Relative motion and the rotation of the axes of reference

  • #1
Holystromboli
21
0
I'm still very early on in my reading, so forgive me if this question isn't coherent. In the "historical introduction" section of the 1920 University of Calcutta translation of the original papers of Einstein and Minkowski available via the MIT online archive, mention is made of the fact that "in the spacetime reality, relative motion is reduced to a rotation of the axes of reference," but no mathematical or graphical representation of this concept is given. Does this imply that the velocity (this isn't a good word for what I mean here but it's the best I could do:) ) of an object relative to an assumed stationary point can be described by a 4 component vector (3 space components and one time) with magnitude c such that the combined magnitude of the 3 spatial components defines the "observed spatial velocity" of the object?
 

Answers and Replies

  • #2
39,036
16,788
Does this imply that the velocity (this isn't a good word for what I mean here but it's the best I could do:) ) of an object relative to an assumed stationary point can be described by a 4 component vector (3 space components and one time) with magnitude c such that the combined magnitude of the 3 spatial components defines the "observed spatial velocity" of the object?

Yes. See here:

http://en.wikipedia.org/wiki/Four-velocity
 
  • #3
Holystromboli
21
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Wow. That is pretty awesome stuff. Thanks for the link. I read through it, but I'm still a bit confused. I took a lot of undergraduate level advanced math classes (impressive, I know :p ), but it's been a loonnngggg time. I can understand the need to delineate "proper time" from time as defined at the stationary observation point, but even with that in mind I'm struggling in my understanding of the units involved. Don't units matter when calculating the component vectors? If the component vector in spatial dimension x is given in in/min and y in m/s, you'd have to convert to common units before the resultant magnitude could be calculated correctly, right? How does this principle apply to the fourth component vector for time? If my assumption were correct, it would imply that there must be a fundamental relation between units of distance and time (e.g. the number of seconds in a meter) in order for us to be able to convert to common units, and this by extension would imply the existence of a unitless value for c, which I've never heard of. What am I missing? Or can we use the value of c itself to calculate an absolute number of seconds in a meter? Also, do you have a reference for a visual and mathematical representation of these concepts for a 3D universe containing 2 space and 1 time dimensions?
 
  • #4
39,036
16,788
even with that in mind I'm struggling in my understanding of the units involved.

In the units used in the Wikipedia page, 4-velocity has the same units as ordinary velocity (see further comments below).

In "natural" units commonly used in relativity, in which ##c = 1## by definition, velocities (4-velocity and 3-velocity) have no units; they are unitless numbers. This amounts to using the same units for time and space (again, see further comments below).

If the component vector in spatial dimension x is given in in/min and y in m/s, you'd have to convert to common units before the resultant magnitude could be calculated correctly, right?

Yes.

How does this principle apply to the fourth component vector for time?

In ordinary units, you multiply it by ##c##. (In the case of 4-velocity, the time component is just ##c## times ##\gamma##, where ##\gamma = 1 / \sqrt{ 1 - v^2 / c^2 }## is the standard relativistic factor.)

there must be a fundamental relation between units of distance and time

Yes, there is.

this by extension would imply the existence of a unitless value for c

No, it implies what you yourself say further on:

Or can we use the value of c itself to calculate an absolute number of seconds in a meter?

Yes, exactly: ##c## is just the conversion factor between ordinary distance units and ordinary time units. If you adopt the "natural" units I referred to above, in which ##c = 1##, you are simply adopting the same units for distance and time: for example, feet and nanoseconds (approximately), or meters and "light-meters", the time it takes light to travel 1 meter (about 3.3 nanoseconds), or years and light-years (the latter is often used in astronomy and cosmology).
 
  • #5
Holystromboli
21
0
Hahaha that's awesome! Thanks for the quick responses!
 
  • #6
Holystromboli
21
0
Also, do you have a reference for a visual and mathematical representation of these concepts for a 3D universe containing 2 space and 1 time dimensions?
Was that a no? :)
 
  • #7
39,036
16,788
Was that a no?

Correct, sorry, I don't have any useful references handy for that.
 

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