With all the chatbot posts in this forum, I decided to test Bing with science and maths questions of various difficulty.
(1) Explain how the Schwarzschild metric when expressed in Droste-Hilbert coordinates results in the speed of light being anisotropic.
Calculate the difference Δv between the radial and peripheral velocity of light at the Earth’s surface.
This is a honours level question in applied mathematics.
Here is the answer.
The Schwarzschild metric in Droste-Hilbert coordinates is;
ds² = c²(1-2MG/ c²r)dt² - dr²/ (1-2MG/c²r) -r²(dθ²+sin²θdϕ²)
One needs to look at the radial and peripheral velocity of light.
Since light travels along null geodesics ds² =0.
For the radial velocity dr/dt, dθ = dϕ = 0 and the metric reduces to;
dr/dt = c(1-2MG/c²r)
For the peripheral velocity rdθ/dt, dr = dϕ = 0 and the metric reduces to;
rdθ/dt = c(1-2MG/c²r)⁰·⁵ ≈ c(1-(1/2)2MG/c²r) as 2MG/c²r is very small for Earth.
The difference between the radial and peripheral velocity is;
Δv = c(2MG/c²)(1/2) = 21 cm/sec.
I tried a much simpler question.
(2) What is the Schwarzschild metric?
It would have done even better if it showed the mathematical form of the metric.
I needed a question of intermediate complexity.
(3) I am driving at a uniform speed of 60 km/hr in one direction and at 80 km/hr per during the return journey, what is my average speed over the entire trip?
This was a "trick" question.
Impulsively one would say the answer is simply the average of the forward and return speeds and equals 70 km/hr.
Bing got the maths right but the calculation wrong!!
The correct answer is.
(2d) / ((d/60) + (d/80)) = 480/7 ≈ 68.6 km/hr.
In the calculations Bing used d instead of 2d for the total distance.
I would give Bing 5/10.
(1) Explain how the Schwarzschild metric when expressed in Droste-Hilbert coordinates results in the speed of light being anisotropic.
Calculate the difference Δv between the radial and peripheral velocity of light at the Earth’s surface.
Oh well this was too tough for the chatbot.I’m sorry, but I couldn’t find any information on how the Schwarzschild metric when expressed in Droste-Hilbert coordinates results in the speed of light being anisotropic or the difference between the radial and peripheral velocity of light at the Earth’s surface. Would you like me to search for something else?
This is a honours level question in applied mathematics.
Here is the answer.
The Schwarzschild metric in Droste-Hilbert coordinates is;
ds² = c²(1-2MG/ c²r)dt² - dr²/ (1-2MG/c²r) -r²(dθ²+sin²θdϕ²)
One needs to look at the radial and peripheral velocity of light.
Since light travels along null geodesics ds² =0.
For the radial velocity dr/dt, dθ = dϕ = 0 and the metric reduces to;
dr/dt = c(1-2MG/c²r)
For the peripheral velocity rdθ/dt, dr = dϕ = 0 and the metric reduces to;
rdθ/dt = c(1-2MG/c²r)⁰·⁵ ≈ c(1-(1/2)2MG/c²r) as 2MG/c²r is very small for Earth.
The difference between the radial and peripheral velocity is;
Δv = c(2MG/c²)(1/2) = 21 cm/sec.
I tried a much simpler question.
(2) What is the Schwarzschild metric?
No surprises here Bing like any search engine would have found the answer.The Schwarzschild metric is a solution to the Einstein field equations of general relativity that describes the gravitational field outside a spherical, non-rotating mass such as a star or planet. It is named after Karl Schwarzschild, who found the solution in 1916. Is there anything else you would like to know about this topic?
It would have done even better if it showed the mathematical form of the metric.
I needed a question of intermediate complexity.
(3) I am driving at a uniform speed of 60 km/hr in one direction and at 80 km/hr per during the return journey, what is my average speed over the entire trip?
To calculate the average speed over the entire trip, we need to know the distance traveled in each direction. Let’s say the distance traveled in each direction is d km.
The time taken to travel from the starting point to the destination at a speed of 60 km/hr would be d/60 hours. The time taken to travel back to the starting point at a speed of 80 km/hr would be d/80 hours. The total distance traveled is 2d km and the total time taken is (d/60) + (d/80) hours.
The average speed over the entire trip is calculated as the total distance traveled divided by the total time taken: (2d) / ((d/60) + (d/80)) = 240/7 ≈ 34.29 km/hr.
So, your average speed over the entire trip is approximately 34.29 km/hr.
This was a "trick" question.
Impulsively one would say the answer is simply the average of the forward and return speeds and equals 70 km/hr.
Bing got the maths right but the calculation wrong!!
The correct answer is.
(2d) / ((d/60) + (d/80)) = 480/7 ≈ 68.6 km/hr.
In the calculations Bing used d instead of 2d for the total distance.
I would give Bing 5/10.
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