Is the 4-velocity always a space-tipe 4-vector?

Re: 4-velocity

No, 4-velocity is always time-like, unless you're talking about tachyons.

 

Re: 4-velocity

4-velocities of ordinary particles are timelike because they travel slower than light.

 

Re: 4-velocity

Show that it's components satisfy t² - x² - y² - z² = 1.

 

Re: 4-velocity

What do you mean why? That's the definition of "timelike". A 4-vector is timelike if t² - x² - y² - z² > 0. All 4-velocites of ordinary particles must satisfy t² - x² - y² - z² = 1 (by definition), which is positive, so 4-velocities are timelike.

BTW, questions like this should be posted in the "Homework & Coursework Questions" forum.

 

Re: 4-velocity

Just calculate its square in the reference frame where the ordinary velocity is equal to zero.
Then only time-component contributes to the square. A time-like four-vector remains such in the other reference frames.

Bob.

 

Re: 4-velocity

Do you know the definition of 4-velocity?

 

Re: 4-velocity

I'm just asking you what the definition of 4-velocity is. What are the components of the 4-velocity of a particle in a reference frame in which it's velocity is (v_x, v_y, v_z)?

If you can answer this question, then I will be able to answer your question:

 

Re: 4-velocity

What's [itex] i [/itex] in the fourth component?

 

Re: 4-velocity

Wow. Might you be able to find a textbook that was written after 1936? Because that was about the last time anyone ever used the ict convention. In that convention, the norm of the 4-velocity will always be -1, and is still timelike, because that is what timelike means in that convention. Get a newer book. Seriously. We've learned a lot since World War II.

 

Re: 4-velocity

No, 4-vectors don't have imaginary components.

The components of the 4-velocity u_µ are

[tex] u_0 = \frac{1}{\sqrt{1-\beta^{2}}} [/tex]

[tex] u_1 = \frac{v_{x}}{c\sqrt{1-\beta^{2}}} [/tex]

[tex] u_2 = \frac{v_{y}}{c\sqrt{1-\beta^{2}}} [/tex]

[tex] u_3 = \frac{v_{z}}{c\sqrt{1-\beta^{2}}} [/tex]

Now, calculate [tex] u_0^2 - u_1^2 - u_2^2 - u_3^2 [/tex]. You should find that it is equal to 1.