What is meant by multiplication in physics?

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Multiplication in physics can vary significantly depending on the context, such as when dealing with vectors, matrices, or scalars. A common interpretation involves using multiplication to calculate quantities like distance or cost, where the area under a graph (e.g., velocity vs. time) represents the product of two variables. For example, traveling at a constant speed of 5 meters per second for 10 seconds results in a distance of 50 meters, illustrating the concept of multiplication in a physical context. Additionally, dimensional analysis can clarify how different units interact, such as dollars per hour multiplied by hours yielding total dollars. Understanding these concepts helps bridge the gap between abstract mathematics and practical physics applications.
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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 sum of velocity 'time' times". But what does this even mean?
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S = v.t
s = v.t
 
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"Multiplication in physics" is rather vague. In school we learn that there is one and only one multiplication. After a few semesters of mathematics or physics one quickly realizes that what multiplication really is strongly depends on the objects your talking about. Vectors? Matrices? Polynoms? p-forms? Spaces?

In your case, I'll assume you talk about the standard multiplication as we learn it for natural numbers. I'ld say the most intuitive understanding comes from looking at integrals:
Think of a graph where you plotted velocity on the y-axis and time on the x-axis. If your velocity is constant, the graph will just be a straight line. We now know that the integral of a function (in this case the function is v(t), i.e. velocity at a given time) is just the area below the graph, i.e. some kind of rectangle. And the area of that rectangle is easily computed (even without doing any integration) as length times height, or in our case velocity times time: Voila.

Okay, you asked for a more physical interpretation. How about this: If you travel at 5 meters per second, it means that (per definition) you travel 5 meters in one second (duh!). For any other period of time, say 10 seconds, you could just might just as well travel 10 times for 1 second each: Thats a product!

Ehm... did that help?
 
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Thanks It helped me a lot.
Can you give me another physical example?
 
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SSG-E said:
Can you give me another physical example?
Suppose you took a job at $20 per hour, and you worked 10 hours. How much money would you expect to be paid and why?
 
PeroK said:
Suppose you took a job at $20 per hour, and you worked 10 hours. How much money would you expect to be paid and why?
20*10 = $200
 
SSG-E said:
20*10 = $200
Why multiply? What's the "physical explanation"?
 
PeroK said:
Why multiply?
As I get $20 for 1 hour so,
for 10 hours I will get
20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20
= 20*10
= $200
 
SSG-E said:
20*10 = $200

It can also be helpful to look at dimensions for justification, ##\text{dollars}\,\text{hr}^{-1} \times \text{hr} = \text{dollars}##

Just in case you're interested, I thought I might add a little note about your example. Quantities like displacement and velocity are more completely described as vectors. In the absence of any acceleration, you can say ##\vec{d} = t\vec{v}##. Since ##\vec{d}## and ##\vec{v}## are really arrows, this just means that the ##\vec{d}## is ##t## times as long as the ##\vec{v}## arrow.

Now you might then choose to establish these vectors in a chosen coordinate system, in which case you can now consider equations pertaining to the components. You can now say things like ##\Delta x = v_x t##, where all of those numbers are ordinary scalars and the multiplication just ordinary multiplication.

You have to be slightly careful when discussing distance, since a distance is the magnitude of a displacement. If ##\Delta \vec{r} = t \vec{v}##, then the distance you go ##|\Delta \vec{r}| = t|\vec{v}|##. Or better still, for non-constant velocities, a distance is an integral of speed (which is itself the magnitude of velocity).
 
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SSG-E said:
As I get $20 for 1 hour so,
for 10 hours I will get
20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20
= 20*10
= $200
How is that different from multiplication in physics? 20 km/h for 10 hours equals 200 km.

Or, if something costs $5 per metre, then 6 metres costs $30. Or, if the density is 5 kg per metre, then the mass of 6 metres is 30 kg.

This is what multiplication is. In terms of numbers or quantities, at least.
 
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PeroK said:
How is that different from multiplication in physics? 20 km/h for 10 hours equals 200 km.

Or, if something costs $5 per metre, then 6 metres costs $30. Or, if the density is 5 kg per metre, then the mass of 6 metres is 30 kg.

This is what multiplication is. In terms of numbers or quantities, at least.
Thanks I finally got the concept
 
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etotheipi said:
It can also be helpful to look at dimensions for justification, ##\text{dollars}\,\text{hr}^{-1} \times \text{hr} = \text{dollars}##

Just in case you're interested, I thought I might add a little note about your example. Quantities like displacement and velocity are more completely described as vectors. In the absence of any acceleration, you can say ##\vec{d} = t\vec{v}##. Since ##\vec{d}## and ##\vec{v}## are really arrows, this just means that the ##\vec{d}## is ##t## times as long as the ##\vec{v}## arrow.

Now you might then choose to establish these vectors in a chosen coordinate system, in which case you can now consider equations pertaining to the components. You can now say things like ##\Delta x = v_x t##, where all of those numbers are ordinary scalars and the multiplication just ordinary multiplication.

You have to be slightly careful when discussing distance, since a distance is the magnitude of a displacement. If ##\Delta \vec{r} = t \vec{v}##, then the distance you go ##|\Delta \vec{r}| = t|\vec{v}|##. Or better still, for non-constant velocities, a distance is an integral of speed (which is itself the magnitude of velocity).
Thanks it is very helpful
 
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