Displacement and distance when particle is moving in curved trajectory

[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!

 

[haruspex]

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.

 

[member 731016]

It is a matter of mathematics, not physics.

Ok thanks for your reply @haruspex !

 

[PeroK]

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

Scalar, vector or tensor generally.

 

[member 731016]

Scalar, vector or tensor generally.

Ok thank you @PeroK!

 

[kuruman]

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.

 

[member 731016]

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.

Thank you very much for your answer @kuruman !

 

[member 731016]

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!