Special relativity problems and answers

Recent questions in Special Relativity
Brody Collins 2022-05-20 Answered

It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume $-+++$ signature. Given ${\gamma }_{\mu }$ as gamma matrices that satisfy $\mathrm{t}\mathrm{r}\left({\gamma }_{\mu }{\gamma }_{\nu }\right)=4{\eta }_{\mu \nu }$, then we have $\mathrm{t}\mathrm{r}\left({\gamma }_{\mu }^{\prime }{\gamma }_{\nu }^{\prime }\right)=4{\eta }_{\mu \nu }$ if we put:${\gamma }_{\mu }^{\prime }=\mathrm{exp}\left(-A\right){\gamma }_{\mu }\mathrm{exp}\left(+A\right)$where $A$ is any matrix.Boosts and rotations use $A$ as a bivector. For example, with $\alpha$ a real number, $A=\alpha {\gamma }_{0}{\gamma }_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma }_{1}{\gamma }_{2}$ gives a rotation around the $z$ axis.Solving for the value of α that gives a boost with velocity $\beta =v/c$ appears to be straightforward. But how to do the rotations?And by the way, what happens when you generalize $A$ to be something other than bivectors?

Jaiden Bowman 2022-05-20 Answered

In history we are taught that the Catholic Church was wrong, because the Sun does not move around the Earth, instead the Earth moves around the Sun.But then in physics we learn that movement is relative, and it depends on the reference point that we choose.Wouldn't the Sun (and the whole universe) move around the Earth if I place my reference point on Earth?Was movement considered absolute in physics back then?

Regina Ewing 2022-05-19 Answered

Looking for specific Relativity exampleThe example had to do with two people walking along a sidewalk in opposite directions, and an alien race on a planet millions of light-years away planning an invasion of the Solar System. The example showed that in one walker's reference frame the invasion fleet had departed, but in the other reference frame the fleet had not.At the time, the explanation made perfect sense, but I have forgotten the details and have never run across this example again.Does anybody know where this was, or have the text of the explanation?

Justine Webster 2022-05-19 Answered

Does a relativistic version of quantum thermodynamics exist? I.e. in a non-inertial frame of reference, can I, an external observer, calculate quantities like magnetisation within the non-inertial frame?

Carley Haley 2022-05-19 Answered

Is there a time + two spatial dimension representation of a Minkowski-space surface which could be constructed within our own (assumed Euclidean) 3D space such that geometric movement within the surface would intuitively demonstrate the “strange" effects of the Lorentz transformation (length contraction, time dilation)?

Matilda Webb 2022-05-19 Answered

It seems odd that entropy is usually only defined for a system in a single 'slice' of time or spacelike region. Can one define the entropy of a system defined by a 4d region of spacetime, in such a way that yields a codimension one definition which agrees with the usual one when the codimension one slice is spacelike?

Matthew Hubbard 2022-05-19 Answered

if $u$ and ${u}^{\prime }$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by$\left(l,m,n\right)=\frac{1}{|u|}\left({u}_{x},{u}_{y},{u}_{z}\right)$and$\left({l}^{\prime },{m}^{\prime },{n}^{\prime }\right)=\frac{1}{|{u}^{\prime }|}\left({u}_{x}^{\prime },{u}_{y}^{\prime },{u}_{z}^{\prime }\right)$are related by$\left({l}^{\prime },{m}^{\prime },{n}^{\prime }\right)=\frac{1}{D}\left(l-\frac{v}{u},m{\gamma }^{-1},n{\gamma }^{-1}\right)$and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.$D=\frac{{u}^{\prime }}{u}\left(1-\frac{{u}_{x}v}{{c}^{2}}\right)={\left[1-2l\frac{v}{u}+\frac{{v}^{2}}{{u}^{2}}-\left(1-{l}^{2}\right)\frac{{v}^{2}}{{c}^{2}}\right]}^{\frac{1}{2}}$Why is that better than the second expression?Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ and $|u|=|\left({u}_{x},{u}_{y},{u}_{z}\right)|=\sqrt{{u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}}$

London Ware 2022-05-17 Answered

The relativistic energy-momentum equation is:${E}^{2}=\left(pc{\right)}^{2}+\left(m{c}^{2}{\right)}^{2}.$Also, we have $pc=Ev/c$, so we get:$E=m{c}^{2}/\left(1-{v}^{2}/{c}^{2}{\right)}^{1/2}.$Now, accelerating a proton to near the speed of light:$\begin{array}{}0.990000000000000& c=>& 0.0000000011& J=& 0.01& TeV\\ 0.999000000000000& c=>& 0.0000000034& J=& 0.02& TeV\\ 0.999900000000000& c=>& 0.0000000106& J=& 0.07& TeV\\ 0.999990000000000& c=>& 0.0000000336& J=& 0.21& TeV\\ 0.999999000000000& c=>& 0.0000001063& J=& 0.66& TeV\\ 0.999999900000000& c=>& 0.0000003361& J=& 2.10& TeV\\ 0.999999990000000& c=>& 0.0000010630& J=& 6.64& TeV\\ 0.999999999000000& c=>& 0.0000033614& J=& 20.98& TeV\\ 0.999999999900000& c=>& 0.0000106298& J=& 66.35& TeV\\ 0.999999999990000& c=>& 0.0000336143& J=& 209.83& TeV\\ 0.999999999999000& c=>& 0.0001062989& J=& 663.54& TeV\\ 0.999999999999900& c=>& 0.0003360908& J=& 2,097.94& TeV\\ 0.999999999999990& c=>& 0.0010634026& J=& 6,637.97& TeV\\ 0.999999999999999& c=>& 0.0033627744& J=& 20,991.10& TeV\end{array}$If the LHC is accelerating protons to $7TeV$ it means they're traveling with a speed of $0.99999999c$.Is everything above correct?

Landon Mckinney 2022-05-13 Answered