciding if you have to use the relativistic formulae to describe the motion of an object. If you know the \(KE\) and the \(E_0\), use Equation (8.23) to get \(\gamma\). If this number is significantly greater than one, then you need to use the relativistic equations, but if it’s very close to one, you can get away with using Newtonian mechanics.

8.5. The Rest Energy#

The rest energy is the energy measured in the rest frame of a particle. It can be determined from the rest mass using Equation (8.16). The elementary particles have fixed rest energies. For the proton – 938.26 MeV, the neutron – 939.55 MeV, the electron – 0.511 Mev, etc. However, not all particles have rest energy. A photon (gamma ray) does not have any rest energy, and a neutrino’s is so small it can barely be measured.

Equation (8.23) would therefore imply that the value for the \(\gamma\) of these particles is infinite. If \(\gamma\) is infinite, then \(\beta= 1\), and these particles must be traveling at the speed of light. Always. This is not surprising for the gamma ray, as it is light. What is surprising is that other particles such as the neutrino, which are not photons, also travel at the same speed.

Also surprising is that all of these particles have momentum even though they do not have any rest mass. You might think, if you think of momentum as mass times velocity, that no mass means no momentum. However, if you look at Equation (8.18), the only way this equation can work for particles that have energy but no mass is if \(p=E/c\). Particles with energy but no mass still have momentum!!!

Note

You can also see this in Equation (8.21): if the rest energy goes to zero, \(p\rightarrow KE/c\). For a photon, the KE is \(hf\), Planck’s constant times the frequency, so the momentum of a photon is \(hf/c\), or \(h/\lambda\), where \(\lambda\) is the wavelength. See also the treatment in the section of Chapter 9 on the Doppler Shift.

On the other hand, for a particle with \(E_0 \neq 0\) to travel at the speed of light, Equation (8.23) implies it must have infinite kinetic energy because \(\gamma\) is infinite for an object traveling the speed of light. No matter how much energy is given to this massive particle, it will always be traveling slower than the speed of light. Consequently, objects with \(E_0 \neq 0\) travel slower than the speed of light and objects with \(E\approx 0\) travel at the speed of light. Nothing that travels at or lower than the speed of light can ever travel faster than the speed of light because it would require an infinite amount of energy, which is not readily available to us, to make the transition from slightly less than the speed of light to slightly more than the speed of light. As the saying goes…

The speed of light. It’s not only a good idea, it’s the law!

8.6. Example of Using the Definitions#

An electron (rest energy = 0.511 MeV) moving in the \(x\)-direction has a kinetic energy of 1.000 MeV (1 eV = \(1.602 \times 10^{-19}\) joules) Find the total energy, the physical momentum, the momentum 4-vector and the speed of this electron.

The kinetic energy is almost twice the rest energy, so \(\gamma\approx3\), and I will have to do this problem using relativistic formulae. The total energy is the sum of the kinetic and rest energy, so \(E_{\rm tot} = 1.511\) MeV. To find the momentum, you use Equation (8.19) to get

\[(cp)^2 = (1.511~{\rm MeV})^2 - (0.511~{\rm MeV})^2 = 2.022~{\rm MeV}^2\]

Take the square root and divide by \(c\) to get

\[\vec{p} = 1.422~\frac{\rm MeV}{c}\hat{x}\]

And yes, MeV/\(c\) is a perfectly good unit for momentum, and very useful for elementary particles. If you really wanted to convert this to the dump truck units (SI) for momentum, kg m/s, use the conversion factor

(8.24)#\[1~{\rm eV}/c = 5.340 \times 10^{-28}~{\rm kg m/s}\]

The momentum 4-vector now can be written (using Equation (8.22)) as:

\[\begin{split}[p_4] = \begin{bmatrix} i1.511~{\rm MeV}/c\\ 1.422~{\rm MeV}/c\\ 0.0\\ 0.0 \end{bmatrix}\end{split}\]

There are a couple ways you could get the speed. The easiest way would be to use Equation (8.23) to get \(\gamma\) and then solve for \(\beta\).

\[\gamma = 1 + \frac{KE}{E_0} = 1 + \frac{1.00}{.511} = 2.957\]

Then use the definition of \(\gamma\) to get

\[\beta = \sqrt{1-\frac{1}{\gamma^2}} = 0.941\]

If you really want that in SI units, that would be \(2.82\times10^8\) m/s. The other way to do it would be to go from the four vector. We have \(p_x = 1.422\times10^6\) eV/c, which by the conversion ((8.24)) above is \(7.593\times10^{-22}\) kg m/s. This momentum is \(\gamma m_0 v_x\), so we divide by \(\gamma = 2.957\) to get \(2.568\times10^{-22}\) kg m/s. Divide by the mass of the electron (\(9.109\times 10^{-31}\) kg) to get \(2.82\times10^8\) m/s, as before. If you forget that momentum has a \(\gamma\) built into it, you would get a speed almost three times faster than light speed.

8.7. Conservation Laws and the Momentum Four-Vector#

If an interaction happens that is isolated from its surroundings, the total energy \(E_{\rm tot}\) (as measured by a particular inertial observer) of that system will be constant before, during, and after the interaction. This is the same \(E\) that appears in the time component of the total momentum 4-vector of this system. If the interaction experienced by each of the pieces of the system is only due to interactions with other pieces of the system, then the net physical momentum (as measured by a particular inertial observer) for this system is constant. The physical momentum is a 3-vector, so both its size and its direction must remain constant. These laws of physics hold for any particular inertial observer.

When a certain physical parameter remains constant while time changes, then that parameter is said to be conserved. Under the conditions stated in the previous paragraph, the sum of spatial components of the momentum 4-vector, when added as 3-vectors, is conserved. In a similar manner, the time component of the momentum 4-vector is also conserved. For a particular inertial observer, the momentum 4-vector is conserved.

If a particular inertial observer measures the total energy of the system, this value will remain constant as long as the system observed is isolated. A second inertial observer, traveling with some relative velocity \(\beta_R\) with respect to the first, will also measure a constant total energy, but the value that that observer measures will, in general, be different from the value measured by the first observer. The Lorentz transformation shows how the two different total energy values are related. What remains constant between two two reference frames is the size of the momentum 4-vector, as it is a Lorentz invariant.

We therefore try to use the words consistently: to be “conserved” is to remain constant within a particular reference frame, during a time interval, but to be “invariant” is to be the same in different reference frames moving with respect to each other. Shifting from one reference frame to another is not a process that conserves energy. An object at rest in one frame will have only its rest energy, but in another frame it will have rest and kinetic energy. Shifting between these frames does not do work upon the object. If I change my walking speed relative to you, your speed changes relative to me, but I have not exerted a force on you to change your speed. The kinds of analysis that depend on conservation of energy or the work-energy theorem are meant to be carried out in a single frame of reference. If you switch to another reference frame, you must recalculate the energy number from scratch – it will not in general be the same.

8.8. Example of Momentum Conservation#

A neutral pion (rest energy \(E_0=135\) MeV) decays into two photons. Show that in the rest frame of the pion, the two photons move in opposite directions. Find the energy of these two photons.

In the rest frame of the pion, the momentum 4-vector for the pion is:

(8.25)#\[\begin{split}[p_4]_{\rm before} = \begin{bmatrix} i135~{\rm MeV}/c\\ 0.0\\ 0.0\\ 0.0 \end{bmatrix}\end{split}\]

The observer in the rest frame of the pion watches the decay take place. The \(x\)-direction is defined as the direction that one of the dautgher photons takes. The second photon travels in some unknown direction. The observer in the rest frame of the pion measures a momentum 4-vector of the two photons as:

(8.26)#\[\begin{split}[p_4]_{\rm after} = \begin{bmatrix} i(E_1+E_2)/c\\ p_{x1}+p_{x2}\\ p_{y2}\\ p_{z2} \end{bmatrix}\end{split}\]

For two 4-vectors to be equal, each of their equivalent components have to be equal. Therefore the second photon cannot have any momentum in the \(y\) and \(z\) directions, and the momentum of the second photon in the \(x\) direction must be equal and opposite to that of the first photon, so that the momentum adds up to equal the zero it was before the collision. This shows that the photons, whichever direction they go, must go opposite to each other.

Note

The ease with which we can set things to zero and cancel them makes this particular frame of reference, in which the momentum was and remains zero, particularly useful. It is referred to as the center of momentum frame, and we will use it in a detailed example in the next chapter.

We can examine the energy of the two photons by looking at the time component. Since the four momentum is conserved, the time component must also be conserved, and therefore the sum of the two photon energies must be 135 MeV/c. Since the photons have no mass, their momenta and their energies must be proportional (by Equation (8.13)), so the energy must be split evenly between them. Each photon has an energy of 67.5 MeV, and therefore a momentum of 67.5 MeV/c, but each in an opposite direction to the other.

8.9. Problems#

  1. Find the physical momentum of a 100 MeV proton, a 100 MeV positive pion and a 100 MeV electron. (the units should be in MeV/c).

  2. Make a plot of the kinetic energy of an electron as a function of \(\beta\) for \(\beta = 0 to 0.999\) (or some maximum number just slightly less than one). On the same graph, plot the classical value of the kinetic energy of this electron (\(KE = 1/2 mv^2= 1/2 Eo \beta^2\)). Describe the difference between these two graphs. What does this graph predict about the chances of an electron to travel the speed of light?

  3. A negative pion is traveling with \(\beta = 0.9900\) in the \(+x\) direction heading straight towards a proton that is traveling with \(\beta = 0.5720\) in the \(-x\) direction. Write down the elements of the momentum 4-vector for this system.

  4. Find the speed and momentum of a negative pion that has kinetic energy 300 MeV.

  5. Calculate the elements of the momentum 4-vector for a 2.00 eV photon. What kind of light is a 2 eV photon?

  6. A proton is moving with a speed of \(\beta = 0.99882\). Calculate the momentum and kinetic energy of this particle.

  7. A 2.00 eV photon is traveling in the \(+x\) direction as seen by observer 1. Observer 2,traveling with \(\beta_R= -.8800\) relative to observer 1 also observes this 2.00 eV photon. Use the Lorentz transformation to determine the elements of the 4-momentum measured by the second observer.

  8. In the lab reference frame, a certain particle has total energy \(E_{\rm tot} = 3.2 \times 10^{-10}\) J and momentum \(p = 9.4 \times 10^{-19}\) kg m/s. Find the speed and rest mass of this particle.

  9. An observer in the lab watches a neutral \(K^0\) (rest energy 497.7 MeV) decay into a positive and negative pion. The kaon is traveling with \(\beta = 0.8666\) before it decays. We write this decay as

(8.27)#\[K^0 \rightarrow \pi^+ + \pi^-\]

Warning

In this context, the zero in \(K^0\) means a neutral particle, not the time component number of a contravariant four vector!

a) Calculate the components of the momentum 4-vector of the kaon as measured by the observer in the lab.

b) Use the Lorentz transformation to find the components of the momentum 4-vector of the kaon in the rest frame of the kaon (hint: the rest frame is moving with \(\beta_R= +0.8666\)).

c) If in the rest frame of the kaon after the decay the negative pion is traveling in the \(+x\) direction, write down the momentum 4 vector of the pair of pions.

d) From the momentum conservation laws, find the kinetic energy and the physical momentum of each of the pions as seen by an observer in the rest frame of the kaon.

e) Write down the momentum 4-vector for each of the pions produced in the decay as seen by an observer in the rest frame of the kaons.

f) Use the inverse Lorentz transformation to transform the momentum 4- vector of each of the pions as measured in the rest frame of the kaons to the momentum 4-vectors as measured in the laboratory reference frame.

g) Show that the sum of the two 4-vectors you calculated in f) is the same as the result in a). This would mean that the 4-momentum is conserved in the lab frame!

  1. A photon is a particle that has zero rest energy. Suppose that \(10^{30}\) photons, each with total energy 0.500 MeV, are bouncing around in a perfectly reflecting container. Assume that other than the photons, the inside of the container is empty. These photons are moving in random directions, so the total physical momentum of the container is zero, because on average the photon’s momenta cancel out.

a) What is the total energy of the inside of the container?

b) What is the rest energy of the inside of the container? Convert this number to rest mass. (hint: it is not zero)

c) Where did this mass come from?

d) At what speed would you have to move the container for the total energy inside to increase by 1 joule?

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