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So we differentiate displacement (from our frame of reference) with respect to proper time of the moving object $\dfrac{dx}{d\tau}$ to get to the momentum equation: $$p=\gamma mv$$ But when we derive the energy equation: $$E=\int\frac{\mathrm{d}p}{\mathrm{d}t}\mathrm{d}x$$ we use our time instead of the object's.

Why do we use our time to get the energy and the object's time to get the momentum? And, why do we differentiate displacement from my frame of reference with respect to the object's (proper) time to get the momentum? (Why don't we use the contracted displacement from the object's reference frame?)

Amit
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Anjan Sharma
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2 Answers2

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You simply got it wrong. $$\tag1\begin{pmatrix}E/c\\\vec p\end {pmatrix}=\begin{pmatrix}\gamma mc\\\gamma m\vec v\end {pmatrix}=mc\begin{pmatrix}\frac{c\,\mathrm dt}{c\,\mathrm d\tau}\\\frac{\mathrm d\vec x}{c\,\mathrm d\tau}\end {pmatrix}$$ i.e. both quantities are defined using the proper time of the moving object in the denominator, and the numerators are all your coördinate functions.

naturallyInconsistent
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@naturallyinconsistent is correct. Additionally, proper time is the closest thing you'll come to "real objective actual time" because it's the only time that's Lorentz-invariant. It's special because it falls along the geodesic. It is unaccelerated.

BUT:

Every point of mass in the universe has in its watch pocket its very own private "right now." These clocks are completely independent and run at different rates.

If they run at the same rate for a while, it's temporary. They'll de-sync as soon as either of the two objects moves any amount in any direction.

That inviolate "right now" which only you own is proper time.

Miss Understands
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