# Displacement and distance when particle is moving in curved trajectory

• NTesla
In summary, the conversation discusses the equations used to solve a problem in Irodov's book. The speaker mentions using the magnitude of the vector ##\vec{v}## to solve the equations and getting the correct answer for displacement. However, the other speaker points out that the equality ##\|\int\vec{v}dt\|\leq\int\|\vec{v}\|dt## only holds for rectilinear motion and not for curved trajectories. They also mention that the equations in the original post are incorrect as they mix scalar and vector quantities. The correct equations should be ##\int_0^T\vec v.dt=l\hat j+\vec u.T## and ##\int_{0}^{\
archaic said:
##||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##.

archaic said:
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
Delta2 said:
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 said:
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.

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NTesla
archaic said:
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.

archaic said:
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.

Delta2
archaic said:
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##.

PeroK said:
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.

PeroK
Steve4Physics said:
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.

NTesla said:
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
PeroK said:
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..

PeroK

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 ##

NTesla said:
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.

haruspex said:
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.

NTesla said:
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.

haruspex said:
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.

NTesla said:
@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.

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..

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NTesla said:
vector from South-west to North east.
No, second quadrant: SE to NW.
NTesla said:
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.
NTesla said:
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.

PeroK said:
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!

Callumnc1 said:
What is the proof of that fact? Many thanks!
It's part of the definition of equality, mathematical or physical.

ChiralSuperfields
PeroK said:
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!

Callumnc1 said:
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.

ChiralSuperfields
PeroK said:
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!

Callumnc1 said:
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.

ChiralSuperfields
haruspex said:
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!

Callumnc1 said:
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!
It is a matter of mathematics, not physics.

ChiralSuperfields
haruspex said:
It is a matter of mathematics, not physics.

Callumnc1 said:
Thanks for your reply @PeroK , what do you mean same 'order'. Many thanks!
Scalar, vector or tensor generally.

ChiralSuperfields
PeroK said:
Scalar, vector or tensor generally.
Ok thank you @PeroK!

To me the "##=##" sign is shorthand notation for "is the same as" so that I don't have to write out the entire locution in English every time I put an equation down.

In mathematics, we write the vector equation ##\mathbf{A}=\mathbf{B}##. Bereft of symbolic notation, this says, in English, "Vector A is the same as vector B". If we want to qualify this some more, we say "This means that the x-component of vector A is the same as the x-component of vector B and the y-component of vector A is the same as the y-component of vector B and the z-component of vector A is the same as the z-component of vector B." All that verbosity can be eliminated by writing "##A_x=B_x;~A_y=B_y;~A_z=B_z.##"

In physics, which uses mathematics as its language, the "##=##' is still shorthand for "is the same as" but more nuanced because it also provides a correspondence between observables. For example, I can use a scale to measure mass ##m##. I can then apply to it a constant force ##F##, which I can measure with a force gauge, and figure out the mass's constant acceleration ##a## by measuring its positions with a ruler and the time at these positions with a clock. Then I can write down the number from the force gauge and, next to it, the scale reading for the mass with the result of my calculation for the acceleration. Lo and behold, the product of the mass and the acceleration is the same as (to within measurement uncertainty) the reading of the force gauge.

In this context, "Net force equals mass times acceleration" says less than "Net force is the same as mass multiplied by acceleration." The former implies that if I give you numbers of the mass and the acceleration, you can find a number for the net force. The latter implies that if you measure all quantities that participate in the equation on both sides separately and then put them together as prescribed, the left-hand side will be the same as the right hand side. We think of the former as a definition and the latter as a "law". One tends to forget that equations in physics labeled "laws" are based on painstaking measurements.

String theory guy and ChiralSuperfields
kuruman said:
To me the "##=##" sign is shorthand notation for "is the same as" so that I don't have to write out the entire locution in English every time I put an equation down.

In mathematics, we write the vector equation ##\mathbf{A}=\mathbf{B}##. Bereft of symbolic notation, this says, in English, "Vector A is the same as vector B". If we want to qualify this some more, we say "This means that the x-component of vector A is the same as the x-component of vector B and the y-component of vector A is the same as the y-component of vector B and the z-component of vector A is the same as the z-component of vector B." All that verbosity can be eliminated by writing "##A_x=B_x;~A_y=B_y;~A_z=B_z.##"

In physics, which uses mathematics as its language, the "##=##' is still shorthand for "is the same as" but more nuanced because it also provides a correspondence between observables. For example, I can use a scale to measure mass ##m##. I can then apply to it a constant force ##F##, which I can measure with a force gauge, and figure out the mass's constant acceleration ##a## by measuring its positions with a ruler and the time at these positions with a clock. Then I can write down the number from the force gauge and, next to it, the scale reading for the mass with the result of my calculation for the acceleration. Lo and behold, the product of the mass and the acceleration is the same as (to within measurement uncertainty) the reading of the force gauge.

In this context, "Net force equals mass times acceleration" says less than "Net force is the same as mass multiplied by acceleration." The former implies that if I give you numbers of the mass and the acceleration, you can find a number for the net force. The latter implies that if you measure all quantities that participate in the equation on both sides separately and then put them together as prescribed, the left-hand side will be the same as the right hand side. We think of the former as a definition and the latter as a "law". One tends to forget that equations in physics labeled "laws" are based on painstaking measurements.

kuruman said:
In this context, "Net force equals mass times acceleration" says less than "Net force is the same as mass multiplied by acceleration."
@kuruman Could you please clarify more how "Net force is the same as mass multiplied by acceleration" says more than "Net force equals mass time acceleration"?

Many thanks!

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