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I'm having a difficult time making the connection between the Newtonian conception of mass and that proposed by special relativity and particle physics. From a Newtonian perspective, mass is that which resists acceleration. If an experimenter were to slide a book across a table, they could measure the required force the book has due to its mass. We will ignore friction here.

From the perspective of special relativity mass can be thought of as $m=E/c^2$, meaning most of the mass of protons and neutrons are due to the binding energy of the quarks.

From the perspective of particle physics and special relativity, where does the resistive force one experiences while sliding a book across a table come from?

Morphyl
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  • Where did the resistive force [I'd prefer 'required force'] come from before you knew about $E=mc^2$ ? Is it any less mysterious in pre-relativity Physics? – Philip Wood Oct 15 '21 at 14:02
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    See the Veritasium video Your Mass is NOT From the Higgs Boson. The earlier video he mentions is Empty Space is NOT Empty – mmesser314 Oct 15 '21 at 14:18
  • When you stop a book from moving at speed $v$, its energy goes from $mc^2 + \frac{1}{2}mv^2$ to just $mc^2$. To lower its energy further, you would need to make it undergo fusion. – Connor Behan Oct 15 '21 at 15:37
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    It was the Veratasium video that got me started on this. I'll clarify my reasoning. Q: Why does the book not travel at the speed of light? A: It has mass Q: Why does the book have mass? A: Mostly from the energy fluctuations in the gluon field. Q: Why do energy fluctuations in the gluon field cause objects not to move at the speed of light? This is the question I am looking to answer. – Morphyl Oct 15 '21 at 20:06
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    Is your underlying question: "how does energy cause inertia"? This might be helpful: https://physics.stackexchange.com/q/525657/123208 – PM 2Ring Oct 15 '21 at 22:13
  • That is basically the mass gap problem. Early evidence for it came from lattice QCD. – Connor Behan Oct 16 '21 at 15:05

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Since $F = dP/dt$ and $E^2 = m_0^2 c^4 + P^2c^2$, we can rearrange and obtain

$$F = \frac{1}{c} \frac{d}{dt}\sqrt{E^2 - m_0^2 c^4}$$

Since $E = mc^2 = \gamma m_0 c^2$ we can replace the contents of the square root with:

$$E^2 - m_0^2c^4 = m_0^2c^4(\gamma^2 -1) = m_0^2c^2\frac{v^2}{1-v^2/c^2}$$

For $v\ll c$, this simplifies to $(m_0cv)^2$

hence we obtain the Newtonian form

$$F \approx \frac{d}{dt} (m_0v)$$

or if you prefer in terms of rest energy,

$$F \approx \frac{1}{c^2} \frac{d}{dt} (E_0v)$$

jng224
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g s
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