Solve E = mc^2 & Time Dilation Equation

[samuel toco]

TL;DR

how to equal the equation: E = mc2 and the Time dilation equation

I think that have something lost on it!!!!!

how to equal the equation: E = mc2 and the Time dilation equation

 

[Ibix]

They aren't equal. I'm not even sure what "equal" would mean between two equations. Why would you think they would be?

 

[berkeman]

Welcome to the PF.

Thread prefix changed from "A" = Advanced/Graduate School level to "I" = Intermediate/Undergraduate level.

Summary:: how to equal the equation: E = mc2 and the Time dilation equation

I think that have something lost on it!

Instead of "equal", maybe you mean "reconcile"? If so, what do you think is inconsistent between them? Could you please show links to the reading you've been doing about each of those equations, and say why that reading has you confused about their compatibility? Thanks.

 

[vanhees71]

I'm also not understanding your question, but note that the correct equation is $E=m c^2 \gamma$, where $\gamma=1/\sqrt{1-v^2/c^2}$ is the same Lorentz factor that also enters the time-dilation formula. Indeed if a clock shows $\Delta t$ in its rest frame an observer moving with velocity $v$ relative to the clock will measure $\Delta t'=\gamma \Delta t$.

The similarity between the equation for energy and the time dilation formula is not a surprise, because it's closely connected. One defines everything in relativity in covariant quantities. Thus having vectors like the time-position four-vector in special relativity and wants to define covariant quantities by time derivatives one uses proper time $\mathrm{d} \tau = \sqrt{1-v^2/c^2} \mathrm{d} t$, because that's a scalar under Lorentz transformations and thus
$$V^{\mu}=\mathrm{d}_{\tau} x^{\mu}$$
is a four-vector, the four-velocity of the particle. In terms of usual velocity (which is NOT a relativistically covariant quantity) this reads
$$(V^{\mu})=\begin{pmatrix} c \mathrm{d}_{\tau} t \\ \mathrm{d}_{\tau} \vec{x} \end{pmatrix},$$
but $\mathrm{d}_{\tau} = \mathrm{d} t (\mathrm{d}_{\tau} t)=\gamma \mathrm{d} t$. Thus you get
$$V^{\mu}=\begin{pmatrix} \gamma c\\ \gamma \vec{v} \end{pmatrix},$$
where $\vec{v}=\mathrm{d}_t \vec{x}$ is the usual three-velocity measured in the lab frame.

Now four-momentum is defined as
$$p^{\mu} = m V^{\mu},$$
where $m$ is the invariant mass. To understand what the time-component means we note that
$$p^0=m c \gamma =m c \left (1+\frac{1}{2} \frac{v^2}{c^2} +\mathcal{O}(v^4/c^4) \right),$$
and this suggests to define $p^0=E/c$ with $E$ the energy of the particle, because for the reason that $p^{\mu}$ is a four-vector it's convenient to include the rest energy $E_0=mc^2$ in the energy of the particle, and then in the non-relativistic limit $E=E_0 + mv^2/2$, and $m v^2/2=E_{\text{kin}}$ in Newtonian mechanics, where $E_0$ is simply an additive constant which is not changed in any way by any physical phenomenon.

Note that this is different in relativistic physics! The rest mass of a composite object is related to the energy of this object in its rest frame by $E_0=m c^2$. E.g., if you have some macroscopic body its invariant mass changes with temperature, because if the body gets hotter the heat-energy gain $\Delta Q$ adds to the rest energy, i.e., the invariant mass of the body changes by the amount $\Delta m=\Delta Q/c^2$.

Thus, while the invariant mass does not change by any physical processes within Newtonian mechanics (except you add or take away some matter to a composite object of course), relativity shows that this is only approximately right. In other words: The invariant mass is not a conserved quantity for a closed system in relativistic physics. Here only the energy of a closed system is conserved but not invariant mass!