elfmotat

If you're using x as the radius, then you're revolving around the y-axis. Which means you're integrating over y. So to evaluate the integral you need x in terms of y, and x(y) will depend on y(x). So I'm not really sure what the problem is.

No, they aren't in a coherent state. Coherent means a collection of particles in the same state, which the Pauli exclusion principle explicitly forbids.

So you're wondering whether or not worldlines terminate? Sure: the particles that make up an object can be created and annihilated, meaning there are endpoints on a particle's worldline. The length of the worldline from one endpoint to another is the particle's lifetime. It depends on what you mean by "objects." If we're talking about point particles, then the particle's worldline goes through each time coordinate exactly once. Particles extend in time the same way they do in space, with the length of their worldline representing their lifetime. Extended objects will be cross each time coordinate at a finite length (i.e. a bunch of points) instead of at a single point, this length is shorter than the object's length than in its rest frame (I accidentally drew that second plot so that they appear longer - disregard this) due to the object's world-volume being rotated. This makes no sense. You can't simultaneously hold t constant and not hold it constant. I see no problems with "double booking" in spacial coordinates. We never stated how the objects we're considering will interact. Let's say they're neutrinos: they could pass right through each other with no interaction.

I don't know what "restricted subcontinua" means, but I made no claims about the dimensions of the space in my post. I made the object in question 1-dimensional so that I could draw it on a 2-d plot. Would you prefer that I make all my diagrams 4-dimensional? What do you mean by "where is your observer?" The word "observer" means "coordinate system," not "material object." I still don't know what your question is. What am I supposed to comment on?

Indeed it's just a clever trick. In canonical QFT it's actually an unnecessary trick. The propagator is just defined as the amplitude for a particle to be created at one point and destroyed at another from the vacuum: [math]iS(x-y)= \langle 0| \, T \left \{ \psi (y) \bar{\psi} (x) \right \} |0 \rangle[/math]. If you expand the field into Fourier modes and go through a tedious calculation you get the propagator. It's the same propagator that you get by introducing anti-commuting "sources" into the Lagrangian.

Then I really have absolutely no idea what you're asking. Could you try rephrasing your question? I never said that spacetime was (1+1)-dimensional in that scenario, I said that you were. This is also completely irrelevant to the rest of the thread.

Maybe you were saying something different than what I interpreted, but what I read was that you wanted to know if we occupied multiple time coordinates. And we do. If that wasn't your point, then I'm not exactly sure what you're trying to say/ask.

Studiot and michel: snap your fingers at the same time. For a moving observer, your snaps are not simultaneous. Instead there is a finite period of time between them. You may feel like you're only occupying one time coordinate, but that's only because you're trapped in your own rest frame! Edit: to expand on this to make it easier to visualize, pretend for a moment that you're a (1+1)-dimensional creature. You have some finite length. In your rest frame, if you were to plot your location as a function of time, it would appear as a series of straight lines with each line occupying exactly one time coordinate: But if a moving observer were to plot your position over time, they would plot something like this: Notice how you occupy a finite interval of time. Your body is "smeared out" over time for a moving observer due to relativity of simultaneity.

The electron field doesn't have a source like the photon (electromagnetic) field does because electron number is conserved (which is why charge is conserved!) while photon number isn't, as a consequence of U(1) symmetry. If you create an electron then you must destroy one somewhere else, and vice versa. You can create as many photons as you want. Something semi-related is that if you get a large number of photons together in coherent states, you get a macroscopic classical electromagnetic field. There's no macroscopic classical version of the electron field because electrons obey the Pauli exclusion principle, which means that you can't get a large collection of electrons in coherent states.

Alright, I'll take your word for it that it's not homework. We have up to first order in h (which is all that's necessary if h->0): [math]f(x+d_i h)=x^2+2d_i hx - d_e x-d_i d_e h+r x +d_e h-rd_e[/math] [math]f(d_e x)=d_e x^2+rx-d_e x - r[/math] where r=de/di. So: [math]\frac{f(x+d_i h)-f(d_e x)}{h}=2d_i x +d_e (1-d_i)+\frac{(x^2+r)(1-d_e)}{h}[/math] So if the equation holds for all x, then the limit as h->0 DNE because the last term goes to infinity. However if d_e=1 then the last term drops out and the limit, and therefore the derivative, exists.

You're right, sorry about that. Evaluate it and see if any contradictions pop up. If they do then the derivative DNE.

This still seems like philosophy. What do "empty coordinates" mean, numerically?

Alright, so they tell you that: [math]a(x)=d_i x+d_e[/math] [math]b(x)=a^{-1}(x)=\frac{x-d_e}{d_i}[/math] [math]f(x) = a(x) b(x) = \frac{(d_i x + d_e) (d_i - d_e)}{x}[/math] This does seem a bit homework-ish, so I'll let you take it from here. Just plug and chug and see what you get. EDIT: corrected a-1.

This seems more like philosophy than physics. Can you describe your idea more concretely? You're right, of course, that x, y, and z are linearly related to t when you're describing light spreading out over time. But that isn't what I was talking about. My point was that T=it for all t, so if you know t then you know T regardless of what physical situation you're trying to describe. That's why it isn't an additional dimension - because it contains no new information. x, y, and z are only linearly related to t in very specific circumstances.

When gravitational fields are very strong (like around a black hole) that equation doesn't work. The relativistic one I posted is accurate for all macroscopic objects regardless of their speed or the strength of the gravitational field.

It doesn't matter what the majority of people believe. What does that have to do with my point? The fact that we're free to redefine time this way is all that matters. Defining it this way sometimes does help, and I've already given you some examples where it does. No different in the sense that the signature is + for all dimensions instead of -+++, which is what your original post was all about.

In the relativistic case, you'll have some spacetime metric [math]g_{\mu \nu}[/math] and Christoffel symbols [math]\Gamma^{\mu}_{\alpha \beta}[/math] constructed from the metric. From these you can use the geodesic equation to get a differential equation for the energy of a small particle with respect to some parameter s: [math]\frac{d E(s)}{ds}=- \frac{1}{m} \Gamma^{0}_{\alpha \beta} \, p^{\alpha} (s) p^{\beta} (s)[/math] If you can solve that equation for E(s), then you'll know the energy of the particle at any value of s. If the metric is independent of time (which it usually is), then conservation of energy holds and dE/ds=0. An alternative approach would be to use [math]g_{\mu \nu} p^\mu p^\nu = const.[/math] when the metric is time-independent, where [math]p^0[/math] is the particle's energy.

What? x, y, and z aren't dependent on the value of t. For example, let's say I want to hang a painting on the wall. I tell you exactly when (t) I'm going to be hanging the painting. Based on this information, can you tell me where (x,y,z) on my wall I'm going to hang it? Of course not, because they are completely independent variables.

Of course it's "true." It's unclear what you mean by "true." All you did was change variables. It still physically represents the same thing. As long as your functions are analytic (which they usually are), you're free to make any variable complex. Both equations are "true." Let's define T=it. Then the metric (in 1+1 dimensions) is: [math]ds^2= -dt^2+dx^2= dT^2+dx^2[/math] We're now completely free to define the variable T as "time." In other words, the numbers we read on clocks are no longer called "time." Instead, we read a number off a clock and we multiply that number by i to get time. If we define things in this way then the metric is positive definite, and time is no different from any of the other dimensions. Plus, we get the added bonus that a lot of physics becomes much simpler. Which variable is more fundamental: t or T? Your instinct might be to say, "well obviously t is more fundamental." But why? If we interpret the numbers on clocks as "distances measured in an imaginary dimension," then T is a real number and t is imaginary. I don't know what you mean by this. Why would our ability to change variables depend on how long after the big bang we waited? it is not independent of t, so it's not an additional dimension. If we know t then we know it.

The total energy is conserved. In the case you're talking about, there are only two relevant types of energy: kinetic and gravitational potential. Their sum must be constant, so changing the potential necessarily changes the kinetic energy. This translates to the equation: [math]\frac{1}{2} m v_1^2 - \frac{GMm}{r_1} = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2}[/math] Edit: I just noticed this is in the relativity section, so if you're looking for the relativistic case then it's a bit different. I'll post the relativistic version of this later, but I'm not at home right now and I'm typing on my phone.

Lots of people have suggested that it is more "fundamental," in a sense, than t, and that all of the strange minus signs and imaginary numbers in relativity and QM can be explained away by the idea that we're "accidentally" using a Wick rotated dimension in all of our equations. The most notable proponent of this idea is Hawking. Semi-related:

You can change variables [math]t \to it[/math] and the metric becomes positive-definite. We're more than free to define the quantity [math]it[/math] as "time." In fact this little trick can enormously simplify lots of problems in physics, for example the integral in my signature.

So your ideas are supported by 'evidence' (personal experience does not constitute evidence by the way), but you didn't 'gather the evidence' so we'll just have to take your word for it?

Looks like we found another explanation: some of them are just really, really high.

On every science board and physics-related video comments section, there is invariably a plethora of crackpots willing to endlessly spew their personal brand of buzzword-salad. They usually have virtually no math, physics, or even general science knowledge. They usually think they're smarter than everyone else. Attempting to correct them is like arguing with a wall. And they're always talking about physics. If there's crackpottery around, you can bet it's physics-related. But why is this? My guess would be that some of it is because of popsci books and documentaries that romanticize "the elusive quest for the theory of everything." That would explain the people who seem to enjoy making up nonsensical diagrams and equations and then passing them off as some deep theory. But that doesn't explain a bunch of other types of crackpot: the relativity deniers, the quantum deniers, the new-age quantum people, and the people who hate math and try to do physics without it. Worst of all is the occasional crackpot with just enough genuine physics knowledge to actually be dangerous. Any thoughts on what draws the loonies to physics in particular?