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GRE Subject Test: Physics : Special Relativity

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Example Question #1 : Special Relativity

The difference in age for the twins in the Twin Paradox occurs during which key moment in the trip?

Possible Answers:

During the time returning to Earth traveling fast.

While approaching a black hole in space during the trip.

None of these

During the period of great acceleration during the changing of directions and return to Earth.

During the beginning of the journey traveling fast.

Correct answer:

During the period of great acceleration during the changing of directions and return to Earth.

Explanation:

While moving clocks do in fact record time moving at different rates, the time dilation works both ways. This means that a stationary person will view a moving clock ticking slower, but at the same time, a person moving alongside the moving clock will see the stationary clock ticking slower. However, clocks experiencing great accelerations will be permanently changed, "losing" time relative to a clock not being accelerated. Thus, the age difference occurs during the portion of the journey when the traveler accelerates at a great rate in order to return to Earth.

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Example Question #1 : Special Relativity

A black hole is an object whose gravitational field is so strong that even light cannot escape. Assuming no change in radius, approximately how much mass would our Sun have to have in order to become a black hole?

Sun's radius: $R=6.963\times10^{8}\,\textup{m}$

Possible Answers:

$1\times10^{35}\,\textup{kg}$

$1.5\times10^{27}\,\textup{kg}$

$5\times10^{35}\,\textup{kg}$

$2\times10^{30}\,\textup{kg}$

$3\times10^{40}\,\textup{kg}$

Correct answer:

$5\times10^{35}\,\textup{kg}$

Explanation:

To derive the Schwarzschild radius of a black hole, set gravitational potential energy equal to kinetic energy at escape velocity:

$\frac{GMm}{r}=\frac{1}{2}mc^2$

Solving for mass of the black hole:

$M=\frac{rc^2}{2G}=\frac{(6.963\times 10^8 \textup{m})(2.998\times 10^8 \textup{m\,s}^{-1})^2}{2(6.674 \times 10^{-11}\textup{m}^3\,\textup{kg}^{-1}\textup{s}^{-2})} =5\times10^{35}\,\textup{kg}$

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Example Question #1 : Special Relativity

At one point in time, two twins are 30 years old. At this time, one of them gets on a rocket and travels at 0.8 c, for what he experiences to be 12 years. How old is the twin that remained on Earth when the traveling twin returns home?

Possible Answers:

37 years old

70 years old

50 years old

42 years old

Correct answer:

50 years old

Explanation:

The equation for time dilation is given by:

$T'=\frac{T}{\sqrt{1-\frac{v^2}{c^2}}}$

In this problem v=0.8c T=12. Using this equation, we get:

$T'=\frac{12}{\sqrt{1-(\frac{4}{5})^2}}=\frac{12}{\sqrt{1-\frac{16}{25}}}=\frac{12}{\sqrt{\frac{9}{25}}} =\frac{12}{\frac{3}{5}}=(12)(\frac{5}{3})=20$

Adding 20 years to the age initial age of 30:

30+20=50

The Earth-twin is now 50.

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Example Question #1 : Length Contraction

A meter stick, at an angle of 30 degrees with the x-axis, is traveling at 0.6c in the direction of the positive y-axis. To a stationary observer, how long does the meter stick appear to be?

Possible Answers:

$\frac{4}{5}\textup{m}$

$\sqrt{\frac{31}{42}}\textup{m}$

$\frac{1}{5}\sqrt{\frac{29}{2}}\textup{m}$

$1\textup{m}$

$\frac{\sqrt{91}}{10}\textup{m}$

Correct answer:

$\frac{\sqrt{91}}{10}\textup{m}$

Explanation:

Length contraction only occurs in the direction of motion. This means that the x component of the length, which is cos(30), does not change; length contraction only occurs the the y component, which is sin(30).

First, we find the Lorentz factor:

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{9}{25}}}=\frac{1}{\sqrt{\frac{16}{25}}}=\frac{5}{4}$

Next, we apply the time dilation equation to the length in the y direction:

$L'_y=L_{y}/ \gamma=\frac{4}{5}\sin(30)=\frac{2}{5}$

Finally, we find the total length by combining the length-contracted y component and the unchanged x component:

$L=\sqrt{L_x^2+L'_y^2}=$ $\sqrt{\frac{3}{4}+\frac{4}{25}}=\frac{\sqrt{91}}{10}$

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Example Question #1 : Length Contraction

A rocket of length 5 meters passes an observer on earth. The observer measures the passing rocket to be 3 meters long. What is the velocity of the rocket in the reference frame of the Earth-based observer?

Possible Answers:

1.25c

0.6c

0.8 c

0.4c

0.75c

Correct answer:

0.8 c

Explanation:

Length contraction is given by

$L_{rest}=\gamma L_{move}$

Where in this case,

$L_{rest}=5\textup{m}, \,\,\,L_{move}=3\textup{m}$

The Lorentz factor is given by:

$\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Combining these two equations, we get:

$L_{rest}=\frac{L_{move}}{\sqrt{1-\frac{v^2}{c^2}}}$

Solving for v:

$v=c\sqrt{1-\left ( \frac{L_{move}}{L_{rest}}\right )^2}= c\sqrt{1-\left ( \frac{3}{5}\right )^2}= c\sqrt{1-\left ( \frac{9}{25}\right )}$

$= c\sqrt{\left( \frac{16}{25}\right )}=\frac{4}{5}c=0.8c$

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Example Question #1 : Energy And Momentum

A relativistic particle of mass m has a total energy 37 times its rest energy. What is the momentum of the particle, in units of mc?

Possible Answers:

144

Correct answer:

Explanation:

The total energy E of a relativistic particle is related to its rest mass energy E_o by:

$E=\gamma E_{o}$

Where gamma is related to the momentum by:

$p=\gamma mc$

Combining the equations and solving for p, we get:

$p=\frac{E}{E_{o}}mc=37 mc$

Which, in the units specified, is 37.

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Example Question #1 : Special Relativity

A particle of mass m traveling at a relativistic speed has a momentum of 50 mc. What is the total energy E of the particle, expressed in units of the rest mass energy E_o?

Possible Answers:

5 E_o

75 E_o

$5\sqrt{2}E_o$

50 E_o

2500 E_o

Correct answer:

50 E_o

Explanation:

For a relativistic particle, momentum is given by:

$p=\gamma mc$

from which we can solve for gamma:

$\gamma=\frac{p}{mc}=\frac{50 mc}{mc}=50$

Total energy of a relativistic particle is given by:

$E=\gamma E_{o}=50E_{0}$

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Example Question #1 : Special Relativity

The rest mass energy E_o of a particle with mass is one quarter of its total energy . What is the of the particle's momentum, in units of ?

Possible Answers:

$\sqrt{3}$

$\sqrt{15}$

Correct answer:

$\sqrt{15}$

Explanation:

The question tells us:

$E=4E_{o}$

Where the rest mass energy of a particle is:

$E_{o}=mc^{2}$

Using the equation for the total energy of a particle, we can substitute:

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

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

Solving this for , we find:

$p=\sqrt{15m^2c^2}=\sqrt{15}\,\,mc$

Which, in units of mc, gives us the correct answer.

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Example Question #1 : Energy And Momentum

The relativistic momentum of a particle with mass is 3mc . What is the total energy of the particle, given in units of the rest mass energy E_o ?

Possible Answers:

$2E_{o}$

$4E_{o}$

$10E_{o}$

$\sqrt{10}E_{o}$

$\sqrt{2}E_{o}$

Correct answer:

$\sqrt{10}E_{o}$

Explanation:

The total energy of a relativistic particle is given by:

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

Substituting the momentum, we get:

$E=\sqrt{9m^2c^4+m^2c^4}=\sqrt{10}mc^2$

Because the rest mass energy of a particle is given by:

$E_{o}=mc^2$

The total energy is:

$E=\sqrt{10}E_{o}$

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Example Question #1 : Energy And Momentum

A scientist measures the spectrum of relativistic jet emitted from a black hole. He finds that the a particular spectral line, which has a stationary wavelength of 212.5 nm, has a Doppler shifted wavelength of 643.7 nm. What is the radial velocity of the relativistic jet?

Possible Answers:

$2.2*10^8 \frac{m}{s}\textup{ away from the Earth}$

$2.2*10^8 \frac{m}{s}\textup{ towards the Earth}$

$2.4*10^8 \frac{m}{s} \textup{ towards the Earth}$

$3.2*10^8 \frac{m}{s} \textup{ towards the Earth}$

$2.4*10^8 \frac{m}{s} \textup{ away from the Earth}$

Correct answer:

$2.4*10^8 \frac{m}{s} \textup{ away from the Earth}$

Explanation:

The relativistic Doppler shift equation is given by:

$\lambda_{moving}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\lambda_{stationary}$

Where $\beta$ is defined as:

$\beta=\frac{v_{source}}{c}$

Because the stationary wavelength is shorter than the moving wavelength, the object must be receding from the Earth, eliminating two answers.

The speed of light is approximately $3*10^8 \frac{m}{s}$ , so $3.2*10^8 \frac{m}{s}$ is not a possible answer.

Making the approximation that

$\frac{\lambda_{m}}{\lambda_{s}}=\frac{643.7}{212.5}\approx3$

Combining this with the first equation:

$\frac{\lambda_m}{\lambda_s}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}=3$

$3\sqrt{1-\beta}=\sqrt{1+\beta}$

$9(1-\beta)=1+\beta$

$9-9\beta=1+\beta$

$10\beta=8,\,\,\,\beta=0.8$

From beta, we can find the velocity:

$\beta=\frac{v_{source}}{c}\rightarrow v_{source}=\beta c=(0.8)(3\times10^8 \frac{m}{s})=2.4\times10^8 \frac{m}{s}$

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GRE Subject Test: Physics : Special Relativity

Study concepts, example questions & explanations for GRE Subject Test: Physics

All GRE Subject Test: Physics Resources

Example Questions

Example Question #1 : Special Relativity

Example Question #1 : Special Relativity

Example Question #1 : Special Relativity

Example Question #1 : Length Contraction

Example Question #1 : Length Contraction

Example Question #1 : Energy And Momentum

Example Question #1 : Special Relativity

Example Question #1 : Special Relativity

Example Question #1 : Energy And Momentum

Example Question #1 : Energy And Momentum

All GRE Subject Test: Physics Resources

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