Show that this vector is timelike?

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In summary, the vector v is timelike but its dot product with spacelike vector X^{\mu\prime} is not zero, so there must be a way to make this vector timelike.
  • #1
Dixanadu
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Hi guys,

So I have a vector which I need to show is timelike. The vector is

[itex] v^{\mu}=t^{\mu}-\frac{t^{\mu}\cdot X^{\mu\prime}}{X^{\mu\prime}\cdot X^{\mu\prime}}X^{\mu\prime} [/itex],

where [itex]t^{\mu}[/itex] is a timelike vector and [itex]X^{\mu\prime}[/itex] is spacelike, however these two vectors are not perpendicular so their dot product does not vanish.

I understand that in order to show that [itex]v^{\mu}[/itex] is timelike, I need to find [itex]v_{\mu}v^{\mu}[/itex] and show that this is greater than 0. So:

[itex]v_{\mu}v^{\mu}=t^{2}-2\frac{(t\cdot X')^{2}}{X^{\prime}\cdot X^{\prime}}+(t\cdot X^{\prime})^{2}[/itex]

So first of all I don't know if this is correct (yes I don't know how to take a dot product obviously lol) and secondly, even if it is, how do I show that this is greater than 0?

Thanks guys!
 
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  • #2
Your last term is wrong. Try to redo the inner product.

You should also look over your use of indices in the original problem statement.
 
  • #3
Yes doctor that is wrong - fixed it and now I get this:

[itex]v_{\mu}v^{\mu}=t^{2}-\frac{(t\cdot X^{\prime})^{2}}{X^{\prime}\cdot X^{\prime}}[/itex] and... is this equal to 0? :O
 
  • #4
No, if it was zero then v would be light-like. I suggest you try to figure out the signs of both of these contributions.
 
  • #5
I don't know how the signs could change. But is the maths right? I mean is that the true expression?
 
  • #6
Nevermind I think i got it -- because X is spacelike, its square is negative right :O so i get something positive overall?
 
  • #7
Correct. Also ##(t\cdot X)^2## is just the square of a number and thus also positive.
 
  • #8
Thank you doctor once again :) while we're at it could you please also confirm / deny the following statement:

[itex](t\cdot X^{\prime})^{2} = t^{2}X^{\prime 2}[/itex]?
 
  • #9
Not true. The left hand side is positive and the right hand side is negative.
 
  • #10
I thought so. Thank you doctor :D you're a life saver lol ! (no pun intended)
 

1. What does it mean for a vector to be timelike?

A vector is considered timelike if it has a positive magnitude and points in the direction of time. This means that the vector can be interpreted as representing a possible path of a particle traveling through time.

2. How do I determine if a vector is timelike?

To determine if a vector is timelike, you can calculate its magnitude using the Pythagorean theorem. If the magnitude is positive, then the vector is timelike. Additionally, you can check if the vector's components have a positive time component and a negative spatial component.

3. What is the significance of a timelike vector in physics?

Timelike vectors are important in physics because they represent possible paths of particles traveling through time. In special relativity, timelike vectors are used to describe the worldlines of massive particles, while lightlike vectors (having zero magnitude) are used to describe the worldlines of massless particles.

4. Can a vector be both timelike and spacelike?

No, a vector can only be either timelike or spacelike. A timelike vector has a positive magnitude and points in the direction of time, while a spacelike vector has a negative magnitude and points in a spatial direction.

5. How does the concept of a timelike vector relate to the concept of spacetime?

The concept of a timelike vector is closely related to the concept of spacetime, as it represents one component of the four-dimensional spacetime continuum. In this continuum, time is considered to be the fourth dimension, and timelike vectors represent possible paths of particles through this dimension. Spacetime is a fundamental concept in both special and general relativity, and timelike vectors are essential for understanding the behavior of particles in these theories.

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