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

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

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  • #2
dx
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No, 4-velocity is always time-like, unless you're talking about tachyons.
 
  • #3
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But |4-velocity|^2 = -1 . So why is it time-lke?
 
  • #4
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How can I show that a 4-velocity is always time-like?
 
  • #5
dx
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4-velocities of ordinary particles are timelike because they travel slower than light.
 
  • #6
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Ok, so if I have a given 4-velocity how can I show in terms of mathematics that it is time-like ?
 
  • #7
dx
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Show that it's components satisfy t² - x² - y² - z² = 1.
 
  • #8
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why?
 
  • #9
dx
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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.
 
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  • #10
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Ok, so if I have a given 4-velocity how can I show in terms of mathematics that it is time-like ?
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.
 
  • #11
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Definition of "timelike" : travel slower than light
WHY a 4-velocity that travel slower than light satisfies t² - x² - y² - z² = 1 ?!!

BTW, I'm trying to understand the meaning of time-like 4-velocity, I have not written any equation, any text of an exercise
 
  • #12
dx
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Do you know the definition of 4-velocity?
 
  • #13
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I wanted to say :

Definition of "timelike" 4-velocity :4-velocity of a PARTICLE THAT travels slower than light (ordinary particles). I wrote in a rush.
(if you're referring to this)
 
  • #14
dx
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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 (vx, vy, vz)?

If you can answer this question, then I will be able to answer your question:
WHY a 4-velocity that travel slower than light satisfies t² - x² - y² - z² = 1 ?!!
 
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  • #15
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([tex]\frac{v_{x}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{v_{y}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{v_{z}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{i}{\sqrt{1-\beta^{2}}}[/tex] )
 
  • #16
dx
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What's [itex] i [/itex] in the fourth component?
 
  • #17
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complex number : i^2=-1
 
  • #18
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([tex]\frac{v_{x}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{v_{y}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{v_{z}}{c\sqrt{1-\beta^{2}}}[/tex] , [tex]\frac{i}{\sqrt{1-\beta^{2}}}[/tex] )
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.
 
  • #19
dx
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No, 4-vectors don't have imaginary components.

The components of the 4-velocity uµ 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.
 
  • #20
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:) very funny, however my prof is able to find a textbook that was written after 1936. Infact I don't find a textbook that uses the ict convention!But I had to study on the prof's notes. The professors should be renovate their lessons, I should learn from them!
 
  • #21
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Hi Martyf,

It is a good exercise to show why the norm of the 4-velocity is always c. If r is the spacetime "position" 4-vector:
r = (ct,x,y,z)
then the 4-velocity is
dr/dtau
Just carry out the differentiation remembering that dt/dtau is the time dilation factor gamma.
 
  • #22
DrGreg
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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!)
 
  • #23
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Ah , I understood why I was confused! In the prof's notes, first there is the definition of timelike 4 vector as the vector such that

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

But then he introduces the ict convention. So when I calculate the norm^2 of the 4-velocity it's =-1, so I thought it should be spacelike!

Thank's DrGreg, you were very clear!
 

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