Can you have velocity without acceleration? My mind is fucking with me right now. How can anything move without any acceleration?

File: 127 KB, 800x640, 1346614682794.jpg [View same] [iqdb] [saucenao] [google]

Anonymous Sat Sep 15 18:06:28 2012 No.5059993 [Reply] [Original]

Can you have velocity without acceleration?

My mind is fucking with me right now. How can anything move without any acceleration?

>>	Anonymous Sat Sep 15 18:07:27 2012 No.5059997 You move at a constant velocity. It's not that hard.

>>	Anonymous Sat Sep 15 18:09:12 2012 No.5060001 >>5059997 Yea I know that... I'm asking to start movement.

>>	lkj Sat Sep 15 18:10:57 2012 No.5060004 >>5060001 relative to what?

>>	Anonymous Sat Sep 15 18:10:57 2012 No.5060005 Light.

>>	Anonymous Sat Sep 15 18:12:27 2012 No.5060009 Acceleration is the change in velocity. Now, please, get out of /sci/, kid.

>>	Anonymous Sat Sep 15 18:17:28 2012 No.5060019 >>5060004 To anything. But I think I have the answer now. >>5060009 No.

>>	Anonymous Sat Sep 15 18:18:43 2012 No.5060022 It would need to start with movement.

>>	Anonymous Sat Sep 15 18:19:27 2012 No.5060024 >>5060019 Yes, it is. > acceleration is the rate at which the velocity of a body changes with time. (Crew, Henry (2008). The Principles of Mechanics. BiblioBazaar, LLC. pp. 43. ISBN 0-559-36871-2.)

>>	Anonymous Sat Sep 15 18:19:44 2012 No.5060025 >>5060009 You mean acceleration is a change in speed.

>>	Anonymous Sat Sep 15 18:20:27 2012 No.5060028 >>5060024 Either he's misquoted, or he made a mistake.

>>	Anonymous Sat Sep 15 18:22:27 2012 No.5060030 >>5060025 speed is scalar. acceleration and velocity are vectors. You are wrong.

>>	Anonymous Sat Sep 15 18:22:27 2012 No.5060031 >>5060025 that actually works too

>>	Anonymous Sat Sep 15 18:23:27 2012 No.5060036 >>5060030 > acceleration > a vector No. There is zero indication of a direction in acceleration.

Anonymous Sat Sep 15 18:28:57 2012 No.5060042 [DELETED]

A path or curve on a manifold M[/spoiler] is a connected set of manifold points, with these path points labeled by a continuous, monotonically increasing real parameter, t[/spoiler].

Each path point has a tangent vector \vec{t}(t)[/spoiler], called velocity, in tangent space T_p M[/spoiler] defined via the directional derivative along the path.

Acceleration is then the directional derivative of \vec{t}(t)[/spoiler].

Please get this thread off the front page now thx

Anonymous Sat Sep 15 18:30:12 2012 No.5060045

>>5060036
There most certainly is, if the velocity is decreasing the acceleration vector points in the opposite direction of the velocity vector. An object in orbit will have an acceleration vector pointing towards the massive body, which keeps the velocity vector tangent to the sphere.

Anonymous Sat Sep 15 18:30:27 2012 No.5060047

A path or curve on a manifold is a connected set of manifold points, with these path points labeled by a continuous, monotonically increasing real paramater, t[/spoiler].

Each path point p[/spoiler] has a tangent vector \vec{v}(t)[/spoiler], called velocity, in tangent space T_p M[/spoiler] defined via the directional derivative along the path.

Acceleration is then the directional derivative of \vec{v}(t)[/spoiler].

Please get this thread off the front page now thx.

Anonymous Sat Sep 15 18:30:43 2012 No.5060049

>>5060036

>Physics A content
>not getting it

Then what happens when a constant velocity is introduced to an acceleration opposite to it? Does that not happen or is it null?

Does a car going at 20 mph not slow down to a stop and speed up if it accelerated backing up?

>>	Anonymous Sat Sep 15 18:31:27 2012 No.5060052 >>5060045 Assuming a uniform orbit.

>>	Anonymous Sat Sep 15 18:31:27 2012 No.5060053 I am at university, but away from physics. Your description of acceleration is pretty interesting. What book do you recommend in such subject?

>>	Anonymous Sat Sep 15 18:32:42 2012 No.5060056 >>5060047 I am at university, but away from physics. Your description of acceleration is pretty interesting. What book do you recommend in such subject?

Anonymous Sat Sep 15 18:33:57 2012 No.5060058

>>5060036
Oh for fucks sake. It would have taken you litterally 5 seconds to check what you just said. Wikipedia entry for Acceleration. The FIRST BLOODY SENTENCE:

In physics, acceleration is the rate at which the velocity of a body changes with time.[1] In general, velocity and acceleration are vector quantities, with magnitude and direction[2][3]

Seriously. At least TRY to do some research before you profess your ignorance.

>>	Anonymous Sat Sep 15 18:34:12 2012 No.5060060 >>5060056 How about a basic physics textbook? Just any old one will do.

Anonymous Sat Sep 15 18:34:43 2012 No.5060062

>>5060047
This is a mathematical definition of velocity and it's not worth the effort to ensure that this conforms to the classical definition of velocity. You are just trying to sound smart by using terms not everybody understands and contributing nothing.

>>	Anonymous Sat Sep 15 18:35:12 2012 No.5060064 >>5060060 My highschool textbook didn't cover manifolds, obviously. I only learned about them in uni

>>	Anonymous Sat Sep 15 18:38:58 2012 No.5060071 >>5060062 >Wants to understand a mathematical term >Doesn't want a mathematical explanation Analogously, we could argue as follows: "I wonder how many bricks there are in this wall" "I will now count them without using numbers"

>>	Anonymous Sat Sep 15 18:39:13 2012 No.5060073 >>5060062 >classical definition of velocity This is the only definition of velocity. Are you 12 or just retarded?

Anonymous Sat Sep 15 18:41:27 2012 No.5060075

>>5060064
For the love of god don't listen to that pretentious faggot. He is just confusing the matter and adding nothing. Why on earth would you even need a manifold or it's tangent space to explain velocity and acceleration? A path is already overkill.

>>	Anonymous Sat Sep 15 18:41:42 2012 No.5060076 >>5060031 No it doesn't. Something can change direction but not speed, and it's still accelerating. Example: a circular orbit.

>>	Anonymous Sat Sep 15 18:41:58 2012 No.5060077 >>5060075 Good to know! Thank you

>>	Anonymous Sat Sep 15 18:45:57 2012 No.5060084 >>5060073 What I was referring to by classical definition is what we all intuitively understand as velocity, which can simply be explained as the change in position. >>5060071 It's a physical term in the context of this thread.

>>	Anonymous Sat Sep 15 18:48:43 2012 No.5060091 >>5060077 It is of great mathematical interest and probably some interest in general relativity. But that is, mostly, outside the scope of this thread.

>>	Anonymous Sat Sep 15 18:49:57 2012 No.5060095 >>5060091 Still, is there a good book? Just out of curiosity right now. You got me going

Anonymous Sat Sep 15 18:50:57 2012 No.5060096

>>5060084
>what we all intuitively understand as velocity, which can simply be explained as the change in position.
Define change. Define position.

See? It's ambiguous. Mathematics allows you to precisely say what you mean while preserving intuition.

If you don't know the definition of anything, you can look it up.

>>5060075
> Why on earth would you even need a manifold or it's tangent space to explain velocity and acceleration?
Because that's what velocity and acceleration are? You sound like a retard. Please explain to me what velocity and acceleration are without using these terms or synonyms.

>>	Anonymous Sat Sep 15 18:52:42 2012 No.5060102 >>5060056 >>5060064 Any freshman undergrad textbook on classical mechanics. Example: Landau/Lifshitz. Sadly, /sci/ is mostly filled with highschoolers who haven't even taken a freshman university physics course so they wouldn't know.

Anonymous Sat Sep 15 18:54:57 2012 No.5060106

>>5060075
I didn't post that, bu I think it comes down to how deeply you want to understand the maths. Being a joint honours student in Maths and Physics, I like to understand the treatment of things like this completely, relating to manifolds and spaces and such, but I realise that most physicists view maths as a means to an end rather than something for it's own sake. It's not necessary to understand a maths concept completely to apply it to problems effectively,but some people find it interesting.

Anonymous Sat Sep 15 18:58:12 2012 No.5060115

>>5060096
I agree, precise mathematical definitions are helpful as long as most of us agree that they represent what we've all been talking about all along.
Velocity is the derivative of the position vector in R^3 with respect to time and the acceleration is it's derivative.
No need for manifolds or their tangent spaces. Why on earth would you need every single tangent vector to a point on a manifold when you are only interested in it's path and the associated tangent vector?

>>	Anonymous Sat Sep 15 18:59:58 2012 No.5060119 >>5060106 I'm studying mathematics and I much prefer rigorous treatment of physics, but there is no need to go overboard.

Anonymous Sat Sep 15 19:02:27 2012 No.5060123

>>5060096
>Define position.

I personally define "velocity" as the integral of acceleration over time, and "position" as the integral of velocity over time. I choose these definitions because, unlike velocity and position, acceleration can be measured. If we use this definition, then there can be no velocity without acceleration.

Of course, "velocity" and "relative velocity between two objects" are two different things; and "position" and "distance between two objects" are also two very different things. Even if you have no acceleration, obviously if two objects have different starting velocities, they will have a relative velocity.

>>	Anonymous Sat Sep 15 19:03:28 2012 No.5060125 >>5060115 This is maths. Nothing is done for "need". What you've described is a specialised case of the former description. Not that it's wrong, and not that it's not perfectly usable for all applicable purposes, it's just not complete :)

Anonymous Sat Sep 15 19:06:13 2012 No.5060133

>>5060115
>Velocity is the derivative of the position vector in R^3
But that's wrong. It's R^4 in classical mechanics, Lorentzian in special relativity, and arbitrary beyond that.

>No need for manifolds
A manifold gives you a generic definition that always works. The laws themselves do not change.

>or their tangent spaces.
How do you define a tangent vector without a tangent space? How do you define coordinate basis vectors (so you know what coordinate system you're dealing with) without tangent space? How do you do vector bundle holonomy (e.g. Newton's laws) without knowing what the tangent space is?

Anonymous Sat Sep 15 19:07:43 2012 No.5060135

>>5060123
>I personally define "velocity" as the integral of acceleration over time, and "position" as the integral of velocity over time. I choose these definitions because, unlike velocity and position, acceleration can be measured. If we use this definition, then there can be no velocity without acceleration.

Well first of all, acceleration is no more or less measurable than position or velocity. Second, in order to have an integral over time and actually get a value, you need initial conditions for position and velocity. So yes, you can certainly have velocity without acceleration. The integral of a=0 over time can be any value, because it depends on v(0).

>Of course, "velocity" and "relative velocity between two objects" are two different things; and "position" and "distance between two objects" are also two very different things

No they're not. Position and velocity are ALWAYS measured relative to something. There's no such thing as position or velocity not relative to something.

>Of course, "velocity" and "relative velocity between two objects" are two different things; and "position" and "distance between two objects" are also two very different things. Even if you have no acceleration, obviously if two objects have different starting velocities, they will have a relative velocity.

Anonymous Sat Sep 15 19:09:12 2012 No.5060136

>>5060123
That's actually a remarkably good point. The definitions using derivatives require there to be something "solid" from which the others are derived. However, in the integral definition, something "solid" is also required. It's keeps going on forever unless you give the fixed point. eg, I'll define accel. as the integral of the rate of change of accel, call it p. then define p as the integral of q and so on.

Or the other way, I'll define position as the derivative of the integral of position, and so on.

Where does it stop? The derivative/integral definition doesn't state which one causes the others.

>>	Anonymous Sat Sep 15 19:09:27 2012 No.5060137 >>5060123 The coordinates of the position are not important since we are interested in the change, the difference in the choice of the origin will disappear in the differentiation.

Anonymous Sat Sep 15 19:12:28 2012 No.5060142

>>5060123
>I personally define "velocity" as the integral of acceleration over time, and "position" as the integral of velocity over time.
There's no difference between your definitions and mine besides added ambiguity. What's time? What is position?

>I choose these definitions because, unlike velocity and position, acceleration can be measured.
So you're telling me I can speed however much I want and I'll never get pulled over? Or a meter stick is useless?

>Of course, "velocity" and "relative velocity between two objects" are two different things; and "position" and "distance between two objects" are also two very different things.
Except there's not. Preferred frames do not exist in physics. There is no "universal reference frame". You're either inertial w.r.t. to something, or not. There's no other observable quantity, and no other quantity that exists. Have you even done high school physics?

>>	Anonymous Sat Sep 15 19:12:59 2012 No.5060140 You're kidding me, right? If you don't have velocity, you stand still.

>>	Anonymous Sat Sep 15 19:13:43 2012 No.5060145 >>5060102 Thank you. Buying Landau soon.

Anonymous Sat Sep 15 19:13:58 2012 No.5060143

>>5060135
>Well first of all, acceleration is no more or less measurable than position or velocity.

Wrong. A lead sphere, in a box, attached to the sides by elastic cords. Use the force on the cords to measure the weight of the sphere, and compare that weight to its known mass. Bam. Right there. Simple accelerometer.

>Second, in order to have an integral over time and actually get a value, you need initial conditions for position and velocity.

No you don't. You just need a starting time, and an ending time. Those are the bounds of the integral.

>Position and velocity are ALWAYS measured relative to something.

Oh? But I just posited a definition of "position" and "velocity" that does NOT require them to be measured relative to anything.

Anonymous Sat Sep 15 19:22:14 2012 No.5060161

>>5060142
>There's no difference between your definitions and mine

What are your definitions?

>besides added ambiguity

There's absolutely zero ambiguity whatsoever. The definite integral of the acceleration of an object, from t=a to t=a+1, is it's velocity at t=a+1. The definite integral of its velocity, from t=a to t=a+1, is its position at t=a+1.

>So you're telling me I can speed however much I want and I'll never get pulled over? Or a meter stick is useless?

My definitions do not deal with relative velocity or distance.

>Preferred frames do not exist in physics.

Yes they do. The subjective frame is the only frame that matters when dealing with a single object. My definitions do not deal with multiple-body systems.

>>	Anonymous Sat Sep 15 19:22:14 2012 No.5060163 >>5060143 You know what I need to measure my position? I need to set the origin of my system at my position and my position is the 0 vector. A bit simpler than yours and works anywhere.

Anonymous Sat Sep 15 19:24:13 2012 No.5060164

>>5060163
>I need to set the origin of my system at my position and my position is the 0 vector.

But if you have velocity, then your system can only be used for measurement when t=a. If t=a+0.0000000000001, you could already have moved away from your origin, at which point your position is once again unmeasureable.

>>	Anonymous Sat Sep 15 19:25:44 2012 No.5060165 >>5060163 How do you set you axis between two points? Set yourself as the origin; where does the other point go? On the x axis? On the y axis? On the z axis, somewhere between two of them? Or all three? In the negative direction? etc. etc.

>>	Anonymous Sat Sep 15 19:27:44 2012 No.5060169 >>5060161 Wait, wait, wait. That integral is v(a+1)-v(a) and you seem to assume that v(a)=0. Your definition seems more complicated bro.

>>	Anonymous Sat Sep 15 19:27:59 2012 No.5060167 >>5060165 I'm not even gonna get on to asking about what happens when your axis aren't orthonormal.

Anonymous Sat Sep 15 19:28:44 2012 No.5060171

>>5060163
>and works anywhere.

Oh, also, how does mine not work everywhere? You're an idiot.

>>5060169
>and you seem to assume that v(a)=0

Yes. I do.

>Your definition seems more complicated bro.

It's the only definition that works in a one-body system.

>>	Anonymous Sat Sep 15 19:31:13 2012 No.5060177 You guys do realise you're arguing in the same manner as: Person 1: "A x B = C" Person 2: "NO! C / B = A" You're both saying the same bloody thing!

>>	Anonymous Sat Sep 15 19:33:43 2012 No.5060180 >>5060164 Dude, I'm now in a coordinate system that has the origin where I used to be and my current position is well defined in it. >>5060165 x front, y left, z up. Come at me bro.

>>	Anonymous Sat Sep 15 19:34:58 2012 No.5060183 >>5060180 Define "front", "left" and "up".

Anonymous Sat Sep 15 19:35:28 2012 No.5060184

>>5060177
>You're both saying the same bloody thing!

No, we're not.
In order to define velocity as the derivative of position, then you must assume that there always exists such a thing as a measurable position. And in a one-body system, there does NOT exist any such thing; you just need to hope that there exists a second body in the system; and if not, you're screwed.

If you define velocity as the integral of acceleration over time however, you can ALWAYS measure your velocity, even in a one-body system.

>>5060180
>Dude, I'm now in a coordinate system that has the origin where I used to be

Where did you use to be? How do you know how far you have traveled? You don't even know your own velocity, because there's nothing else in the system with you by which to measure it. And even if there were something else in the system with you, you could never know that that thing didn't *itself* have some velocity with respect to the origin.

>x front, y left, z up.

>implying you can tell which way is up when you have no other objects in the system with you to orient yourself with

>>	Anonymous Sat Sep 15 19:36:58 2012 No.5060185 >>5060171 Yeah, your contraption had something to do with weight, which changes depending on where you are. You also seemed to read something of a forcemeter and elastic cords don't function very well in the vacuum of space.

Anonymous Sat Sep 15 19:37:59 2012 No.5060187

>>5060161
>What are your definitions?
See >>5060047

>There's absolutely zero ambiguity whatsoever. The definite integral of the acceleration of an object, from t=a to t=a+1, is it's velocity at t=a+1. The definite integral of its velocity, from t=a to t=a+1, is its position at t=a+1.
I'm not asking you the boundary you are integrating over, I'm asking you WHAT you are integrating over. Also, what is time? What is position? What kind of values does it take? What does it map to in terms of coordinates?

>My definitions do not deal with relative velocity or distance.
But they do? You have to do a Galilean transformation between observers.

>Yes they do.
No, they do not. http://en.wikipedia.org/wiki/Principle_of_relativity

>The subjective frame is the only frame that matters when dealing with a single object.
What the fuck is a subjective frame?

>My definitions do not deal with multiple-body systems.
This has fuck all to do with multiple-body definitions.

>>	Anonymous Sat Sep 15 19:38:43 2012 No.5060188 >>5060184 Why is one of you assuming position and the other acceleration then? Why not the integral of position? Why not velocity? Why not the derivative of acceleration? And so on and so forth.

>>	Anonymous Sat Sep 15 19:39:29 2012 No.5060190 Oh god, too much crackpottery in this thread and conceptual difficulties with high school physics. I'm out, sorry for starting this argument over a simple freshman definition of velocity/acceleration.

Anonymous Sat Sep 15 19:39:58 2012 No.5060189

>>5060185
>Yeah, your contraption had something to do with weight, which changes depending on where you are.

Wrong. Your ignorance of the definition of the word "weight" betrays your lack of an education in physics.

Protip – when you are in an elevator that is accelerating upward at 9.8 meters per second, you weigh twice as much as you will weigh when the elevator stops. When you are in an elevator that is accelerating downward at 9.8 meters per second, you have zero weight.

Anonymous Sat Sep 15 19:40:43 2012 No.5060192

>>5060184
... my head is up, most of the time. And it's completely irrelevant if I can measure it or not, I have a new position in my coordinate system. You seem to be bogged up in the practical issues of measuring these things which is obviously completely irrelevant.

>>	Anonymous Sat Sep 15 19:43:28 2012 No.5060195 >>5060189 Weight is the gravitational force and is very much dependent on where you are, your weight will be about 1/6 on the Moon compared to Earth.

>>	Anonymous Sat Sep 15 19:44:59 2012 No.5060196 >>5060189 Weight is not what your scale says, be it on an elevator or otherwise. Wait, u troll me?

Anonymous Sat Sep 15 19:45:59 2012 No.5060197

>>5060187
>A path or curve on a manifold is a connected set of manifold points, with these path points labeled by a continuous, monotonically increasing real paramater, t.

Your definition requires that there exist a stationary cartesian grid by which to measure the positions of your "manifold points". There does not exist any such grid.

>Also, what is time?

Time is time, as experienced by the object whose acceleration you are measuring, and whose velocity and position you are calculating. Stop asking this question. It is inane.

>What is position?

I've already defined it. It is the integral of the integral of acceleration.

>You have to do a Galilean transformation between observers.

There is absolutely zero dependence on foreign observers, in my definitions.

>What the fuck is a subjective frame?

Are you a moron? Time as measured by a clock onboard the object. Acceleration as measured by an accelerometer onboard the object. How could you not deduce this? Why must I explain this to you, as if you are a child?

>This has fuck all to do with multiple-body definitions.

Your definitions require multiple bodies.

Anonymous Sat Sep 15 19:48:14 2012 No.5060205

>>5060192
>my head is up, most of the time.

So what you're saying is that your definition cannot be used, A, when the object is not anthropomorphic, and B, when the object *is* anthropomorphic, but not oriented away from the nearest source of gravity? That sounds like a pretty shitty definition. My definition is applicable no matter what the object, and no matter what its orientation with respect to local gravitational fields.

>You seem to be bogged up in the practical issues of measuring these things which is obviously completely irrelevant.

If you can't measure anything except acceleration, and if your definition of velocity requires things other than acceleration to be known, then your definition is unusable, and therefore sucks.

>>	Anonymous Sat Sep 15 19:50:58 2012 No.5060208 >>5060195 >Weight is the gravitational force Wrong. Take a physics class. >>5060196 >Weight is not what your scale says, be it on an elevator or otherwise. Yes, it is. >Wait, u troll me? Take a physics class.

>>	Anonymous Sat Sep 15 19:51:28 2012 No.5060211 >>5060205 I can easily measure my position, it's 0. And assigning axes to my coordinate system is also a practical triviality, assign them randomly for all I care and make sure they're orthogonal for simplicity.

>>	Anonymous Sat Sep 15 19:53:13 2012 No.5060213 >>5060208 http://en.wikipedia.org/wiki/Weight and check the references 1 and 2 which deal with this issue.

>>	Anonymous Sat Sep 15 19:53:13 2012 No.5060214 >>5060211 >I can easily measure my position, it's 0. Is your position still zero after one second?

>>	Anonymous Sat Sep 15 19:53:43 2012 No.5060215 >>5060208 You're out of your depth son.

Anonymous Sat Sep 15 19:54:43 2012 No.5060217

>>5060213
>http://en.wikipedia.org/wiki/Weight

Looks like someone stopped reading after the first sentence.

"In the 20th century, the Newtonian concepts of gravitation were challenged by relativity. Einstein's principle of equivalence put all observers, accelerating in space far from gravitating bodies, or held in place against gravitation near such a body, on the same footing. This led to an ambiguity as to what exactly is meant by the 'force of gravity' and (in consequence) by weight. The ambiguities introduced by relativity led, starting in the 1960s, to considerable debate in the teaching community as how to define weight for their students. The choice was a Newtonian definition of weight as the contact reaction-force against the force of gravity, for an object at rest on the ground, or an operational definition defined by the act of weighing. In the operational definition, weight becomes zero in conditions of weightlessness such as Earth orbit or free fall in vacuum. In such situations, the Newtonian view is that there remains a force due to gravity which is not measured (thus causing an apparent weight of zero), while the Einsteinian view is that there never does exist a measurable force due to gravity, even in everyday experience. Instead, weight and all sensation of weight are always produced by contact forces (push or pull) from the ground, or a scale. In free-fall, no force is measured simply because the force due to gravity is (still) never felt, and the floor (or the scale) now fails to exert the mechanical force that is what is always observed as 'weight.'"

>>	Anonymous Sat Sep 15 19:54:43 2012 No.5060218 >>5060214 If I'm not moving it is.

>>	Anonymous Sat Sep 15 19:56:43 2012 No.5060221 >>5060215 >You're out of your depth son. And you probably think that gravity is a force. >>5060218 >If I'm not moving it is. How do you measure whether you are moving or not?

>>	Anonymous Sat Sep 15 19:57:43 2012 No.5060223 >>5060217 Still doesn't define weight as anything else, just presents a problem with the current definition.

Anonymous Sat Sep 15 19:58:59 2012 No.5060225

>>5060223
>Still doesn't define weight as anything else, just presents a problem with the current definition.

Boy, you suck at reading.

"The choice was a Newtonian definition of weight as the contact reaction-force against the force of gravity, for an object at rest on the ground, or an operational definition defined by the act of weighing. In the operational definition, weight becomes zero in conditions of weightlessness such as Earth orbit or free fall in vacuum."

>>	Anonymous Sat Sep 15 19:59:28 2012 No.5060227 >>5060226 >observe that my position is changing How?

>>	Anonymous Sat Sep 15 19:59:59 2012 No.5060226 >>5060221 I don't measure anything, observe that my position is changing.

>>	Anonymous Sat Sep 15 20:02:59 2012 No.5060231 >>5060225 Lol, so you want to propose this even fuzzier definition to fix a non-problem with a clear definition?

Anonymous Sat Sep 15 20:05:28 2012 No.5060234

>>5060231
>Lol, so you want to propose this even fuzzier definition to fix a non-problem with a clear definition?

How is it fuzzy? Your acceleration, multiplied by your mass, is your weight. This is literally the simplest and most cut-and-dry formula in all of physics.

>>	Anonymous Sat Sep 15 20:05:59 2012 No.5060233 >>5060227 Magic? This is still an irrelevant practical matter that I am completely uninterested in. How do you measure acceleration in the vacuum of space where your rubber chords won't do anything?

>>	Anonymous Sat Sep 15 20:08:28 2012 No.5060239 >>5060234 The only other proposed definition of weight was an operational definition based on the act of measuring weight, which is very fuzzy.

Anonymous Sat Sep 15 20:08:58 2012 No.5060238

>>5060233
>Magic? This is still an irrelevant practical matter that I am completely uninterested in.

Physics is the study of the real world. If your definitions are inoperable in the real world, then they suck and should be abandoned.

>How do you measure acceleration in the vacuum of space where your rubber chords won't do anything?

You fail physics forever. I wish I could slap you.

The lead sphere has mass. Therefore it resists the acceleration of the spacecraft. Therefore it exhibits force on the cords holding it in place. This is so simple that a child could understand it; yet you have failed to grasp this basic idea. I pity you.

Anonymous Sat Sep 15 20:09:14 2012 No.5060241

>>5060197
Sorry, I just had to respond to this. It made me too mad. You fail at elementary physics. Be sure to publish your findings on viXra! I'm sure they'll be glad to read it. You do not understand simple classical mechanics. You have a set of coordinate basis vectors, you need an observer to define a coordinate system. If you do not have this, you cannot have physics. A linear transformation of the vector components of said observer gives you a different reference frame.

I think you should read this:
>http://en.wikipedia.org/wiki/Tests_of_special_relativity#Basic_experiments
>There is no preferred inertial frame under Newtonian mechanics or special relativity
>Any uniformly moving observer in an inertial frame cannot determine his "absolute" state of motion by a co-moving experimental arrangement.

>There is absolutely zero dependence on foreign observers, in my definitions.
LMFAO. Your definitions fail in preserving the laws of physics in every frame of reference, then, which means that causality does not work, the speed of light is not constant, and Lorentz invariance fails.

>There is absolutely zero dependence on foreign observers, in my definitions.
No, there is not. How do you measure acceleration, velocity, or position without an observer? How does the speed of light remain constant in every frame of reference?

> Time as measured by a clock onboard the object.
I asked you how you define time in mathematical terminology, not how to measure it. How does time evolution work?

> Acceleration as measured by an accelerometer onboard the object.
Obviously you do not understand what F=ma means. If you have two inertial objects and they both experience the same amount of acceleration, it's impossible to tell that acceleration exists.

You're up to almost 80 points on http://math.ucr.edu/home/baez/crackpot.html, good job!

>>	Anonymous Sat Sep 15 20:10:28 2012 No.5060245 >>5060241 Also, acceleration, velocity, and position of what? What is this object you speak of? How do you calculate things?

Anonymous Sat Sep 15 20:10:44 2012 No.5060246

>>5060239
>an operational definition based on the act of measuring weight, which is very fuzzy.

It isn't fuzzy in the slightest. In the presence of a force, an object with mass will resist that force, by pushing in the opposite direction. This resistance can be measured; the measured force is the weight of the object. Simple as that.

>>	Anonymous Sat Sep 15 20:11:58 2012 No.5060247 >>5060238 My definitions work fine in the real world, it's just not my problem to obtain the data. And in the vacuum of space at, what... 0,3K no part of your contraption will measure anything.

>>	Anonymous Sat Sep 15 20:15:43 2012 No.5060258 >>5060246 Seems like the only thing you're doing is dealing with the mass of the object, you have an arbitrary force and you see how an object reacts to it.

Anonymous Sat Sep 15 20:16:15 2012 No.5060259

>>5060241
>you need an observer to define a coordinate system. If you do not have this, you cannot have physics.

I have an observer. He is a passenger onboard a spacecraft. He is trying to measure his spacecraft's velocity. There are no other objects in the system, except for him, and the spacecraft he is in.

How does he measure his velocity, with respect to his starting velocity? How does he measure his position, with respect to his starting position? He cannot, unless he employs MY definitions.

The fact that you cannot see this is proof positive that it is you, and not I, who fail at elementary physics.

>How do you measure acceleration, velocity, or position without an observer?

I said "foreign observer". As in, an observer substantively outside of the frame of reference of the object being measured.

In addition to elementary physics, you also fail at reading comprehension.

>I asked you how you define time in mathematical terminology, not how to measure it.

All we are concerned with here is measurement.

>If you have two inertial objects and they both experience the same amount of acceleration

I'm going to make this really simple for you, so that you can understand.

1. Cord is attached to wall of box.
2. Box is attached to floor of spacecraft.
3. Lead sphere is attached to cord.
4. Cord is elastic.

When spacecraft accelerates, lead sphere pulls backward on cord. Force exerted on cord is measurable. Therefore, acceleration is measurable.

>>	Anonymous Sat Sep 15 20:17:13 2012 No.5060261 >>5060259 Fuck off, Carl

Anonymous Sat Sep 15 20:18:28 2012 No.5060263

>>5060247
>it's just not my problem to obtain the data.

It's not a question of whether you personally are the one obtaining the data. It's a question of whether the data you are requesting can even be obtained at all.

>And in the vacuum of space at, what... 0,3K no part of your contraption will measure anything.

>implying my contraption could not be mounted onboard a heated spacecraft

>>5060261
>Fuck off, Carl

My name is Jacob.

Anonymous Sat Sep 15 20:22:28 2012 No.5060272

>>5060263
Yes, by magic. And your contraption only works where you have an unlimited power source to heat your equipment? You want to be able to measure the acceleration always and everywhere, right? Seems awfully restrictive to me. To be honest, I don't care how you get the acceleration, from where I'm sitting it's just as easy as getting the position.

>>	Anonymous Sat Sep 15 20:27:43 2012 No.5060275 That's why experimental physicists and mathematicians shouldn't talk.

>>	Anonymous Sat Sep 15 20:28:58 2012 No.5060276 >>5060272 >Yes, by magic. I hate to break this to you, sir, but that ain't real. (Continued – I have to break my post into pieces because the system thinks something in it is spam; and I need to find the problem.)

>>	Anonymous Sat Sep 15 20:29:28 2012 No.5060279 (Continued from >>5060276.) >And your contraption only works where you have an unlimited power source to heat your equipment? Who says my power source needs to be unlimited? I'm not measuring my acceleration for all of eternity. (Continued.)

Anonymous Sat Sep 15 20:30:58 2012 No.5060280

(Continued from >>5060279.) I'm just measuring it over a finite interval of time.

>You want to be able to measure the acceleration always and everywhere, right?

"Starting at any time and in any place" would be a better way of setting the parameters.

Your attempt at a reductio-ad-absurdum is kind of pathetic.

>from where I'm sitting it's just as easy as getting the position.

In physics, we seek universal laws and definitions. Your definitions are only applicable where you are sitting (on the earth). Mine are applicable anywhere.

Anonymous Sat Sep 15 20:30:59 2012 No.5060281

>>5060259
>I have an observer. He is a passenger onboard a spacecraft. He is trying to measure his spacecraft's velocity.
If he is inertial wrt the spacecraft, he cannot. Again, "Any uniformly moving observer in an inertial frame cannot determine his "absolute" state of motion by a co-moving experimental arrangement."

>He cannot, unless he employs MY definitions.
Your definitions involve integrals. An integral requires coordinate basis vectors, you must employ a coordinate system. Let me remind you that an integral is over the exterior derivative of a differential form. A differential form is defined a=\frac{1}{2}a_{ij}dx^i dx^j[/spoiler] whereas dx^{i...j}[/spoiler] are coordinate basis forms.

>The fact that you cannot see this is proof positive that it is you, and not I, who fail at elementary physics.
Inertial observers cannot measure acceleration. Are you fucking 5?

>As in, an observer substantively outside of the frame of reference of the object being measured.
There will always exist a coordinate transformation if the frames are not identical. "Substantively" is irrelevant.

>In addition to elementary physics, you also fail at reading comprehension.
I think you do. You made a claim. You stated your definitions are fundamentally different from the ones standard in physics. I asked you to define time in your framework. This means give me a framework that allows you to calculate the time evolution of a physical system. Can you do this?

>>	Anonymous Sat Sep 15 20:31:28 2012 No.5060282 >>5060275 >That's why experimental physicists and mathematicians shouldn't talk. What do mathematicians even do? It seems to me that the entire profession is a relic of the pre-computer era.

Anonymous Sat Sep 15 20:31:28 2012 No.5060283

>>5060281
>All we are concerned with here is measurement.
Can you explain to me how measurements work? Can you explain to me using some actual mathematical machinery how your philosophical bullshit works or are you just going to rant on?

>I'm going to make this really simple for you, so that you can understand.
Again, you fail to understand simple physics. Your thought experiment fails because the lead sphere is in a different frame of reference than the spacecraft. It experiences a different force because of its mass. Mass sets inertial resistance. You are in a different coordinate system.

>>	Anonymous Sat Sep 15 20:36:58 2012 No.5060285 >>5060282 You are a fucking idiot. The mathematics done right now is not computation. Get out of this conversation right now because you have nothing to contribute.

Anonymous Sat Sep 15 20:38:28 2012 No.5060287

>>5060280
Fuck it, I want to measure the acceleration of a hydrogen atom in the center of the sun. Your contraption is fucked.
>>5060276
Who says magic isn't real? Just because you haven't observed it? That's just because it only gives ME my position in my coordinate system.

Anonymous Sat Sep 15 20:38:44 2012 No.5060288

>>5060281
>If he is inertial wrt the spacecraft, he cannot.
>"absolute"

I am not speaking of absolutes; and you know it.

Do I really need to say "velocity with respect to his starting velocity" and "position with respect to his starting position" instead of "velocity" and "position"? Are you really going to be that pedantic?

>An integral requires coordinate basis vectors, you must employ a coordinate system.

A vector of acceleration is obtainable via accelerometer. I've already described how such an accelerometer would work. Do I really need to go into deeper detail for you?

>Inertial observers cannot measure acceleration.

Explain why my accelerometer would not work. I've already explained why it would. You have not presented a counter-argument.

>"Substantively" is irrelevant.

The lead sphere is approximately within the same reference frame as the remainder of the accelerometer inside of which it rests. It is not exactly, because it has free range of motion within. If you are going to object to my accelerometer having moving parts, then get the fuck out.

>I asked you to define time in your framework. This means give me a framework that allows you to calculate the time evolution of a physical system. Can you do this?

If we are dealing with a one-body system, evolution is irrelevant; only passage is necessary to measure. And there are atomic clocks that will work perfectly fine in that respect.

Anonymous Sat Sep 15 20:43:34 2012 No.5060292

>>5060283
>Can you explain to me how measurements work? Can you explain to me using some actual mathematical machinery how your philosophical bullshit works or are you just going to rant on?

What are you even asking, here? I've already described how my accelerometer works.

>Your thought experiment fails because the lead sphere is in a different frame of reference than the spacecraft.

Yeah? So?

>>5060285
>You are a fucking idiot. The mathematics done right now is not computation.

Name something mathematicians do.

>>5060287
>Fuck it, I want to measure the acceleration of a hydrogen atom in the center of the sun. Your contraption is fucked.

The contraption is one theoretical way of measuring what needs to be measured for my definition's sake. Obviously we can imagine all sorts of different methods for individual scenarios, as long as all the required variables are being filled accurately.

Anonymous Sat Sep 15 20:44:33 2012 No.5060294

>>5060288
>I am not speaking of absolutes; and you know it.
You said
>Your definitions require multiple bodies.
and
>The subjective frame is the only frame that matters when dealing with a single object.
and
>My definitions do not deal with relative velocity or distance.
Which implies the exact opposite. You are a pathetic crackpot confusing himself.

>A vector of acceleration is obtainable via accelerometer. I've already described how such an accelerometer would work. Do I really need to go into deeper detail for you?
>Explain why my accelerometer would not work. I've already explained why it would. You have not presented a counter-argument.
I've already told you why the accelerometer is in a different frame of reference and experience a difference force.

>The lead sphere is approximately within the same reference frame as the remainder of the accelerometer inside of which it rests. It is not exactly, because it has free range of motion within.
You contradicted yourself. You now imply there exists a coordinate transformation between the ship and the lead sphere. This is correct and does not violate any physical laws.

>If we are dealing with a one-body system, evolution is irrelevant; only passage is necessary to measure.
No, because again the laws of physics remain constant in every frame of reference. Newton's laws, -\nabla V = \dot{p}[/spoiler] say particles move along geodesics when not acted on by any potential. Reproduce this with your definitions or you have no argument.

>>	Anonymous Sat Sep 15 20:46:03 2012 No.5060295 >>5060292 >Yeah? So? You said preferred frames exist in >>5060161 Let me quote you again. >Yes they do. The subjective frame is the only frame that matters when dealing with a single object. My definitions do not deal with multiple-body systems.

Anonymous Sat Sep 15 20:47:33 2012 No.5060296

>>5060292
In other words, you can't measure the acceleration of the hydrogen atom... so it has none?
By the way, mathematicians are generally a few hundred years ahead of their times, Hilbert spaces was just a mathematical concept for a hundred years until some physicists figured they could use that shit for something.

Anonymous Sat Sep 15 20:48:19 2012 No.5060299
File: 68 KB, 500x330, 1344465146681.jpg [View same] [iqdb] [saucenao] [google]

Is Who wants to be a millionaire just a game of luck?

I mean, there's 33.34% chance of being right unless the person happen to know the answer, and lifelines can only get you so far into the game

Anonymous Sat Sep 15 20:49:19 2012 No.5060303

>>5060299

Let's say there are 15 questions with 4 choices each. Let's also say there are 3 lifelines: call-a-friend, 50-50, and ask-the-audience.

For each of the 15 questions, there is only 1 correct answer, and 3 incorrect answers, so immediately your chances are 1/3 = 33.34%. You can use each lifeline once. Using phone-a-friend, let's say gives you a 95% chance the person you call is correct. So 1 question is 95%. Similarly, we can say ask-the-audience gives you a 95% chance. Finally, 50-50 will remove 2 wrong answers, so you have 1 right answer and 1 wrong answer, so you basically know the answer. To recap:

Question 1: 33.34%

Question 2: 33.34%

Question 3: 33.34%

Question 4: 95% <--- ask-the-audience

Question 5: 33.34%

Question 6: 33.34%

Question 7: 33.34%

Question 8: 33.34%

Question 9: 95% <--- phone-a-friend

Question 10: 33.34%

Question 11: 33.34%

Question 12: 33.34%

Question 13: 33.34%

Question 14: 33.34%

Question 15: 100% <--- 50-50

Total: 690.08%

Divided by 15 questions = 46.00%

So basically, it's more a less a game of chance, but more than half the people on the show will lose, so it's not very fair. That show is making tons of money off of people.

>>	Anonymous Sat Sep 15 20:50:03 2012 No.5060305 And this kids is what happens when autistics try and agree on something.

>>	Anonymous Sat Sep 15 20:50:49 2012 No.5060307 >>5060305 shut up fuck face

Anonymous Sat Sep 15 20:51:34 2012 No.5060308

>>5060294
>Which implies the exact opposite.

I also said "velocity with respect to starting velocity" and "position with respect to starting position". Which does not imply anything absolute.

>I've already told you why the accelerometer is in a different frame of reference and experience a difference force.

I haven't refuted this. Yes, the lead sphere is in a slightly different frame of reference. That's obvious. What's your point?

>You contradicted yourself. You now imply there exists a coordinate transformation between the ship and the lead sphere.

I never meant to imply that there didn't. You are the one perceiving a contradiction that isn't there.

>No, because again the laws of physics remain constant in every frame of reference.

How does this contradict anything I've said?
All I need to do here is measure the passage of time, and the force exerted by the subtle movement of a lead ball. These are things which can be accomplished all within the laws of physics.

>>5060295
>You said preferred frames exist

And then specified that by "preferred" I meant "subjective".
How can you fail at such simple reading comprehension?

>>5060296
>In other words, you can't measure the acceleration of the hydrogen atom... so it has none?

My definition does not require that humans be able to measure the acceleration; it only requires that the acceleration be known. And for all practical scenarios, the acceleration can be known. That's my only point.

Anonymous Sat Sep 15 20:56:04 2012 No.5060312

>>5060308
You keep telling other people they have horrible reading comprehension - you yourself do not. You should not use terms when you have no idea what they mean. Your definitions are exactly identical to >>5060047 except you seem to not understand what they really mean.

http://en.wikipedia.org/wiki/Preferred_frame

"In theoretical physics, a preferred or privileged frame is usually a special hypothetical frame of reference in which the laws of physics might appear to be identifiably different (simpler) from those in other frames."

http://wordnetweb.princeton.edu/perl/webwn?s=inertial%20frame

"Inertial reference frame: a coordinate system in which Newton's first law of motion is valid."

Anonymous Sat Sep 15 20:56:04 2012 No.5060313

>>5060307
Lol, I'm one of the autists and that shit's just funny.
>>5060308
Hahaha, there is no reason for you to believe that the acceleration can be known if you need to measure everything. That was your entire point, I have to be able to measure it to know it while I said that I don't care how you know it.

Anonymous Sat Sep 15 21:04:18 2012 No.5060322

>>5060312
Fine. I'm guilty of misusing a term. I'm sorry.

>>5060313
You don't seem to understand the difference between "feasible" and "possible".

"Feasible" means there is a way that humans could accomplish it through technological means.

"Possible" means that there is no physical, logical, or mathematical reason why it couldn't be accomplished in some conceivable way.

It is always possible, in any system, to measure the acceleration of any object larger than a point particle, even if it is not feasible to measure the acceleration of anything smaller than the smallest device humans can invent to accomplish said measurement.

It is NOT always possible, in every system, to measure position, because not all systems will contain foreign objects whose distance and direction from you can be measured.

Anonymous Sat Sep 15 21:05:48 2012 No.5060324

One more thing to add (I'm >>5060312)

People ITT are confusing two things from what I see

coordinate acceleration - the second time derivative of the position wrt some system of coordinates

proper acceleration - the acceleration measured by an accelerometer (i.e. the acceleration "felt" by an observer)

Note that the coordinate acceleration is a frame-variant quantity while the proper acceleration is frame-invariant.

The coordinate acceleration of anything in its own rest frame is always 0. If that object is accelerating relative to an inertial coordinate system then the object's rest frame is non-inerital and the proper acceleration of that object will be non-zero.

>>	Anonymous Sat Sep 15 21:07:18 2012 No.5060328 >>5060324 Then, in this thread, when I have said "acceleration", please assume I have meant "proper acceleration".

>>	Anonymous Sat Sep 15 21:09:34 2012 No.5060331 >>5059993 Well yes, you can change velocity without acceleration. How? Move around a corner at a constant speed. Velocity is a vector, speed and direction.

>>	Anonymous Sat Sep 15 21:14:18 2012 No.5060339 >>5060331 Not true, changing directions is also acceleration.

>>	Anonymous Sat Sep 15 21:15:33 2012 No.5060341 >>5060339 Mind you, it's centripetal acceleration.

>>	Anonymous Sat Sep 15 21:40:33 2012 No.5060367 faggots, all of you

light superobserver Sat Sep 15 22:01:33 2012 No.5060407

Velocity is an observable, not an operator. A change in velocity is measured as a change in momentum and can be calculated as a change in the frequency for some constant.

Velocity is equivalent to the absorption of a photon.

A decent book.

http://www.amazon.com/Maxwell-Pragmatic-Quantum-Relativity-ebook/dp/B002WPZVFA

Anonymous Sat Sep 15 22:02:04 2012 No.5060408

lol wow, this thread.

I don't think anyone really understood what OP meant either.

In order to have some velocity you need to apply to it an acceleration, and in order to have some acceleration you need some jerk, you can keep going on this perpetually.

Think about it with differentials.

>>	superobserver Sat Sep 15 22:04:48 2012 No.5060412 You only need a measured difference in momentum and this is always transmitted in units of hf (plancks constant times some frequency) and hence the observation of velocity is relative to a local frame and is observable only via a photon absorption.

>>	Anonymous Sat Sep 15 22:07:48 2012 No.5060416 >>5060412 What is jerk then? or for that matter jounce/jilt/jouse/jolt/delta jerk, snap, crackle, pop, etc..?

brownian motion superobserver Sat Sep 15 22:11:48 2012 No.5060420

All changes in state are accompanied by a change in energy measured in units hf. So a jerk can be quantified as some change (be it positive or negative) in photon energy for a given material object.

This is not relate f=ma to e=hf (yet).

>>	brownian motion superobserver Sat Sep 15 22:22:34 2012 No.5060439 File: 455 KB, 240x240, Brownian_motion_large.gif [View same] [iqdb] [saucenao] [google] And so it goes that every collision transmits energy in units hf.

>>	Anonymous Sat Sep 15 22:43:49 2012 No.5060474 >>5060439 Didn't Einstein write one of his early papers on the subject of brownian motion?

>>	Anonymous Sat Sep 15 22:51:49 2012 No.5060489 >Can you have velocity without acceleration? >122 posts and 1 image reply omitted. Click here to view. Now I know why I swore off this fucking cesspool of retardation. Are you faggots serious?

Einstein superobserver Sat Sep 15 22:53:19 2012 No.5060491

Yes he did. While I will not claim to have read it I will say that Brownian-scale forces can be calculated as changes in photon energy (at the level of electron valence and at the level of simple momentum or temperature).

An object in motion tends to stay in motion until acted upon by an opposing (or additive) force - this force is quantified in units hf.

>>	Anonymous Sat Sep 15 22:54:03 2012 No.5060493 >>5060489 To be fair, it's because it's one of the few concepts that everyone that's finished high school will know about. Hence lots of replies.

Anonymous Sat Sep 15 22:56:48 2012 No.5060498

>>5060489
>Now I know why I swore off this fucking cesspool of retardation. Are you faggots serious?

We got into an argument over whether acceleration was the second derivative of position, or whether position was the second integral of acceleration. The debate was ended when >>5060324 pointed out that, in physics, there are two types of acceleration – coordinate acceleration, which *is* simply the second derivative of position, and "proper" acceleration, which is indeed simply the acceleration as measured by an accelerometer. Thus, everyone was right.

>>	Anonymous Sat Sep 15 22:58:03 2012 No.5060500 >>5060491 lol. wat

acceleration superobserver Sat Sep 15 23:00:18 2012 No.5060503

There is of course the situation where a material objects has no position (or it is distributed to Unity) when it has no velocity.

This is a problem that crops up using Schrodinger's Equation. The probability that an electron has a location is zero when the electron has zero velocity. Or, I should say, the electron can appear at any photon in the universe.

This is quite obviously wrong. But it persists as a result of using Time as an independent parameter, rather than tying time to something like "number of photon events".

Anonymous Sat Sep 15 23:00:33 2012 No.5060506

>>5060407
>>5060420
>>5060412
>>5060439
>Velocity is an observable, not an operator.
I don't think you know what an operator is. Velocity is an operator in QM.

>A change in velocity is measured as a change in momentum
That's no necessarily true. There are many cases where velocity can change and momentum does not.

>Velocity is equivalent to the absorption of a photon.
No, it is not. You are spewing nonsensical crackpot drivel.

Momentum is not the same as energy, and that's not even relativistic. This question has nothing to do with the de Broglie relations.

>All changes in state are accompanied by a change in energy
No, they are not. States are rays in Hilbert space which contain all the information in the density matrix. Changes in the density matrix do not always imply changes in energy.

> measured in units hf.
No. That's nonrelativistic momentum.

>So a jerk can be quantified as some change (be it positive or negative) in photon energy for a given material object.
Jerk is an entirely different concept from photon energy. You have no idea what you're talking about.

>>	Anonymous Sat Sep 15 23:01:48 2012 No.5060509 >>5060498 which type is gravity most like?

Anonymous Sat Sep 15 23:04:48 2012 No.5060516

>>5060509
>which type is gravity most like?

Technically, gravity isn't even a true force. Freefall is inertial. So "acceleration" from gravity is really just coordinate acceleration, and not "proper" acceleration at all; much like the "centrifugal" acceleration observed in objects within a rotating reference frame.

velocity operator superobserver Sat Sep 15 23:06:49 2012 No.5060518

States are rays in a Hilbert space, true. Mutually orthogonal states can have the same energy, true.

I should say that you cannot observe an event without observing a photon or a change in energy. And so a change in state is a change in energy inasmuch as you cannot encode that an event has actually occurred until you have absorbed a photon. Or, until you have absorbed energy quantized in photon units.

Velocity is an observable, not an operator, in QM. That is an argument, not a statement of fact. I am arguing that you cannot operate upon a mass to provide acceleration if you cannot observe the energy added during acceleration. The observation of the energy is the observation of a frequency differential over some period of time (time being a tricky subject here).

So velocity is an operator inasmuch as it is quantized in units hf.

Anonymous Sat Sep 15 23:09:18 2012 No.5060527

>>5060489
This is actually a (fairly) difficult question in physics, believe it or not. Proper acceleration is frame independent with the Rindler metric, and this is very paradoxical. Accelerating frames, besides being very trivial physics, are still not completely understood.

For example, an accelerating charge produces electromagnetic radiation, but if acceleration is frame independent this implies a charged particle in a gravitational field can produce an infinite energy source.

See, eg: http://xxx.lanl.gov/pdf/gr-qc/0006037v1.pdf

rate of change in acceleration = jerk superobserver Sat Sep 15 23:10:18 2012 No.5060529

the rate of change is simply the difference in absorbed energy.

So if you want to measure the change imparted by having a velocity, you must measure the velocity. And so what units do you think velocity is transmitted in when it is observed?

>>	Anonymous Sat Sep 15 23:10:48 2012 No.5060532 >All this shit I'm so glad I went into mathematics instead of physics.

>>	Anonymous Sat Sep 15 23:11:34 2012 No.5060535 >>5060518 >Velocity is an observable, not an operator, in QM. So <span class="math">\langle \dot{x} \rangle = \frac{i}{\hbar}[H,x][/spoiler] isn't correct? Any observable is an operator, btw.

>>	Anonymous Sat Sep 15 23:11:34 2012 No.5060536 >>5060532 Why? How is that even useful?

observable operator superobserver Sat Sep 15 23:17:18 2012 No.5060553

Yes but rather than using the standard notion of classical velocity you want to quantize velocity in units of photon energy. You can replace velocity altogether if you simply say that the ***observable difference*** in energy for an object in motion is calculated in units hf.

Anonymous Sat Sep 15 23:30:03 2012 No.5060583

>>5060553
Velocity is quantized. The momentum eigenstate is \hat{p} | \psi \rangle = p | \psi \rangle[/spoiler], if we're dealing with nonrelativistic physics where canonical momentum = regular momentum, velocity is just \hat{v}=\frac{\hat{p}}{m}[/spoiler], so \hat{v} | \psi \rangle = v | \psi \rangle[/spoiler].

It's an observable \langle v \rangle = \langle\psi|\hat{v}|\psi\rangle[/spoiler]

And has uncertaintly \Delta x\; \Delta p \ge \frac{\hbar}{2}\;\; \to \;\; \Delta x\; \Delta v \ge \frac{\hbar}{2m}[/spoiler]

Not sure what you're implying. The physical significance can be seen in the group velocity of a free wavepacket.

If velocity did not exist, particles would not move across their target manifold. Interaction would never happen.

Target manifold superobserver Sat Sep 15 23:35:03 2012 No.5060591

http://books.google.com/books?id=sdVrBM2w0OwC&pg=PA206&lpg=PA206&dq=relativistic+velocit
y+operator&source=bl&ots=MYFkasVkqy&sig=NC7VCy6_q3Px901znpO2YOmv7vI&hl=en#v=onepage&
amp;q=relativistic%20velocity%20operator&f=false

I am stating that you cannot observe a velocity without registering an event. The event is encoded as a change in energy. The chane in energy is quantized in units h. I am not speaking about determining the location (or, spread of possible locations) for a particle having a given energy (or momentum) I am specifying that the observation of velocity requires an event and that such an event is measured in units hf=E.

So a specific observation that transmits information regarding velocity is quantized in units of hf.

I am not sure I understand how you are deriving that relativistic velocity is quantized. It (the velocity operator in QM) is p/m and this is the definition of a classical relation, not a quantum one.

Anonymous Sat Sep 15 23:40:19 2012 No.5060603

Hokay so to the guy who was talking about the intergral form of accelertaion.
Beilieve it or not you have at one point implied a coridiante system. When you mesure an accleration, you defined it as when the lead mass was not moving. that is when you defined it. You defined it relitive to that mass at a time t. If someone was walking by the space ship at time t, how do you rectify that he see's the acceleration from him different? If you can't then your system does not hold to observations made, and does not hold with physics.
How do you convey this acceleration to other people? lets assume you are measuring the blox moves in meters. You see it move 1m directly twords the aft of the ship. But me standing outside the ship sees it move 1.1m directly aft. how do you rectify these two different observation of the same thing?

Also you do not need a second observer to define a cordinate system. you simply define it. Does not matter how so your one body system does not change with this fact. You definitions do 'work' however you are trying to add something to a feild of physics that has its shit sorted out pretty well, and has easier methods to aquire your answers by simply defining a cordinate system.

Anonymous Sat Sep 15 23:42:18 2012 No.5060607

>>5060603
The device containing the lead weight was simply a basic accelerometer design that I proposed off the cuff. There are other methods by which such devices can work. Does it matter whether they involve moving parts, which would naturally have different reference frames? The point is that proper acceleration is measurable.

>>	Anonymous Sat Sep 15 23:43:33 2012 No.5060612 >>5060591 OK, that makes much more sense. I thought you were implying you can replace the physical interpretation of velocity with energy.

coordinate systems superobserver Sat Sep 15 23:44:48 2012 No.5060617

There are no coordinate systems, there are only events.

That sounds argumentative but bear with me....

When you mention time, you are talking about a series of events. At time t an event takes place. At time t+1 another event happens. This is not the correct way of looking at the situations.

The object being observed is emitting photons. You will observe that these photons are red or blue shifted or remain constant. Velocity (relativistic sense) is the registration of these differences. It does not matter if you are moving or the observed object is moving the only detectable event is a photon, and this photon has a red or blue shift.

Anonymous Sat Sep 15 23:46:18 2012 No.5060624

>>5060607
You asked why this would not work I did. Your results can not always be comunicated to someone else. This is why it would not work. While yes it is measurable, in order for it to be meaining full you define a cordinate system. Your argument that you don't is now no longer true. Other wise your integration is now leaving 2 arbitrary constants that can only be solved by defining an origin, which defines a coordinate system.
And these other stystems with out defineing anything still leave this paradox of two different observations made for the same thing.

Anonymous Sat Sep 15 23:48:48 2012 No.5060630

>>5060624
You seem to be putting a lot of words in my mouth.

I only claimed that, because position is not measurable without external references, it is best to define position (with respect to starting position) as the second integral of proper acceleration. I never said this could or should be done without the use of a coordinate system.

coordinates superobserver Sat Sep 15 23:51:33 2012 No.5060637

The "two different observations" are equal to "two observers detecting separate photon(s)." So they are different events because they involve different information.

While one observer registers a redshifted photon series another registers a blue shifted photon series. The blue shifted photons are more frequent (more per unit local time) than the redshifted photons. Thus the idea of distance traveled is different for different observers. But there is no coordinate system, or if there is it is comprised only of photons separating observers.

Anonymous Sat Sep 15 23:51:33 2012 No.5060638

>>5060617
ok, but then how do you define this red and blue shift?
All you really said there was that there is no absolute reference system, but even from these photons you defined a reference frame. And in order to quantify this red and blue shift you define a cordinate system, it may only be depending on the amount of red and blue shift, but it is still defined.

>>	Anonymous Sat Sep 15 23:52:18 2012 No.5060641 >>5060630 my aplogies, I may now be confusing you with >>5060637

>>	Anonymous Sat Sep 15 23:54:33 2012 No.5060649 >>5060637 How can you define a notion of distance with out attaching a 0 point? even with this red and blue shift, you still define no difference as a 0, wich now you attach a cordiante system.

red and blue superobserver Sat Sep 15 23:57:03 2012 No.5060654

You define a local energy. IF you have an opportunity to compare results, you will notice that one result indicates light of a higher frequency then the other recorded result.

So I am saying that in the frame of the observer you cannot determine whether the sequence of photons is red or blue shifted unless there is a change from one photon event to the next. Or,

h*1 @t1 , h*2 @t2, h*3 @t3 .... etc indicates that the point source for the photons is accelerating.

The same is true in reverse. But, when you have,

h*1 @t1, h*1 @t2, h*1 @t3....... etc., you cannot say if the point source has a velocity. Or, you can say that if the point source has a velocity is has no acceleration.

>>	Anonymous Sat Sep 15 23:59:18 2012 No.5060657 >>5060009 You are right, thread ends here. >>5060019 Fuck you. No, you are wrong. Fuck. Trying to back up your argument with a "no"? Get the fuck out.

Distance superobserver Sun Sep 16 00:00:33 2012 No.5060659
File: 41 KB, 320x400, awesome.jpg [View same] [iqdb] [saucenao] [google]

Distance is the number of photon events between you and a point source. Alpha Centari is 4 light years away. But is that measured in miles or photon events?

It is measured in photon events. There is no distance without an event.

Anonymous Sun Sep 16 00:01:48 2012 No.5060663

>>5060654
While that is still dandy, you still define a 0. forgive be if in the 100+ posts I started to confuse arguments, but the whole point was to find a defintion of acceleration that was independent of a refrence frame yes? And the agrument has been made that no matter how you measure it you still, at some point, defined a 0. Your red and blue shift also implies you know the staring frequency.

>>	Anonymous Sun Sep 16 00:02:03 2012 No.5060665 >>5060657 >You are right, thread ends here. The question is much deeper than this. If you have two comoving spacecraft that are accelerating with respect to an inertial observer, do the individual spacecraft see acceleration with respect to each other?

>>	Anonymous Sun Sep 16 00:04:03 2012 No.5060672 >>5060665 Actually, yes. The signals between them would have to cross longer and longer distances. They would perceive each other as moving farther away from each other; and each would perceive the other, paradoxically, as moving ahead of them.

Anonymous Sun Sep 16 00:05:33 2012 No.5060677

>>5060659
while yes there is no distance with out an 'event' meaning an observation, distance is not measured in photons. I think you need to go back and do a unit analysis.
Photons are measured by energy. When you mesure a red or blue shift you are actually measuring a difference in energy. distance is well, distance. Given a red or blue shift you can not obtain distance, only a difference in energy.

superobserver Sun Sep 16 00:05:33 2012 No.5060678

If the spacecraft are very far apart, and they wait a long time until they can both see each other (start getting photons from each others ships) then we can define a span of space equal to teh number of photons of a given frequency that can populate a space.

So, we have a distance that is "filled" with a given number of photons of a given frequency (and wavelength). When the ships start to accelerate toward each other, it will take ***some time*** for the new signal (the blueshifted photon) to reach the distant ship.

>>	superobserver Sun Sep 16 00:07:03 2012 No.5060680 If there are no photons between Earth and Alpha Centari how can an observer experience distance?

>>	Anonymous Sun Sep 16 00:07:48 2012 No.5060682 >>5060672 Sure, and for the case v << c proper acceleration is the same as acceleration. That's a triviality. But what is happening to their wordlines?

>>	Anonymous Sun Sep 16 00:07:48 2012 No.5060683 >>5060672 how do you rectify this paradox?

superobserver Sun Sep 16 00:10:03 2012 No.5060691

If you skip the photons between you and a distant point you do not have to travel the distance. Paradox solved (if I understand it correctly).

Distance exists so long as events exist. If you can avoid the events, you can go straight to Alpha Centari. In an instant.

Anonymous Sun Sep 16 00:10:18 2012 No.5060692

>>5060678
but the amount of photons that can fill space is infinite. you can send an infinite number of photons though any region of space. So you can not mesure distance by this notion. Also if they are moving at the same velocity, you no longer have a red or blue shift to measure anything off of.

Anonymous Sun Sep 16 00:13:18 2012 No.5060702

>>5060691
I think your not understanding an event. What I defined as an event was simply an observation. However that does not mean becouse I do not make an observation that the measurment is imposible. What you are trying to say is that because I have my eyes closed I can not measure distance, therefore I can be anywhere. This is not true. The distance will still exist regradless if you look.

superobserver Sun Sep 16 00:14:33 2012 No.5060704

You can easily send a finite (precise) number of photons through space.

http://iopscience.iop.org/0034-4885/68/5/R04

Using a precise number of photons (in an experimental setting) you can arrange the experiment as explained. Also there are not an infinite number of photons or there would be an infinite amount of heat generated upon every material surface.

superobserver Sun Sep 16 00:17:34 2012 No.5060714

Closing your eyes is not the same as having no velocity. The absence of velocity spreads the probability of location to Unity (infinity). That makes the zero-v state a very interesting one from the point of view of Schrodinger's eq.

However, if two ships are moving away from each other and sending signals out toward each other then we must assume at some point in time they were closer to each other. And by closer I mean to say that fewer photons separated them.

Anonymous Sun Sep 16 00:24:04 2012 No.5060730

>>5060704
the number of photons emited is not dependant on distace, only the number recieved. so the only way this would work would be if you know the number of photons emited, and then the number recieved. However in doing this the cordiante system is not dependent on the photons itself but rather you defined a radial one centered at the emision source, and the photons really become a tool of measurment rather then the measurement itself

superobserver Sun Sep 16 00:25:03 2012 No.5060732

To be more precise, let me say that space exists, but space is defined as the number of photons of a given wavelength that can "fit" within a finite length.

Then if we say that two objects are separated by 10 red photons of space we are saying that two objects are separated by 550 nanometers * 10.

Anonymous Sun Sep 16 00:29:33 2012 No.5060743

>>5060714
ok I don't think your really understand that equation. That equation does not contridict with classical physics. I mean that you can't suddenly apear somehwere else then where you started. untill you observe where you are you can think of yourself as everywhere, but in doing so you are also no where.

the number of photons is not dependent on distance, you can measure a distance by using a photon in one of two ways. Using the way I mentioned in >>5060730
or by measureing the time it takes to emit a photon from a to be, and then back from b to a, however eiather way you still define a radial system

>>	Anonymous Sun Sep 16 00:30:48 2012 No.5060746 >>5060732 You can't say that though. Even in an experimental there is no way to limit the number of photons in a given length.

superobserver Sun Sep 16 00:31:48 2012 No.5060748

Distance must be an event. Without the existence of 4 light years worth of photons between earth and alpha centari there is no distance. The apparent separation is a result of the photons, not a result of some other property of space. You do not need to separate the observable quantities from the metric, and say that the metric has a reality that supersedes the observable quantities within the metric. The only thing that can be measured is an event, or a sequence of events, and distance follows from this, it does not precede it.

Anonymous Sun Sep 16 00:36:48 2012 No.5060755

>>5060748
Ok when we say light year we do not mean anything to do with actual light. Well we do only in what we define as a light year. A photon interaction is not required for you to be able to measure a light year. It is simply ~9.46X10^(15) meters. it is a convinent way of measurement in astronomical terms. Saying something like light years worth of photons is meaning less. Again the notion that you can limit the number of photons in a given space is simply not true.

also you still don't seem to understand an event. No matter what interpetation of shordigers eqation you use, 100% of the time you will stay exactly where you are

superobserver Sun Sep 16 00:37:04 2012 No.5060757

The metre is the length of the path travelled by light in vacuum during a time interval of 1⁄299,792,458 of a second

We already use light to fix concepts regarding distance. I am merely stating that distance cannot be experienced absent an event (or a series of events). So to say that two points are separated in space is to say that there exists a sequence of photons between two points.

superobserver Sun Sep 16 00:40:18 2012 No.5060767

I am not saying that you cannot have multiple photons in the same space. Lasers are a perfect example of stimulated emission. I am saying that in a very specific way of talking we can say that a distance contains photons, that the relativistic experience of length contraction and time dilation require photon measurements, and that objective distance is not a relational structure, photons are relational structures.

Anonymous Sun Sep 16 00:41:48 2012 No.5060771

>>5060757
Photons are not required for a measurement of distance. We use light to fix the lenght of a meter, however that is only becuase it is conistant. That is only a recent development. For a long time people measured feet based on the measurment of the lenghts of a kings foot! ultimatly what you measure distance in is arbitrary, however what it represents is not.

Distance is not an event. To say that there exists a sequence of photons between two points is to give a photon a dimenstion, wich they simply do not have.

>>	superobserver Sun Sep 16 00:43:33 2012 No.5060775 Photons have wavelength and this is the dimension of length. Are you saying that you can become aware of velocity or acceleration without measuring a photon (or a sequence of photons)?

>>	Anonymous Sun Sep 16 00:44:34 2012 No.5060779 >>5060767 why must a photon exist between two points? There might be a point in the univerise where photons have yet to travel! so this point must not exist?

superobserver Sun Sep 16 00:46:48 2012 No.5060786

You cannot measure an event without a measurable quantity!

Yes, an area of space that has no photons is not observable. And there is no distance between two points that are not separated by photons. What would you be be observing if there is no photon?

Anonymous Sun Sep 16 00:46:48 2012 No.5060787

>>5060775
A wavelenght is not a dimension, but rather an energy conversion. They do not ocupy physical space in the same sense. I can close my eyes, blocking out all photons, while sitting in a car where the driver steps on the cause causing an acceleration, because I have my eyes closed I can not feel this acceleration?

>>	Anonymous Sun Sep 16 00:49:03 2012 No.5060793 >>5060786 how do we observe black holes? we do not see any photons emited from them. But yet we are certian they exist! we can even measure distacne from the earth to these black holes!

Anonymous Sun Sep 16 00:49:03 2012 No.5060794

>>5060787
Not the person you're responding to, but you really ought to stop using that "closing your eyes" analogy. It doesn't work here. Photons are still hitting your eyelids, and thus interacting with you.

A proton "closing its eyes" would be a proton which suddenly becomes incapable of interacting with anything, at which point it exists simultaneously everywhere in the universe, until it interacts with something, somewhere, at which point it is there.

>>	superobserver Sun Sep 16 00:49:03 2012 No.5060795 To be clear, the distance between two points is entirely determined by photon observations. There is no distance event aside from observing photons.

Anonymous Sun Sep 16 00:54:48 2012 No.5060809

>>5060795
If I put a stick between me and a wall and I say I am one stick distance between me and the wall in what sense am I using photons?
>>5060794
while you can not isolate photons itself, you can isolate all, for lack of a better word, inertial photons. That is to say i am only interacting with photons that are in my refrence frame. with no external photons, I can still 'feel' this acceleration.

>>	superobserver Sun Sep 16 00:57:03 2012 No.5060817 You see the stick, do you not?

>>	Anonymous Sun Sep 16 00:58:03 2012 No.5060821 I will agree to the notion that photons are the most usefull method of derning distance, but they are not the only means. And you can not define a photon distance, you can only simply use them as a tool

>>	Anonymous Sun Sep 16 00:59:19 2012 No.5060826 >>5060817 Why do I have to see the stick? Holding the stick is enough to know it is there, and not being able to push it farther in front of me is enough to know it is up against something.

superobserver Sun Sep 16 00:59:48 2012 No.5060830

You are at rest with respect to the stick. Or, are you in motion? your estimation of the length of the stick requires you to use a photon measurement if you want to be precise regarding distance.

But if you want to create an arbitrary length and say that it is the span between you and some other point, that is fine. But you will know that you are somewhere relative to some other somewhere because you are monitoring photon events. And without those events, time itself has no meaning.

Anonymous Sun Sep 16 01:03:33 2012 No.5060845

>>5060830
I'm not trying to define time. I am simplly trying to define distance with out using a photon, which has been my argument all along. If I stand up and proclaim to the world this stick that I am holding is exactly equal to one stick length. I am not intereseted in reltivistic effects on a stick length. I have defined this with out using photons or a photon event.

superobserver Sun Sep 16 01:04:03 2012 No.5060846

The stick pushing against you and another object is registered as photon energy. Or, heat.

The stick has mass. Agreed. It has length. The length changes slightly with temperature. This is true of all materials.

But I see your point. you are saying that you can get a measuring rod and that proves that length exists. But the rod can be observed from different points of view. Here, it is not the rod that is being observed, it is the light that scatters from the rod that is being observed.

Your pushing against the rod is an event, one that cannot exist or have a reality of any kind without photon mediation.

There is no distance event without a photon event.

Anonymous Sun Sep 16 01:08:49 2012 No.5060864

>>5060846
heat is more then just photons. heat radation is photons yes, but heat is more then just that. Again if I say that this stick is one stick lenght, I am not defining it over time, so its effect from temperature is simply not their. why am I observing photon scatered from it. I am talking about a physical reaction caused from the friction in my hand, the rigidity of the stick, and the friction of the stick on the wall.

superobserver Sun Sep 16 01:08:49 2012 No.5060866

But, ok, let's say that we do not need light to define distance. Then we say that alpha centari is 4 light years away we are really saying that it is (4*365*24*60)*186,000 miles away. So we fix this distance.

Now let's say that no light at all exists over this distance. Does the distance still exist if there are no events separating the earth from alpha centari.

Because each photon is an event. And without those, there is no distance (nor time, but that is another discussion).

>>	Anonymous Sun Sep 16 01:10:18 2012 No.5060871 File: 37 KB, 403x598, jimmies2.jpg [View same] [iqdb] [saucenao] [google] >>5060866

Anonymous Sun Sep 16 01:12:18 2012 No.5060877

>>5060866
yes distace is still their. how we measure it is now a different story. And I did agree that a photon is usually the best method, it is not the only method of measuring distance. You can still define it with out the use of photons. I am not talking about measurement. This brings us back to the black hole. We have not observed any photons from black holes. how do we know their there? How do we know how far away we are from it?

>>	Anonymous Sun Sep 16 01:16:48 2012 No.5060886 >>5060866 I guess my point being is we can use photons to measure the diameter of a black hole, however because no photons exist between each side, is the distance between those two points undefined?

superobserver Sun Sep 16 01:18:03 2012 No.5060890

We know black holes are there because we see the effect they have on photons. We call them black holes because they are in a thermodynamic cycle whereby they absorb energy but they do not seem to re-radiate any of it (a kind of self-referential heat sink).

But if you want to travel a distance you will be measuring events and these events are photon events.

>>	Anonymous Sun Sep 16 01:20:03 2012 No.5060893 >>5060890 so how do you overcome the fact that now the measurement of the diameter of a black hole is now undefiend?

superobserver Sun Sep 16 01:24:48 2012 No.5060901

Distance is normally thought of a span of physical space. More often than not people refer to distance as a length. The length matters because you can fit things into it.

A mathematical idea, length. Physics deals with objects, and it says that objects of finite energy take up a finite space. We measure things by registering absolute changes (0,1) or by measuring changes of degree (1, 1.2, 1.6).

Space is not a physical object. It is however filled with physical objects. So we grab an arbitrary object and say that it has a length. But we are measuring this object. The measurements require photons.

Maybe I am wrong and we don't need photons to measure things. But if I am wrong about that then qM is also entirely wrong. As a matter of fact, almost all physics would have to be revised.

A math concept is not always easily converted into a QM concept. Distance is one of those classical ideas that gets blown to bits by the reality underpinning the "observable" universe.

My two cents.

Anonymous Sun Sep 16 01:25:03 2012 No.5060903

>>5060890
I think I may have initially misunderstood your definiton of a photon event. I'm not even sure I'm arguing with the same person at this point. But the notion of a cordinate system based soley on a photon I do not agree with. If you wish to define a photon event as an observation then go right ahead, but that really is as arbitrary as my stick distance.

superobserver Sun Sep 16 01:26:49 2012 No.5060906

Hmm... Distance is normally thought of a span of physical space. More often than not people refer to distance as a length. The length matters because you can fit things into it.

A mathematical idea, length. Physics deals with objects, and it says that objects of finite energy take up a finite space. We measure things by registering absolute changes (0,1) or by measuring changes of degree (1, 1.2, 1.6).

Space is not a physical object. It is however filled with physical objects. So we grab an arbitrary object and say that it has a length. But we are measuring this object. The measurements require photons.

Maybe I am wrong and we don't need photons to measure things. But if I am wrong about that then qM is also entirely wrong. As a matter of fact, almost all physics would have to be revised.

A math concept is not always easily converted into a QM concept. Distance is one of those classical ideas that gets blown to bits by the reality underpinning the "observable" universe.

My two cents.

We also know the "size" of a black hole because we can see the radius compared with the background temperature. So we can see them indirectly, or better yet we can compare temperatures of the background and the object and discern a size.

>>	superobserver Sun Sep 16 01:29:49 2012 No.5060913 Sauce. http://www.sciencenews.org/view/feature/id/63612/title/Black_hole_silhouettes

Anonymous Sun Sep 16 01:33:03 2012 No.5060922

>>5060906
I will admit I am thinking more of the mathmatical concept rather then the physical. I am still developing my knowledge of physics, and there is a high chance that I'm completly talking out of my ass, in wich case I hope you had fun watching me do so. I was really trying to point out that defining velocity as the intergral of acceleration still implies a cordianate system, or you leave a constant that is unsolvable.

At any rate I think that we can both agree that wining an argument on the internet (espcially 4chan) is like wining a retared race, win or loose your still retarted. I will admit I am.

Anonymous Sun Sep 16 01:34:49 2012 No.5060924

holy shit /sci/. holy shit.

what the hell is with all you people talking about Hilbert spaces and momentum and quantum mechanics???

The answer to OP's question is simple.

>How can anything move without any acceleration?

it doesn't. Newton's first and second laws address this exact question. 1st law: an object moving at a constant velocity will keep moving at that velocity forever unless an external force acts on it. 2nd law: if an external force acts on it, the object accelerates to a different velocity, with its mass as a proportionality constant. in other words F=ma.

OP, it's simple: objects don't move unless a force acts on them, which gives them an acceleration which changes the velocity to another value, and then it will keep moving at that velocity until some force again acts on it to change it's velocity again (accelerates it). you're basically right, nothing moves without being accelerated.

the real mystery here is why /sci/ is making this so much more complicated than it is. the fuck??

>>	superobserver Sun Sep 16 01:35:33 2012 No.5060925 Yeah. I am here for strange reasons. Never been on sci thread before. Just started this: www.piratenode.net Unrelated. TV. Science of control. More fun.

Anonymous Sun Sep 16 01:39:18 2012 No.5060930

>>5060925
I fail to see the point of this site.
>>5060924
Welcome to the internet. His question was answered several hours ago and OP has since left with his answer, however we still have discusions. This is the first non high school physics thread we have had in a long time. let us have our twisted fun

superobserver Sun Sep 16 01:40:18 2012 No.5060931

Quantifying the force that acts upon an object is the difficult part.

I mean, classical mechanics is simply the mechanics of infinitesimal change and Quantum Mechanics is the mechanics of quantized change (multiples of the unit h).

So acceleration is the addition of a finite amount of energy and this exactly corresponds to a photon energy (otherwise it could not be observed).

Distance and related concepts follow from observable quantities. So we have to identify what the actual observable quantities are.

And the only observable quantities we have are photons.

So we reduced the discussion to one regarding quantizing velocity in units of photon energy. Then we tried to quantize distance in terms of photon events and created a loggerhead.

Fail.

>>	superobserver Sun Sep 16 01:44:18 2012 No.5060938 File: 62 KB, 800x532, Loggerhead_sea_turtle.jpg [View same] [iqdb] [saucenao] [google] And so it is loggerheads all the way down....

Anonymous Sun Sep 16 01:53:18 2012 No.5060957

Using photons as observable quantities is nothing new (that's what we usually always use), but space-time still exists independently of the photon. We could scatter neutrinos, composite neutral particles, or even gravitons hypothetically to recover observable quantities.

The real underlying laws (Lorentz invariance) are present regardless what you use, for example the photon still takes light-like paths in space-time and classically all massless particles have an action functional
<div class="math">\int_{gamma}\sqrt{-g_{\mu\nu}\dot{z}^{\mu}\dot{z}^{\nu}}</div>

If we were to strip off the U(1) gauge field from space-time, meaning electrodynamics no longer exists, there would still be Lorentz invariance.

Anonymous Sun Sep 16 02:00:48 2012 No.5060977

>>5060682

I would say that proper acceleration cannot be taken as "absolute" because it is defined in terms of the metric. The metric gives the intrinsic geometry of spacetime given by the gravitational curvature from every other particle in the universe. So if we measure the proper acceleration, we are measuring against every other particle in the universe.

faith in metrics superobserver Sun Sep 16 02:06:18 2012 No.5060985

Space-time exists independently of photons but motion upon the metric cannot be measured without reference to a photon (or several photons).

Now, we can certainly detect a neutrino or a proton or an electron but we do not see these objects rather we see the momentum (photon energy) that they transfer to another object when undergoing an interaction.

WE would not say that a perturbation was created where no energy was exchanged. Of course, we can create a positron-electron pair in the strong field of an atom with gamma ray annihilation. Now, if the atom is in motion then we do not need a gamma ray - a less energetical photon will do the trick. So long as the *total energy* of the reaction is sufficient.

Space-time may exist independently from photons but you cannot measure its presence without measuring photons - is my statement. So to say it exists is a bit like saying God exists - more a matter of faith than demonstration.

Anonymous Sun Sep 16 02:24:48 2012 No.5061011
File: 5 KB, 300x183, ef0d008d39.jpg [View same] [iqdb] [saucenao] [google]

>>5060985
But a neutrino doesn't interact electromagnetically. This means it doesn't emit or absorb photons - it can only interact with W or Z bosons. See the pic. It still exchanges momentum/energy in these events, but you can't measure anything without a change in momentum or energy, is this what you mean?

>So to say it exists is a bit like saying God exists - more a matter of faith than demonstration.
Well, every theory has to make some sort of assumptions. If this weren't true we couldn't predict anything in science. Although we can't measure space-time directly (at least for now, until we can detect gravitons), there's lots of evidence for its existence with the predictions of special and general relativity - http://relativity.livingreviews.org/Articles/lrr-2006-3/

Really, it all boils down to a model which can reproduce all of our existing experiments in one theoretical framework and simultaneously make predictions. The Einsteinian concept of "space-time" has been the only thing so far that's able to do it in a competent manner.

neutrino detection superobserver Sun Sep 16 02:42:04 2012 No.5061043

Cherenkov radiation is still light radiation. And that is how neutrinos are detected.

Or, that is what the detectors are set to record. And so even a neutrino is reduced to being observed by its impacts on other objects, and the resulting photon production.

Anonymous Sun Sep 16 02:54:18 2012 No.5061072

>>5061043
But that's an intermediate process after the neutral current from the neutrino occurs; it's not what actually triggers the detection. Still, we'd sure be blind without electromagnetism. ;) (Now that I think about it though, not really because of dark matter/dark energy, we'd only be 4.6% less observable than what we can see now)

This is getting quite philosophical, though. I think the best question would now be if it's possible to reproduce a framework without reference to the metric that is capable of reproducing all known physical laws/experiments.

>>	unobserved photons superobserver Sun Sep 16 03:04:49 2012 No.5061099 File: 15 KB, 198x226, terminator.jpg [View same] [iqdb] [saucenao] [google] The metric is unnecessary, it can be eliminated without any loss to our ability to make prediction. But that is a much larger discussion....

Anonymous Sun Sep 16 06:59:18 2012 No.5061547

>>5060826
>Why do I have to see the stick? Holding the stick is enough to know it is there, and not being able to push it farther in front of me is enough to know it is up against something.

The contact force between your hand and the stick is caused by the electric repulsion between the electrons in the atoms of your hand and the electrons in the atoms of the stick. The electromagnetic force is propagated by photons. Likewise, the contact force between the stick and the wall is also caused by electromagnetism.

Anonymous Sun Sep 16 07:03:34 2012 No.5061556

>>5060977
>I would say that proper acceleration cannot be taken as "absolute" because it is defined in terms of the metric.

It is defined subjectively. It is possible, for example, to have a proper acceleration of 1022 meters per second per second, for 293229 seconds, which would logically mean you were traveling at exactly lightspeed at the end of that interval. An external observer will not observe you traveling at lightspeed; but that's irrelevant to the case of trying to determine your subjective, or proper, acceleration.

Absolute acceleration superobserver Sun Sep 16 13:21:18 2012 No.5062216

You cannot change your motion without a force. A force is going to be computed as a change in energy. A change in energy is equivalent to a change in frequency (or the addition or subtraction of units h) in QM and thus is quantized or, in Classical Mechanics, we say that your velocity is changed.

Velocity is a classically continuous curve, which means that changes in energy have no defined minimum. Now, changes in photon energy are quantized in units h.

If we restrict ourselves to saying that acceleration can be determined locally, what we are saying is that a local measurement detects some change in a local condition. This local condition is a blue shift inside the vessel that is accelerating (and this correlates to the experience of "weight" (think accelerating from a dead stop in an elevator). Once the change has occurred, and a terminal velocity has been achieved, no experiment conducted inside the moving frame can detect motion. Only by comparing events happening outside the moving frame to events happening inside the moving frame can we come to recognize the presence of accelerated motion.

This detection *always* gets transmitted by photons, or is always reduced to an analysis of photon frequencies.

>>	Anonymous Sun Sep 16 13:34:03 2012 No.5062251 >>5062216 >implying energy is in all systems quantized in QM >implying velocity needs to be continuous in classical mechanics >implying this means changed in energy have no defined minimum. >implying you know wtf you're talking about

>>	superobserver Sun Sep 16 13:48:48 2012 No.5062287 Velocity is a classically continuous evaluation, is it not? p/m How do you quantize p? Stating that you cannot observe velocity without an interaction. Stating that all interactions are based on photon exchanges.

Anonymous Sun Sep 16 15:20:34 2012 No.5062549

>>5062287
1. velocity doesn't need to be continuous, take a particle in a box
2. you quantize p if there's boundary conditions, a free particle doesn't have energy, momentum or position quantized
3. photons aren't the only particles that carry force y'know

Anonymous Sun Sep 16 15:42:59 2012 No.5062624

velocity is speed of movement and means literally nothing if it isn't relative to something.

acceleration is the process of using energy to change your velocity relative to everything else in the universe. the closer you get to c when moving away from something, the more energy it takes to increase your speed relative to that object (and this is exponential growth). However this being relative, your velocity is changing at a different rate depending entirely on what you are measuring against.

Think about how when you are driving a car, and how objects close to you appear to be moving faster relative to you than objects farther away. Switch distance with velocity and you've got a very basic idea of the concept.

Advanced search
Text to find
Subject [?]Search by post subject. Leave empty for any.
Username [?]Search for user name. Leave empty for any user name.
Tripcode [?]Search for tripcode. Leave empty for any.
Email [?]Search by email. Leave empty for any.
Filename [?]Search by image filename. Leave empty for any.
From Date [?]Enter what date to start searching from. Format is YYYY-MM-DD
To Date [?]Enter what date to start searching until. Format is YYYY-MM-DD
Image hash
Search in	All Posts OPs Only
Deleted posts	Show all posts Show only deleted posts Only show non-deleted posts
Internal posts	Show all posts Show only internal posts Show only archived posts
Order	New posts first Old posts first
Capcode	All Posts Only by Users Only by Mods Only by Admins Only by Developers
Results	Posts Threads
Action	[ Simple ]

/sci/ - Science & Math