Solving for Vector u in Terms of Vector η

In summary, the homework statement states that the proper velocity is equal to the square root of the product of the proper distance and the proper time. The relation between proper velocity and rapidity is that the tanh of theta is equal to the eta divided by the square root of c+eta^2.
  • #1
ehrenfest
2,020
1

Homework Statement

[tex] \mathbf{\eta} = 1/ \sqrt{1 - u^2/c^2} \mathbf{u} [/tex] represents the proper velocity in terms of the ordinary velocity. Find vector u in terms of vector eta.

Then find the relation between proper velocity and rapidity.

Homework Equations


The Attempt at a Solution



Scalar u is the magnitude of the vector u, so I cannot just bring the denominator to the other side, right? Do I have to get scalar eta in terms of scalar u?
 
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  • #2
ehrenfest said:
Do I have to get scalar eta in terms of scalar u?

The other way round. Find scalar u in terms of scalar eta.
 
  • #3
George Jones said:
The other way round. Find scalar u in terms of scalar eta.

Yes. That is what I meant. Those velocity magnitude's so I am not sure how to do that...
 
  • #4
Remember that both [itex] \mathbf{u} [/itex] and [itex] \mathbf{\eta} [/itex] are ordinary vectors in R^{3} in which the norm of a vector is given by a scalar product. So compute [itex] \left \langle \mathbf{\eta},\mathbf{\eta}\right\rangle [/itex].
 
  • #5
dextercioby said:
[itex] \left \langle \mathbf{\eta},\mathbf{\eta}\right\rangle [/itex].

[itex] \left \langle \mathbf{\eta},\mathbf{\eta}\right\rangle = \eta_x^2 + \eta_y^2 + \eta_z^2[/itex]

But we have

[tex] \sqrt{1 - u^2/c^2} \mathbf{\eta} = \mathbf{u} [/tex] so how do you get that scalar u in terms of the scalar eta?

Let me check my understanding of what u and eta are. The usage of these two terms implicitly uses an inertial reference frame S and an object in motion with respect to S. vector u is the displacement measured in S per unit time measured in S and vector eta is the displacement measured in S per unit time measured by a clock attached to our object. All good?
 
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  • #6
[tex]\left\langle \vec{\eta},\vec{\eta}\right\rangle =\left\langle \vec{u},\vec{u}\right\rangle \frac{1}{1-\frac{u^{2}}{c^{2}}}\Rightarrow \eta ^{2}=\frac{u^{2}}{1-\frac{u^{2}}{c^{2}}}\Rightarrow u^{2}=\frac{\eta ^{2}}{1+\frac{\eta ^{2}}{c^{2}}}[/tex]

[tex]\vec{\eta}=\vec{u}\frac{1}{\sqrt{1-\frac{u^{2}}{c^{2}}}}\Rightarrow \vec{u}=\vec{\eta}\sqrt{1-\frac{u^{2}\left(\eta \right)}{c^{2}}}=\vec{\eta}\sqrt{1-\frac{1}{c^{2}}\frac{\eta ^{2}}{1+\frac{\eta ^{2}}{c^{2}}}}=\vec{\eta}\frac{c}{\sqrt{c^{2}+\eta ^{2}}} [/tex]
 
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  • #7
Thanks. But is that paragraph I wrote correct? I am very confused about proper velocity. It is supposedly proper distance divided by proper time. But what is proper distance? Isn't the distance an object moves in a reference frame attached to it always 0?
 
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  • #8
Let me check my understanding of what u and eta are. The usage of these two terms implicitly uses an inertial reference frame S and an object in motion with respect to S. vector u is the displacement measured in S per unit time measured in S and vector eta is the displacement measured in S per unit time measured by a clock attached to our object. All good?

I think I got it. Also, is the relation between proper velocity and rapidity the following:

tanh theta = eta/ sqrt(c^2 + eta^2)

?
 
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1. What is the purpose of solving for vector u in terms of vector η?

The purpose of solving for vector u in terms of vector η is to find the relationship between these two vectors. This can help in simplifying equations and solving for unknown variables in a vector equation.

2. How do you solve for vector u in terms of vector η?

To solve for vector u in terms of vector η, you can use the dot product or cross product of the two vectors. You can also use algebraic manipulation and substitution to isolate vector u in the equation.

3. Can vector u and vector η be parallel?

Yes, vector u and vector η can be parallel if they have the same direction or are scalar multiples of each other. In this case, solving for vector u in terms of vector η would result in a scalar multiple of vector η.

4. What is the difference between solving for vector u in terms of vector η and finding the magnitude and direction of vector u?

Solving for vector u in terms of vector η gives the relationship between the two vectors, while finding the magnitude and direction of vector u gives information about the size and direction of vector u. Solving for vector u in terms of vector η is a more general approach, while finding the magnitude and direction is more specific.

5. Are there any other methods for solving for vector u in terms of vector η?

Yes, there are other methods such as using the components of the vectors in a coordinate system, using geometric interpretations, or using matrix operations. The method used may depend on the specific problem and the information given.

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