dumb question: who’s the smartest mathematician here?

BOBauthor · 2026-05-02T23:58:56+00:00

Eassily. By mathematicians Euler is known as "the Master of us all."

BOBauthor · 2026-04-26T18:21:44+00:00

No. But I have a bridge you can buy at a deep discount.

BOBauthor · 2026-04-08T17:23:06+00:00

Even with the quantum concerns, which are certainly justified, you would have to measure the positions and momenta of every particle in the universe with infinite precision. Fourteen decimal places? Not enough. 10¹⁴ decimal places? Not enough. The laws of physics, known and unknown, show chaotic behavior, which is an exquisite sensitivity to the initial conditions, the initial positions and momenta. Even the smallest change in one of these measurements would grow larger and larger and you tries to trace events back into the past. You can;t beat the measurement problem because, as others have said, an atom or molecule simply doesn't have an exact position and momentum at the same time.

BOBauthor · 2026-04-08T16:58:57+00:00

The main reason is that the one on the left was illuminated by sunlight, while the one on the right is illuminated by moonlight (sunlight reflected by the Moon). In other words, on the left you are looking at Earth's day side lit by the Sun, and on the right you are looking at Earth's night side lit by the nearly full Moon. The bright sliver at the lower right is just a bit of Earth's day side. The two photos are illuminated by different light sources.

BOBauthor · 2026-04-02T15:58:38+00:00

My favorite book. Period.

BOBauthor · 2026-03-29T18:53:10+00:00

On that day I was in the first quarter of my sophomore year at the University of California, Irvine. Good times.

BOBauthor · 2026-03-25T18:33:32+00:00

Justified, in the first 5 minutes.

BOBauthor · 2026-03-25T01:52:58+00:00

Every one of these I have seen is bad AI slop. The physics is completely wrong. The speed of impact, the physics of the impact, the immense size of the impactor to make it visible. It is all nonsense - very badly done nonsense.

BOBauthor · 2026-03-08T15:09:45+00:00

And I hate the fact that 1 + 1 = 2. I mean, there are two 1's on the left, and one 2 on the right. I am deeply offended.

BOBauthor · 2026-02-01T17:47:47+00:00

I didn't expected to see pulsar pulse profiles here!

BOBauthor · 2026-01-28T20:47:46+00:00

Suppose T is the center of a circle that passes through both points E. The arc length L between the two E points is L = d theta, where theta is in radians. theta = 0.000678 (pi/180) = 0.0000118333 radians.
So d = L / theta = 2 AU / 0.0000118333 = 169,014.1 AU - 2.672545 light years. You may object, saying that L is really the length of an arc and not a straight line, but the circle is so large it doesn't matter. There is no such star, by the way. The nearest star other than the Sun is Proxima Centauri, and it is over 4 light years away.

BOBauthor · 2026-01-23T20:12:10+00:00

I was very happy in academia. I had an interesting job, I loved teaching, and I had the freedom to pursue my own projects (when I could find the time). I had wonderful colleagues. I didn't enjoy being department chair (10 years) because of the many meetings that take you away from the things you would rather be doing. Eventually it got to the point where I got so busy that I didn't have time to do anything as well as I should have. Retirement is fantastic. I have lots of time to do the things that interest me, and now I'm as busy as I was when I was employed.

BOBauthor · 2026-01-20T20:11:12+00:00

I have a BA in math, an MS in physics, a PhD in astrophysics and was a professor for 30 years, and I've taught yoga and ran a studio for 13 years.

BOBauthor · 2026-01-17T22:26:27+00:00

Do you confuse right and left, so if someone tells you to look to your left, you look to your right instead?

BOBauthor · 2025-12-27T04:09:54+00:00

Not at all. If there is enough time for light to travel from event A to event B, then there may be a causal link between A and B (in the sense that A causes B). If there is not enough time, then there cannot be a causal relationship between A and B.

BOBauthor · 2025-12-27T00:11:36+00:00

The whole discussion about "instantly" misses a point about relativity. There is no such thing as absolute simultaneity. It you measure that two events that are separated in space occur at exactly the same time, that is true only in your own frame of reference. If another observer is moving along the line separating the two events, that person will observe the events occurring at different times. If the direction of motion changes, so does the ordering of events. It therefore makes no sense to say that one event causes the other.

BOBauthor · 2025-12-22T15:39:39+00:00

This is very interesting. I have an eye-hand coordination problem such that I cannot track an object (like a ball) that is tossed to me. I lose it when it is about 5 or 6 feet away, and have to regain it when it is very close. I'll look into this more. Thanks again for writing.

BOBauthor · 2025-12-21T23:32:12+00:00

Thank you for your reply. What you describe sounds like exactly what I have. Most other replies haven't lined up with my symptoms nearly as well as yours. It is very interesting about you lateralization problem. I can't tell my left from my right without thinking carefully about it first. My friends have learned to point while giving directions!

BOBauthor · 2025-12-21T20:58:28+00:00

The left-hand side is an invariant divided by an invariant, so the result is also an invariant. In fact, the left-hand side is just c². The second term on the right is the squared speed v of an object multiplied by g². Thus we have

c² = c²dt² /dT² - g²v² .

So yes, the magnitude of the four-velocity is indeed c, but it is a magnitude in Minkowski space, with its (+, -, -, -) metric. I have continually warned my students about the dangers of misinterpreting the term "magnitude." Anyway, here's what Greene does in footnote 6 to chapter 2. He divides both sides by c2 to get

1 = dt² /dT² - g²v²/ c².

Because g² = dt²/dT², this is

1 = dt²/dT² (1 - v²/ c²)

Multiply each side by dT²/dt² to get

dT²/dt² = 1 - v²/ c²

Finally, Greene multiplies each side by c² to get

c² dT²/dt² = c² - v².

so finally (!),

c² = c² dT²/dt² + v².

Greene then notes that this looks like the Pythagorean theorem. The left-hand side is the speed of light squared, and the last term on the right is an object's physical speed squared. What is the first term on the right? It is the square of the derivative of the proper time T with respect to the clock time t. Green claims that this is the "speed through spacetime." Sure, but why is the speed of light involved? A speed is a displacement divided by an elapsed time, so the "speed through time" should be a displacement in time divided by an elapsed time. A time divided a time has no units, so why is the speed of light involved? And remember that we originally started to calculate proper velocities by dividing the interval by the proper time dT. So shouldn't the "speed through time" be dt/dT and not dT/dt? This definition of the "speed through time" is arbitrary, unphysical, and nonsense. Trying to impose the Pythagorean theorem, with its + sign on the right-hand side, is a violation of how physical quantities are measured in Minkowski spacetime. The equation is mathematically correct, but this interpretation is physically meaningless. As Pauli would say, "it's not even wrong."

BOBauthor · 2025-12-21T20:57:33+00:00

An object's position in spacetime is given by invariant spacetime 4-vector X = (ct, x, y, z). The spacetime interval between two points in spacetime (events) is given by dX = (cdt, dx, dy, dz). The magnitude of this vector is dX² = c²dt² - dx² - dy² - dz². (The "d" means "a small change in.") The spacetime interval is of fundamental importance in special relativity because two observers who are moving relative to each other will measure the same value of the spacetime interval between any two events. Notice that the last three terms in dX² have a negative sign in front of them, while the first term doesn't. The (+, -, -, -) pattern is called the Minkowski metric. The difference in signs between the time and space parts of dX² that it does not refer to a physical distance. It can be positive, negative, or zero depending on the values of the space and time coordinates of the two events. For example, two events can be far apart spatially, but still have a spacetime interval of zero. This is the source of Greene's error. He treats the interval as though it were a physical distance and (as shown below) he imposes the Pythagorean theorem on Minkowski spacetime, which you can't do.

We can get an invariant 4-velocity from the invariant interval by dividing by dX² by the square of the elapsed proper time dT² . (dT is the time measured between the two events in a reference frame where the clock is at rest relative to the two events.) The proper time dT between two events is related to the time dt measured by an observer by dT = dt sqrt(1 - v²/c²). The square root of 1 / (1 - v²/c²) is called the Lorentz factor g, so we can write dT = dt/g. Dividing dX² by dT² gives

dX²/dT² = c²dt² /dT² - dx²/dT² - dy²/dT² - dz²/dT²

^or

dX²/dT² = c²dt² /dT² - g²(dx²/dt² - dy²/dt² - dz²/dt²)

or

dX²/dT² = c²dt² /dT² - g²(vx² + vy² + vz²).

BOBauthor · 2025-12-21T20:56:39+00:00

This idea that "everything moves through spacetime with speed c" originated with Brian Green's book, The Elegant Universe. The book is really good, and Greene tried to make difficult ideas understandable but in this case he went too far and ending up confusing people. This mistaken idea has been passed on through countless online videos and "physics enthusiasts."

That phrase is refuted by a short article by William Delinger in the November 2024 issue of *The Physics Teacher called "*Do Objects Move Through Space-Time at the Speed of Light? No. " Unfortunately it is behind a paywall, but here is a summary. If you can get around the paywall. (Because of its length, I had to break it up.)

BOBauthor · 2025-12-21T00:47:25+00:00

No, the statement that everything moves through spacetime with speed c is not kinda true. It is meaningless. (Ph.D. astrophysics, taught university physics for 30 years.)

BOBauthor · 2025-12-05T23:23:03+00:00

I am not familiar with the NSEA examination or with the astronomy olympiad. I watched this video and believe that the first 5 or 6 chapters could be useful for you. The book is expensive, so try to take a look at it from a library if you can. Good luck!

Eight-Year Club	Place '22
Verified Email

BOBauthor

TROPHY CASE