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Gravity is not a force – free-fall parabolas are straight lines in spacetime

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Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#441

Post by CamperBob2 »

pdonis wrote: post #430
CamperBob2 wrote: post #414

Earlier quoted context omitted.

Dude, this is basic theory at the high school level. Here's how it works: spacetime is 4-dimensional, and a vector in spacetime is always c units long. You can restrict your travel solely to X, Y, Z, or time if you like. If you do that, the other three components are going to be zero. Photons put it all into the X, Y, and Z components, leaving nothing for the t component. They experience a change of position in space…

> Dude, this is basic theory at the high school level. No, Tyson's claim is not "basic theory at the high school level". It is a particular interpretation of a theory (Special Relativity) that does not work, for the reasons I gave. > Photons put it all into the X, Y, and Z components, leaving nothing for the t component. Wrong. The spacetime vector that describes a photon's trajectory does have a t component. > Seems…

Wrong. The spacetime vector that describes a photon's trajectory does have a t component.

Sounds interesting. Where can I read more about this t component?

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#442

Post by comboy »

ithkuil wrote: post #440
comboy wrote: post #434

Earlier quoted context omitted.

I think it does if they do not occupy the exact same space. It's only objective that they have the same velocity from their point of view (but then parallel lines concept hardly makes any sense).

are you talking special relativity or general?

Special.

I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view.

By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consisted just of 3 bodies, 1 being you and 2 being rest of the world, then the way of thinking you described still doesn't make sense in context of relativity and may lead to some confusion (apart from it being, to me, incoherent in context of special relativity).

But maybe I'm missing something from your picture, I'm happy to read and learn.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#443

Post by Dylan16807 »

comboy wrote: post #442
ithkuil wrote: post #440

Earlier quoted context omitted.

are you talking special relativity or general?

Special. I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view. By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consiste…

Special relativity definitely won't be a problem. Everyone see timings and speeds a little differently, but everyone can also do the calculation from anyone else's point of view. Objects that are in the same reference frame are objectively in the same reference frame in special relativity; everyone agrees. The reference frame is determined only by velocity. And the distortion is determined only by the observer's velocity, so both objects will have the same distortion.

GR I'm less familiar with.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#444

Post by CamperBob2 »

phizy wrote: post #432
CamperBob2 wrote: post #204

Earlier quoted context omitted.

I wouldn't say it's incorrect at all. From the point of view of a photon, no time elapses between its the origin and destination endpoints.

There is no "point of view of a photon." Photons do not have frames of reference. So you are completely incorrect insofar as what you are saying is physically nonsense.

Again, take it up with Tyson and others with doctoral-level credentials who've made a career out of explaining these subjects to the unwashed laity. I'm not one of those people. Nobody posting on Hacker News is, as far as can be discerned.

If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly? Or perhaps with the appropriate faculty committees at their institutions?

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#445

Post by pdonis »

lopmotr wrote: post #435
pdonis wrote: post #408

Earlier quoted context omitted.

> It still doesn't sound physically distinct If you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction. If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero. There is no continuum between tho…

> Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above. I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited…

> velocities don't add like that relativistically and I think you'd still see it as travelling at 1.0c

Not indefinitely. Sure, if you accelerated for a short enough time in the direction of the neutrino, you might still be within your measurement error and so not have learned anything. But that just means you need to accelerate for a longer time. For any given measurement accuracy, you will be able to calculate how long you need to accelerate, by your clock, to definitely distinguish the two cases. Relativistic velocity addition does not change that fact. All it changes is the details of that calculation; yes, for a given measurement accuracy, you need to accelerate for a longer time, by your clock, to definitely distinguish the cases than you would if velocity addition were linear. But that doesn't mean relativistic velocity addition makes it impossible to distinguish the cases at all, ever. It doesn't.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#446

Post by pdonis »

lopmotr wrote: post #436
ymolodtsov wrote: post #316

Earlier quoted context omitted.

With neutrinos we might find that it's not but it'd be impossible to catch a photon as it would always have the same speed of c in our reference frame. > I'd like to add that even photons have a nonzero upper bound to their possible rest mass. At least they used to. Not sure what you're talking about, their momentum? No object with mass can reach the speed of light and we know they're travelling at that exact speed.

How do we know they're travelling at exactly c? That's my concern. Last I heard, a couple of decades ago, physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.

> How do we know they're travelling at exactly c? That's my concern.

We don't, strictly speaking. The measurements you refer to aren't even measuring the speed of photons. They're measuring their rest mass.

> physicists would occasionally measure a new maximum possible rest-mass for photons. It would be very tiny, of course, but they couldn't say it's exactly zero.

Based on just those measurements, no. The most they can say is that the photon rest mass is zero to within some error bar, and the size of the error bar keeps getting smaller. (The current error bar, IIRC, is 10^-52 grams, or about 24 orders of magnitude smaller than the electron mass.)

However, we have a ton of indirect evidence that photons are massless; the most extensive body of such evidence is all the evidence for the gauge invariance of electromagnetism. If photons had a nonzero rest mass, that would break electromagnetic gauge invariance. So photons having a nonzero rest mass would be a huge issue for our current theories, in the way that neutrinos having a nonzero rest mass would not; there is no important symmetry coresponding to electromagnetic gauge invariance that is broken by neutrinos having a nonzero rest mass.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#447

Post by pdonis »

lopmotr wrote: post #435
pdonis wrote: post #408

Earlier quoted context omitted.

> It still doesn't sound physically distinct If you try what I described with a light ray, it will be moving away from you at c no matter how much you accelerate in its direction. If you try it with a massive object, even a neutrino with a very, very tiny invariant mass, that will not be the case; its speed relative to you will decrease as you accelerate after it, eventually to zero. There is no continuum between tho…

> Yes, you will know, because you will know if the neutrino's speed relative to you has decreased or not. If it has, it's possible to bring it to rest relative to you. If it hasn't, it's not. See above. I don't think this quite works because of relativistic addition of velocities. Naively, it seems to. For example, if the object was travelling at 0.999999c relative to you (appears to be 1.0c according to your limited…

> Maybe I've got my relativistic velocity addition wrong?

For the specific case you give, that depends on what accuracy you are assuming. The relativistic velocity addition would be:

v_new = (0.999999 - 0.5) / (1 - 0.999999 * 0.5) = 0.999997.

So if your accuracy is, say, 1 part in 100,000, you wouldn't be able to see the difference. But with an accuracy of 1 part in 500,000, you would, even though you wouldn't have been able to see the difference before the acceleration.

Also, suppose you accelerated for a second increment equal to the first; you would get

v_new = (0.999997 - 0.5) / (1 - 0.999999 * 0.5) = 0.999991.

And one more increment:

v_new = (0.999991 - 0.5) / (1 - 0.999999 * 0.5) = 0.999973.

As you can see, the differences in velocity grow fairly quickly for each equal increment of acceleration; the growth is not at all linear. And, as I said in my other post just now, for any given measurement accuracy, it would be simple to calculate how much acceleration you would need to be able to distinguish moving at exactly c from moving at 0.999999c (or any other speed less than c that you choose) to that accuracy.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#448

Post by pdonis »

CamperBob2 wrote: post #441
pdonis wrote: post #430

Earlier quoted context omitted.

> Dude, this is basic theory at the high school level. No, Tyson's claim is not "basic theory at the high school level". It is a particular interpretation of a theory (Special Relativity) that does not work, for the reasons I gave. > Photons put it all into the X, Y, and Z components, leaving nothing for the t component. Wrong. The spacetime vector that describes a photon's trajectory does have a t component. > Seems…

Wrong. The spacetime vector that describes a photon's trajectory does have a t component. Sounds interesting. Where can I read more about this t component?

If you're actually serious, try any textbook on SR that uses the four-vector formalism.

However, this comment has significantly raised my estimate of the probability that you are just trolling.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#449

Post by pdonis »

CamperBob2 wrote: post #444
phizy wrote: post #432

Earlier quoted context omitted.

There is no "point of view of a photon." Photons do not have frames of reference. So you are completely incorrect insofar as what you are saying is physically nonsense.

Again, take it up with Tyson and others with doctoral-level credentials who've made a career out of explaining these subjects to the unwashed laity. I'm not one of those people. Nobody posting on Hacker News is, as far as can be discerned. If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with…

> If you and others in the thread feel that these popular authorities are spreading misinformation or using inappropriate analogies, doesn't it behoove you to raise an objection with them directly?

If they want to make money by getting "the unwashed laity" to buy their books, why should I stop them? I simply don't buy them myself. If other people want to get told comforting nonsense, that's their problem. Caveat lector.

> Or perhaps with the appropriate faculty committees at their institutions?

Which would be pointless and absurd, since, as I have already said, the claims in question are not being made in textbooks and peer-reviewed papers.

I suspect you are trolling.

Re: Gravity is not a force – free-fall parabolas are straight lines in spacetime

#450

Post by ithkuil »

comboy wrote: post #442
ithkuil wrote: post #440

Earlier quoted context omitted.

are you talking special relativity or general?

Special. I mean the whole way looking at it seems wrong to me. There is no "rest of the world" in relativity. Assigning some objective vector to everybody doesn't work. These only make sense from some specific point of view. By the question you asked now I'm assuming, you meant "but hey, without GR..", but even without GR, ignoring that the universe is expanding, assuming flat space time etc. If the universe consiste…

I don't understand.

You can certainly imagine a scenario where two objects measure their mutual distance as being constant in time. You can also imagine other scenarios, but I'm asking to imagine the scenario where two objects do measure their distance to be D and then measure it again and it's still D, and measure it again and again and it's still D. That's just the definition of standing still with respect of each other, and when you plot a space time diagram, their space curve is parallel (because their spacial distance doesn't change).

Special relativity doesn't make that scenario impossible. It doesn't force things to move. It just describes what happens when things do move (through spacetime).

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