Olivier5
 
  0  
Reply Tue 28 Apr, 2015 02:59 pm
@maxdancona,
Thanks, I did try to make it funny... :-) But what I meant was: you have no reason to assume your training superior to mine; nor any reason to assume that, just because some teacher of yours used a particular turn of phrase, that this turn of phrase is un-problematic.

The idea of anything being absolutely stationary is problematic in relativity, and hence any mention of a stationary object, PERIOD, is problematic in that context.

If by that phrase one simply means a shortcut to: 'stationary as compared to the frame of reference X or Y which is otherwise determined as being inertial', then that's fine, provided the implicit frame of reference is obvious to all and not ambiguous.
maxdancona
 
  0  
Reply Tue 28 Apr, 2015 03:16 pm
@Olivier5,
First of all, I have an advanced degree in Physics. If you don't have an advanced degree in Physics, then my training in Physics is superior to yours. That is nothing against you... it is just a fact. Second, I have worked in Physics with other Physicists and I can tell you quite confidently, from experience solving problems and communicating with other trained people, that you are wrong.

I don't know what your training in "special relativity" in college consisted of. Generally a week or two of something called "special relativity" is included in the second semester of undergraduate physics in the US. You obviously have been exposed to some of the terminology and ideas... but you are also missing something in your understanding.

But I can tell you quite assuredly that you are wrong (and the simple google search I suggested to find professional Physicists using the term you say they never use would show you that).

I could try to explain to you why you are wrong, but apparently you are so interested in showing how right you are that this is quite impossible.
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 03:21 pm
@maxdancona,
Quote:
I could try to explain to you why you are wrong, but apparently you are so interested in showing how right you are that this is quite impossible.

Not at all, in fact I have been waiting for your argument. I'd gladly consider it. Shoot.
maxdancona
 
  -1  
Reply Tue 28 Apr, 2015 03:25 pm
@Olivier5,
Do you understand that an object in circular motion at any instant in time is travelling in a straight line in any inertial frame.... I probably could prove this to you, but hopefully we can skip this part.
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 03:34 pm
@maxdancona,
Are you are trying to say that a circular motion can be considered in straight line, as a approximation, over the course of a very short period?
layman
 
  -1  
Reply Tue 28 Apr, 2015 03:35 pm
@maxdancona,
Quote:
...at any instant in time is travelling in a straight line...


By "at any instant in time" do you mean if time is stopped? If you take a picture, for example?
layman
 
  -1  
Reply Tue 28 Apr, 2015 03:38 pm
@layman,
Or do you mean that in GR (in stark contrast to SR) an object in "free fall" (such as a satellite orbiting a planet) is deemed to be "inertial?"
0 Replies
 
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 03:41 pm
@layman,
If time is stopped, there is no movement anymore, by definition of what is movement.
layman
 
  -1  
Reply Tue 28 Apr, 2015 03:43 pm
@Olivier5,
Quote:
If time is stopped, there is no movement anymore, by definition of what is movement.


Yes, of course.
0 Replies
 
maxdancona
 
  -1  
Reply Tue 28 Apr, 2015 03:44 pm
@Olivier5,
No. I am saying that at any given instant, a particle in circular motion can be considered to be traveling in a straight line.

In other words... at any given point in time, a particle has a velocity (a speed and a direction) which can be represented as a vector.


(There is a difference as apparently you see).
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 03:53 pm
@maxdancona,
If 'instant' means an interval of time = 0 seconds, then there is no movement whatsoever.

If 'instant' means 'a very very short period of time', then I understand you. You're saying that the arc of a circle tends to resemble a straight line when the length of the arc is very small as compared to the circle radius. That's why the apparatus in the michelson-morley experiment can be said to be inertial: because the experiment only lasts a very short time during which the apparatus' orbiting around the sun and its rotation around the earth axis don't really matter: its trajectory over a few seconds can be approximated as a movement in straight line at constant speed, hence inertial.
0 Replies
 
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 03:54 pm
@maxdancona,
Quote:
at any given point in time, a particle has a velocity (a speed and a direction) which can be represented as a vector.

Alright, that's clear enough.
maxdancona
 
  -1  
Reply Tue 28 Apr, 2015 04:27 pm
@Olivier5,
Ok. The velocity vector (speed and directly) for that particle at that instant of time will be different for each inertial frame. You with me so far?
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 04:37 pm
@maxdancona,
Ok.
maxdancona
 
  -1  
Reply Tue 28 Apr, 2015 04:40 pm
@Olivier5,
And for this particle at this instance of time, there is an inertial frame of reference where this velocity vector is zero.

Now we have defined a mathematically valid inertial frame of reference in which this particle has a zero velocity for an instant of time.
layman
 
  -1  
Reply Tue 28 Apr, 2015 04:51 pm
@maxdancona,
I'm on a train and I order a beer. After guzzling most of it, I set the can down on the tray in front of me. It is now in a "co-moving" frame with me. Neither of us are moving with respect to each other. That the idea? So what?
0 Replies
 
Olivier5
 
  0  
Reply Tue 28 Apr, 2015 04:52 pm
@maxdancona,
Sure, there is even an infinite number of frames of reference in which that vector has a null value.
0 Replies
 
usery
 
  1  
Reply Tue 28 Apr, 2015 10:45 pm
@Olivier5,
Quote:
as anyone opening a sink can testify (the water swirls around the sink due to the rotation of earth, aka Coriolis force).


lol, not quite.

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

The vector cross product can be evaluated as the determinant of a matrix:

http://upload.wikimedia.org/math/b/0/1/b013b03f3c334748a0eb29f32cfde225.png
Olivier5
 
  0  
Reply Wed 29 Apr, 2015 06:12 am
@usery,
Why, not quite?
Olivier5
 
  0  
Reply Wed 29 Apr, 2015 06:16 am
@maxdancona,
Hhhhm...

Max? Where's your demonstration going, baby step by baby step?

I guess at some point, you realized you were going nowhere, hence you dropped from the thread... Better late than never.
0 Replies
 
 

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