Can someone state clearly a sufficient condition for a 4-tuple of

[Kalidor]

Can someone state clearly a sufficient condition for a 4-tuple of quantities to be a four vector? For instance I saw the naive definition of speed four vector is not but I couldn't really understand when a 4-tuple is or is NOT a four vector.

 

[Bill_K]

When (v_x, v_y, v_z, v_t) transform under Lorentz transformations the same way that (x, y, z, ct) do.

 

[Dale]

It is a four vector if it transforms like a four vector. So take a given four-tuple, transform it via the Lorentz transform and see if it is still the same four-tuple in the boosted frame.

 

[bcrowell]

So take a given four-tuple, transform it via the Lorentz transform and see if it is still the same four-tuple in the boosted frame.

Wait, don't you mean to transform it the way you know it physically transforms (based on the transformation properties of the things its defined in terms of) and then transform it with a Lorentz transformation, and make sure the results are the same?

 

[jfy4]

In a separate thread I started a little bit ago about commutation relations, It was pointed out that $x^{\alpha}$ does not in general transform as a vector...

If you defined a vector based on how it transforms in comparison to the coordinate "4-vector", wouldn't that only be valid for flat-space time? Since in curved spaces the coordinate vector looses meaning except in local cases?

I'm just starting to come to grasps with loosing $x^{\alpha}$... so bear with me...

 

[Bill_K]

Sorry, I thought we were talking special relativity. In general relativity one compares the transformation of v to the transformation of dx rather than x, and distinguishes between covariant and contravariant vectors. But I still thought the question was about special relativity?

 

[Dale]

Wait, don't you mean to transform it the way you know it physically transforms (based on the transformation properties of the things its defined in terms of) and then transform it with a Lorentz transformation, and make sure the results are the same?

Yes, sorry about the sloppiness, that is what I meant by "same". In other words, see if it represents the same physical quantity in the new frame. Which is determined in the way you mention.

 

[robphy]

Here's a 4-tuple that is not a 4-vector:
assemble these components of the electric and magnetic fields into a 4-tuple
Q=(B_x,E_x, E_y,E_z).

That doesn't transform like a 4-vector.
In particular, B_x*B_x - E_x*E_x - E_y*E_y - E_z*E_z (square-norm) is not Lorentz-invariant,
and v_t*B_x - v_x*E_x - v_y*E_y - v_z*E_z (dot-product with a 4-vector) is not Lorentz invariant...
i.e. g( Lv , LQ ) =/= L ( g(v,Q) ), where L is a boost and g is the metric.

 

[Kalidor]

Thanks for all your answers. I'm trying to understand these brief notes and all I have there is the standard boost that you can find e.g. first thing on this page
http://en.wikipedia.org/wiki/Lorentz_transformation

How do I show from those equations that , say, (E/c,p_x,p_y,p_z) is a four vector?

 

[DrGreg]

How do I show from those equations that , say, (E/c,p_x,p_y,p_z) is a four vector?

Assuming you are talking about a particle with mass, do you understand that the thing you wrote could also be written as

$$\left( mc\frac{dt}{d\tau}, m\frac{dx}{d\tau}, m\frac{dy}{d\tau}, m\frac{dz}{d\tau} \right)$$​

That may be enough of a hint to complete this.