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October 10

Why not travel to Mars in a straight line?

My understanding is that when we send a space craft to Mars, the ship goes into an orbit around the Sun like this and the ship's orbit eventually intersects with Mars' orbit. Why not go in a straight line from Earth to Mars? That would certainly be the shortest distance. I'm guessing that takes more energy? Pealarther (talk) 21:57, 10 October 2022 (UTC)[reply]

Going in a proper straight line would take very much energy indeed. And you would need to aim very carefully not to miss. But more generally, for any plausible spacecraft, such an approach does not work. Remember that you don't only have to match position, you also need to match speed. In other words, both when departing Earth and when arriving at Mars you must necessarily have the same position and the same speed as the corresponding planet. Usually, the most efficient way to do that is a Hohman transfer orbit. --Stephan Schulz (talk) 00:19, 11 October 2022 (UTC)[reply]

Yes, the path shown uses the least energy. Bubba73 ^{You talkin' to me?} 05:44, 11 October 2022 (UTC)[reply]

Hello, Pealarther. Start by reading Orbital mechanics several times until you understand it. Then read the references in that article until you understand them. Then you will understand the answer to your question. Cullen328 (talk) 05:49, 11 October 2022 (UTC)[reply]

The ideal path to Mars is the one that follows a geodesic, which is (in some sense) a "straight" path through curved spacetime, which has been curved through gravity (see General relativity for how this works). The analogy of a geodesic in terms we can visualize is a great circle path on a globe. The most efficient path between two points on earth is along a great circle. This path is not straight and curves when viewed in three dimensions, but in a sense it is straight along the 2D curved manifold that forms the surface of a sphere. In the same way, the most efficient path between two points in 4D spacetime is a geodesic. If you want to think about it in non-GR terms (i.e. not thinking about curved spacetime), then the most efficient path to Mars (the one that uses the least fuel) is the one where you spend the least amount of energy fighting against the gravity. Since the earth, the sun, and other planets all exert gravitational forces on your spacecraft, fuel spent fighting against those forces unnecessarily is wasted fuel. The greatest effect is going to be that of the sun; since the spacecraft leaving the Earth already has the earth's momentum, it's going to want to continue to orbit the sun in an ellipse under the sun's gravity. The most efficient path to Mars is one that doesn't fight against that tendency, and so will be a curved path. --Jayron 32 12:29, 11 October 2022 (UTC)[reply]

Pealarther, here's my best take for explaining why... If you just go on a straight line, you will need a ton of energy to overcome the Sun's gravitational influence. We already need a lot propellant just to get to Earth's orbit, and we want to save the remaining propellant as much as humanly possible. So instead, it would be more clever if we just boost the spacecraft's orbit and make use of the Sun's gravity, just enough so that close to the furthest point in the orbit (apogee) the spacecraft would reach Mars. This is a transfer orbit that requires the least amount of energ and it is called the Hohmann transfer orbit.

In practice, unless you are really, really short on fuel, the spacecraft will be boosted more than what the Hohmann transfer requires to reach Mars faster. And yes, if you are in a hurry and want to move around space like the Expanse, see Space travel under constant acceleration, where the first half of the journey is used to accelerate the spacecraft, and the second half is used to decelerate. CactiStaccingCrane (talk) 15:43, 16 October 2022 (UTC)[reply]

To chime in and hopefully add some help in inuiting this: the problem is that you, Earth, and Mars are all constantly moving—as observed from a Sun-centered reference frame—which makes the notion of "going in a straight line" somewhat incoherent. Your goal is to start at Earth and "hit" Mars. To do this you have to aim at where Mars will be when you intersect its orbit. This takes some calculus to work out, and since you don't control the positions of Mars and Earth you are forced to accommodate their trajectories. The thing you do control is when you leave, and people sending probes to Mars often wait for favorable times when Mars is close to Earth in their orbits to reduce fuel and travel time. Our intuition, "designed" for life as African plains apes, misfires here, because to us it looks like the heavens are sitting there nice and stately, but in reality we're whizzing at fantastic speeds around a nuclear fireball. Try to stand on a moving platform, throw a ball, and hit something on a different platform moving at a different speed. ‐‐47.147.118.55 (talk) 03:58, 17 October 2022 (UTC)[reply]

Oh, also, and looking at the linked image again, I think I see another element of the confusion. Most spacecraft only fire their engines briefly to change trajectory. The rest of the time they're just coasting. Space is empty, nothing to stop you, so you just keep moving the way you already were. Spacecraft are heavily limited on the amount of fuel they can carry, because there are no fuel stations in space. Fuel has mass and every bit of mass requires more fuel to push it, and soon your fuel requirements balloon into something unworkable, in what's known as "the tyranny of the rocket equation". There are some ways to work around this, such as ion drives that use very little propellant, or a light sail, which eliminates the problem of carrying fuel. If you can get your craft to go really fast, you can indeed make it from Earth to Mars in what is practically a "straight line". Constructing a pulsed laser array to kick a light sail craft up to relativistic velocity left as an exercise for the reader. --47.147.118.55 (talk) 12:02, 17 October 2022 (UTC)[reply]

"Constructing a pulsed laser array to kick a light sail craft up to relativistic velocity left as an exercise for the reader" LMAO CactiStaccingCrane (talk) 13:33, 17 October 2022 (UTC)[reply]

Photon duration

Today we know how to generate photons individually or a few units. So a photon is emitted at a time t1 and from a time t2 it is considered to have left (like a car with a front and a rear), if so, does its length correspond to its wave length ?

In this Nature article ^[1] they used "pulse of light", so a pulse have a Start time t1 and end time t2, right ? Malypaet (talk) 22:06, 10 October 2022 (UTC)[reply]

References

^ https://www.nature.com/articles/nnano.2017.218

A photon exhibits wave–particle duality. Viewed as a wave, it has no definite extension in space, and viewed as a particle, it has no definite wavelength. The wavelength of a photon is directly proportional to the inverse of the magnitude of its momentum. Position and momentum are complementary variables. So, by the uncertainty principle, the more precisely we determine the wavelength, the less we know about its position. --Lambiam 05:54, 11 October 2022 (UTC)[reply]

The photon either has left, or it has not. There's no time when it's in the process of leaving. PiusImpavidus (talk) 09:11, 11 October 2022 (UTC)[reply]

This is like saying, either Schrödinger's cat has died, or it has not. Welcome to the world of quantum weirdness. --Lambiam 15:04, 11 October 2022 (UTC)[reply]

When you take a measurement, the photon has either left or not. Otherwise, it can be in a superposition of having left and not having left (in which case the photon hasn't been created), but the photon is never in the state of leaving, which is the state OP implied. Schrödingers cat however can be dying, in addition to being dead and alive at the same time. PiusImpavidus (talk) 15:13, 12 October 2022 (UTC)[reply]

Photons, insofar as they are particles, are point particles. As such, they have no dimensions, and take no time leaving. They don't have sides or edges as such. --Jayron 32 12:21, 11 October 2022 (UTC)[reply]

Viewed as wave packets, however, photons do have a spatial extension. --Lambiam 14:56, 11 October 2022 (UTC)[reply]

(edit conflict)Kinda-sorta-in-a-way-but-probably-not-really-depending-on-how-you-think-of-it. The wave in this sense is a probability distribution of some property of the particle in question (its wave function), and while "location" (Δx in one dimension) is one of those properties, the "wave packet" in this case is a complex function, and while it is common to treat the "real part" of this function as something like "location", eh? It's really that the wave packet represents the uncertainty in the location of the photon itself; when you do any measurement of the photon, either at emission or absorption, it acts as a point particle. It doesn't have any dimensions when you measure it, so it doesn't take any "time" to absorb or emit; it doesn't behave as though it has a front or back edge. --Jayron 32 16:25, 11 October 2022 (UTC)[reply]

(post EC comment) Or what Nemur said below, which is much better than my comment. --Jayron 32 16:25, 11 October 2022 (UTC)[reply]

Or, to resolve the condundrum, (if we're being really rigorous), we have to stomp on the notion and shout loudly in to the abyss: "photons don't work like that!" And then we can rest easy knowing that we don't really know...

Really, the issue comes down to a semantic problem; an issue of trying to use natural English language to describe something for which natural English language is a poor representation.

The original post used an interesting word-choice - it asked about when a photon had left. What exactly does that mean? What exactly does it mean for a photon to "leave"?

If we can use some physics-ese to define precisely what "leaving" means for a photon, we can probably answer with scientific rigor. Has a photon "departed" or "left" from some well-defined point in space only when the expectation value for the electric field falls below some threshold? If this is what "departure" means for a photon, we can answer precisely - albeit, couched in a statistical, predictive sort of manner - by solving for some quantum-mechanical wave equation, or something. The details of this boring equation, I leave to the physicists, who already know how to solve it (it's trivial!)

In my experience, most of the conundrums of quantum mechanics stem from trying to force a vague, plain-language description onto the system. In large, every-day macroscopic systems, the vagueness is not quite so apparent - but when we try to translate plain language descriptions onto the microscopic scales that we care about for most of the text-book problems of quantum mechanics, those ambiguous, plain-language words make for ill posed problems. This is why many physicists quip: "shut up and calculate."

Now, to first order, that sounds like we're being dismissive about a sophisticated question; we're saying "stop asking that!" And that's not a very nice or educational thing to do. But... there's nuance, here. So let's rephrase "shut-up-and-calculate" into something that is both more polite, and more aligned to what we're trying to say: "The words you're using, in the English language, are not a good way to describe photons. We'd rather teach you to use a different language - mathematics - and then we will politely ask you to phrase your question using this language - because this language will be better at expressing questions and answers relating to the way photons really behave."

And that's why if we pull out any of our books-about-photons - I mean, take your pick! I've got a shelf full of them, if you'd like to read them! - as soon as they get into the nitty-gritty details about photon behavior, they jump out of English and codeswitch into physics-ese mathematical language. We don't talk about "when photon has left." Rather, we reference Equation 26.3 in Section 12, and in order to understand it, you need to have learned how to read it in its native language.

Fundamentally, this is not so different from any other academic pursuit. We can teach an introductory class on Roman history in English, but an advanced study of the topic invariably requires learning at least a little Latin. To English, certain things, well, do not translate, well.

Nimur (talk) 16:14, 11 October 2022 (UTC)[reply]

I did mention some books, for further reading -

The Griffiths' series -

Jackson's famous book -

Introduction to Electrodynamics

...

Of course, As I've said many times before, in many years of responding on the reference desk - don't underestimate this stuff.

I quote myself, here:

Normally, people read this book after they have completed a full four or five years of prior full-time preparation as full-time undergraduate physics students, because many very smart people find this mathematical content to be just at the limit of their mental capacity for comprehension after four or five years of full-time preparation. If you are not presently writing and solving wave equations for electrodynamics in conventional cases, you probably are not going to have great success writing and solving wave equations for nontrivial relativistic cases; and if you plan to understand general relativity without solving any equations, you're not going to get very far.

So - and I really don't want to be dismissive - the answer is, it ain't easy to answer questions about quantum mechanical behaviors of photons - not correctly and concisely, at least. Very smart people typically spend years trying to understand the complexity.

Nimur (talk) 16:30, 11 October 2022 (UTC)[reply]

A radar which emits a pulse of an electromagnetic beam of a certain energy for one microsecond between a time t1 and t2 is very real and trivial. Either we have n waves (E=hv) at // and there we know the beginning and the temporal end of each wave (t1 and t2), or we have a flux of n photons (E=hv) at // and there they must each be framed within an interval < t2 - t1? Sorry, but a point is in the realm of math, not physics. This impulse, when it encounters an obstacle, comes back to us at a time t3, it reacts like classical mechanics, simple ballistics. There's a very old cartoon from the beginning of computing where engineers disassemble a huge computer looking for a fault, next to them is a cleaning lady leaning on her broom and showing them the plug unplugged: "Is that what you're looking for?" Malypaet (talk) 20:41, 11 October 2022 (UTC)[reply]

But you asked about a single photon, not a pulse. A photon has no size and cannot be known precisely in location and momentum simultaneously. Also it might not even exist if you don't look at it, according to this year's Nobel prize winners (or something like that) Rmhermen (talk) 04:59, 12 October 2022 (UTC)[reply]

Exactly! The reformulated question posed an "either/or" scenario - and Lambian already described earlier (by bringing a cat into the mix) that this kind of "binary" yes/no certainty is at least inapt.

Even if we restrict to macroscopic systems, simply determining whether we did receive a signal devolves into a game of statistics and probability and confidence.

Every real signal has noise. Every signal! This is a stronger statement than it might initially appear. Noise isn't some mere annoyance that we can will into oblivion. It's not even some "practical detail" that we can usefully ignore in theoretical physics. We cannot construct a perfect (frictionless, while we're at it!) machine in our lab that observes signals with zero noise - not even in a purely theoretical construction. Some noise is an annoyance of unrefined engineering; we can hire better designers or purchase better parts to take this noise out of our machine (real or theoretical). But... other noise is inherent to the physics, and we may not ignore it!

At the most fundamental level - the regime of quantum mechanics - we have quantization noise - the inherent uncertainty involved in correctly counting a small integer number of physical events. If the number of events is "either zero or one," the best we can say is that we "either did or did not observe one single event, with some level of confidence."

How can we boost our confidence? We can re-take the measurement - but you see, that would literally entail observing a second time, for which the event either did or did not occur! If the first event had uncertainty and never even arrived, the second experiment might be the the first observance of the single event; or it might be a second repeated observance of two identical events, or maybe we failed to observe either, or maybe we saw the event only on one of two tries (first, or second attempt?...) and so ... this is noise. It cannot be avoided. As I said, even if we define "photon present" as equal to some measured level of electric field ... how do you know that electric field came from your photon and not from my other thermal photon that I (a notoriously troublesome nearby emitter of white noise) fired toward your detector at exactly the instant to interfere with your finely-crafted imported artisinal photon detector?

And this bizarreness - this "quantum weirdness" - as Lambian said, and as Rmhermen said, is really important stuff. Great minds have spent a lot of cycles stewing on it. As Rmhermen also said - this has something to do with some prestigious development in modern physics.

What we can do here is point our readers toward resources, and maybe inspire them to be enthusiastic about the miraculous complexity of even the most simple small particle of our universe.

We cannot, however, completely, correctly, concisely explain "the photon." It is too mysterious an animal to describe in a few words.

Nimur (talk) 15:37, 12 October 2022 (UTC)[reply]

The radar pulse is a group of photons and is a very concrete example contradicting the answers given to me. And for the photon that doesn't exist if you don't look at it, that's science fiction. So for a blind person, the moon does not exist? And yet it exists, regardless of whether we can see it or not. It is the same for a photon in a radar pulse, we are sure that it travels in this pulse. Here we know its speed, its wavelength and its position at least somewhere in the pulse. Malypaet (talk) 21:29, 12 October 2022 (UTC)[reply]

What Rmhermen said. If you change your question midpost than the answer will of course change too. One photon is a point particle and has no dimensions, and takes no time to interact with what is absorbing it. A group of photons has a spatial component, and does take an amount of time to interact with what is absorbing them. --Jayron 32 10:50, 12 October 2022 (UTC)[reply]

The easiest way to simplify the description of one complicated thing is to add lots more of them!

Yeah, we treat individual photons differently than we treat groups of photons, ... but I'm already feeling that the original question has become less coherent as it progresses forward along its light cone...

Nimur (talk) 15:50, 12 October 2022 (UTC)[reply]

Less coherent? Do you frequently make such apt puns? --Jayron 32 11:34, 13 October 2022 (UTC)[reply]

[1] ttps://www.nature.com/articles/nnano.2017.218

[1]