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

[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.

 

[Meir Achuz]

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]

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

 

[anantchowdhary]

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]

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.

 

[anantchowdhary]

Yea i get that,but in his DIRECTION of motion wouldn't he be measuring the velocity of B as zero?

 

[Hootenanny]

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

 

[anantchowdhary]

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

 

[Hootenanny]

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

Yes, so what is the magnitude of this velocity?

 

[anantchowdhary]

umm..
a + bcos(theta)

 

[Hootenanny]

Perhaps, the hypotenuse of your triangle...

 

[anantchowdhary]

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

 

[Hootenanny]

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...?

 

[anantchowdhary]

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]

the positive direction of the x axis.

No it isn't.

 

[anantchowdhary]

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

 

[Hootenanny]

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.)

 

[anantchowdhary]

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?

 

[Hootenanny]

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.

 

[anantchowdhary]

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?

 

[robphy]

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

 

[anantchowdhary]

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

 

[Hootenanny]

anant;

http://mathworld.wolfram.com/VectorAddition.html
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/vectors/u3l1b.html

 

[anantchowdhary]

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

 

[Hootenanny]

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

 

[Meir Achuz]

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).

 

[country boy]

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]

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<br />

r_y = a_y + b_y<br />

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}

 

[country boy]

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]

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&#039; \right),

and thus

g \left(u_b , u_a \right) = \gamma.[/itex]<br /> <br /> Here, the relative speed v between A and B is used in \gamma.<br /> <br /> But from C&#039;s spacetime split,<br /> <br /> 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]&lt;br /&gt; &lt;br /&gt; Consequently,&lt;br /&gt; &lt;br /&gt; \gamma = \gamma_a \gamma_b,&lt;br /&gt; &lt;br /&gt; which can be used to find easily the relation between v, a, and b.

 

[country boy]

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?

 

[Meir Achuz]

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.

 

[country boy]

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.

 

[Meir Achuz]

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.
How could this lead to a thread with 33 posts?

 

[country boy]

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.
How could this lead to a thread with 33 posts?

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

 

[George Jones]

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?

For speeds sufficiently high that Lorentz transformations are significantly different than Galilean transformations, as implied by the original post in this thread, then relativity really is needed.

From the the last equation in my previous post (#30),

(1 - V^2)^(-1/2) = (1 - u^2)^(-1/2) (1 - v^2)^(-1/2),

which, after restoring the c's, leads to

V^2 = u^2 + v^2 - (u^2 v^2)/c^2.

Without relativity, the last term disappears. This is to be expected, since the original post sums perpendicular velocities, so, non-relativistically, the speeds satisfy the Pythagorean theorem.

 

[country boy]

V^2 = u^2 + v^2 - (u^2 v^2)/c^2.

Without relativity, the last term disappears. This is to be expected, since the original post sums perpendicular velocities, so, non-relativistically, the speeds satisfy the Pythagorean theorem.

Thanks for the detailed 4-vector derivations you have posted. It is interesting to follow your reasoning. However, the relation above is only correct for the case where A and B were at the origin at the same time. That constraint was not stated in the original post.

In reading the original post and early exchange, it still seems that the conceptual problem here is with the contruction of a relative velocity. Once that is understood, one can move on to the Lorentz trasformations.

 

[Mentz114]

Anant, if you're still following this, look at post 29. Country Boy has given the method.

Start with D^2 = x^2 + y^2, differentiate wrt time and you're there.

 

[George Jones]

Anant, if you're still following this, look at post 29. Country Boy has given the method.

Start with D^2 = x^2 + y^2, differentiate wrt time and you're there.

I don't think that this gives the relative speed. The relative speed is the magnitude of the derivative of the relative position vector, not the derivative of of the magnitude of the relative position vector. Post #29 and the above use the part of the relative velocity that is (edit) parallel to the relative position vector, but neglect the part of the relative velocity that is perpendicular to the relative position vector.

Let \vec{D} be the relative position of B with respect to A. Then,

\vec{V} = \dot{\vec{D}} = \dot{D} \hat{D} + D \dot{\hat{D}}.

The relative of B with respect to A is

\vec{D} = \vec{r_B} - \vec{r_A};

differentiating gives

\vec{V} = \vec{v_B} - \vec{v_A};

dotting this with itself gives

\vec{V} \cdot \vec{V} = \left( \vec{v_B} - \vec{v_A} \right) \cdot \left( \vec{v_B} - \vec{v_A} \right).

Finally,

V^2 = v^2_A + v^2_B,

since \vec{v}_A is perpendicular to \vec{v}_B.

However, the relation above is only correct for the case where A and B were at the origin at the same time.

The above non-relativistic stuff is: modified by relativity; not dependent on whether the observers go throught the spatial origin. I think my relativistic version also is independent of spacetime origin, but I could be wrong. A good check would be a derivation using methods similar to those used in the standard derivation of the sums of parallel and anti-parallel (in C's frame) velocities.

 

[country boy]

Reply to George Jones:

I see the difference in our two approaches to relative speed. [There has been some confusion between speed and velocity in this thread, which I probably added to.] You derive the relative speed between the two reference frames of A and B, which is the same everywhere, while I derive the speed between the moving points A and B. These are, of course, different. I thought from reading the early posts that the latter was what was asked for, but now I believe that your derivation is what is needed when using the Lorentz transformation. The relative speed between A and B, as I defined it, is what the two observers see as they watch each other. It relates to the doppler effect, for instance. But I agree that it is better to approach the problem through the velocity between two moving reference frames. Relativity can then be applied without confusion and everything can be derived, including the doppler effect.

Thanks very much for sticking with me on this.