Four Vectors - GRE Subject Test: Physics

Card 1 of 8

Didn't Know

Knew It

1 of 87 left

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?

Tap to reveal answer

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'}${c^2$}}$

$v\approx0.6c$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

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

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

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}$\cdot 2)=5.25$

← Didn't Know|Knew It →

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?

Tap to reveal answer

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'}${c^2$}}$

$v\approx0.6c$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

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

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

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}$\cdot 2)=5.25$

← Didn't Know|Knew It →

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?

Tap to reveal answer

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'}${c^2$}}$

$v\approx0.6c$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

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

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

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}$\cdot 2)=5.25$

← Didn't Know|Knew It →

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?

Tap to reveal answer

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'}${c^2$}}$

$v\approx0.6c$

← Didn't Know|Knew It →

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?

Tap to reveal answer

Answer

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

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

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}$\cdot 2)=5.25$

← Didn't Know|Knew It →

Knew It

Four Vectors - GRE Subject Test: Physics

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?

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately To find the relative speed, use the following equation:

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?

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

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?

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately To find the relative speed, use the following equation:

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?

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

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?

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately To find the relative speed, use the following equation:

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?

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

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?

The speed of light through water in a stationary frame is found from the index of refraction of water, which is approximately To find the relative speed, use the following equation:

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?

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

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'}${c^2$}}$

$v\approx0.6c$

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

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

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}$\cdot 2)=5.25$

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'}${c^2$}}$

$v\approx0.6c$

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

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

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}$\cdot 2)=5.25$

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'}${c^2$}}$

$v\approx0.6c$

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

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

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}$\cdot 2)=5.25$

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'}${c^2$}}$

$v\approx0.6c$

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

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

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}$\cdot 2)=5.25$