Special Relativity - GRE Subject Test: Physics

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Question

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 Eo?

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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?

Answer

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 $310^8 \frac{m}{s}$ , so $3.210^8 \frac{m}{s}$ 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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Question

A particle in an accelerator has a total energy and a relativistic momentum $p=4 \frac{GeV}{c}$ . What is the rest mass of the particle, in units of $\frac{1}{c^2}$ ?

Answer

The total energy of a relativistic particle is given by:

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

Solving for the rest mass :

$m=\frac{\sqrt{E^2-p^2c^2}}{c^2}=\frac{\sqrt{25-16}}{c^2}=\frac{3}{c^2}$

Which, in the units specified, gives the answer 3.

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Question

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?

Answer

The total energy E of a relativistic particle is related to its rest mass energy Eo 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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Question

A container of water is carried on a space ship traveling at 60% the speed of light relative to the Earth. From the perspective of an observer on Earth, what is the speed of the light in the water?

$n_{water}=1.33$

Answer

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately $\frac{4}{3}$

$v'=\frac{c}{n}=\frac{3}{4}c$

To find the relative speed, use the following equation:

$v_{p}=\frac{v'+v}{1+\frac{vv'}{c^2}}$

$v\approx0.6c$

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Question

A reference frame S' is moving at 0.6_c_ in the z direction with respect to a stationary frame S. An event occurs in S' with the coordinates (x', y', z', ct')=(1, 0, 2, 3). What are the coordinates (x, y, z, ct) of the event with respect to the stationary frame S?

Answer

To apply a Lorentz transformation, we need gamma and beta:

$\beta=\frac{v}{c}=\frac{3}{5}$

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

Then, apply the Lorentz transformation:

$z=\gamma(z'+\beta ct')=\frac{5}{4}(2+\frac{3}{5}\cdot 3)=4.75$

$ct=\gamma(ct'+\beta z')=\frac{5}{4}(3+\frac{3}{5}\cdot 2)=5.25$

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Question

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: $R=6.963\times10^{8},\textup{m}$

Answer

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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Question

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

Answer

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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Question

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?

Answer

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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Question

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

Answer

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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Question

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?

Answer

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:

The Earth-twin is now 50.

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Question

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 Eo?

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

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?

Answer

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 $310^8 \frac{m}{s}$ , so $3.210^8 \frac{m}{s}$ 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:

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

Compare your answer with the correct one above

Question

A particle in an accelerator has a total energy and a relativistic momentum $p=4 \frac{GeV}{c}$ . What is the rest mass of the particle, in units of $\frac{1}{c^2}$ ?

Answer

The total energy of a relativistic particle is given by:

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

Solving for the rest mass :

$m=\frac{\sqrt{E^2-p^2c^2}}{c^2}=\frac{\sqrt{25-16}}{c^2}=\frac{3}{c^2}$

Which, in the units specified, gives the answer 3.

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Question

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?

Answer

The total energy E of a relativistic particle is related to its rest mass energy Eo 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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Question

A container of water is carried on a space ship traveling at 60% the speed of light relative to the Earth. From the perspective of an observer on Earth, what is the speed of the light in the water?

$n_{water}=1.33$

Answer

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately $\frac{4}{3}$

$v'=\frac{c}{n}=\frac{3}{4}c$

To find the relative speed, use the following equation:

$v_{p}=\frac{v'+v}{1+\frac{vv'}{c^2}}$

$v\approx0.6c$

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