DaveC426913
Valued Senior Member
What does any of this have to do with ChatBots?
This went off the rails at post 34.
This went off the rails at post 34.
ChatGPT is a gibbering idiot (this is not news)
- Thread starter DaveC426913
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DaveC426913
Valued Senior Member
Sure. But now it's becoming a physics discussion. It's hard enough to Write4U to stay on-topic as it is.
Anyway, carry on. It was a one-off thread topic anyway.
exchemist
Valued Senior Member
I know. It was just that Write4U had contrived to forget why Copilot was wrong on that occasion, so I reminded him. There seems no doubt that Copilot invented the wrong answer, as it was not to be found in any of the references it cited.
My only doubt, or suspicion, now is whether it did that because of the way Write4U asked the question of it. It could be an example of a chatbot telling the client what it thinks he wants to hear.
DaveC426913
Valued Senior Member
I wonder what Copliot thinks of Write4U spamming it with a dozen "you might find this interesting" articles every day.
Do you think Copilot is capable of rage-quitting?
exchemist
Valued Senior Member
Haha, I wonder. But actually what I meant was if you ask one of these chatbots a leading question, the bot may pick up the hint and tailor its answer to support the insinuation in the question.
DaveC426913
Valued Senior Member
Pinball1970
Valued Senior Member
Co-pilot has told me I am wrong on some a lot of stuff, politely. "Not quite" or simply "no" and as you pointed out references are usually given.
Where it has been wrong (I test it) is on maths problems and last week Colorectal polyp statistics.
Write4U
Valued Senior Member
I also asked if a wave is physical object.
Copilot:
1. https://www.askamathematician.com/2017/05/q-if-light-is-a-wave-then-whats-doing-the-waving/
2. https://www.physicsclassroom.com/class/waves/Lesson-1/What-is-a-Wave
Question: Is the curve in a line a physical object?
But even if a wave is not a physical object, can it have a "value" ?
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exchemist
Valued Senior Member
It’s the word “object” that is problematic here. A wave is most certainly a physical entity, but calling it an object presents problems when it is extended over an indeterminate distance.
But this is a distraction from the thread topic so let’s not pursue it further here.
I use ChatGPT as a guide - to kickstart some marketing ideas I might have at work. It does make mistakes; there's even a disclaimer to that effect when you use the free version. But, if you use it when you're having writer's block or you just need potential insights on a specific subject, it can be really helpful. Depending on the topic (for now), it might seem like a 'gibbering idiot,' but I suspect it will continue to grow in popularity, because of the speed with which you can find answers. If you don't already know the subject matter well, however, you might not know if ChatGPT is accurate or not and it's a little scary if people are using it as a fountain of knowledge. I don't think it "should" be used like that.
DaveC426913
Valued Senior Member
Precisely.
To clarify my stance: It has its uses. It is useful as a prompt. But it cannot answer questions that you can't just verify for yourself some other way. That's why some people call it a "glorified search engine". (I mean it will be happy to pretend it can, but you have no way of knowing it's got it right).
But it is being egregiously misused, because most people don't understand it. If it sounds like a human, they assume it thinks like a human, and that they can treat it like a human as far as its "knowledge" goes. This is extremely dangerous and will only get moreso as it is adopted. People will die.
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Write4U
Valued Senior Member
It depends on the subject.
For instance, in the case of law and precedence, a lawyer with access to AI will always have an advantage over a lawyer doing research manually.
Copilot:
DaveC426913
Valued Senior Member
That is what I said: an AI can't create an answer that you couldn't have gotten yourself by other means, but it can look up an answer faster. That's the advantage it provides. That's still just a glorified search engine.
Niccodeamus
Registered Member
I copied and pasted the question into ChatGPT, model o3
This is the entire output after its ‘thinking time’ statement which is clickable and shows what it did
Result
How to get that number
Rule-of-thumb check
A handy approximation for 1 g “flip” trips is
T_{\text{Earth}}\![\text{yr}] \approx \text{distance [ly]} + 1\; \text{yr} .
For 20 ly that gives ≈ 21 yr, very close to the exact 21.9 yr we found—good confirmation the calculation is on track.
Caveats
That’s the trip in a nutshell: about 22 years pass on Earth, while the crew ages only 6 years.
This is the entire output after its ‘thinking time’ statement which is clickable and shows what it did
Result
- Time seen from Earth (or the destination): ≈ 21.9 years
- Time experienced on board: ≈ 6.0 years
- Peak speed at the midpoint: ≈ 0.996 c (γ ≈ 11.3)
How to get that number
- Set up the half-trip.
With 1 g proper acceleration a=9.81\;\text{m s}^{-2} you cover half the distance (10 ly) while accelerating and the other half while decelerating. - Relativistic constant-acceleration equations (Earth’s inertial frame)
\begin{aligned} x(\tau) &= \frac{c^{2}}{a}\Bigl(\cosh\!\frac{a\tau}{c}-1\Bigr),\\ t(\tau) &= \frac{c}{a}\;\sinh\!\frac{a\tau}{c}, \end{aligned}
where- x = distance seen from Earth after the traveller’s proper time \tau,
- t = Earth’s clock time that has passed in the same interval.
- Solve for the proper time to the midpoint.
Half-distance x_{1}=10\;\text{ly}=9.461×10^{16}\,\text{m}.
\cosh\!\frac{a\tau_{1}}{c}=1+\frac{a\,x_{1}}{c^{2}} \quad\Longrightarrow\quad \tau_{1}=\frac{c}{a}\,\cosh^{-1}\!\Bigl(1+\frac{a\,x_{1}}{c^{2}}\Bigr)
Numerically, \tau_{1}\approx9.53×10^{7}\,\text{s}=3.02\;\text{yr}. - Earth’s time to the midpoint.
t_{1}=\frac{c}{a}\,\sinh\!\frac{a\tau_{1}}{c}\approx3.45×10^{8}\,\text{s}=10.93\;\text{yr}. - Full trip (accelerate + decelerate).
T_{\text{Earth}}=2t_{1}\approx6.90×10^{8}\,\text{s}=21.9\;\text{yr}.
\tau_{\text{ship}}=2\tau_{1}\approx1.91×10^{8}\,\text{s}=6.0\;\text{yr}. - Peak speed.
Rapidity at the midpoint \eta = a\tau_{1}/c ⇒
v=c\tanh\eta\approx0.996\,c .
Rule-of-thumb check
A handy approximation for 1 g “flip” trips is
T_{\text{Earth}}\![\text{yr}] \approx \text{distance [ly]} + 1\; \text{yr} .
For 20 ly that gives ≈ 21 yr, very close to the exact 21.9 yr we found—good confirmation the calculation is on track.
Caveats
- The destination must be at rest with respect to Earth; otherwise you must add any initial relative velocity.
- Continuous 1 g thrust for 22 years is far beyond our present engineering (fuel, shielding, life-support).
- General-relativistic effects from the Galaxy’s gravitational field are negligible compared with the special-relativistic ones here.
That’s the trip in a nutshell: about 22 years pass on Earth, while the crew ages only 6 years.
DaveC426913
Valued Senior Member
Great. Run it again. Do you get the same answer?
As I said: I got six different answers (for the trip in Earth's frame of reference):
22 years
3.4 years
4.4 years
665 years
40 years
44.3 years
The fact that there's a two-orders-of-magnitude spread in its guessing is amusing, but the fact that it appears to have discovered faster-than-light travel (making a 22ly journey in a mere 3.4 years) is laughable.
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Niccodeamus
Registered Member
Relativistic 1-g “half-accel / half-brake” trip
(distance = 20 ly in Earth’s rest frame)
Because the crew hold a constant proper acceleration a for the first half-distance and the same proper deceleration for the second half, the motion in the Earth frame obeys
x(t)=\frac{c^{2}}{a}\!\left(\sqrt{1+\left(\frac{a t}{c}\right)^2}-1\right),\qquad v(t)=\frac{a t}{\sqrt{1+\left(\frac{a t}{c}\right)^2}} .
1. Time to the midpoint (10 ly)
Set x=\frac{D}{2}:
\frac{D}{2}= \frac{c^{2}}{a}\left(\gamma_{\!}-1\right),\; \gamma_{\!}=1+\frac{aD}{2c^{2}}\;, \qquad t_{\frac{1}{2}} =\frac{c}{a}\,\sqrt{\gamma_{\!*}^{2}-1}.
Numerically,
t_{\frac{1}{2}} = 10.93\ \text{yr}.
2. Total Earth-frame flight time
t_{\text{Earth}} = 2\,t_{\frac{1}{2}} = 6.896\times10^{8}\,\text{s} \approx \boxed{21.85\ \text{yr}} \;(\text{about }21\text{ yr}\,10\text{ mo}\,6\text{ d}).
Extra insight (not asked but handy)
Why Newtonian maths fails here
If one naively kept a=g constant in Earth’s frame, the kinematic result would be only 8.8 yr—implying speeds well above light speed. Special relativity corrects this, stretching the Earth-frame duration to ≈ 21.9 yr while limiting v<c.
(distance = 20 ly in Earth’s rest frame)
| symbol | meaning | value |
| D | total distance | 20 ly |
| a | proper acceleration | g = 9.81\text{ m s}^{-2} |
| c | speed of light | 2.9979\times10^{8}\text{ m s}^{-1} |
x(t)=\frac{c^{2}}{a}\!\left(\sqrt{1+\left(\frac{a t}{c}\right)^2}-1\right),\qquad v(t)=\frac{a t}{\sqrt{1+\left(\frac{a t}{c}\right)^2}} .
1. Time to the midpoint (10 ly)
Set x=\frac{D}{2}:
\frac{D}{2}= \frac{c^{2}}{a}\left(\gamma_{\!}-1\right),\; \gamma_{\!}=1+\frac{aD}{2c^{2}}\;, \qquad t_{\frac{1}{2}} =\frac{c}{a}\,\sqrt{\gamma_{\!*}^{2}-1}.
Numerically,
- D/2 = 9.461\times10^{16}\,\text{m}
- \gamma_{\!*}=1+\frac{gD}{2c^{2}} = 1+11.2787 = 12.2787
- t_{\frac{1}{2}} =\frac{c}{g}\sqrt{\gamma_{\!*}^{2}-1}=3.4479\times10^{8}\,\text{s
t_{\frac{1}{2}} = 10.93\ \text{yr}.
2. Total Earth-frame flight time
t_{\text{Earth}} = 2\,t_{\frac{1}{2}} = 6.896\times10^{8}\,\text{s} \approx \boxed{21.85\ \text{yr}} \;(\text{about }21\text{ yr}\,10\text{ mo}\,6\text{ d}).
Extra insight (not asked but handy)
- Peak speed (at midpoint)
\beta_{\max}=\frac{at/c}{\sqrt{1+(at/c)^{2}}}=0.996,\quad v_{\max}\approx0.996\,c. - Time felt on board (integrating proper time)
\tau_{\text{crew}} = \frac{2c}{a}\sinh^{-1}\!\left(\frac{at_{\frac{1}{2}}}{c}\right) \approx 6.0\ \text{yr}.
Why Newtonian maths fails here
If one naively kept a=g constant in Earth’s frame, the kinematic result would be only 8.8 yr—implying speeds well above light speed. Special relativity corrects this, stretching the Earth-frame duration to ≈ 21.9 yr while limiting v<c.
Niccodeamus
Registered Member
How do I format equations in the chat to make them render properly. They’re really hard to read like this
To answer your question, it got the correct answer twice although clearly answered it completely differently, or in a different way
To answer your question, it got the correct answer twice although clearly answered it completely differently, or in a different way
DaveC426913
Valued Senior Member
I don't really dabble in LaTeX much. Every site seems to implement it differently. Not sure if SciFo supports it.
You need some sort of opening/closing tags to identify it. In this case, I'm trying '$' and '\begin' but obviously that's not correct.
$\beta_{\max}=\frac{at/c}{\sqrt{1+(at/c)^{2}}}=0.996,\quad v_{\max}\approx0.996\,c.$
\begin{align*}\beta_{\max}=\frac{at/c}{\sqrt{1+(at/c)^{2}}}=0.996,\quad v_{\max}\approx0.996\,c.\end{align*}
You could always screen grab the formulae directly from the chatbot and paste them. Here, I used https://www.quicklatex.com/ :
Twice is better than my track record but still not a high degree of confidence. See how many times it gets the correct answer before - or if - it fails.
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Niccodeamus
Registered Member
A lot will depend on what engine or model version you’re using. It’s improving very fast. Our lunches are in danger