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.

 

Re: 4-velocity

martyf,

You are unlucky to have a professor who is really out of date.

He is using an outdated convention that you write a 4-vector as X = (x, y, z, ict). With that convention, a 4-vector is timelike if ||X||² < 0 (where ||X|| denotes the Euclidean norm of a 4D vector).

However, the modern convention is to write a 4-vector as X = (ct, x, y, z). The condition for it to be timelike may be written as either

[tex]c^2t^2 - x^2 - y^2 - z^2 > 0[/tex]​

or as

[tex]x^2 + y^2 + z^2 - c^2t^2 < 0[/tex]​

(Different authors might express it either way, but they are equivalent, of course.)

You have just learned the hard way that different authors use different conventions, so if you want to read lots of different books on a subject, you need to be able to understand more than one convention and to translate between them. (As if relativity wasn't difficult to learn already!)