# How can we accurately measure the relative velocity between two moving bodies?

• anantchowdhary
In summary, the two bodies are moving perpendicular to each other with velocities 'a' and 'b'. If A measures the velocity of B in his direction of motion, it comes out to be zero. To compare their relative velocity, A makes a LT with velocity 'b' and uses 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
anantchowdhary
Suppose two bodies A and B are moving perpendicular to each other with velocities 'a' and 'b'.

A with 'a'
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---------------------------------------------->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.

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

anantchowdhary said:
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 wouldn't he be measuring the velocity of B as zero?

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

anantchowdhary said:
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)

Perhaps, the hypotenuse of your triangle...

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

anantchowdhary said:
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?

anantchowdhary said:
the positive direction of the x axis.
No it isn't.

y isn't 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 isn't it?the answer is positive

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 haven't done the above stated question.Im n grade 10.

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

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.

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

so
r=b i+a j

So what's the problem if theyre in the positive direction of the x and y axes?

IMHO, in spite of the subject matter, this thread seems more like a homework-type question.

Hehe,but i can't get y I am wrong:S

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

sry but i still can't get it

When I say resultant vector, what does that mean to you?

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).

masudr said:
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.

anantchowdhary said:
r(x) is positive and so is r(y)

so
r=b i+a j

So what's 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 wouldn't 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 $$cos\theta^{-1}$$

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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.

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 $b.$

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I may be missing something, but it seems that the problem here of understanding relative velocity is not a special relativity problem. Can it be treated classically as a first step?

country boy said:
I may be missing something, but it seems that the problem here of understanding relative velocity is not a special relativity problem. Can it be treated classically as a first step?
This is the relativity forum. Ask your question in the appropritate forum.

Meir Achuz said:
This is the relativity forum. Ask your question in the appropritate forum.

That is why I asked. The original question was posed in the context of relativity, but upon examination it may not be a relativity question. The difficulty in understanding the relative velocity between A and B is not a result of the magnitude of the velocities. It can be dealt with at low velocities.

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country boy said:
That is why I asked. The original question was posed in the context of relativity, but upon examination it may not be a relativity question. The difficulty in understanding the relative velocity between A and B is not a result of the magnitude of the velocities. It can be dealt with at low velocities.
What puzzled me is that this is taught in the first week of physics 101.
V_x=u_x-v_x
V_y=u_y-v_y.

Meir Achuz said:
What puzzled me is that this is taught in the first week of physics 101.
V_x=u_x-v_x
V_y=u_y-v_y.

34 ... or 35 ... :-)

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