1

Physics, Astronomy & Cosmology / Re: How does special relativity explain dimensional components ...

« on: 25/06/2022 13:02:01 »

Hi.

Does time and space have separate components like that?

   Yes and No.

The yes bit:
    4-vectors are what are important in spacetime.    These have 4 components,   3 of them are called spatial components and the other component is called the time component.  You can write the spatial components first and the time component last but it's more common to write the time component first.   It's also fairly common to start numbering the components from 0 and not from 1.   The final slightly confusing thing you might see is that if you had a 4-vector  X   then you may see the components written as  X⁰,  X¹,  X²  and  X³.    Superscripts instead of subscripts like  X₀,  X₁  can be used.
    For example    r =    ( ct,  x,  y,  z)  is the usual way of writing  the position 4-vector of an object.    It has a time component  ct  =    c (the speed of light)  multiplied by the position of the object on the time axis, t.    It has spatial components   x, y, z   which are the position of the object along the x, y and z axis respectively.    Now you could use t as the time component instead of ct but the algebra turns out to be much easier if you use  c (the speed of light) multiplied by t   as the time component.

The "no" bit
1.   There's an unusual way of determining the magnitude of a 4-vector.  You might see it called the "norm" or "Minkowski norm" of a 4-vector.    For simple vectors used in Euclidean space or Newtonian mechanics,  whenever you increase the size of one component the overall magnitude of the vector would increase.   For 4-vectors that's not always the situation,  you can increase the size of one component and end up reducing the overall magnitude of the 4-vector.    The Minkowski metric is described in various places  [For example,   https://phys.libretexts.org/Bookshelves/Modern_Physics/Book%3A_Spiral_Modern_Physics_(D%27Alessandris)/3%3A_Spacetime_and_General_Relativity/3.1%3A_Minkowski_Metric  ].

2.   You are mainly discussing velocities and motion in your posts, rather than just positions.   In ordinary Newtonian mechanics, velocities are just a rate of change of position with respect to the time co-ordinate.   For 4-velocities we can't determine rates of change with respect to a fixed time co-ordinate,  instead we must determine rates of change with respect to what is called the proper time for that object undergoing the motion.
   What this boils down to is that spatial components of the 4-velocity are NOT exactly the spatial components of the ordinary Newtonian or 3-velocity that the object might have.   Instead the spatial components are a multiple of the spatial velocity you would have assigned the object in Newtonian mechanics.  Also it's not a fixed multiple,  the multiple changes according to the Newtonian 3-velocity of the object.    Specifically, the  spatial components of the 4-velocity are given by   γ  (the gamma factor) multiplied by the spatial components of the 3-velocity.

- - - - - - - -
     That might be more detail than you were after.   Overall there is a lot of similarity between  4-velocities used in Special relativity and more conventional velocity vectors you might have seen in Newtonian mechanics.    I've mentioned the differences because, in my limited experience, if we don't then it's human nature to run away with the idea that it's all exactly like Newtonian mechanics and ordinary 3-velocity vectors.   You'll soon hit problems if you do that.
     For example,  it can be useful to consider the magnitude of a 4-velocity vector.  An ordinary object with some positive rest mass always has a 4-velocity vector of magnitude c (the speed of light).   That magnitude can be shared out between the time component and the spatial component of the objects 4-velocity.   An object at rest (in a given frame which we will use to assign the velocity vector) has all of its velocity in the time component while the spatial components would have the value 0.     Meanwhile, an object in motion (in the given frame) has a non-zero value in the spatial components of the 4-velocity and a correspondingly lower value* in the time component.

Best Wishes.

* LATE EDITING:  I don't like this on a second reading.   It's precisely one of those examples where you could have a larger numerical value for the time component but that is actually reducing the overall magnitude of the 4-velocity and not increasing it (because the Minkowski metric subtracts the time component instead of adding it).  It's fair to say the object has less velocity through time and many Pop Sci articles will do this - but it's not correct to imply that the numerical value you find written in the time component of the 4-velocity has to be smaller.

2

Physics, Astronomy & Cosmology / Re: Is this a correct spacetime diagram with world lines?

« on: 21/01/2022 13:42:01 »

Thanks, I did not know that all 3 rocks are moving in the primed frame.  Then what is the point of reference of the primed frame?

The worldline of a statioary object in the primed frame would be parallel to ct' axis, just like the worldlines of the two rocks stationary in the unprimed frame are parallel to ct.

Why this specific choice of primed frame was chosen for the picture is something only the author of the picture can answer.

And I did not know that the two rocks are on the world lines from the unprimed frame.

The worldline of any object is a set of objective events, and events are not frame specific. You can transform the line to a different frame (say one in which the rocks are not stationary), but all the events are still the same events. Only in the one frame in which they are stationary do they map to the same location in space, so only in that one frame are they parallel to the time axis.

This is related to my original comment about the angles of some of the lines, ct' in particular.
These are not necessarily wrong since the scales of the two axes are not labeled.
They're both in distance (meters say), but maybe the long x axis goes from 0 on the left to 10 meters on the right, and the shorter ct axis goes from 0 at the bottom to 20 meters at the top, in which case the axes, while both in meters, are simply not to the same scale. So the drawing is not necessarily wrong in that way unless the scales are similar.
If they are scaled similarly, then your diagonal yellow 'high speed rock' is moving at considerably faster than c since it goes to the right faster than it goes up.

3

Physics, Astronomy & Cosmology / Re: Is this a correct spacetime diagram with world lines?

« on: 21/01/2022 00:47:29 »

Hi.

   Here's an annotated diagram with some of the points that Halc raised (see attachment).

I also put the rocks on the x' axis just to remind me what is happening in the frame of reference of the rock that is traveling.

   That's a little bit worrying.  I'm not sure I've understood but I'm a little concerned that you might be thinking that the primed frame is one where the rock passing through the origin is actually at rest.   
   All of the 3 rocks are moving in the primed frame.

Best Wishes.

4

Physics, Astronomy & Cosmology / Re: Is this a correct spacetime diagram with world lines?

« on: 20/01/2022 23:17:09 »

The one rock (diagonal yellow line) seems to be moving dang fast in both frames, not just the primed frame. Faster in the unprimed frame.

About the angles:
You've not shown a 'light speed' line originating from the origin, but it would go at a 45° angle and the x' and ct' lines would be symmetric about that (equidistant), and they're not. The ct' line as you've drawn it is too close to the light line.

As for the two rocks, they're shown at one set of locations in the unprimed frame (intersections of their worldline with the grey x' line), and another location (dots) in the primed frame. These are objective events and the rocks cannot be in two locations at the same time in the primed frame as shown. So the red dots belong at the intersections, not where you've drawn them. Their worldlines will tilt to non-vertical if you do the Lorentz transform where the primed frame has perpendicular axes.