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What is the meaning behind multiplication in physics? Is multiplication in physics purely mathematical or there is a physical explanation to it? How do we explain the product for example, $s = v·t $? Is there any meaning behind this? For example, I can say that "Distance is defined as the product of velocity 'times' time"? But what does this even mean?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z 2 days ago
  • $\begingroup$ Note that the nomenclature $s = v . t$ looks like a dot product, which is different than multiplication. The correct way to write this equation is $s=vt$, where people who know physics recognize that "v" is velocity and "t" is time (in other words, physics has a standard nomenclature). $\endgroup$ – David White yesterday
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Many, albeit far from all, "multiplications" in physics are simplified versions of integrals. For example, in your case, saying $\Delta s = v \Delta t$ comes from the definition of velocity $$ \vec{v} = \frac{d\vec{s}}{dt} $$ for which we can easily see this has come from an integral: $$ \vec{s} - \vec{s}_0 = \int_{t_0}^t \vec{v}(t)dt. $$ This equation -- as well as $\Delta s = v\Delta t$ -- tell you that the change in position of an object from $\vec s_0$ to $\vec s$ is the result of a velocity $\vec{v}$ that occurs over time period $t-t_0$. During that time period, the velocity $\vec v$ uses every infintesimal moment $dt$ to change the position by a tiny amount $d\vec s$, until all the little $d\vec s$'s have accumulated up to $\vec{s}-\vec{s}_0$.

There are numerous other places where this occurs: $F = -\frac{dU}{dx}$, $I = \frac{dq}{dt}$, $\Delta V = - \frac{d\Phi_B}{dt}$, $P = - \frac{dU}{dV}$, and many more.

Another example is Newton's second law -- $F=ma$. If only conservative forces act on your object then you can see it is equivalent to $$ \frac{dU}{dx} = -m\frac{dv}{dt}.$$ That is to say -- the changes in potential energy over displacement (force) cause the velocity to change over time in proportion to mass.

In the context of vectors, scalar and vector multiplication also contribute the important directions of products, e.g. in Maxwell's equations we have

$$ \nabla \cdot E = \frac{\rho}{\epsilon_0} $$ and $$ \nabla \times \vec{B} = \mu_0\left[\vec{J}+\epsilon_0\frac{\partial \vec{D}}{\partial t}\right]$$

The first tells you that electric field diverges from a source in a way consistent to the distribution of source charge. The second tells you that the magnetic fields produced from real or displacement currents curl around the source and form a new vector.

All multiplication in physics has physical meaning. Often this can be found by looking at the underlying definitions of the quantities. After all, the fundamental equations in physics -- Newton's 2nd law, the Schrodinger equation, Maxwell's equations, and so forth -- are each differential equations that link the interactions and energy of/between objects to their motion through spacetime using multiplication.

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  • $\begingroup$ Can you provide an image/ a figure in my case ? $\endgroup$ – Omer Farooq 2 days ago
  • $\begingroup$ See faculty.utrgv.edu/emmett.tomai/courses/3370/mod1/refs/img/… $\endgroup$ – zhutchens1 2 days ago
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    $\begingroup$ An example of a multiplification that is not a simplified version of an integral is $E=\hbar\omega$. $\endgroup$ – J.G. 2 days ago
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    $\begingroup$ This is one of those answers that sounds good, but that I'm actually somewhat skeptical is fundamentally correct. Specifically this claim: "All multiplication in physics has physical meaning". It's hard for me to see how you can justify this statement without it just becoming tautological - i.e. it has physical meaning because it appears in a physics equation. $\endgroup$ – aquirdturtle 2 days ago
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    $\begingroup$ @aquirdturtle fair, but I would say that all physics equations have physical meaning because they are mathematical models build to explain the behavior we observe or predict in reality. It’s not tautological because the equations are models, the meaning is experimental. Probably worth moving this to chat for more discussion. $\endgroup$ – zhutchens1 yesterday
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Basically the multiplication in physics is the same as in mathematics.

But there is one more important thing you need to keep in mind. Unlike in mathematics, in physical formulas the values have a number and a unit. Therefore, when multiplying two physical values, then you need not only to multiply their numbers. You also need to multiply their units.

Let's take for example a car driving with speed $v = 50 \text{ miles/hour}$, during a time $t = 2 \text{ hours}$.

Then you can calculate the distance driven by the car by $$s = v \cdot t$$

Using the values from above and doing the multiplication you get $$\begin{align} s &= v \cdot t \\ &= 50 \text{ miles/hour} \cdot 2 \text{ hours} \\ &= (50 \cdot 2)\ (\text{miles/hour} \cdot \text{hours}) \\ &= 100 \text{ miles} \end{align}$$

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  • $\begingroup$ can u provide a figure/ image $\endgroup$ – Omer Farooq 2 days ago
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    $\begingroup$ @OmerFarooq The image would exactly look like the image in Krishna's answer, just only with different numbers. $\endgroup$ – Thomas Fritsch 2 days ago
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    $\begingroup$ then why does v + v + v + v + v............t does not make sense $\endgroup$ – Omer Farooq 2 days ago
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    $\begingroup$ @OmerFarooq I guess you first need to understand multiplication of real numbers, which is an extension of integer multiplication, i.e. not just repeated addition/counting. $\endgroup$ – Paŭlo Ebermann 2 days ago
  • $\begingroup$ @OmerFarooq v = s/t. So v + v = (s/t) + (s/t) = (2s/t) = 2v. Thus (v + v) * t = ((s/t) + (s/t))*t = (2s/t) * t = 2s. It all works out correctly, even if doing pure maths. $\endgroup$ – Polygnome yesterday
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Physics can be thought of as the subset of mathematics which happens to make correct predictions about the Real World (tm) [at least roughly speaking]. As such in my view it's probably a mistake to read too much into the "physical meaning of multiplication" beyond the mathematical definitions.

Of course there are lots of friendly-sounding examples to give people of intuition about multiplication - e.g. counting groups of something. $4\times5$ can be thought of as 4 groups of 5 things each, and this remains true for most physical quantities in some sense. However, the math doesn't care about this interpretation of the expression - the math stands by itself. So it's perfectly reasonable and self-consistent to drop the fluff and focus on the cold-hard math.

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Another way of thinking about this intuitively is to first understand the meaning of speed. If a car has a speed of 60 miles per hour, this means that every hour the car traverses 60 miles. So, if 2 hours have elapsed, the car traverses 120 miles which is 60*2. Abstracting this now to variables, if a car is moving with a speed $v$, and a time $t$ has elapsed, the distance the car traverses is $v*t$.

This is precisely the same interpretation multiplication has in mathematics. If your teacher hands 3 cookies per student, and they have 30 students, how many cookies do they have to hand out? It's 30*3=90.

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Perhaps the simplest way multiplication can be understood in linear relations such as in your example is: “is proportional to with this coefficient.”

Distance is proportional to time. The coefficient of proportionality is called “velocity.”

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  • $\begingroup$ This is very misleading and fails even in very simple cases. Displacement is the integral of velocity over time, and there is no 'proportionality'. The mass of a solid homogenous sphere is not proportional to its radius $r$, but to the cube of its radius, and there is no meaningful sense in which what you said can be used to understand the product $r·r·r$. $\endgroup$ – user21820 yesterday
  • $\begingroup$ @user21820 I disagree. You appear to have a narrow conception of proportionality. “Displacement = Rate * Time” is exactly a statement of proportionality. If you wanted a more nuanced expression, capturing the possibility that velocity might not be constant, then you would have used a different formula. $\endgroup$ – Gilbert yesterday
  • $\begingroup$ You fail to acknowledge your explanation's failure in anything beyond linear relations. I wonder whose conception is narrow... $\endgroup$ – user21820 yesterday
  • $\begingroup$ @user21820 Read OP’s question again. Think about where he/she is asking the question from. Do you really think that my answer is unhelpful? If you wish to improve it, you have the power. $\endgroup$ – Gilbert yesterday
  • $\begingroup$ I read the question carefully, and I think there are many much better answers than yours that have already been posted (which I upvoted). The bottom line is that your answer is based on too narrow conceptualizations of multiplication in physics. So there is no point for me to attempt to improve it. If I were to give an answer, it would be roughly along the lines of aquirdturtle's, though zhutchens1's answer addresses the fact that many instances of multiplication are just special cases of the actual physically meaningful integrals. Or perhaps, more accurately, $v = \frac{dx}{dt}$. $\endgroup$ – user21820 21 hours ago

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