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B From dE=dm*c^2 to E=m*c^2: Integration constant?

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  1. Jun 8, 2016 #1
    Hi,

    I've read some high school "derivations" of ##E=m\cdot c^2## that all considered single photons with momentum ##p=E/c## that are absorbed or emitted from some massive object, changing its mass. So they actually only showed the incremental
    $$\Delta E=\Delta m\cdot c^2 .$$

    Most of those books simply generalized this to ##E=m\cdot c^2##, only one author said that this formula is obtained by integration and dropping constants. It's unclear to me why the integration constants simply can be set to zero. Maybe they don't matter in SR, but wouldn't they change things in GR? Or could this integration constant be absorbed in Einstein's cosmological constant?

    Also, one of those books used ##\Delta E=\Delta m\cdot c^2## to compute that the mass of a spring with spring constant ##k## increases by
    $$\Delta m=\frac{\Delta E_S}{c^2}=\frac{\frac{1}{2}k\cdot s^2}{c^2}$$
    when compressed by a distance ##s##. How is such a generalization possible? The derivations critically relied on ##p=E/c## for photons which are not present in this example.
     
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  3. Jun 8, 2016 #2

    PeroK

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    In answer to your first question. If a small amount of energy can be converted to a small amount of mass, related by a factor of ##c^2##, then that's it. There's no need to integrate.

    If you do integrate in this situation, you would have a definite integral of some sort, so no constant of integration would arise.
     
  4. Jun 8, 2016 #3
    Well of course constants can arise in definite integrals. It would look like
    $$\int_{E_0}^E dE'=E-E_0=\int_{m_0}^m dm'\cdot c^2=(m-m_0)\cdot c^2 .$$
    One of the derivations I mentioned assumes an object gaining mass by absorbing photons. Let's say we start with an initial mass of zero, ##m_0=0##, there could still be a nonzero energy ##E_0##. Differentials are about changes, not absolute values.
     
  5. Jun 8, 2016 #4

    PeroK

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    ##E_0## and ##m_0## are not constants of integration, they are initial values.

    If you imagine removing mass/energy from an object, then ##E_0 = m_0 = 0## eventually.

    Nevertheless, integration is unnecessary. Not least because in fact what you have is a large sum, the definite integral being an approximation.

    ##\Delta E## is not a differential, it is a small finite change.

    Finally, constants of integration do not arise in definite integrals.
     
  6. Jun 8, 2016 #5
    Well you name it, they are inital values, and hence CANNOT be derived from ##dE=dm\cdot c^2##.
    Why shouldn't it be possible that at some point we end up with no mass but there's still energy?
    That doesn't change anything. Say ##a_n=\sum_{i=1}^n 1=n## and ##b_n=10+\sum_{i=1}^n 1=10+n##, then ##\Delta a_n:=a_n-a_{n-1}=1## and ##\Delta b_n:=b_n-b_{n-1}=1##, so ##\Delta a_n=\Delta b_n## without being ##a_n=b_n##.
    I just wrote it as integration because I didn't want to make assumptions about energy or mass being quantized.
     
  7. Jun 8, 2016 #6

    PeroK

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    A particle cannot decay into a single photon, owing to conservation of momentum, but it can decay into two photons. Essentially that's your scenario. All the mass is gone and there is still energy in the form of two or more massless particles.
     
  8. Jun 8, 2016 #7
    Why not just derive it using kinematics and generic initial and final values? (taking note to use relativistic momentum in place of p=mv)

    For example (only worrying about x direction)

    1. X = (dx, cdt) (c there to match units)

    2. V = (ɣv, ɣc) (dividing by proper time; ɣ = [1 - (v/c)2 ]-1/2, of course)

    3. Follow normal kinematics for x. I.e. p = mɣv, F = m d(ɣv)/dt, W = m∫[d(ɣv)/dt]dx from x0 to x1. If you do those steps, T = mc2(ɣ-1) will pop up without fail, and at that point it will be very obvious that rest energy = mc2 and is always the same regardless of inertial reference frame.​

    When you do it that way it is clear there is no possible way for E to ever be zero unless m is zero, nor is there any way for energy to still be there if m = 0.

    As for additional potential energy contribution from height, if I'm not mistaken in the Lorentz transformation equations y=y' and z=z', so it should be the same in all inertial reference frames.
     
  9. Jun 11, 2016 #8

    vanhees71

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    Well this is the usual confusion about the meaning of "##E=m c^2##". One should stress that this is an outdated formula from the ancient days where physicists used "relativistic mass" to confuse students of relativity ;-).

    The modern argument goes as follows. We like to work with covariant four-vector quantities. That brings us to the definition of four-momentum by defining it as in Newtonian mechanics but using the four-vector of time and position of a particle and the proper time, whose increment is defined by
    $$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2}.$$
    Then one defines
    $$p^{\mu} =m_0 \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
    where ##m_0## is the invariant mass of the particle (which is usually labeled ##m## in modern times, but you have unfortunately used ##m## for the outdated relativistic mass here). Now plugging everything in, leads to
    $$(p^{\mu})=m_0 \frac{\mathrm{d}}{\mathrm{d} \tau} \begin{pmatrix} c t \\ \vec{x} \end{pmatrix} = m_0 \frac{\mathrm{d} t}{\mathrm{d} \tau} \frac{\mathrm{d}}{\mathrm{d} t} \begin{pmatrix} c t \\ \vec{x} \end{pmatrix} = \frac{m_0}{\sqrt{1-\vec{v}^2/c^2}} \begin{pmatrix}c \\ \vec{v} \end{pmatrix}.$$
    So we have the relatistic momentum
    $$\vec{p}=\frac{m_0 \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}.$$
    To interpret also the time component of the four-vector we just defined we note that for small velocities ##|\vec{v}|\ll c## we can approximate
    $$c p^0=\frac{m_0 c^2}{\sqrt{1-\vec{v}^2/c^2}}=m_0 c^2 \left (1+\frac{\vec{v}^2}{2 c^2} \right) = m_0 c^2 + \frac{m_0}{2} \vec{v}^2,$$
    i.e., we interpret
    $$E=c p^0$$
    as the energy of the particle since up to the constant ##m_0 c^2## it's just the usual non-relativistic kinetic energy. The advantage to keep the additive constant in the definition is just that it gives together with the relativistic momentum a four vector ##p^{\mu}##.

    Now that all this makes physical sense is due to observation. In scattering experiments with elementary particles it's observed with very high accuracy that the total four-momentum in the above defined sense is indeed conserved, and there are processes, where the incoming particles are destroyed and other particles with different masses are produced, but the four-momentum as defined above is still observed, i.e., you can change a system's total rest mass by reactions, and this change of mass is related to ##\Delta E/c^2##.

    The example with the absorbed photon is a very nice one. We only have to generalize the above considerations to massless particles, i.e., particles for which ##m_0=0##. Then you can only work with momentum and energy obviously, because it doesn't make sense to write the Lorentz factor ##1/\sqrt{1-\vec{v}^2/c^2}## for a massleess particle for which always ##\vec{v}^2=c^2##, and this makes the Lorentz factor diverge. So we have to use only energy and momentum. The trick is to use an invariant expression. Out of the four-momentum we can make only one invariant expression, namely ##p_{\mu} p^{\mu}##. Now for a massive particle, using the definitions above, we find
    $$p_{\mu} p^{\mu}=m_0^2 c^2.$$
    Writing this out in time and space-components we get
    $$E^2/c^2-\vec{p}^2=m_0^2 c^2 \; \Rightarrow \; E=c \sqrt{m_0^2 c^2+\vec{p}^2},$$
    and the latter equation makes sense for the limit ##m_0 \rightarrow 0##. For a massless particle we thus have the energy-momentum ("dispersion") relation
    $$E=c |\vec{p}|.$$

    Now we can treat the example in your high-school book in a modern way: You have some body of (invariant) mass ##m_0## at rest and a photon running towards it with some momentum ##\vec{p}## the four-momentum vectors before the photon hits the body thus is
    $$(p_1^{\mu})=\begin{pmatrix} m_0 c \\ 0\end{pmatrix}, \quad (p_2^{\mu})=\begin{pmatrix} |\vec{p}| \\ \vec{p} \end{pmatrix}.$$
    Now after the photon is absorbed by the body (heating it up a bit absorbing the photon's energy). After this "collision" we have a body which is a little hotter than before. Now we have four-momentum conservation, i.e.,
    $$(p_3^{\mu})=\begin{pmatrix} m_0 c+|\vec{p}| \\ \vec{p}\end{pmatrix},$$
    and we can get the invariant mass by using
    $$p_3 \cdot p_3=m_0'^2 c^2=(m_0 c+|\vec{p}|)^2-\vec{p}^2=m_0^2 c^2 +2 m_0 c |\vec{p}|,$$
    i.e.,
    $$\Delta m = \sqrt{m_0^2+2 m_0 |\vec{p}|/c}-m_0 = m_0 (\sqrt{1+2 |\vec{p}|/(m_0 c)}-1 \simeq |\vec{p}|/c = E_{\gamma}/c^2.$$
    Note: This above "naive" result is only an approximation, because in reality part of the photon's energy gets into the motion of the combined body. The reason is that energy AND momentum must be conserved, i.e., the body after the collision is not strictliy at rest. However, for a macroscopic body ##m_0 c^2 \gg E_{\gamma}##, and thus we could make the above approximation, which neglects the tiny velocity of the body after the collision.
     
  10. Jun 12, 2016 #9
    Thanks for those last two responses, they seem more rigorous to me, though still making some assumptions and interpretations.
    One of my books actually avoided this problem (I think) by considering two photons from opposite sides, cancelling out their momentum in the rest frame of the body:
    Capture.PNG

    However, I'm still wondering if you can derive ##E=m\cdot c^2## from ##\Delta E=\Delta m\cdot c^2## in a more "basic" way. An example: After my book derived time dilation by looking at a light clock and using Pythagoras, I asked myself why this can be generalized to all clocks, i.e. time itself. Then I came up with following thought experiment: Say we construct a clock with two minute hands directly above each other that conduct a current if and only if their tips both are at the 12 mark. If we now attach a device to that clock (say a bomb) and start both clocks at the same time, the minute hands both reach the 12 mark after the same time interval and the bomb explodes.

    If we now watched this from a moving frame and time dilation only applied to the light clock, the hands wouldn't reach the 12 mark simultaneously and the bomb wouldn't explode from this frame. This is a logical absurdity and would lead to a whole lot of causality problems. Hence, time dilation applies to all clocks, or time itself.

    Can we apply a similar, "simple and logical" thought experiment to show that ##\Delta E=\Delta m\cdot c^2## necessarily implies ##E=m\cdot c^2## and everything else leads to absurdities or causality problems? To be clear: I'm only talking about massive body at rest, for which ##E=m\cdot c^2## applies.
     

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  11. Jun 12, 2016 #10

    vanhees71

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    Sure, that was Einstein's original thought experiment: A body is emitting two photons back-to-back. I've no simpler way to explain it than Minkowski vectors. In fact it's the most simple way to derive relativistic formulae of all kind. For an introduction, see

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

    So here's the explanation of this case: Before the emission of the two photons we have one body at rest. It's four-momentum is given by
    $$p_1=\begin{pmatrix}M c \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
    After the emission you have a body with four-momentum ##p_1'## and two photons with four-momenta ##p_2'## and ##p_3'##. By assumption they are emitted back-to-back with the same magntitude of three-momentum, i.e., we have
    $$p_2'=\begin{pmatrix} |\vec{p}_{\gamma}| \\ \vec{p}_{\gamma} \end{pmatrix}, \quad p_3'= |\vec{p}_{\gamma}| \\ -\vec{p}_{\gamma}.$$
    From energy-momentum conservation you get
    $$p_1'=p_1-p_2'-p_3'=\begin{pmatrix} Mc-2|\vec{p}_\gamma| \\0 \\0 \\0 \end{pmatrix}.$$
    But the invariant mass ##M'## after the decay is given by
    $$ p_1' \cdot p_1'=M'^2 c^2=(Mc-2 |\vec{p}_{\gamma}|)^2 \; \Rightarrow \; M'=M-2 |\vec{p}_{\gamma}|/c=M-2 E_{\gamma}/c^2.$$
    So here the "naive" picture is exact.

    So "##E=m c^2##" follows from the assumption that with defining energy and momentum of a particle such that these quantities form a four-vector in the sense explained in my previous section and the validity of energy-momentum conservation, which in fact is a consequence of the special-relativistic space-time symmetries (Noether's theorem).
     
  12. Jun 12, 2016 #11
    Well since the Lorentz transformation is linear there isn't anything to derive. If it applies to E it applies to delta E and dE.
     
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