# Relative velocity problem

Suppose two bodies A and B are moving perpendicular to each other with velocities 'a' and 'b'.

A with 'a'
^
|
|
|
|
|
|
|
|
|
---------------------------------------------->B with 'b'

Now how would they measure their relative velocitites?As A's velocity in the direction of motion of B = 0 ,B would feel that A is moving away(linearly) from it with a velocity'b'
And A would feel that B is moving away with velocity 'a'.

Now how do we compare their relative velocity? This is in order to aply the Lorentz transfomation.

Meir Achuz
Homework Helper
Gold Member
To find A's velocity with respect to B, make a LT with velocity b, using the transformation laws for velocity a (separate law for a perp and a parallel to b).

Or rotate the system so that A and B are moving away from each other along an axis.

But the velocities of A and B have components that hav the value of 0 in their respective axes.

Like if A measures velocity of B in his direction of motion,it comes out to be zero :(
Im confused

Hootenanny
Staff Emeritus
Gold Member
Like if A measures velocity of B in his direction of motion,it comes out to be zero :(
No it doesn't. Remember that A is moving upwards in your diagram.

Yea i get that,but in his DIRECTION of motion wouldnt he be measuring the velocity of B as zero?

Hootenanny
Staff Emeritus
Gold Member
Okay, try it this way. How fast is B moving away from A from A's rest frame?

with velocity 'a'? and with veolcity 'b' in the direction of the perpendicular axis

Hootenanny
Staff Emeritus
Gold Member
with velocity 'a'? and with veolcity 'b' in the direction of the perpendicular axis
Yes, so what is the magnitude of this velocity?

umm..
a + bcos(theta)

Hootenanny
Staff Emeritus
Gold Member
Perhaps, the hypotenuse of your triangle...

ok ok sry its the resultant of the two vectors sry :p.How foolish of me

Hootenanny
Staff Emeritus
Gold Member
ok ok sry its the resultant of the two vectors sry :p.How foolish of me
No problem, and the direction of the resultant vector is...?

the positive direction of the x axis.BBut one more question..wudnt we do resultantcos(theta) to get the magnitude of the vector in the reference frame's exact direction?

Hootenanny
Staff Emeritus
Gold Member
the positive direction of the x axis.
No it isn't.

y isnt it ??????

r(x)=a(x) +b(x)
r(y)=a(y) +b(y)

r=r(x)i + r(y)j

r(x)=b
r(y)=a

so y isnt it?the answer is positive

Hootenanny
Staff Emeritus
Gold Member
Think about it for a minute. What is the direction of the resultant velocity (think about the 'poeple swimming across a river' questions that I'm sure you've done.)

umm i havent done the above stated question.Im n grade 10.

but wats wrong with the resultant as i said in my previous post?

Hootenanny
Staff Emeritus
Gold Member
I asked for the direction of the resultant velocity and you haven't given it, you've just given two components, which is what we were given at the start of the problem.

Last edited:
r(x) is positive and so is r(y)

so
r=b i+a j

So whats the problem if theyre in the positive direction of the x and y axes?

robphy
Homework Helper
Gold Member
IMHO, in spite of the subject matter, this thread seems more like a homework-type question.

Hehe,but i cant get y im wrong:S

err..thnx
but still please tell me if both r(x) and r(y) are positive wats wrong?

sry but i still cant get it

Hootenanny
Staff Emeritus
Gold Member
When I say resultant vector, what does that mean to you?

Meir Achuz
Homework Helper
Gold Member
To find A's velocity with respect to B, make a LT with velocity b, using the transformation laws for velocity a (separate law for a perp and a parallel to b).

Or rotate the system so that A and B are moving away from each other along an axis.

Why not go back and consider what masudr suggested. The axes can be drawn any way you want, so if an axis is drawn directly between the two, and a shift is made to a reference frame in which that axis is not moving, then it is clear that A and B are moving away from each other. Go from there.

r(x) is positive and so is r(y)

so
r=b i+a j

So whats the problem if theyre in the positive direction of the x and y axes?

Is there anything wrong here(above)?
$$r_x = a_x + b_x$$

$$r_y = a_y + b_y$$

But
$$r_x = 0 + b_x$$

$$r_y = a_y + 0$$

So $$r_x$$ and $$r_y$$ are positive.So wouldnt the resutlant also be in the poitive direction of the ' x' axis and in the positive direction of the 'y' axis?

And to find out the angle it makes with the 'x' axis we can find out then $$cos\theta^{-1}$$

Last edited:
It might help to start with the expression for the distance D between the two bodies in terms of x and y coordinates (y for A and x for B), and then take the derivative dD/dt to get the relative velocity. The velocities 'a' and 'b' can then be inserted for dy/dt and dx/dt. The result shows that the relative velocity is the sum of the projections of the two velocities on the line between the two bodies: dD/dt = a cosQ + b sinQ, where Q is the angle at A between the origin and B. This corresponds to intuition.

George Jones
Staff Emeritus
Gold Member
In the spirit of chapter 15, Example: Minkowski Vector Spaces, from Mathematical Physics by Robert Geroch, I'm going to have a go at this problem using 4-vector methods.

Let me make sure I understand the problem.

There is an inertial observer C, such that in C's frame: A moves with constant (spatial) velocity $\vec{a}$; B moves with constant velocity $\vec{b}$; $\vec{a} \cdot \vec{b} = 0.$

Now, let the 4-velocities of A, B, and C be $u_a$, $u_b$, and $u_c$, respectively. The 4-velocity of any inertial observer is the unit timelike vector used by that observer to splt spacetime into space and time. Let C do this. Then,

$$u_a = \gamma_a \left(u_c + \vec{a} \right)$$
$$u_b = \gamma_b \left(u_c + \vec{b} \right),$$

where, in each expression, the first term lies in C's time and the second term lies in C's space.

But A can also split spacetime into space and time. Then,

$$u_b = \gamma \left(u_a + b' \right),$$

and thus

$$g \left(u_b , u_a \right) = \gamma.[/itex] Here, the relative speed $v$ between A and B is used in $\gamma.$ But from C's spacetime split, [tex]g \left(u_b , u_a \right) = g \left( \gamma_b \left(u_c + \vec{b} \right) , \gamma_a \left(u_c + \vec{a} \right) \right) = \gamma_a \gamma_b.[/itex] Consequently, [tex]\gamma = \gamma_a \gamma_b,$$

which can be used to find easily the relation between $v,$ $a,$ and $b.$

Last edited: