Is the Equation Vy' = Vy * sqrt(Vc^2 - Vx^2) Valid for Velocities Less Than Vc?

[sha1000]

Hi everyone. I'll be grateful if someone can help me with this problem.

1. Homework Statement

I have a closed system composed of one particle. The maximal velocity that this particle can have is equal to V_c.

Here we consider only 2D space: X and Y direction. The particle velocity is V (which is limited to V_c) and have two velocity components V_x and V_y, each cab vary from 0 to V_c.

The existence of V_c creates the relation between V_x and V_y. For example if V_x = V_c then V_y = 0.
We can write this relation as follows: V_x² + V_y² ≤ V_c².

I want to express the component V_y as a function of V_x.

2. The attempt at a solution

In order to resolve this, first I studied the case when the overall velocity V is equal to V_c. In this case:

V_x² + V_y² = V_c².

Then,

V_y = sqrt( V_c² - V_x²) . (1)

Can I conclude that this last relation (1) is correct for all cases (and not only when V = V_c)? My problem is that I don't know if it is mathematically justified to take a special case: V = V_c and then conclude that the final relation (1) between V_y and V_x works for general case for any value of V.

 

[Ray Vickson]

Hi everyone. I'll be grateful if someone can help me with this problem.

1. Homework Statement

I have a closed system composed of one particle. The maximal velocity that this particle can have is equal to V_c.

Here we consider only 2D space: X and Y direction. The particle velocity is V (which is limited to V_c) and have two velocity components V_x and V_y, each cab vary from 0 to V_c.

The existence of V_c creates the relation between V_x and V_y. For example if V_x = V_c then V_y = 0.
We can write this relation as follows: V_x² + V_y² ≤ V_c².

I want to express the component V_y as a function of V_x.

2. The attempt at a solution

In order to resolve this, first I studied the case when the overall velocity V is equal to V_c. In this case:

V_x² + V_y² = V_c².

Then,

V_y = sqrt( V_c² - V_x²) . (1)

Can I conclude that this last relation (1) is correct for all cases (and not only when V = V_c)? My problem is that I don't know if it is mathematically justified to take a special case: V = V_c and then conclude that the final relation (1) between V_y and V_x works for general case for any value of V.

No, it is not correct. Your $v_x,v_y$ satisfy $v_x^2+v_y^2 \leq v_c^2$, but are not otherwise restricted. That means that $(v_x,v_y)$ lies inside on on the edge of a circular disc of radius $v_c$. You have $v_y = \sqrt{v_c^2 - v_x^2}$ or (a case you forgot) $v_y = -\sqrt{v_c^2 - v_x^2}$ only for points $(v_x,v_y)$ that lie on the boundary of the disc. For example, $v_x = v_y = 0$ satisfies the restriction, as does $v_x = 0, v_y = \frac{1}{2} v_v$ or $v_x = \frac{1}{2} v_c, v_y = \frac{3}{4} v_c$, etc. There are infinitely many possible solutions!

 

[sha1000]

No, it is not correct. Your $v_x,v_y$ satisfy $v_x^2+v_y^2 \leq v_c^2$, but are not otherwise restricted. That means that $(v_x,v_y)$ lies inside on on the edge of a circular disc of radius $v_c$. You have $v_y = \sqrt{v_c^2 - v_x^2}$ or (a case you forgot) $v_y = -\sqrt{v_c^2 - v_x^2}$ only for points $(v_x,v_y)$ that lie on the boundary of the disc. For example, $v_x = v_y = 0$ satisfies the restriction, as does $v_x = 0, v_y = \frac{1}{2} v_v$ or $v_x = \frac{1}{2} v_c, v_y = \frac{3}{4} v_c$, etc. There are infinitely many possible solutions!

Hi. thank you for the response.

Actually I made a mistake, the problem is somehow more complicated. I'll do my best to describe it in detail.

The condition with the limit velocity is the same. But, in order to accelerate our particle to the speed V_c we need to apply energy equal to some precise value, let's call it E_c. Even If we give more energy to the particle will never go beyond the limit velocity V_c.
(This is somehow similar to the special relativity. Example: the limit speed for the electron is the speed of light, even if you give it an infinetly high amount of energy its velocity will never go beyond the speed of light and will only converge to it).

Lets return to our problem. If we take for example the case when the particle has already an initial velocity in the X direction V_x = 0.9*V_c. Now, if we will try to accelerate this moving particle in the Y direction using the energy = E_c, the overall velocity will be still equal to V_c (V = V_c). We can calculate V_y' = sqrt(V_c² - V_x²); since we know that the component V_x = 0.9*V_c;

IMPORTANT: we need to make the difference between V_y and V_y'

We call V_y the velocity that our particle should have when initially V_x is equal to 0. And V_y' is the reduced velocity due to the initial non-zero V_x component.

we can write: V_y' = V_y*sqrt(V_c² - V_x²) (1)

This surely works when we use the energy E_c in order to accelerate the moving particle, since the overall velocity V will always be equal to V_c (as you said we are on the edge of a circular disc). BUT, can we use the equation (1) in order to calculate V_y' even when the overall velocity V is inferior to V_c (when we are inside of a disc)?

If we finally write V² = V_y'² + V_x² (2)

with V_y' (see equation (1))

We can see through the equation (2) that we will never obtain a velocity higher than V_c, so this equation suits the answer very well. But how to justify that the relation (1) can be used not only on the edge of a circle but also inside of it.

Can we use this argument: in order to find the general relation between V_y and V_y' we resolve the equation for the special case V_y = V_c?

i'm really sorry for my English.