Constant scalings of 4-vectors

  • #1

Summary:

Constant scalings of a vector do not matter to decide if a vector is timelike or spacelike.
Could anybody prove this statement ?

Main Question or Discussion Point

If anybody has studied the book:

A First course in String Theory - Barton Zweibach - 2nd edition

This statement is present in 6th chapter of book on pg 110
 

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  • #2
PeterDonis
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Could anybody prove this statement ?
I assume "constant scalings" just means multiplying all components by the same constant factor. What effect would this have on the sign of the vector's squared norm?
 
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  • #3
I assume "constant scalings" just means multiplying all components by the same constant factor. What effect would this have on the sign of the vector's squared norm?
IMG_5565.jpg
IMG_5566.JPG
 
  • #4
I assume "constant scalings" just means multiplying all components by the same constant factor. What effect would this have on the sign of the vector's squared norm?
If it was about multiplying all of the components by the same constant factor, then the answer was intuitively clear, but the problem here is that, we are not multiplying all of the components with a constant vactor, instead we are multiplying just one component with a constant factor. Let me post the complete problem.
IMG_5567 2.jpg
IMG_5567 2.jpg
IMG_5568 2.jpg
IMG_5569 2.jpg
 

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  • #5
Ibix
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I think it's saying that ##\frac{\partial X^\mu}{\partial \tau}+\lambda\frac{\partial X^\mu}{\partial \sigma}## defines tangent vectors in every direction in the worldsheet, but not every vector. To get every vector you would need to write ##A\left(\frac{\partial X^\mu}{\partial \tau}+\lambda\frac{\partial X^\mu}{\partial \sigma}\right)##, where ##A## is just a constant scaling of the vector. Presumably the modulus of the vector isn't important to the argument they're making (just the sign of its square), so they arbitrarily set ##A=1## and simplify the maths a bit.
 
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  • #6
George Jones
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Constant scalings of a vector ...

Consider the following concrete example.

Since ##\frac{\partial X^\mu}{\partial \tau}## and ##\frac{\partial X^\mu}{\partial \sigma}## are linearly independent, an arbitrary tangent vector ##u## can be expressed as

$$u = a \frac{\partial X^\mu}{\partial \tau} + b \frac{\partial X^\mu}{\partial \sigma}$$

for real constants ##a## and ##b##.

Let

$$v\left(\lambda\right) = \frac{\partial X^\mu}{\partial \tau} + \lambda \frac{\partial X^\mu}{\partial \sigma},$$

and suppose ##a=2## and ##b=3##, so that

$$u = 2 \frac{\partial X^\mu}{\partial \tau} + 3 \frac{\partial X^\mu}{\partial \sigma}.$$

Then, for every ##\lambda##, we have ##v\left(\lambda\right) \ne u##, but when ##\lambda = 3/2##, we have ##v = u/2##. In other words, ##v\left(\lambda\right)## is never equal to this specific tangent vector ##u##, but there is a value of ##\lambda## such that ##v\left(\lambda\right)## is proportional to ##u##.
 
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  • #7
PeterDonis
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Consider the following concrete example
So in this example, ##u## would be a "constant scaling" of the vector you get by setting ##\lambda = 3/2##, i.e., ##v(3/2)##; specifically, we have ##u = 2 v(3/2)##. And the statement "constant scalings do not matter to decide if a vector is timelike or spacelike" just means that both ##u## and ##v(3/2## have the same sign of their squared norm, i.e., they are either both timelike or both spacelike (or both null). And the same would be true for any vector that could be obtained from ##u## or ##v(3/2)## by multiplying both components by the same constant factor.
 
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