ChatGPT is a gibbering idiot (this is not news)

[DaveC426913]

The first answer looked a little suss, so I asked again. And then I kept asking...



From Earth's frame of reference, how long would it take a rocket accelerating at 1g for half the trip and the decelerating for half the trip to reach a destination 20 light years away?


Every time I ask it this same question I get different results (I am literally copy-pasting the question):


1. Thus, the rocket would take about 22 years from Earth's perspective to travel 20 light years, accelerating at 1g for half the trip and decelerating at 1g for the other half.

2. In Earth's frame of reference, the rocket would take approximately 3.4 years to travel 20 light years,...

3. From Earth's frame of reference, the rocket would take approximately 4.4 years to reach a destination 20 light-years away...

4. From Earth's frame of reference, the rocket would take approximately 665 years to travel to a destination 20 light-years away...

5. From Earth's frame of reference, it would take approximately 40 years ...

6. ...it would take approximately 44.3 years...

 

[DaveC426913]

Here's ChatGPT showing some of its work. I haven't yet checked how it's getting different answers, let alone wrong answers.



22 years:

To calculate how long it would take for a rocket to travel 20 light years, accelerating at 1g1g1g for half the trip and decelerating at 1g1g1g for the second half, from Earth's frame of reference, we need to use the relativistic equations of motion. The process involves breaking the trip into two phases: acceleration and deceleration, with both phases involving constant proper acceleration.

Key details:​

• Distance to the destination: D=20 light yearsD = 20 \, \text{light years}D=20light years
• Constant proper acceleration: a=1g≈9.8 m/s2a = 1g \approx 9.8 \, \text{m/s}^2a=1g≈9.8m/s2
• The rocket accelerates for half the trip and decelerates for the second half, maintaining the same constant proper acceleration.

We use relativistic kinematics to solve this problem.

Step-by-step solution:​

1. Total Time in Earth's Frame:
   When the rocket accelerates at 1g1g1g for half the trip and decelerates at 1g1g1g for the other half, we can use the following equation for the total time (TTT) in the Earth's frame:
   T=2DvavgT = \frac{2D}{v_{\text{avg}}}T=vavg2D
   Where vavgv_{\text{avg}}vavg is the average speed of the rocket in Earth's frame.
2. Relativistic Rocket Motion:
   The distance traveled during constant proper acceleration is governed by the relativistic equation:
   x=c2a(cosh⁡(atc)−1)x = \frac{c^2}{a} \left( \cosh\left(\frac{a t}{c}\right) - 1 \right)x=ac2(cosh(cat)−1)
   where xxx is the distance traveled, aaa is the proper acceleration, ccc is the speed of light, and ttt is the proper time in the rocket’s frame.
   However, in this case, we are interested in the distance in the Earth's frame, and the total time is split into two parts (acceleration and deceleration). For simplicity, we use an approximation based on the fact that the proper acceleration is 1g1g1g, and the trip is symmetric.
3. Simplified Time Formula:
   The total time TTT in the Earth's frame for a round trip of a given distance, where the rocket accelerates at 1g1g1g for half the distance and decelerates for the other half, is approximately:
   T≈Dc⋅(2 ln(1+vc))T \approx \frac{D}{c} \cdot \left( 2 \, \text{ln}(1 + \frac{v}{c}) \right)T≈cD⋅(2ln(1+cv))
   For a trip to 20 light years with constant acceleration, we find that the total travel time in Earth's frame is around about 22 years when considering relativistic effects.

665 years:

To solve this problem, we need to determine how long it would take a rocket, accelerating at 1g1g1g (where 1g≈9.8 m/s21g \approx 9.8 \, \text{m/s}^21g≈9.8m/s2) for half the trip and decelerating at 1g1g1g for the second half, to travel to a destination 20 light years away, from Earth's frame of reference. This involves relativistic motion since the rocket will be traveling at speeds close to the speed of light.

Step-by-step approach:​

The problem involves constant acceleration for half of the journey and constant deceleration for the second half. The time in the Earth's frame will be affected by relativistic effects, so we need to consider the relativistic equations of motion for constant acceleration.

Key Information:​

• Distance to destination: D=20 light-yearsD = 20 \, \text{light-years}D=20light-years
• Acceleration: a=1g=9.8 m/s2a = 1g = 9.8 \, \text{m/s}^2a=1g=9.8m/s2
• The rocket accelerates at 1g1g1g for the first half of the trip, then decelerates at 1g1g1g for the second half.
• We need to find the total time TTT from Earth's frame.

Relativistic Time for Constant Acceleration​

In relativistic motion under constant acceleration, the time in Earth's frame for a round trip (where the rocket accelerates and decelerates symmetrically) can be approximated by the following formula:
T=2Dc⋅ln⁡(1+vmaxc)T = \frac{2D}{c} \cdot \ln\left(1 + \frac{v_{\text{max}}}{c}\right)T=c2D⋅ln(1+cvmax)
where:

• D=20 light-yearsD = 20 \, \text{light-years}D=20light-years is the total distance,
• c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s is the speed of light,
• vmaxv_{\text{max}}vmax is the maximum velocity reached by the rocket at the end of the acceleration phase.

To solve this, we first need to determine the maximum velocity reached during the acceleration phase. For constant acceleration aaa, the velocity reached at the end of the acceleration phase can be found using:
vmax=ctanh⁡(atc)v_{\text{max}} = c \tanh\left(\frac{a t}{c}\right)vmax=ctanh(cat)
where ttt is the time in the rocket's frame of reference.

Key Approximation for Time in Earth's Frame​

For the case of constant acceleration 1g1g1g and relativistic speeds, we can use the well-known approximation for the total travel time TTT in Earth's frame for the trip, derived from relativistic rocket equations.
The total time TTT for a one-way journey where the rocket accelerates at 1g1g1g for half the trip and decelerates at 1g1g1g for the second half is approximately given by:
T=Dc⋅(ca)⋅ln⁡(1+aDc2)T = \frac{D}{c} \cdot \left( \frac{c}{a} \right) \cdot \ln\left( 1 + \frac{a D}{c^2} \right)T=cD⋅(ac)⋅ln(1+c2aD)
Where:

• D=20 light-yearsD = 20 \, \text{light-years}D=20light-years (convert to meters),
• a=9.8 m/s2a = 9.8 \, \text{m/s}^2a=9.8m/s2,
• c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}c=3×108m/s.

Step 1: Convert Distance to Meters​

1 light year = 9.461×10159.461 \times 10^{15}9.461×1015 meters, so:
D=20 light-years=20×9.461×1015 m=1.8922×1017 mD = 20 \, \text{light-years} = 20 \times 9.461 \times 10^{15} \, \text{m} = 1.8922 \times 10^{17} \, \text{m}D=20light-years=20×9.461×1015m=1.8922×1017m

Step 2: Calculate Time​

Now we can substitute the values into the formula:
T≈1.8922×10173×108⋅(3×1089.8)⋅ln⁡(1+9.8×1.8922×1017(3×108)2)T \approx \frac{1.8922 \times 10^{17}}{3 \times 10^8} \cdot \left( \frac{3 \times 10^8}{9.8} \right) \cdot \ln\left( 1 + \frac{9.8 \times 1.8922 \times 10^{17}}{(3 \times 10^8)^2} \right)T≈3×1081.8922×1017⋅(9.83×108)⋅ln(1+(3×108)29.8×1.8922×1017)
First, calculate the terms:
1.8922×10173×108≈6.307×108 seconds\frac{1.8922 \times 10^{17}}{3 \times 10^8} \approx 6.307 \times 10^8 \, \text{seconds}3×1081.8922×1017≈6.307×108seconds3×1089.8≈3.06×107 seconds\frac{3 \times 10^8}{9.8} \approx 3.06 \times 10^7 \, \text{seconds}9.83×108≈3.06×107seconds
Now, calculate the logarithmic term:
9.8×1.8922×1017(3×108)2≈2.04\frac{9.8 \times 1.8922 \times 10^{17}}{(3 \times 10^8)^2} \approx 2.04(3×108)29.8×1.8922×1017≈2.04
Thus:
ln⁡(1+2.04)≈ln⁡(3.04)≈1.11\ln(1 + 2.04) \approx \ln(3.04) \approx 1.11ln(1+2.04)≈ln(3.04)≈1.11
Now, multiply all the terms together:
T≈6.307×108×3.06×107×1.11≈2.1×1016 secondsT \approx 6.307 \times 10^8 \times 3.06 \times 10^7 \times 1.11 \approx 2.1 \times 10^{16} \, \text{seconds}T≈6.307×108×3.06×107×1.11≈2.1×1016seconds

Step 3: Convert Seconds to Years​

Convert the time from seconds to years:
T≈2.1×10163.155×107 years≈665 yearsT \approx \frac{2.1 \times 10^{16}}{3.155 \times 10^7} \, \text{years} \approx 665 \, \text{years}T≈3.155×1072.1×1016years≈665years

Conclusion:​

From Earth's frame of reference, the rocket would take approximately 665 years to travel to a destination 20 light-years away, accelerating at 1g1g1g for half the trip and decelerating at 1g1g1g for the second half.

 

[DaveC426913]

44.3 years:

To solve the problem of calculating how long it would take a rocket accelerating at 1g1g1g (approximately 9.8 m/s29.8 \, \text{m/s}^29.8m/s2) for half the trip and decelerating at 1g1g1g for the second half to reach a destination 20 light-years away, we need to consider relativistic effects, as the rocket will travel at high speeds approaching the speed of light.

Key Assumptions:​

• Distance to destination: D=20 light-yearsD = 20 \, \text{light-years}D=20light-years
• Acceleration: a=1g=9.8 m/s2a = 1g = 9.8 \, \text{m/s}^2a=1g=9.8m/s2
• The rocket accelerates at 1g1g1g for half the trip and decelerates at 1g1g1g for the second half.

Relevant Relativistic Formula:​

The total time in Earth's frame of reference for a journey under constant acceleration, where the rocket accelerates for half the distance and decelerates for the other half, is approximately given by the formula:

T=2Dc⋅ln⁡(1+aDc2)T = \frac{2D}{c} \cdot \ln\left( 1 + \frac{aD}{c^2} \right)T=c2D⋅ln(1+c2aD)
Where:

• DDD is the total distance to the destination,
• ccc is the speed of light (3×108 m/s3 \times 10^8 \, \text{m/s}3×108m/s),
• a=9.8 m/s2a = 9.8 \, \text{m/s}^2a=9.8m/s2 is the constant proper acceleration.

Step 1: Convert Distance to Meters​

We need to convert the distance from light-years to meters.

1 light-year=9.461×1015 m1 \, \text{light-year} = 9.461 \times 10^{15} \, \text{m}1light-year=9.461×1015m
Thus, the total distance D=20 light-yearsD = 20 \, \text{light-years}D=20light-years is:

D=20×9.461×1015 m=1.8922×1017 mD = 20 \times 9.461 \times 10^{15} \, \text{m} = 1.8922 \times 10^{17} \, \text{m}D=20×9.461×1015m=1.8922×1017m

Step 2: Substitute Values into the Formula​

Now we substitute the known values into the formula:

T=2×1.8922×10173×108⋅ln⁡(1+9.8×1.8922×1017(3×108)2)T = \frac{2 \times 1.8922 \times 10^{17}}{3 \times 10^8} \cdot \ln\left( 1 + \frac{9.8 \times 1.8922 \times 10^{17}}{(3 \times 10^8)^2} \right)T=3×1082×1.8922×1017⋅ln(1+(3×108)29.8×1.8922×1017)

Step 3: Simplify the Expression​

First, calculate the terms:

2×1.8922×10173×108=1.2615×109 seconds\frac{2 \times 1.8922 \times 10^{17}}{3 \times 10^8} = 1.2615 \times 10^9 \, \text{seconds}3×1082×1.8922×1017=1.2615×109seconds
Next, calculate the logarithmic term:

9.8×1.8922×1017(3×108)2=1.85×10199×1016=2.04\frac{9.8 \times 1.8922 \times 10^{17}}{(3 \times 10^8)^2} = \frac{1.85 \times 10^{19}}{9 \times 10^{16}} = 2.04(3×108)29.8×1.8922×1017=9×10161.85×1019=2.04
So:

ln⁡(1+2.04)=ln⁡(3.04)≈1.11\ln(1 + 2.04) = \ln(3.04) \approx 1.11ln(1+2.04)=ln(3.04)≈1.11
Now, multiply everything together:

T=1.2615×109×1.11≈1.4×109 secondsT = 1.2615 \times 10^9 \times 1.11 \approx 1.4 \times 10^9 \, \text{seconds}T=1.2615×109×1.11≈1.4×109seconds

Step 4: Convert Time to Years​

To convert seconds to years, divide by the number of seconds in a year (3.155×107 seconds/year3.155 \times 10^7 \, \text{seconds/year}3.155×107seconds/year):

T=1.4×1093.155×107 years≈44.3 yearsT = \frac{1.4 \times 10^9}{3.155 \times 10^7} \, \text{years} \approx 44.3 \, \text{years}T=3.155×1071.4×109years≈44.3years

Final Answer:​

From Earth's frame of reference, it would take approximately 44.3 years for the rocket to reach a destination 20 light-years away, assuming it accelerates at 1g1g1g for half the trip and decelerates at 1g1g1g for the second half.

 

[exchemist]

There has been some discussion of the limitations of LLMs on the .net forum. The view seems to be that as they are language models they cannot reason and will make stuff up, concocted from the sources they have scanned. Your exercise seems to illustrate this. That forum now forbids people to use results or passages of text from LLMs in support of arguments or claims.

It certainly seems they are worthless for science discussions, except perhaps as a form of advanced search engine, if one makes use of the primary sources that some of them cite in their responses.

Nevertheless, we can expect -indeed we already see - the effects of some people outsourcing their thinking to a chatbot.

 

[Seattle]

How is this?

What factors would you need to know to figure out how long it would take a rocket to launch from Earth and land on a planet 20 light years away from Earth?

ChatGPT said:​

To determine how long it would take a rocket to travel from Earth to a planet 20 light-years away, you would need to consider several key factors:

1. Rocket's Maximum Speed – The speed of the spacecraft relative to Earth, typically expressed as a fraction of the speed of light (e.g., 0.1c or 10% of the speed of light).
2. Acceleration and Deceleration – If the spacecraft cannot instantly reach its cruising speed, you must account for the time it takes to accelerate to maximum speed and then decelerate upon arrival.
3. Propulsion System – The type of propulsion determines the achievable speed. Chemical rockets are too slow, while nuclear, ion, or hypothetical fusion or antimatter propulsion systems could allow for relativistic speeds.
4. Relativistic Time Dilation – At speeds approaching a significant fraction of the speed of light, time dilation (from special relativity) means that time onboard the spacecraft would be much shorter than time experienced on Earth.
5. Interstellar Medium and Resistance – The spacecraft may encounter dust, gas, or other obstacles in interstellar space, which could affect its speed or require shielding.
6. Gravity Assists or Other Speed Enhancements – The use of gravity assists from massive objects (e.g., the Sun or nearby planets) could help increase velocity.
7. Life Support and Fuel Constraints – If a crew is onboard, life support duration and the feasibility of carrying enough fuel for acceleration and deceleration must be considered.
8. Course Corrections and Navigation – The need for mid-course corrections due to gravitational influences or unexpected obstacles.

Rough Estimate​

• If the rocket traveled at 0.1c (10% of the speed of light), it would take 200 years to reach the destination (ignoring acceleration and deceleration time).
• At 0.5c (50% of the speed of light), the trip would take 40 years.
• At 0.9c (90% of the speed of light), it would take about 22 years (accounting for relativistic effects, onboard time would be much shorter).

Would you like a specific calculation based on a chosen speed and acceleration model?

4o

 

[Sarkus]

I can't parse the calculations to see what it's doing wrong, but it seems to suggest it understands the assignment... and then goes mad. Interesting stuff, though.

FYI, the correct answer is the first in your initial list. It's about 21.85 years.

 

[Seattle]

I can't parse the calculations to see what it's doing wrong, but it seems to suggest it understands the assignment... and then goes mad. Interesting stuff, though.

FYI, the correct answer is the first in your initial list. It's about 21.85 years.

The deal with the LLM's, in general, is that you need to know the ballpark answer as a check and if the answers are "odd" then you need to improve the question a bit more with more specificity (and realize that math isn't their strong suit).

With Dave's examples you can mention that you've been given multiple differing answers and ask what's going on? Or you could be more specific with more details in the initial question or just ask what all the factors are and then be more specific with those.

If you ask a few follow up questions you generally get a pretty good answer with a lot less legwork. Look at it as a tool that you need to learn how to use rather than as an infallible "genius".

It's not accurate, IMO, to conclude that LLM's are just "gibbering idiots".

 

[TheVat]

The fact that it would give some answers less than 20 years does suggest a major confusion between the Earth FoR and the ship's (where Lorentz factors are applicable). A confusion that can only happen with an AI that understands nothing. It cannot grasp that it is the ship's reference frame that is non-inertial.

 

[Seattle]

If the confusion or question or point of discussion is whether LLM's are truly "AI", they aren't. They are light years (pun intended) ahead of a Google search.

If the implication is that LLM's aren't useful. Obviously, they are.

 

[DaveC426913]

The fact that it would give some answers less than 20 years does suggest a major confusion between the Earth FoR and the ship's (where Lorentz factors are applicable). A confusion that can only happen with an AI that understands nothing. It cannot grasp that it is the ship's reference frame that is non-inertial.

That was what made me suss. The first answer was a plausible ~22 years. But when I changed the g-accel, it gave me a wildly different answer of, like, 4 years.

I actually started this post to be more about "Wow! ChatGPT has invented FTL!!" before I realized got much more silly the farther down the rabbit hole I went.

 

[DaveC426913]

... it seems to suggest it understands the assignment... and then goes mad.

Indeed. Recall the post in kingiyk's thread where I asked it to arrange a darts tourney.
1. It explicitly told me, spec-by-spec, it understood exactly what I wanted.
2. It explicitly gave me a result that was exactly not what I wanted.
3. It summarized by lying to me, explicitly, spec-by-spec, that its results exactly matched my requirements.
4. No matter how many times I corrected it, it did steps 1-3 anyway, explicitly lying each and every time.

FYI, the correct answer is the first in your initial list. It's about 21.85 years.

Yeah. I would never have questioned it if I hadn't rerun the same query by happenstance.

 

[ольга]

The first answer looked a little suss, so I asked again. And then I kept asking...



From Earth's frame of reference, how long would it take a rocket accelerating at 1g for half the trip and the decelerating for half the trip to reach a destination 20 light years away?


Every time I ask it this same question I get different results (I am literally copy-pasting the question):


1. Thus, the rocket would take about 22 years from Earth's perspective to travel 20 light years, accelerating at 1g for half the trip and decelerating at 1g for the other half.

2. In Earth's frame of reference, the rocket would take approximately 3.4 years to travel 20 light years,...

3. From Earth's frame of reference, the rocket would take approximately 4.4 years to reach a destination 20 light-years away...

4. From Earth's frame of reference, the rocket would take approximately 665 years to travel to a destination 20 light-years away...

5. From Earth's frame of reference, it would take approximately 40 years ...

6. ...it would take approximately 44.3 years...

1.Какова масса вашего корабля?
2. Может ли ваш корабль быть безмассовым?
3. При каких условиях ваш корабль может стать безмассовым, если изначально он таким не был?
4. Почему у безмассовых объектов есть ограничение по скорости?

 

[Sarkus]

The fact that it would give some answers less than 20 years does suggest a major confusion between the Earth FoR and the ship's (where Lorentz factors are applicable). A confusion that can only happen with an AI that understands nothing. It cannot grasp that it is the ship's reference frame that is non-inertial.

The time experienced by passengers would be c.6.04 years, if my calcs are correct. So even those answers by ChatGPT seem to be inaccurate, if one assumes that is what it is doing.

 

[exchemist]

If the confusion or question or point of discussion is whether LLM's are truly "AI", they aren't. They are light years (pun intended) ahead of a Google search.

If the implication is that LLM's aren't useful. Obviously, they are.

Do LLMs always give you the references they have consulted? If so, that would save time which would certainly be valuable. (I've never used one and don't intend to, hence my question.)

 

[Sarkus]

Okay, so I asked ChatGPT the same question as in DaveC's OP.

It gave me the answer of 1.04 years - clearly incorrect. But it also gave a breakdown of the formulae it was using, and came up with:
x(t) = (c^2/a) (cosh (at/c) -1)
where x(t) is the distance at a given time t, and a is the acceleration, and c is the speed of light.

This is not the correct formula for the question, though, which should be:
x(t) = (c/a) (sinh (at/c)

I pointed this out to ChatGPT, and it agreed that the one it initially used was not the right formula. I asked it why it made this mistake, and it said it incorrectly referenced a more complex formula derived from non-constant acceleration, even though the OP's enquiry stated "accelerating at 1g for half the trip" etc. So is it failing to see that as referencing "constant acceleration"?

Or maybe it looks up the initial idea of what it is being asked, then sees a number of formulae in all the places it looks, and tries to identify the one that looks most correct. I mean, it's not that far off.

I then asked it to give me the relativistic rocket equation for an object experiencing "constant proper acceleration", and it did then at least give me the correct formula.

 

[exchemist]

That was what made me suss. The first answer was a plausible ~22 years. But when I changed the g-accel, it gave me a wildly different answer of, like, 4 years.

I actually started this post to be more about "Wow! ChatGPT has invented FTL!!" before I realized got much more silly the farther down the rabbit hole I went.

I have another example, this time from Copilot, to which Write4U had outsourced his mind: https://www.sciforums.com/threads/write4us-wobbly-world-of-word-salad-woo.166165/page-22 See post 439.

Copilot claimed photons were not physical entities, based on a load of references in none of which such an idea was suggested. Copilot made it up. And it’s wrong of course. But Write4U goofily trusted it to be correct.

 

[Seattle]

Do LLMs always give you the references they have consulted? If so, that would save time which would certainly be valuable. (I've never used one and don't intend to, hence my question.)

It depends on the model. If they use real-time web info they generally do. If it's a model that doesn't use real-time info and is just based on their training, they don't.

I remember a guy (long ago) who didn't use email or computers because that was what his secretary was for.

 

[exchemist]

It depends on the model. If they use real-time web info they generally do. If it's a model that doesn't use real-time info and is just based on their training, they don't.

I remember a guy (long ago) who didn't use email or computers because that was what his secretary was for.

Yes I once had a boss like that, an American.

But I have yet to see what use a chatbot is, for somebody like me. We’ve established you can’t rely on the veracity of its output. There are plenty of things I don’t use, such a TV, or a virtual reality headset. What do you use chatbots for?

 

[DaveC426913]

What do you use chatbots for?

It makes a pretty good search engine. I've used it when I want to know a very specific detail about some app or device that would otherwise be buried ten pages deep in documentation. "How do I paste text without the formatting in a Mac?", or "How do I rescan my QX-47 Explosive Space Modulator?"

The other day I actually used it to help me write a couple of lines of code: "Attach a listener to the Submit button that returns false if the length of 'student_id' field is not exactly 9 digits".

It still doesn't always get those right, but it's a good jumping off point. In these cases, the consequence of it getting the answer wrong are negligible, since I'm about to verify them myself anyway.

Finally, I found my my "killer app" - the one I can't live without. While not specifically a chatbot, it is still AI-powered: Content-Aware fill in Photoshop is now indispensible for me to alter the backgrounds of photos. This saves me oodles of time.

 

[exchemist]

It makes a pretty good search engine. I've used it when I want to know a very specific detail about some app or device that would otherwise be buried ten pages deep in documentation. "How do I paste text without the formatting in a Mac?", or "How do I rescan my QX-47 Explosive Space Modulator?"

The other day I actually used it to help me write a couple of lines of code: "Attach a listener to the Submit button that returns false if the length of 'student_id' field is not exactly 9 digits".

It still doesn't always get those right, but it's a good jumping off point. In these cases, the consequence of it getting the answer wrong are negligible, since I'm about to verify them myself anyway.

Finally, I found my my "killer app" - the one I can't live without. While not specifically a chatbot, it is still AI-powered: Content-Aware fill in Photoshop is now indispensible for me to alter the backgrounds of photos. This saves me oodles of time.

Seems fair enough, though these are all IT examples. Does it give you the references, so you can check it is not talking crap?