How does any sign off the big bang, light, microwaves etc., make it back to earth for us to detect?

ccupini · 9ヶ月前

So Hubble parameter is decreasing but because it's not decreasing fast enough objects accelerating away from us.

Just so. More specifically, in the standard model of cosmology, it turns out that as long as the ratio of recessional acceleration to recessional velocity is less than the Hubble parameter, the Hubble parameter will be decreasing.

So what is the rate Hubble parameter is decreasing?

I'm not sure of the value.

ccupini · 9ヶ月前

There's a subtlety here that needs to be addressed. First, a note on terminology. When we talk about the rate of expansion, we need to be careful. The important quantity, called the Hubble parameter, is a speed per distance. That is, it tells you how fast an object at a given distance is receding from us. In particular, closer objects are receding more slowly than distant objects.

Now, on the one hand, cosmological expansion is "accelerating" in the sense that if you look at a specific object right now, the apparent recessional speed (the rate at which the distance from us to that object changes) is increasing. That is, we can say that the object is accelerating away from us.

But when I said

Over time, as the parameter that determines the expansion "rate" decreased, the light found itself in a region where the recession speed was lower than the speed of light

I was talking about how the Hubble parameter changes. This is actually decreasing, even now. This means that while a fixed object sitting in space is accelerating away from us, the recessional speed at a fixed distance from us is decreasing. To make that a little more clear, imagine that we look at two distant galaxies, called A and B. For each of them, we measure how far away they are and how fast they're receding, and we find that galaxy A is farther away and receding more quickly than galaxy B. Now, since they're receding, we can wait a bit and measure again. Let's say we do this and find that galaxy B is now as far away as galaxy A was the first time. We note the following:

Both galaxies are farther from us than they were the first time.
Both galaxies are moving faster than they were the first time.
Galaxy B is moving slower than galaxy A was moving the first time, even though it's just as far away.

We conclude that both galaxies are accelerating (their speeds increased) away from us (their distances increased), but that the Hubble parameter has decreased (the recessional speed at a given distance has gone down).

So how does that allow us to see galaxies that are receding faster than the speed of light? Well, while the galaxy is receding at some rate, the light it emits is receding at that speed minus the speed of light (you can picture it like swimming against the current). So, at first, both the galaxy and the light get carried away by the expansion, but the distance between the light and the source galaxy increases. Since the Hubble parameter is decreasing, the extra distance between the light and the galaxy means that the recessional speed of the light (even before subtracting the speed of light) will be lower than the speed of the source galaxy. If the Hubble parameter decreases quickly enough, then it can happen that the recessional speed at the light's position actually drops below the speed of light. When that happens, the effective direction of the light reverses and it begins moving toward us. But then it will never turn back around, since its speed is only being increased (decreasing distance, decreasing Hubble parameter).

ccupini · 9ヶ月前

I read enough about detections from the edge of space

Any such reference must be to the edge of the observable universe. So far as we can determine, the universe, taken as a whole, has no edge.

I have no idea how those indicators made it back to earth which isn't at the edge of the universe.

The same way light from anything gets anywhere: light moves. You can, I suppose, detect objects that are on the other side of the room despite the fact that you are not on the other side of the room. This is because light from those objects has traveled across the room to reach you. Similarly, we can detect distant galaxies (even ones on the edge of the observable universe) because light from them has been traveled across the intervening space to reach us. They began emitting light long ago, and (probably) will continue emitting light for a long time. Some of that light has long since passed us, some of it is just now arriving, and some of it has yet to arrive. When we direct a telescope toward these objects, the light that is currently reaching us is captured by the telescope and recorded (the precise mechanism of capture and recording depends on the telescope). We can then analyze the recording to infer information about the source of the light (much as you can determine the approximate size of objects across the room from you).

Technical note: contrary to popular belief, the most distant galaxies are currently receding from us at speeds well above the speed of light. By this I mean that the distance between us and those galaxies is increasing at a rate that exceeds the speed of light. We are able to detect them because the light has been moving toward us since they were much closer. These objects emitted light in our direction, but the expansion of space originally resulted in that light traveling away from us. There is, however, a crucial difference between the light and the source: the source galaxy is basically just being carried along by expansion, whereas the light was "struggling" against it. Over time, as the parameter that determines the expansion "rate" decreased, the light found itself in a region where the recession speed was lower than the speed of light. From that moment on, the distance between us and the light only decreased, until it finally reached us. The source galaxy, on the other hand, has been carried ever farther away by expansion and will never again enter our causal range. To wit, while we can see that galaxy, we're seeing it as it was in the past and we will never see it as it is "right now". Similarly, we could never send a signal to that galaxy; an alien living there might be able to detect evidence of our galaxy's ancient past, but any signal we sent out now would never get there.

Kooko1212 · 1年前

That what seem to be fundamentally quantum-mechanical interactions can give rise to an apparently classical universe.

RelativisticMechanic · 1年前

I don't know about /r/science, but in /r/askscience we have a policy of removing top-level responses that are speculative, anecdotal, or otherwise don't represent the results of peer-reviewed science.

Our more popular threads tend to attract a lot of "my experience has been...", "I've always thought...", and "this youtube video says..." type comments. In order to maintain the standards we've set, such comments (and, typically, all subsequent responses to such comments) are nuked. Of course, this generates a host of "what happened here?" comments, which are removed along with other off-topic comments.

supergalactic · 1年前

If we want to talk about moving "only in time" and not "moving through space", then we really need to specify which locally inertial reference frame we're using to define "stationary".

It's always seemed to me that the most "intuitive" reference frame would be the one in which the time machine finds itself at the moment of activation. If that reference frame happens to be moving around the Sun synchronously with Earth's center of mass, I would expect that the time machine would carry on as such. There might be complications from the fact that Earth is rotating, but one imagines those wouldn't be too hard to correct for.

supergalactic · 1年前

The question then becomes why your time machine is tied to that rest frame and not, for example, the same locally inertial reference frame it was in (i.e., moving along with Earth) at the time of activation.

baseballs200 · 1年前

An infinite number of mathematicians walk into a bar. The first asks for a beer. The second asks for two beers. The third asks for four beers. The fourth tries to ask for eight beers, but the bartender stops them and says, "Just give me my beer already."

poprocksandfolk · 1年前

The Newtonian limit applies when the speeds involved are small, the field is weak (so it can be approximated closely as a perturbation on flat space), and the field is essentially static. It becomes less useful as one deviates further from these approximations. As with all approximations, it stops being useful when you need more accuracy than it can provide.

For example, the Newtonian approximation is quite sufficient to get an astronaut to the moon. However, it doesn't quite get Mercury's precession right, it won't suffice if you need to calibrate clocks on a GPS satellite, and it would be wholly inaccurate if you wanted to know how the distance between us and distant galaxies will be changing a few billion years from now.

poprocksandfolk · 1年前

It's covered in pretty much any decent introductory text on relativity, many of which can be found online for free.

The basic idea is that "objects in motion continue in a straight line at constant speed unless acted on by an outside force" is too tight a constraint. Relativity replaces this with "the worldline through spacetime of an object is a time-like geodesic unless acted on by an outside force", where the spacetime metric is determined by Einstein's field equation.

In an appropriate limit, the geodesic equation for a 3+1 dimensional manifold reduces to Newton's second law, while Einstein's equation reduces to Newton's universal gravity.

poprocksandfolk · 1年前

A so-called fictitious force that arises from living in an accelerated reference frame.

HerpesAunt · 1年前

What I meant by this, was I couldn't understand if time itself operates at different speeds depending on gravitational pull, or if it is just perceived differently depending on gravitational pulls.

It's a real, physical difference. Each observer experiences time as passing at the perfectly normal rate of one second per second; it's only when you compare their clocks that you will find one twin to be aged differently. Note that "clock" here is a common shorthand for "anything at all that varies with time and can therefore be used to show that different amounts of time have passed"; this includes biological "clocks", like a human being.

So does my twin literally age differently?

Yes.

Or has he just experienced a different amount of time?

Again, I don't understand the difference. If your twin experiences 20 years, he ages 20 years. That's what experiencing 20 years means.

HerpesAunt · 1年前

Time dilation is absolutely real. If two identical twins begin together, one moves to a significantly greater magnitude gravitational field and then returns after some time, and we then compare their age, the one who spent time in the stronger field will be younger than the one who remained distant.

That said, you need a ridiculously strong gravitational field to get an effect that would be noticeable over ten years. To make the calculations simple, let's consider a non-rotating mass. You and your identical twin start out in deep space very, very far from the mass. Now your twin descends toward the mass, spends some time there, and returns after exactly 10 years have passed on your clock. If we ignore complicating effects during the actual descent and return, we can ask how the amount by which your twin ages depends on how close he gets to the mass. For example, if you want the dilation effect to make him age only 5 years, he would have to get within four-thirds of the Schwarzschild radius from the center of mass.

By way of comparison, Earth's surface radius is around 718 million times its Schwarzschild radius. This means that if you remain in deep space for 10 years while your twin descends to Earth's surface, he would be only about 1/5 of a second younger than you.

When I talk about our relative ages, I don't mean the temporal duration that has passed in our life. I am talking about our Biological age if you will. As in the physical difference between a 20 year old young man approaching the physical prime of his life, and a full grown 40 year old who has graying hair and back pain

I'm not clear on the distinction here. How could your biological age differ from "the temporal duration that has passed in your life"?

OlejzMaku · 1年前

Consider two people, Alice and Bob, who spend their lives in inertial reference frames in deep space.

In Alice's reference frame, Alice is always at the origin, (0,0,0), while Bob's position is forever changing and never takes on the same value twice.

In Bob's reference frame, Alice is the one that's always moving while Bob spends forever at (0,0,0).

PassthePaper · 1年前

A neutrino walks through a bar.
An infinite number of mathematicians walk into a bar. The first asks for a beer. The second asks for two beers. The third asks for four beers. The fourth tries to ask for eight beers, but the bartender stops them and says, "Just give me my beer already."

marklove039 · 1年前

Short version:

Real numbers are complex numbers and i is a complex number. Since we can add and multiply complex numbers, it follows that if we pick two real numbers a and b, the number a + b*i is a complex number. It turns out that every complex number can be written in this way. That is, if you pick any complex number, z, there exist real numbers a and b such that z = a + b*i.

Long version:

Complex numbers are basically just pairs of real numbers along with special rules for addition and multiplication. Specifically, if we take the complex number (a,b) and the complex number (c,d), we have

(a,b) + (c,d) = (a + c, b + d).
(a,b)*(c,d) = (ac - db, ad + cb).

Note a few things. First, (1,0) multiplied by any other complex number just give that number back:

(a,b)*(1,0) = (a*1 - 0*b, a*0 + 1*b) = (a,b).

Second, if I take two complex numbers of the form (a,0) and add them, I get another number of that form:

(a,0) + (b,0) = (a + b, 0).

Similarly, (0,a) + (0,b) = (0, a + b).

Finally, we see that

(0,1)*(0,1) = (0*0 - 1*1, 0*1 + 0*1) = (-1,0).

All of this leads to the fact that for any complex number (a,b), we have

(a,b) = (a,0) + (b,0)*(0,1).

Introducing the notation i = (0,1), and using "a" as a shorthand for (a,0), this becomes

(a,b) = a + bi,

which is the common notation.

To your other questions:

We can represent complex numbers in other ways. For example, every complex number can be written as the product of a real number, r, and the exponential of a purely imaginary number, it. That is, every complex number z can be written as

z = r*e^it

for real numbers r and t. This representation is really rather useful, but it has the somewhat unnatural feature of being non-unique: if z = r*e^it, then also z = r*e^{i[t + 2}^*^k*π], for any integer k.

Double-decker_trams · 1年前

Probably something like "the 11111th Busy Beaver Number" is the biggest well defined (although in this case unknown) number you could write down very quickly.

According to the rules, I have fifteen seconds. I write:

S^G(G), where G is graham's number, exponentiation is functional iteration, and S(m) is the m-th busy beaver number.

Does that count?

Double-decker_trams · 1年前

If you're allowing ↑⁹, you might as well allow ↑^{9↑^9↑⁹}, which is going to be much, much larger than your ↑^{999999999999999999999999999}, and nevertheless still much smaller than, for example, Graham's number.

Double-decker_trams · 1年前

Not even close.

For example, your number is ridiculously (like, literally unimaginably) smaller than Graham's number, defined by taking G = g(64), where g(1) = 3↑⁴3 and g(n) = 3↑^g(n-1⁾ 3 .

You could do better using Conway's chained arrow notation. With this notation, if we call Graham's number G, one has

3 → 3 → 64 → 2 < G < 3 → 3 → 65 → 2,

so it's very easy to make numbers much, much larger than Graham's number. Moreover, if we introduce a new notation like

n ⇒ m = n → n → … → n → n,

where there are m total arrows, we can go way, way higher. For example, consider G ⇒ G.

And we can, of course, keep doing this, defining ever bigger numbers by simply iterating the above processes.

If you switch over to allowing pre-defined functions, things get even more messy as you can consider iterated busy beaver functions, which will simply swamp anything you could get using the above method.

Shmallies · 1年前

I've removed this question because it's a very frequently asked question here. For example, see this question from 14 hours ago.

RelativisticMechanic · 1年前

It comes into play when explaining why C measures a distance increase of 3 light-seconds over a time of 2 seconds while neither B nor a measure an equal change in distance over the same time. In order to figure out what either A or B observes, you need to account for both the time dilation effect and the length contraction effect.

Dissimulate · 1年前

Yes, absolutely.

Let's say that in your reference frame right now Earth is 9 light-years away coming toward you at 90% the speed of light. Let's further assume, to make my life easier, that you have been watching Earth approach you at this rate for as long as you can remember. Now, in 10 years, it will reach you. If it's 9 light-years away, the light it's emitting right now won't reach you for another 9 years. This means that you will get all of the light that it emits between right now and its arrival in the last year of your wait. How much time do you see pass on an Earth clock during that final year? Well, if it's moving at 90% the speed of light, then time on Earth is passing at roughly 0.436 times your time. So in 10 years for you, Earth clocks pass roughly 4.36 years. This means that in the last year of your wait, you see Earth's clocks run through 4.36 years of time.

RelativisticMechanic · 1年前

Yes, absolutely, in the sense that if C measures the distance between A and B at some time and then again two seconds later, C will find that the separation has increased by three light-seconds.

RelativisticMechanic · 1年前

But the distance between them is NOT increasing by 1.5c.

Yes, it is. Specifically, in the sense specified explicitly by Olog:

However, from your point of view in the middle of the two objects, the distance between A and B indeed increases at a rate of 1.5 times the speed of light.

If you determine the distance between them in your reference frame at some moment, and then determine the distance between them in your reference frame two minutes later, you will find that their separation has increased by three light-minutes: the distance between them in your reference frame is changing at a rate of 1.5 times the speed of light.

The objects are only moving through space at .75c relative to some other point. so they are experiencing time at .25 of that other point.

This is wrong. There is a way to model time-dilation and relative velocities in such a way that "the speed through space" and "speed through time" (by which we mean the number of seconds that a stationary observer would see to pass for the moving observer for every one second that the stationary observer experienced) generate a total "speed through spacetime" of c, but you have to combine them as legs of a right-triangle. That is, if we call the "speed through time" s and "speed through space" v, one has

v² + (cs)² = c².

If you plug in v = 0.75c, you find that one must have s² = 0.4375. That is, an observer moving at 3/4 the speed of light relative to you appears to be ageing at sqrt(7/16) the rate as you.

There is no point where the distance is increasing faster then 1c.

Neither of the observers sees the other moving away faster than c, but our stationary observer (the one making the claim that they're each moving in opposite directions at 0.75c) unequivocally sees the distance between them increasing at 1.5c.

fillingtheblank · 1年前

The global geometry is actually distinct from the topology (which can also be discussed locally or globally). In fact, as I discussed here, it's not at all trivial to associate topological properties to specific spacetimes.

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