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All Solutions of Type: Special Relativity

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GR8677 #20

Special Relativity $\Rightarrow$ }Rest Energy

If one remembers the formulae from special relativity, arithmetic would be the hardest part of the problem.

The problem is to solve $\gamma m_k c^2 = m_p c^2$ , where $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ and $\beta=\frac{v}{c}$ .

The rest-mass of both the kaon and the proton are given in the problem. Thus, the equation reduces to $\gamma=\frac{938}{494}$ .

Now, because the range of velocities vary significantly between $1.x$ and $2$ , one can't directly approximate that as 2. Boo. So, long-division by hand yields approximately $1.9=\gamma=\frac{1}{\sqrt{1-\beta^2}}$ .

The author of this site prefers to use fractions instead of decimals. Thus $\gamma^2=1.9^2=\left(1+\frac{9}{10}\right)^2= 1+\frac{81}{100} +\frac{180}{100} =\frac{262}{100}$ . Express $\gamma$ in terms of $\beta$ to get $1-\beta^2=\frac{100}{262}\approx\frac{1}{3} \Rightarrow \beta^2=\frac{2}{3}$ .

$8\times 8$ is about 64, so the velocity has to be greater than 0.8c. The only choice left is (E).

If one has the time, one might want to memorize the following:

$\begin{eqnarray} \gamma(\beta=0.1)&=&1.005\\ \gamma(\beta=0.25)&=&1.033\\ \gamma(\beta=0.5)&=&1.155\\ \gamma(\beta=0.75)&=&1.51\\ \gamma(\beta=0.9)&=&2.29<br /> \end{eqnarray}$

, or perhaps a more elaborate list of $\gamma-\beta$ correlations.

If one knew that before-hand, then one would immediately arrive at choice (E).

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GR8677 #21	Special Relativity $\Rightarrow$ }Metric Recall that the metric $dS^2=-dt^2+dr^2$ (equivalently, $dS^2=dt^2-dr^2$ ) $dr^2 = dx_i dx_i= (dx)^2+(dy)^2+(dz)^2=4+0+4=8$ and $dt^2=4$ . Thus $dS^2=4$ , and $S=2$ , as in choice (C). Click here to jump to the problem!

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GR8677 #68

Special Relativity $\Rightarrow$ }Relativistic Energy

The relativistic energy relation is,

$E= \sqrt{p^2 c^2 + mc^4.}$

As the speed of a particle $v \rightarrow c$ , $E\rightarrow pc$ , since by the de Broglie relation, $p=h/\lambda=h\nu/c=E/c$ . Plug in the requirement into the energy relation to find that $m\rightarrow 0$ , as in choice (A).

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GR8677 #69

Special Relativity $\Rightarrow$ }Length Contraction

Things moving relative to a particular rest frame appear shorter, i.e., contracted. Knowing that $\gamma > 1$ , one can deduce this relation without remembering much about special relativity (other than the hackneyed phrase length contraction) $L_{rest}=\gamma L_{moving}$ . So, one has $5=\gamma 3$ for the car, where it is seen to be moving in the reference frame of the garage. This implies that $\gamma=5/3\approx 1.6=\frac{1}{\sqrt{1-(\beta)^2}}$ . Recall the following useful table to memorize, for relating $\beta=v/c$ and $\gamma$

$\begin{eqnarray} \gamma &\Leftrightarrow& \beta \\ 1.005 &\Leftrightarrow& 0.1\\ 1.033 &\Leftrightarrow& 0.25\\ 1.155 &\Leftrightarrow& 0.5\\ 1.51 &\Leftrightarrow& 0.75\\ 2.29 &\Leftrightarrow& 0.9<br /> \end{eqnarray}$

One has $\gamma \approx 1.6 \Rightarrow \beta > 0.75c$ (but less than $c$ ). Choice (C) works best.

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GR8677 #70	Special Relativity $\Rightarrow$ }Length Contraction One deduced the approximate value of $\gamma$ in Problem 69 (as well as the equation for length contraction). Thus, $4=\gamma L_{moving}\Rightarrow L_{moving}=4\times 3/5=12/5=2.4$ . This is choice (A). Click here to jump to the problem!

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GR8677 #71

Special Relativity $\Rightarrow$ }Relativity!

The order of the door opening depends on which reference frame one is in. Whether or not one actually put the relativistic cake in one's mouth to have performed the motion of eating the cake... is all relative.

Observe this table summarizing the results of some trivial calculations of rest lengths,

$\begin{eqnarray} ........... own frame & & other's frame\\ car........ 5 m & & 3m \\ garage..... 4 m & & 2.4m<br /> \end{eqnarray}$

So, the car fits in the garage in the garage's frame (the car is $3m$ and the garage is $4m$ ). But, in its own frame, the car is much too big to fit in the garage (car is $5m$ and the garage is merely $2.4m$ ). Which frame is right? Relativity puts equal footing to either, and thus one shall never know. Be an agnostic, and one just might live a happy carefree life.

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GR8677 #36

Special Relativity $\Rightarrow$ }Conservation of Energy

The rest mass for each mass is 4kg. They collide head-on with identical speeds pointing in opposite directions. This implies that the composite mass is at rest. Thus, recalling that the total energy is given by $E=\gamma mc^2$ and that the rest mass is given by $E=mc^2$ , one has
$2\gamma mc^2 = Mc^2$ , where M is the composite mass.

The particle travels at $v=3c/5$ , which yields $\gamma = 5/4$ . Plug this in to get $M=10/4 \times 4 = 10kg$ .

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GR8677 #37

Special Relativity $\Rightarrow$ }Addition of Velocities

Recall that $u' = \frac{v+u}{1+vu/c^2}$ . The problem gives $u=0.6c$ , $v=0.3c$ . Thus, $u'=\frac{0.9c}{1+0.18}$ , which is slightly less than choice (E). Choose choice (D).

If one forgets the addition of velocity formula, one can always derive it from taking derivatives of the Lorentz Transformations, which are easier to remember, with $x'=\gamma(x+vt)$ and $t'=\gamma(t+vx/c^2)$ .

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GR8677 #38	Special Relativity $\Rightarrow$ }Energy The problem gives $\gamma mc^2 = 10$ and $\gamma mv = 5$ . Divide the two to get $v/c^2 = 0.5 \Rightarrow v=c/2$ , as in choice (D). The hardest part of the problem is remembering the definition of relativistic energy in terms of just the rest mass and $\gamma$ . Click here to jump to the problem!

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GR8677 #48

Special Relativity $\Rightarrow$ }Half Life

The halflife of the mesons is given. Since only half of the mesons reach point B 15 meters away, one presume that it takes 1 halflife of proper time to get there.

The proper time is $t_0=2.5E-8$ . The length $L=15=vt$ is in the lab frame, and since the time dilation equation gives $t=t_0\gamma$ , one has $L = 15 = vt_0\gamma=vt_0/\sqrt{1-(v/c)^2}\Rightarrow L/t_0 = \frac{v}{\sqrt{1-(v/c)^2}}$ .

Now, the gory arithmetics. No calculators allowed; 12 years of American public school mathematics wasted! $15/2.5E-8=15/25E-9=3/5E9=6E8=\kappa=\frac{v}{1-(v/c)^2}=\frac{vc}{\sqrt{c^2-v^2}}$ . Multiply things out to get $\kappa^2(c^2-v^2)=v^2c^2 \Rightarrow \kappa^2 c^2 = v^2 (c^2+\kappa^2)$ . Plug in numbers to get $v=\frac{6E8c}{\sqrt{9E16+36E16}}=\frac{6E8c}{\sqrt{45E16}}=\frac{6c}{3\sqrt{5}}$ , which is choice (C). Whew!

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GR8677 #50	Special Relativity $\Rightarrow$ }Spacetime Interval The spacetime interval is defined by the metric that negates spatial and time variables as $dS^2 = (cdt)^2 - (dx)^2$ . $dS$ is invariant. One has thus $dS^2 = dS'^2 \Rightarrow (3c)^2=(5c)^2-(ct)^2\Rightarrow ct = 4 c$ minutes, as in choice (C). Click here to jump to the problem!

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GR8677 #96

Special Relativity $\Rightarrow$ }Maximal Velocity

Conservation of momentum yields $p = 2\gamma m_e v$ , where p is the momentum of the photon.

Conservation of energy yields $pc=2\gamma m_e c^2$ . Plug in the above equation for momentum to get $2\gamma m_e vc = 2\gamma m_e c^2$ . This occurs when $v=c$ . Since v is the velocity of the electron, and since according to relativity, only a photon (to wit: a massless particle) can move at the speed of light---one of the conservation laws is not conserved!

This is choice (A).

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GR8677 #32

Special Relativity $\Rightarrow$ }Momentum

The total energy of a particle of mass m is $E=2mc^2=\gamma mc^2$ , since the problem supplies that it is equal to twice the rest mass. This means that $\gamma =2$ .

The relativistic momentum is $p=\gamma mv=2mv$ . However, $v\neq c$ .

Knowing $\gamma$ , one can quickly solve for $\gamma=2 = 1/\sqrt{1-\beta^2}$ . One finds that $\beta^2 = 3/4$ . Thus, the velocity is $v=\sqrt{3}/2 c$ . Plug this into the momentum equation to get, choice (D).

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GR8677 #33	Special Relativity $\Rightarrow$ }Frames The distance the pion must travel is in the lab frame, and thus $L=30=vt$ . The decay time is given in its proper time, and thus $t=\gamma t_0 \Rightarrow 30=v\gamma t_0$ . Solving for v, one finds that choice (D) works. Click here to jump to the problem!

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GR8677 #34

Special Relativity $\Rightarrow$ }Spacetime Interval

The spacetime interval for the convention $ds^2 = c^2 dt^2 - dx^2$ has regions where the slope is greater than that of the lightcone line, i.e., $ct=x$ , to be timelike. Slopes that are less than the lightcone slope on the plot of x (horizontal) vs ct (vertical) correspond to spacelike phenomena. In this region, one can have two observers disagree on whether an event happens before the other. Thus, one wants $ct<x$ , as in choice (C).

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GR8677 #71

Special Relativity $\Rightarrow$ }Doppler Shift

(The Doppler Shift is derived earlier on in the current author's solutions for GR9677.)

Since the light received has a higher wavelength than the incident wavelength, one concludes that this is a red-shift, that the object is moving away from the Earth.

The Doppler Effect equation gives $\lambda = \sqrt{\frac{1+\beta}{1-\beta}} \lambda_0$ , where $\lambda_0$ is the wavelength of the source. Thus, $\lambda/\lambda_0 =\sqrt{\frac{1+\beta}{1-\beta}}\approx 6$ . Solving, one gets $\beta = 35/37 \Rightarrow v=35/37 c$ , which is, by inspection of possible choices, closest to choice (D).

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GR8677 #79	Special Relativity $\Rightarrow$ }Rest Mass The total relativistic energy is $10GeV=\gamma mc^2=\sqrt{p^2c^2+m^2c^4}$ . The total relativistic momentum is $p=8GeV/c = \gamma mv$ . Plugging the momentum into the first equation, one has $(8^2+m^2c^2)c^2=10^2 \Rightarrow m^2c^4 = 100-64 = 36 \Rightarrow m = 6GeV/c^2$ , as in choice (D). Click here to jump to the problem!

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GR8677 #80

Special Relativity $\Rightarrow$ }Addition of Velocities

If one forgets the addition of velocity formula, one can quickly derive it from the Lorentz transformation (which one really ought to remember),

$x=\gamma(x'+vt')$

$t=\gamma(t'+vx'/c^2)$

where the unprimed system is the rest frame.

One finds that $u_x=dx/dt = \frac{d}{dt}\gamma(x'+vt')=\gamma \frac{d}{dt'}(x'+vt')\frac{dt'}{dt}=\gamma(u_x^{'}+v)$ .

From the time transformation, one gets $\frac{dt}{dt'}=\gamma(1+vu_x^{'}/c^2)$ , where $u_x^{'}$ is the velocity of the particle moving in the primed coordinate system. Plugging the inverse of that into the velocity in the unprimed system, one gets

$u_x=\frac{u_x^{'}+v}{1+vu_x^{'}/c^2}$ .

The problem supplies the following values: $v=c/2$ , $u_x^{'}=c/n=3c/4$ (since $n=4/3$ ). Plugging them in the $u_x$ equation, one finds that $u_x = \frac{3c/4+c/2}{1+3/8}=10c/11$ , as in choice (D).

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GR8677 #99

Special Relativity $\Rightarrow$ }Conservation of Energy

The tricky part of this problem is to note that the momentum of the photon is shared equally between all three final particles. (This insight was supplied by Felipe Birk). Thus, $p_e = p/3$ , where p is the momentum of the photon and $p_e$ is the momentum of the final electron or photon.

The initial energy before the photon strikes the electron is $E_0=pc+m_e c^2$ , which is just the energy of the photon plus the rest energy of the electron.

The final energy after the collision is $E_f = 3 \sqrt{(pc/3)^2+(m_e c^2)}$ , which is the sum total energy of all three final particles, i.e., the positron and two electrons. (A positron is the electron's antiparticle, and thus they have the same mass.) Note that the momentum split relation mentioned above is used to equate the final particle momentum with the initial photon momentum.

Conjure up the good conservation of energy idea. Equating $E_0=E_f$ , one gets $(pc)^2+(m_ec^2)^2 + 2pcm_ec^2 = 9((pc/3)^2+(m_ec^2)^2)$ . Canceling the $(pc)^2$ terms on both side, then solving, one arrives at $pc=4m_ec^2$ , which is choice (D).

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GR8677 #37

Special Relativity $\Rightarrow$ }Maximal Velocity

The maximal velocity of any object, even light itself, is the speed of light. Moreover, light always travels at light speed (c). This is true in all frames, and in fact, it is one of the two postulates of Special Relativity (the other being the equivalence of inertia frames).

There's no need to chunk out the addition of velocity formula for this. The only possibilities are choices (A) and (D). Since $\gamma_2$ is emitted backwards, according to the coordinate system in the diagram, its velocity would be $-c \hat{k}$ , as in choice (A).

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GR8677 #38

Special Relativity $\Rightarrow$ }Time Dilation Formula

The time dilation formula is given by $t = \gamma t_0 = \frac{\Delta t_0}{\sqrt{1-\beta_{ij}^2}}$ , where time is dilated (lengthened) in all but the frame at rest (proper-time $t_0$ ). Note that $\beta_{ij} = v_{ij}/c$ .

So, from that alone, one can deduce the following relations (without looking at the choices yet):

$\Delta t_2 = \frac{\Delta t_1}{\sqrt{1-\beta_{12}^2}}$

$\Delta t_3 = \frac{\Delta t_1}{\sqrt{1-\beta_{13}^2}}$

The latter deduction is just choice (B).

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GR8677 #70	Special Relativity $\Rightarrow$ }Gamma $E=\gamma mc^2 = 100mc^2$ , where ETS supplies the total energy to be 100 times the rest energy. Thus, $p=\gamma mv = 100 mv$ , but since $\gamma = 100 = \frac{1}{1-\beta^2}$ , where $\beta=v/c$ , one has $v\rightarrow c$ , as in choice (D). Click here to jump to the problem!

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GR8677 #85

Special Relativity $\Rightarrow$ }Momentum

Given a total energy of $\gamma mc^2 = 1.5MeV$ and the rest mass of the electron to be $m_e = .5MeV/c^2$ , one can figure out $\gamma =3$ .

The momentum is given by $p=\gamma mv = 3mv=3v/2 (MeV/c^2)$ .

Solve for the velocity from $\gamma= \frac{1}{\sqrt{1-\beta^2}} \Rightarrow \gamma^2 = \frac{1}{1-\beta^2} \Rightarrow 1 = \gamma^2(1-\beta^2) \Rightarrow \beta^2 = 1-\gamma^{-2} \Rightarrow \beta^2=8/9$ . Thus, the velocity is $v=2c/3$ .

Plugging this into the equation for momentum, one gets $p=3/2\times \sqrt{8/9}=\sqrt{2}$ , and thus its momentum is about 1.4, as in choice (C).

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GR8677 #94	Special Relativity $\Rightarrow$ }Lorentz Transformation Lorentz transformations are given by $x^{'} = \gamma (x-vt)$ $t^{'} = \gamma (t - vx/c^2)$ Factoring out the terms, choice (C) is $x=5/4(x-3/5t)$ , and thus $\gamma=5/4$ and $v=3/5$ . Since the equation for t fits the form above, this is a valid Lorentz Transformation. Click here to jump to the problem!

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This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.