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what is 4th dimension? is it just 1 dimensional space? how come time is described as the 4th dimensional path?
thanks..
thanks..
Depends on how you define the 4th. In relativity we say fourth simply because this theory (in its most basic version) only includes 3 spatial dimensions so the natural addendum is to put time as the 4th. In reality, we treat it more like its the zeroth dimension, which is probably more correct. To deal with this time mathematically, here is a snippit:
In flat 3-dimensional space, the infinitesimal distance between any two points is given (in cartesian coordinates) by:
[tex]ds^2=dx^2+dy^2+dz^2[/tex]
Which you might recognize as simply the infinitesimal form of the familiar Pythagorean theorem. In a flat 4-dimensional spacetime, the new expression is:
[tex]ds^2=-dt^2+dx^2+dy^2+dz^2[/tex]
Note how this is not the simple Pythagorean theorem because of the sign change! (The negative on the time part is merely convention. What is important is the time component has a different sign from the spatial component) So time enters into the calculation of the distance between any two points, so that the resulting path is made up of the temporal and spatial displacements according to the above form.
[tex]ds^2=-dt^2+dx^2+dy^2+dz^2[/tex]
Note how this is not the simple Pythagorean theorem because of the sign change! (The negative on the time part is merely convention. What is important is the time component has a different sign from the spatial component) So time enters into the calculation of the distance between any two points, so that the resulting path is made up of the temporal and spatial displacements according to the above form.
I'm not sure where you want to go with this line of questioning, but you are asking about something that is simply a word we use to describe a mathematically useful relationship. It's just a definition.i take this chance to ask another similar question as well.. why C has to be squared in E=MC^2 ... and when you calculate kinetic energy.. E=half MV^2...... why squared?
Why is it important that ds^{2} be invariant? And also why the c^{2} is attached to the dt^{2}. As effectively the product of this is a speed and a time, so hence a distance. Which does make sense in it's self. But why not just right this product as a distance?As for why the time component has an opposite sign, do you want a rigorous derivation? A hand-wavy argument is that we really want ds^{2} to be an invariant quantity.
What we are looking for is something in spacetime geometry (aka Minkowski geometry) which is analogous to distance in Euclidean geometry. In Euclidean geometry distance is invariant under rotations and translations. The equivalent class of operations in spacetime is spatial translations, spatial rotations, time translations, and boosts (relative velocity). So we are looking for something which is invariant under those operations, this thing will be the spacetime analog to spatial distance.Why is it important that ds^{2} be invariant?
I'm not quite sure what you mean "why not just write this product as a distance", but the point is it makes no sense to add quantities with different units. When you actually derive the form of the interval you see the c^{2} there, but in relativity we generally use units where c=G=1, so I tend to forget about factors of c and G and insert them back so at the end. Can seem a bit haphazard to someone not familiar with using geometric units, so I apologize for that.And also why the c^{2} is attached to the dt^{2}. As effectively the product of this is a speed and a time, so hence a distance. Which does make sense in it's self. But why not just right this product as a distance?
https://www.physicsforums.com/showthread.php?t=386951 [Broken]PS. Can you direct me to where you have all the nifty codes for writing squared and other signs? Much appreciated.