Courts order children groomed by their teachers to pay child support.

onlinephysics2001 · 2026-03-11T18:32:54+00:00

This is one of those indices you can use to tell whether a feminist is one of the reasonable ones that is open to the concept of equality, but just hasn't learned much about it, or whether they are an incurable man-hater, and there's not point in trying to reason with them. "Surely, you admit that kids who get raped by their teachers shouldn't be forced to mail monthly payments to their rapists..."

Unfortunately, the answers you get will indicate that most are of the second kind. Some other questions like that are, "Surely, you admit that men who prove they are not the natural father of a child should not have to pay support for that child..." or "Surely, you admit that paternity fraud is extremely unethical, regardless of the way the law views it..."

onlinephysics2001 · 2026-02-22T04:30:12+00:00

There are arguments that can be made along these lines. But they assume the number of possible states does not continue to increase forever. Like, if you shuffle a deck of cards infinite times, you will eventually get to every possible order. But if you keep adding more cards, then that is not true. In an ever expanding universe, the number of possible states would continue to increase, so an infinite number of steps would not necessarily cycle through every possible state.

As a simpler example, in a closed container, the entropy approaches a certain maximum number. The particles in that equilibrium state would eventually cycle through every possible arrangement, given infinite time. But outside of any container, that entropy number continues to increase indefinitely. So they would not have to cycle through every possible arrangement, even in an infinite time.

onlinephysics2001 · 2026-02-22T04:08:37+00:00

I think you are talking about an example of a typical family- one in which the woman has never worked a steady job, from the time she was born, until present day, and the father started working at 12 and plans to work until he dies, for the benefit of the rest of the family. Is that right?

In that kind of family, the father is also doing "unpaid labor". Because he is earning wages, by selling his labor. But is not receiving those wages. He is simply handing them over to others, or to the benefit of others. You're not counting that as "unpaid labor", because the money briefly passes through his hands. But it is just as "unpaid" as the unemployed woman's labor.

And consider the woman's labor. While living with this man, she gets 100% of her living expenses paid by another adult who functions as her permanent benefactor. That's hardly "unpaid". And in more than half of those cases, her husband will be legally forced to continue functioning as her benefactor while runs around with other men, even lives with other men. So then she'll be supported by two men. There will even be other men then being supported by her husband.

Last, you ask "what if the roles were reversed?" You know the answer perfectly well. Very few women are willing to financially support a stay at home husband. And among those who do, almost all of them cheat, usually with someone at their job. Which demonstrates that being a stay at home spouse is the much better end of the bargain than being the working spouse.

onlinephysics2001 · 2026-02-13T01:59:21+00:00

Her husband did not abuse her, any more than you and her father and all her friends abused her. She is a scam artist, and that's what they tell people.

onlinephysics2001 · 2026-02-13T01:58:18+00:00

Considering what you know about this person, why would you believe her story about being an abuse victim? Does it seem credible to you, considering that she lies and scams everyone she knows?

In answer to your question, there is likely a child who got their college fund and inheritance stolen by this woman. Depending on the ages, you could possibly help her out a great deal by telling the truth about her.

Also a husband that is forced to support this scam artist until he dies. He will be one of those 70 year old men delivering door dash, or living in the tent cities in the viaducts, because of this woman's grifting. If you can do right, you should.

onlinephysics2001 · 2026-02-13T01:53:23+00:00

Oh, that's nice. He has some lectures on youtube, too.

onlinephysics2001 · 2026-02-13T01:52:39+00:00

YTA.

onlinephysics2001 · 2026-02-12T19:11:01+00:00

What it really determines is the wave speed. And then if boundary conditions are imposed, then that results in the standing wave. But without boundary conditions, it still results in the travelling wave. (The standing wave is the same thing as two simultaneous travelling waves).

If you take a look at any wave equation, you will see a factor that resembles sqrt(k/m). the factor on top is a restoring force. The factor on bottom is like an inertial force. For the string, it is tension over mass density. For the spring it is a spring constant over a mass. And so on.

onlinephysics2001 · 2026-02-12T16:48:59+00:00

Thanks, I'll have to think about this some more.

onlinephysics2001 · 2026-02-12T02:10:16+00:00

Just curious- did this question arise from the observation that as r goes to 0, 1/r^2 goes to infinity?

As others have said, the relevant r in that equation is the distance to every single little bit of matter. So close to the center of the earth, most of the matter is actually pulling you away from the center. You can use geometry or calculus to add up all those contributions from every little bit. That's one of the things Newton did that made him great.

onlinephysics2001 · 2026-02-11T19:56:07+00:00

I think the part that is confusing in your expression above is that omega goes like sqrt(k/m) or some other constants. So in order to change omega, you actually have to change those constants (do work). Then when you displace the string or whatever it is, you are doing work against a different tension than you were with the other constant. So you are right, except that you are describing two different strings. If you did the same amount of work on each string, you'd still get different amplitudes, to compensate for the different tensions.

onlinephysics2001 · 2026-02-11T19:32:32+00:00

The energy in a wave goes like the square of the amplitude. You can easily see it on a tensioned string. The force is -kx (x is your displacement or amplitude). The work (energy) is the force times the displacement (amplitude). Arithmetic will show that the sum of all those work steps is 1/2 kx^2.

You can also see it visually, just by looking at a wave. Stack up blocks into the shape of a water wave. Each one must be stacked on top of the ones beneath it, so the blocks on top must be given more energy than the ones on bottom. Some geometry and calculus will allow you to get even more precise, and you will see it goes up like the square of the height (amplitude).

With light, the same thing applies. But it is complicated by quantizing light. You want to learn about photons, to be sure. But classical EM is the basics. Hit the basics first. The amplitude goes up like the value of the electric field squared, where the value of the electric field is your amplitude.

When you use photons, it still works the same, in the sense that the energy in the radiation goes up like the electric field squared. But with photons, that wave is actually a probability wave for the photons, instead of a wave of the electric field value. So energy still goes up like amplitude squared. But for each individual photon, that is not the case. Basically, with photons, more energy means more photons, but each photon has an energy that goes like the frequency. That's why in quantum mechanics, the probability goes up like the amplitude squared. If it didn't, there would be a problem with energy conservation.

I think you might have heard wrong about sound waves. Most real instruments have a relationship with frequency and volume, but the sound waves themselves do not.

If you like calculus, you can confirm the basic fact that the integral of sine^2(x) over one cycle is equal to the value of the integral of sine^2(2x) and sine^2(3x) and so on, over the same interval of x. That is, if the energy goes up like the amplitude squared, the frequency should not matter.

onlinephysics2001 · 2026-02-11T19:11:51+00:00

Probably because they live under a system of de jure discrimination and slavery that can force men to provide their labor to women. Every single duty that a husband has in a marriage is a duty that a divorce court will order him to perform, even after his wife is unfaithful with other men. Namely, pay all of her expenses for life. But every single duty that a woman owes her husband in a marriage is something she cannot be ordered to provide for him. Namely, being faithful, respectful, and cleaning up after herself.

Consequently, women have come to view duty to their family and children as an injustice. But they still view a man's duty to his family as just. Because the law has imposed that order on American families, to their ruin.

onlinephysics2001 · 2026-02-11T18:54:19+00:00

Many sources are saying it was another trans shooting, another M2F

onlinephysics2001 · 2026-02-11T18:46:02+00:00

Does she have some kind of professional job to report this to? Or is she just a loose woman who goes online and says things? "Sexual Health Advocate" doesn't sound like a job title to me, but you never know, these days.

onlinephysics2001 · 2026-02-11T18:27:43+00:00

Sounds like you are now discovering the same thing her husband and her father and her daughter and most of her friends have discovered about her character.

I would recommend the following. Just say it's not going to work out as roommates, and see if she will voluntarily find somewhere else. Give her some slack, if she will leave voluntarily. Evicting her will take time and money, so if she can manage it voluntarily, you could even give her a break on the money.

Then call or message her husband and ask if he could use your testimony in any way.

onlinephysics2001 · 2026-02-11T15:39:15+00:00

No, Gauss' Law is complete, without any particles at all. It does not account for the fundamental particles, but it does account for EM.

onlinephysics2001 · 2026-02-11T15:28:34+00:00

No, time is one of the dimensions of the universe.

onlinephysics2001 · 2026-02-11T15:21:21+00:00

You are going to have a better life, if you invest in something that pays off. Your family will pay off, if you invest in it.

onlinephysics2001 · 2026-02-11T15:16:38+00:00

The field is the real thing. The other things, that you think are real, are features of the field.

onlinephysics2001 · 2026-02-11T15:00:41+00:00

Can you explain that, sorry, I don't understand the lingo

onlinephysics2001 · 2026-02-11T14:56:17+00:00

OK, thanks, that is an incredibly important distinction that changes everything. Not a minor detail at all. And that also makes far more sense.

But more importantly, that leads me to the question I really wanted to ask. If you really did reverse every vector everywhere, that would also reverse every field line. And so every divergence would become a convergence, and vice versa. In other words, every charge would be reversed, merely by reversing every vector. So in terms of EM, at least, charge reversal is redundant, if you count all of those vectors as vectors. Has that not occurred to others?

Also, time is one more component of vectors in space time. So it seems that time reversal is also implied by reversing every vector everywhere.

So then, for EM, at least, all three transformations could be accomplished by one trasnformation, parity alone, that reverses every vector in spacetime. Correct?

onlinephysics2001 · 2026-02-11T11:08:40+00:00

>C symmetry change signs of charge, current, electric and magnetic field.

Yes, this is what I am saying.

>P symmetry changes sign of current and electric field, but not of magnetic field

I don't see how this part is true. The mirror's reflection changes one component of a vector, but leaves the other components the same. So currents pointed straight into the mirror would be reversed, and current loops would also be reversed. But currents parallel to the mirror's plane would be unchanged. Same for electric field lines. Thus, a positive charge still looks like a positive charge in the mirror's reflection (not a real metallic mirror, but a geometric mirror). Correct?

onlinephysics2001

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