Displacement and distance when particle is moving in curved trajectory

[Delta2]

@PeroK, @haruspex, @Steve4Physics: I've tried to solve it once again. Please let me know if it is correct(Is the reasoning written in lower half of the first pic correct ?):
$\vec{v_{AG}}$ is the velocity of particle A wrt Ground. Similarly others.
View attachment 276023

Something doesn't look quite right to me with the above, because $\hat v$ depends on time, hence you can't take it out of the integrals.

 

[haruspex]

Something doesn't look quite right to me with the above, because $\hat v$ depends on time, hence you can't take it out of the integrals.

I agree, and I very much doubt the validity of the preceding line, where the integral of the component of $\vec u$ orthogonal to $\vec v$ is taken to be zero. That component starts out as directly East (say), then veers to the SE as A swings to the NE. It never has a westward component, so cannot integrate to zero.

 

[haruspex]

The equation that you wrote: $\int_0^T\vec v.dt=l\hat j+\vec u.T$ is right. I understand it.
But I couldn't understand why the 2nd equation(in the original post) was wrong. $\vec{v}cos\theta$ is always in the direction of displacement $u\tau$.
$\therefore$ atleast this equation should be right: $\int_{0}^{\tau}\vec{v}cos\theta dt=u\tau$ should be right. Isn't it ?

Yes, sorry ... it was late here, should have slept on it. Post corrected.

 

[archaic]

Something doesn't look quite right to me with the above, because $\hat v$ depends on time, hence you can't take it out of the integrals.

$||v||$ and $||u||$ are given as constants.

 

[Delta2]

$||v||$ and $||u||$ are given as constants.

Yes but the direction of $\vec{v}$ (which is the same as the direction of $\hat v$) varies with time, don't you think?

 

[haruspex]

$||v||$ and $||u||$ are given as constants.

Look at the fourth line from the end in the first image in post #28.
$\int_0^\tau vdt\hat v$ is turned into $v\tau \hat v$.

 

[haruspex]

I can think of two ways to go about this:
First, by remembering that for the vector $x$ representing the distance between the two points$$||x||'=\frac{x\cdot\dot x}{||x||}\text{ (for any vector)}\Leftrightarrow \int_0^\tau||x||'dt=-L=\int_0^\tau\frac{x\cdot\dot x}{||x||}dt$$but this is too cumbersome to workout, I think.
Second, using a nice diagram showing the configuration at two instances separated by a small period of time, and the law of cosines to find $||x||'$.
View attachment 276032
$$L^2(t+\Delta t)\approx\left(L(t)-u\Delta t\right)^2+(v\Delta t)^2-2\left(L(t)-u\Delta t\right)(v\Delta t)\cos\varphi$$Keeping in mind that $\cos\varphi=-\cos\theta$, we can develop the expression, and manipulate it to get $\frac{\Delta L^2}{\Delta t}=...$, then compute the limit as $\Delta t\to0$, and find $2L(t)L'(t)=...\Leftrightarrow L'(t)=...$
Finally, $\int_0^\tau L'(t)dt=-L(0)=\int_0^\tau...dt$

The Irodov method is clearly superior. The result follows swiftly from the two equations posted.

 

[archaic]

Yes but the direction of $\vec{v}$ (which is the same as the direction of $\hat v$) varies with time, don't you think?

Look at the fourth line from the end in the first image in post #28.
$\int_0^\tau vdt\hat v$ is turned into $v\tau \hat v$.

To get to what PeroK mention in #24, i.e that the relative velocity is $(v - u\cos \alpha,-u\sin \alpha)$ in the basis $(\hat v,\hat v_\perp)$, we applied a change of basis $(\hat\imath,\hat\jmath)\to(\hat v,\hat v_{\perp})$. In the new basis, $\hat v$ and $\hat v_{\perp}$ are constant.

 

[Delta2]

To get to what PeroK mention in #24, i.e that the relative velocity is $(v - u\cos \alpha,-u\sin \alpha)$ in the basis $(\hat v,\hat v_\perp)$, we applied a change of basis $(\hat\imath,\hat\jmath)\to(\hat v,\hat v_{\perp})$. In the new basis, $\hat v$ and $\hat v_{\perp}$ are constant.

Interesting you define a frame of reference where the basis vectors are time dependent. This I think means that this frame of reference is not inertial, but I am not familiar at all with all these so I will not comment further.

 

[haruspex]

To get to what PeroK mention in #24, i.e that the relative velocity is $(v - u\cos \alpha,-u\sin \alpha)$ in the basis $(\hat v,\hat v_\perp)$, we applied a change of basis $(\hat\imath,\hat\jmath)\to(\hat v,\hat v_{\perp})$. In the new basis, $\hat v$ and $\hat v_{\perp}$ are constant.

Maybe, but that is not the meaning of $\hat v$ in post #28.

 

[NTesla]

To get to what PeroK mention in #24, i.e that the relative velocity is $(v - u\cos \alpha,-u\sin \alpha)$ in the basis $(\hat v,\hat v_\perp)$, we applied a change of basis $(\hat\imath,\hat\jmath)\to(\hat v,\hat v_{\perp})$. In the new basis, $\hat v$ and $\hat v_{\perp}$ are constant.

In post#28, in the 2nd pic, I've drawn the apparent path of particle A that particle B will observe. $\hat{v}$ and $\hat v_{\perp}$ are constant.
I visualised it as two particles in deep space where there is no gravity, particle A is always targeting particle B therefore $\hat{v}$ is constant and $\hat v_{\perp}$ is also constant. However, as seen in pic 2 of post#28, the path will not be a straight line, but will be curved towards the left, because of the component $-usin\theta$.

 

[NTesla]

Okay, I see what's he's done. Let's look at the initial situation. A is moving towards B at speed $v$ and B is moving in a perpendicular direction at speed $u$. The initial separation is $(l, 0)$. After a short time $dt$, we have the new displacement of B from A is $(l - vdt, udt)$ and the new distance of B from A is: $$\sqrt{l^2 + v^2dt^2 - 2lvdt + u^2dt^2}$$ Ignoring higher powers of $dt$ gives $$l - vdt$$ In other words, the distance between A and B is reducing at a rate $v$.

In general, the relative velocity of A with respect to B is $(v - u\cos \alpha, u\sin \alpha)$ and the same applies. The rate of change of distance is the first component only.

In post#24, you've written:

In other words, the distance between A and B is reducing at a rate $v$.

And, you've also written:

The rate of change of distance is the first component only$(v - u\cos \alpha)$.

Both of these can't be right at the same time.
My understanding is that the rate of change of distance given by $(v - u\cos \alpha)$ is the right one, since it doesn't include approximations.

 

[NTesla]

OK, I'm with you. Maybe the equation comes from consideration of relative velocity. In A's frame, B has an instantaneous relative velocity of $v - ucos\theta$. That would produce the required integral.

In A's frame of reference, B will have velocity given by: $(ucos\theta-v)$. The relative speed $(v - ucos\theta)$ will be relative speed of particle A from B's frame of reference.

 

[PeroK]

In post#24, you've written:
And, you've also written:
Both of these can't be right at the same time.
My understanding is that the rate of change of distance given by $(v - u\cos \alpha)$ is the right one, since it doesn't include approximations.

They can be (and are) both right. The first case was the special case where for the initial motion $\alpha = \pi/2$.

 

[NTesla]

They can be (and are) both right. The first case was the special case where for the initial motion $\alpha = \pi/2$.

Yes..you are right. I should have observed that.. my bad..

 

[NTesla]

But can this be right: $\left |\int_{0}^{\tau}(v-ucos\theta) dt\hat{v} \right | = \left | l\hat{j} \right |$
$\Rightarrow \int_{0}^{\tau}(v-ucos\theta) dt = l$

 

[haruspex]

View attachment 276087

But can this be right: $\left |\int_{0}^{\tau}(v-ucos\theta) dt\hat{v} \right | = \left | l\hat{j} \right |$
$\Rightarrow \int_{0}^{\tau}(v-ucos\theta) dt = l$

No. See post #32.

 

[NTesla]

It never has a westward component, so cannot integrate to zero.

@haruspex, I read your post#32 again and again. I understand what you've written which is what one might see from Ground frame of reference. But when seen from B's frame of reference(pic2 in post#28), particle A has a curved trajectory, it starts going westward (due to component $usin\theta$), but one component of velocity of particle A is always along $\hat{v}$ which is always directed towards particle B. Therefore, I think that integral of $usin\theta$ must be zero. Isn't it.? Please let me know your reasoning. I couldn't derive any other meaning from post#32.

 

[haruspex]

when seen from B's frame of reference(pic2 in post#28), particle A has a curved trajectory, it starts going westward (due to component usinθ), but one component of velocity of particle A is always along v^ which is always directed towards particle B. Therefore ...

Therefore?! I see no logical connection there. How does the component of v directed towards B as tell you that the other component integrates to zero?
Maybe you are confusing it with taking a component towards the final position of B and a component orthogonal to that? In that case, yes, the orthogonal component would integrate to zero.

 

[NTesla]

Therefore?! I see no logical connection there. How does the component of v directed towards B as tell you that the other component integrates to zero?
Maybe you are confusing it with taking a component towards the final position of B and a component orthogonal to that? In that case, yes, the orthogonal component would integrate to zero.

@haruspex, The component $usin\theta$ tells that the path of particle A while going towards particle B will not be a straight line. From particle B's frame of reference, particle A will have a curved trajectory as depicted in pic 2 of post#28. The trajectory will be curved solely because of component $usin\theta$ of velocity of particle A. Since both the particles meet after some finite time $\tau$, therefore, in my opinion integral of $usin\theta$ should be zero. This is what I wanted to convey in my post#48.
If this is so, then please explain the reason why you wrote no in post#47.

 

[haruspex]

@haruspex, The component $usin\theta$ tells that the path of particle A while going towards particle B will not be a straight line. From particle B's frame of reference, particle A will have a curved trajectory as depicted in pic 2 of post#28. The trajectory will be curved solely because of component $usin\theta$ of velocity of particle A. Since both the particles meet after some finite time $\tau$, therefore, in my opinion integral of $usin\theta$ should be zero. This is what I wanted to convey in my post#48.
If this is so, then please explain the reason why you wrote no in post#47.

$u\sin(\theta)$ is indeed the magnitude of the component normal to the line joining AB, but its direction keeps changing, so it makes little sense to integrate it.

We can make it a vector by introducing $\bar v$ to mean the unit vector 90 degrees anticlockwise from $\hat v$. So now we can integrate $u\sin(\theta)\bar v$. This points somewhere in the second quadrant at all times, so will integrate to produce a vector pointing in that quadrant.

Meanwhile, the component of relative velocity in the $\hat v$ direction integrates to a vector pointing somewhere in the first quadrant.

I see no reason why these might not add to form a vector in the +y direction, that being the relative displacement over the trajectory.

 

[NTesla]

If I've understood your post#51, then $\int_{0}^{\tau}usin\theta\bar{v}$= a vector from South-west to North east.

While integrating $usin\theta\bar{v}dt$, my understanding was that it will give the displacement that the particle A will undergo in the direction perpendicular to $\hat{v}$. But since from particle B's FoR, particle A has not undergone any displacement in the direction perpendicular to the initial line joining particle A and B, thus, $\int_{0}^{\tau}usin\theta\bar{v}dt$ should be equal to 0. I don't understand how this is wrong. Please let me know..

 

[haruspex]

vector from South-west to North east.

No, second quadrant: SE to NW.

While integrating $usin\theta\bar{v}dt$, my understanding was that it will give the displacement that the particle A will undergo in the direction perpendicular to $\hat{v}$.

The direction of to $\hat{v}$ varies during the integration, so it doesn't make sense to say that.

from particle B's FoR, particle A has not undergone any displacement in the direction perpendicular to the initial line joining particle A and B,

What do you mean by "the initial line joining particle A and B" in B's reference frame?
As far as B is concerned, A started at a point distance L to the S, so in B's view that is still a point distance L to B's S. In B's view, A has made an excursion out to the West from that line.
But this has nothing to do with our definition of $\bar v$. B defines $\hat v$ as pointing from the present position of A to itself, and $\bar v$ as perpendicular to that.
If you mean the line joining A's actual initial position to B's final position, B sees that point as way over to the W (ut to the W, where t is the time from start to finish), so indeed A is seen as having gone some displacement in that direction.

 

[member 731016]

That equation is manifestly wrong. The LHS is vector, the RHS is a scalar. Correct would be: $$\int_{0}^{\tau}v \cos\theta(t) dt=u\tau$$ where $\theta(t)$ is a function of $t$, and $v = |\vec v|$.

Is it true for all equations that both sides must be the same physical quantity (i.e LHS = Scalar = RHS, or LHS = vector = RHS) @PeroK . What is the proof of that fact? Many thanks!

 

[PeroK]

What is the proof of that fact? Many thanks!

It's part of the definition of equality, mathematical or physical.

 

[member 731016]

It's part of the definition of equality, mathematical or physical.

Ok thank you, out of curiosity, when did you learn that @PeroK ? Would you be able to guide me to any physics textbooks? Many thanks!

 

[PeroK]

Ok thank you, out of curiosity, when did you learn that @PeroK ? Would you be able to guide me to any physics textbooks? Many thanks!

I did pure mathematics at university. People forget that equals $=$ has a precise meaning. It's simplest to take it as an axiom of mathematical physics that you can only compare things of the same order and physical dimensions.

 

[member 731016]

I did pure mathematics at university. People forget that equals $=$ has a precise meaning. It's simplest to take it as an axiom of mathematical physics that you can only compare things of the same order and physical dimensions.

Thanks for your reply @PeroK , what do you mean same 'order'. Many thanks!

 

[haruspex]

Is it true for all equations that both sides must be the same physical quantity (i.e LHS = Scalar = RHS, or LHS = vector = RHS) @PeroK . What is the proof of that fact? Many thanks!

For two things to be equal, you must be able to do the same things with each and with the same consequence. I can add a scalar to a scalar because there are defined rules for that, but there are no rules allowing me to add a vector to a scalar.

 

[member 731016]

For two things to be equal, you must be able to do the same things with each and with the same consequence. I can add a scalar to a scalar because there are defined rules for that, but there are no rules allowing me to add a vector to a scalar.

Thanks for your reply @haruspex , I guess the reason why there are no rules for adding a vector to a scalar is because it is not physically meaningful yet. Maybe in the future of physics!