$14.1.9.$ "...A space object was approaching Earth. The fastest space laboratory was rushing to meet it. 'What is the approach velocity of the object and the laboratory?' the general in charge of the encounter requested from Earth.
'In the Earth's system or our laboratory's?' the laboratory operator replied. 'What difference does it make,' the general answered. 'These velocities already differ by 0.01%,' came the voice from space. 'We have now reached an approach velocity of exactly 100,000 km/s in our system and we are not changing it anymore.' 'How do you measure the velocity?' the general asked. 'Just like you, we established a passive link with the object. A radar pulse constantly travels between us and the object, reflecting alternately from our laboratory and the object. The approach velocity is determined by the change in the pulse's return time.' 'That is true when the radar pulse both moves away and approaches the laboratory at a speed equal to the speed of light,' the general thought. 'Then the approach velocity of the object is determined only by the ratio of two successive times. But it is not like that for them. When they chase the reflected pulse, the speed of the pulse $c$ decreases by the magnitude of the laboratory's speed, and it increases by the same amount when the pulse flies towards them.' Unexpectedly to himself, the general asked the operator: 'What approach velocity would you get if we reported from Earth the pulse speeds relative to the laboratory that we observe, and you used these values to calculate the object's speed based on the pulse's return time? Probably the exact same one we saw from Earth.' 'Yes, general,' the answer flew to Earth at the speed of light. A thought flashed through the general's mind: 'Physicists are dissembling. They simply cannot measure the speed of the pulse. There is no scale. So they assume it to be equal to the speed of light. Hence all the discrepancies.' "
This excerpt from a yet unpublished sci-fi story raised the following questions. How right is the general? What are the velocities of the object and the laboratory in the "Earth" frame of reference?
### Solution
#### 1. Analysis of the radar pulse and the general's paradox
Physically, the general is wrong: according to the second postulate of SR (Special Relativity), the speed of light in a vacuum is strictly equal to $c$ for any inertial observer. The scientists in the laboratory measure the approach velocity $v$ via the pulse return times $\tau_1$ and $\tau_2$ absolutely correctly:
$$ v = c \frac{\tau_1 - \tau_2}{\tau_1 + \tau_2} = c \frac{k - 1}{k + 1}$$
where $k = \tau_1/\tau_2$ is the ratio of times.
However, the general's intuition leads to a remarkable result. If one applies the classical velocity addition law to light (the pulse moves away at $c-u$ and returns at $c+u$, where $u$ is the laboratory's velocity relative to Earth), the velocity $v'$ calculated this way is expressed as:
The amazing fact is that this formula $v'$ exactly matches the true approach velocity of the object and the laboratory in the "Earth" reference frame!
Why is this not a coincidence?
The secret lies in the parameter $k$. When transitioning from the laboratory frame to the Earth frame, both time intervals ($\tau_1$ and $\tau_2$) undergo the same relativistic time dilation (multiplied by the Lorentz factor $\gamma$). Since both times are stretched proportionally, their ratio $k$ remains invariant. The general uses incorrect absolute values of classical velocities, but during the calculations, these errors cancel out perfectly. The mathematics of spacetime geometry covertly corrects the general's mistake, yielding the correct relativistic result.
#### 2. Calculation of the laboratory's velocity
According to the condition, the difference between the Earth's approach velocity $v'$ and the laboratory's $v$ is $0.01\%$, i.e., a fraction $\alpha = 10^{-4}$. Let's write their difference:
$$ v' - v = \frac{v\left(1 - \frac{u^2}{c^2}\right)}{1 - \frac{uv}{c^2}} - v = v \frac{u(v - u)}{c^2 - uv} = \alpha v $$
We cancel $v$ and express the laboratory's velocity $u$. Since $\alpha = 10^{-4}\ll 1$, the desired velocity of the laboratory is extremely small ($u \ll v$). Neglecting the magnitude of $u$ compared to $v$ in the numerator and the term $uv$ compared to $c^2$ in the denominator, we simplify the equation:
$$\frac{u(v - u)}{c^2 - uv} = \alpha\implies u \approx\frac{\alpha c^2}{v}$$
Substituting the known values ($c = 300\,000$ km/s, $v = 100\,000$ km/s):
This velocity belongs to the fastest Earth space laboratory (which is a colossal value for human technology).
#### 3. Calculation of the object's velocity
@@ -36,9 +36,13 @@3. Calculation of the object's velocity
The object's velocity relative to Earth $v_0$ is related to the laboratory's velocity $u$ and their relative approach velocity $v$ by the relativistic velocity addition law as:
$$ v_0 = \frac{v - u}{1 - \frac{vu}{c^2}}$$
−
Given that $vu \ll c^2$ (the denominator is close to one), we can use the approximation with tremendous accuracy:
1. The general is physically wrong (the speed of light is invariant) but mathematically right: due to the invariance of the time ratio $k$, the calculation using his classical formulas paradoxically yields the true approach velocity in the "Earth" frame.
−
2. The laboratory's velocity is $u \approx 90$ km/s; the object's velocity is $v_0 \approx 99\,910$ km/s.
+
2. The laboratory's velocity is $u \approx 90$ km/s; the object's velocity is $v_0 \approx 99\,920$ km/s.
@@ -1,4 +1,4 @@
### Statement
### Statment
$14.1.9.$ "...A space object was approaching Earth. The fastest space laboratory was rushing to meet it. 'What is the approach velocity of the object and the laboratory?' the general in charge of the encounter requested from Earth.
$14.1.9.$ "...A space object was approaching Earth. The fastest space laboratory was rushing to meet it. 'What is the approach velocity of the object and the laboratory?' the general in charge of the encounter requested from Earth.
'In the Earth's system or our laboratory's?' the laboratory operator replied. 'What difference does it make,' the general answered. 'These velocities already differ by 0.01%,' came the voice from space. 'We have now reached an approach velocity of exactly 100,000 km/s in our system and we are not changing it anymore.' 'How do you measure the velocity?' the general asked. 'Just like you, we established a passive link with the object. A radar pulse constantly travels between us and the object, reflecting alternately from our laboratory and the object. The approach velocity is determined by the change in the pulse's return time.' 'That is true when the radar pulse both moves away and approaches the laboratory at a speed equal to the speed of light,' the general thought. 'Then the approach velocity of the object is determined only by the ratio of two successive times. But it is not like that for them. When they chase the reflected pulse, the speed of the pulse $c$ decreases by the magnitude of the laboratory's speed, and it increases by the same amount when the pulse flies towards them.' Unexpectedly to himself, the general asked the operator: 'What approach velocity would you get if we reported from Earth the pulse speeds relative to the laboratory that we observe, and you used these values to calculate the object's speed based on the pulse's return time? Probably the exact same one we saw from Earth.' 'Yes, general,' the answer flew to Earth at the speed of light. A thought flashed through the general's mind: 'Physicists are dissembling. They simply cannot measure the speed of the pulse. There is no scale. So they assume it to be equal to the speed of light. Hence all the discrepancies.' "
'In the Earth's system or our laboratory's?' the laboratory operator replied. 'What difference does it make,' the general answered. 'These velocities already differ by 0.01%,' came the voice from space. 'We have now reached an approach velocity of exactly 100,000 km/s in our system and we are not changing it anymore.' 'How do you measure the velocity?' the general asked. 'Just like you, we established a passive link with the object. A radar pulse constantly travels between us and the object, reflecting alternately from our laboratory and the object. The approach velocity is determined by the change in the pulse's return time.' 'That is true when the radar pulse both moves away and approaches the laboratory at a speed equal to the speed of light,' the general thought. 'Then the approach velocity of the object is determined only by the ratio of two successive times. But it is not like that for them. When they chase the reflected pulse, the speed of the pulse $c$ decreases by the magnitude of the laboratory's speed, and it increases by the same amount when the pulse flies towards them.' Unexpectedly to himself, the general asked the operator: 'What approach velocity would you get if we reported from Earth the pulse speeds relative to the laboratory that we observe, and you used these values to calculate the object's speed based on the pulse's return time? Probably the exact same one we saw from Earth.' 'Yes, general,' the answer flew to Earth at the speed of light. A thought flashed through the general's mind: 'Physicists are dissembling. They simply cannot measure the speed of the pulse. There is no scale. So they assume it to be equal to the speed of light. Hence all the discrepancies.' "
This excerpt from a yet unpublished sci-fi story raised the following questions. How right is the general? What are the velocities of the object and the laboratory in the "Earth" frame of reference?
This excerpt from a yet unpublished sci-fi story raised the following questions. How right is the general? What are the velocities of the object and the laboratory in the "Earth" frame of reference?
### Solution
### Solution
#### 1. Analysis of the radar pulse and the general's paradox
#### 1. Analysis of the radar pulse and the general's paradox
Physically, the general is wrong: according to the second postulate of SR (Special Relativity), the speed of light in a vacuum is strictly equal to $c$ for any inertial observer. The scientists in the laboratory measure the approach velocity $v$ via the pulse return times $\tau_1$ and $\tau_2$ absolutely correctly:
Physically, the general is wrong: according to the second postulate of SR (Special Relativity), the speed of light in a vacuum is strictly equal to $c$ for any inertial observer. The scientists in the laboratory measure the approach velocity $v$ via the pulse return times $\tau_1$ and $\tau_2$ absolutely correctly:
$$ v = c \frac{\tau_1 - \tau_2}{\tau_1 + \tau_2} = c \frac{k - 1}{k + 1}$$
$$ v = c \frac{\tau_1 - \tau_2}{\tau_1 + \tau_2} = c \frac{k - 1}{k + 1}$$
where $k = \tau_1/\tau_2$ is the ratio of times.
where $k = \tau_1/\tau_2$ is the ratio of times.
However, the general's intuition leads to a remarkable result. If one applies the classical velocity addition law to light (the pulse moves away at $c-u$ and returns at $c+u$, where $u$ is the laboratory's velocity relative to Earth), the velocity $v'$ calculated this way is expressed as:
However, the general's intuition leads to a remarkable result. If one applies the classical velocity addition law to light (the pulse moves away at $c-u$ and returns at $c+u$, where $u$ is the laboratory's velocity relative to Earth), the velocity $v'$ calculated this way is expressed as:
The amazing fact is that this formula $v'$ exactly matches the true approach velocity of the object and the laboratory in the "Earth" reference frame!
The amazing fact is that this formula $v'$ exactly matches the true approach velocity of the object and the laboratory in the "Earth" reference frame!
Why is this not a coincidence?
Why is this not a coincidence?
The secret lies in the parameter $k$. When transitioning from the laboratory frame to the Earth frame, both time intervals ($\tau_1$ and $\tau_2$) undergo the same relativistic time dilation (multiplied by the Lorentz factor $\gamma$). Since both times are stretched proportionally, their ratio $k$ remains invariant. The general uses incorrect absolute values of classical velocities, but during the calculations, these errors cancel out perfectly. The mathematics of spacetime geometry covertly corrects the general's mistake, yielding the correct relativistic result.
The secret lies in the parameter $k$. When transitioning from the laboratory frame to the Earth frame, both time intervals ($\tau_1$ and $\tau_2$) undergo the same relativistic time dilation (multiplied by the Lorentz factor $\gamma$). Since both times are stretched proportionally, their ratio $k$ remains invariant. The general uses incorrect absolute values of classical velocities, but during the calculations, these errors cancel out perfectly. The mathematics of spacetime geometry covertly corrects the general's mistake, yielding the correct relativistic result.
#### 2. Calculation of the laboratory's velocity
#### 2. Calculation of the laboratory's velocity
According to the condition, the difference between the Earth's approach velocity $v'$ and the laboratory's $v$ is $0.01\%$, i.e., a fraction $\alpha = 10^{-4}$. Let's write their difference:
According to the condition, the difference between the Earth's approach velocity $v'$ and the laboratory's $v$ is $0.01\%$, i.e., a fraction $\alpha = 10^{-4}$. Let's write their difference:
$$ v' - v = \frac{v\left(1 - \frac{u^2}{c^2}\right)}{1 - \frac{uv}{c^2}} - v = v \frac{u(v - u)}{c^2 - uv} = \alpha v $$
$$ v' - v = \frac{v\left(1 - \frac{u^2}{c^2}\right)}{1 - \frac{uv}{c^2}} - v = v \frac{u(v - u)}{c^2 - uv} = \alpha v $$
We cancel $v$ and express the laboratory's velocity $u$. Since $\alpha = 10^{-4}\ll 1$, the desired velocity of the laboratory is extremely small ($u \ll v$). Neglecting the magnitude of $u$ compared to $v$ in the numerator and the term $uv$ compared to $c^2$ in the denominator, we simplify the equation:
We cancel $v$ and express the laboratory's velocity $u$. Since $\alpha = 10^{-4}\ll 1$, the desired velocity of the laboratory is extremely small ($u \ll v$). Neglecting the magnitude of $u$ compared to $v$ in the numerator and the term $uv$ compared to $c^2$ in the denominator, we simplify the equation:
$$\frac{u(v - u)}{c^2 - uv} = \alpha\implies u \approx\frac{\alpha c^2}{v}$$
$$\frac{u(v - u)}{c^2 - uv} = \alpha\implies u \approx\frac{\alpha c^2}{v}$$
Substituting the known values ($c = 300\,000$ km/s, $v = 100\,000$ km/s):
Substituting the known values ($c = 300\,000$ km/s, $v = 100\,000$ km/s):
This velocity belongs to the fastest Earth space laboratory (which is a colossal value for human technology).
This velocity belongs to the fastest Earth space laboratory (which is a colossal value for human technology).
#### 3. Calculation of the object's velocity
#### 3. Calculation of the object's velocity
@@ -36,9 +36,13 @@3. Calculation of the object's velocity
The object's velocity relative to Earth $v_0$ is related to the laboratory's velocity $u$ and their relative approach velocity $v$ by the relativistic velocity addition law as:
The object's velocity relative to Earth $v_0$ is related to the laboratory's velocity $u$ and their relative approach velocity $v$ by the relativistic velocity addition law as:
$$ v_0 = \frac{v - u}{1 - \frac{vu}{c^2}}$$
$$ v_0 = \frac{v - u}{1 - \frac{vu}{c^2}}$$
Given that $vu \ll c^2$ (the denominator is close to one), we can use the approximation with tremendous accuracy:
Here, we cannot neglect the denominator, as it is exactly what accounts for the $0.01\%$ relativistic effect. Let's calculate the exact value:
1. The general is physically wrong (the speed of light is invariant) but mathematically right: due to the invariance of the time ratio $k$, the calculation using his classical formulas paradoxically yields the true approach velocity in the "Earth" frame.
1. The general is physically wrong (the speed of light is invariant) but mathematically right: due to the invariance of the time ratio $k$, the calculation using his classical formulas paradoxically yields the true approach velocity in the "Earth" frame.
2. The laboratory's velocity is $u \approx 90$ km/s; the object's velocity is $v_0 \approx 99\,910$ km/s.
2. The laboratory's velocity is $u \approx 90$ km/s; the object's velocity is $v_0 \approx 99\,920$ km/s.
