There is no evidence to support that time is quantized. So wouldn't the use of discrete values like $dt$ in calculus suggest time is quantized and comes in discrete durations of $dt$?
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$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– David ZCommented Aug 10, 2020 at 6:30
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$\begingroup$ What qualifies as "extended discussion" ? $\endgroup$– Blue5000Commented Aug 10, 2020 at 15:00
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1$\begingroup$ It's an autogenerated comment so don't read too much into that wording, but off the top of my head if an exchange reaches more than four comments we'll probably at least think about moving it. If it looks like it's about to wrap up within the next few comments, we might leave it, but otherwise there's a decent chance of it getting migrated to chat. Of course it's always a case-by-case determination based on several factors, and I'd encourage asking on Physics Meta if you want a more detailed explanation. $\endgroup$– David ZCommented Aug 10, 2020 at 19:55
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4 Answers
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The use of $dt$ is precisely to avoid discrete time. Suppose speed of something at time $t$ equals $t^2$. This scenario clearly involves a continuously changing speed. What we could do to calculate the approximate distance the particle travels from $t=0$ to $t=1$ is:
Divide the interval $t=0$ to 1 into, say, 100 smaller intervals, and assume the speed to be constant throughout each interval.
So the approximate distance traveled in the first 0.01 seconds= speed*time=$0^2*0.01$
In the next 0.01 seconds = $0.01^2*0.01$ .....
In the last 0.01 seconds=$(0.99^2*0.01)$
And finally we could add up these distances.
But the above calculation doesn't exactly capture the continuously changing nature of the speed. It assumes speed to be constant in small finite intervals.
So instead, what we actually do is calculate the limit of the distanced traveled as we make this interval smaller and smaller (the $h \to 0$ in calculus). That's what the integration method achieves (when you calculate $\int_0 ^1 t^2 dt$). It perfectly captures what would've happened if speed was changing continuously.
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$\begingroup$ When this integral is done there are other terms that go to zero as you say. these get left out of the result t^3/3 and intern are left out of the value of 1/3 for t=1 t=0 , however if the value of dt is not zero then no matter how small these extra terms become that is no excuse to exclude them from the final result. $\endgroup$– Blue5000Commented Aug 9, 2020 at 12:46
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1$\begingroup$ Some formatting hints: $h\to0$ can be entered as
$h\to0$; multiplication can be typeset as $\times$ via$\times$or as $a\cdot b$ via$a\cdot b$. $\endgroup$– RuslanCommented Aug 9, 2020 at 12:46 -
$\begingroup$ @Blue5000 in the limit they are zero: $$\lim\limits_{x\to0}x=0.$$ $\endgroup$– RuslanCommented Aug 9, 2020 at 12:47
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$\begingroup$ I have seen this before , but what exactly do you mean by "in the limit" ? $\endgroup$– Blue5000Commented Aug 9, 2020 at 12:49
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$\begingroup$ @Blue5000 On the calculator, you can try calculating $\int_0^1 t^2 dt$ by dividing the interval $t=0$ to $1$ into smaller and smaller subintervals (and assuming the speed to be constant throughout each subinterval). You'd find that you can make the distance traveled as close to $\frac{1}{3}$ as you want if you only increase the number of sub-intervals (this can be proved too). So we say that $\frac{1}{3}$ is the limit of the distance traveled as the number of sub-intervals tends to infinity $\endgroup$ Commented Aug 9, 2020 at 14:28
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No. The use of "dt" in calculus has no correspondence with the passage of time in the physical world; it is a mathematical construct.
answered Aug 9, 2020 at 3:42
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$\begingroup$ "any slice of time shorter than the Planck time (10^-43 second) has no physical meaning", can you please elaborate on that? $\endgroup$ Commented Aug 9, 2020 at 3:57
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$\begingroup$ But doesn't dt represent a non zero duration of time ? $\endgroup$– Blue5000Commented Aug 9, 2020 at 4:28
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1$\begingroup$ @nielsnielsen There is no widely-accepted theory in which times shorter than the Planck time have no physical meaning. You are (reasonably) presupposing what some eventually-accepted theory of quantum gravity will say about time. $\endgroup$– G. SmithCommented Aug 9, 2020 at 4:53
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2$\begingroup$ The thing is that dt is not a value. It is part of the notation of the derivative or integral, and has no meaning on its own. If you want to see the actual value that has a non-zero value, but becomes arbitrarily small as the limit goes to zero, you have to expand the derivative or integral into its limit-based definition, in which the value becomes visible. With the derivative, that small but non-zero value is typically denoted with a h, and not dt. $\endgroup$ Commented Aug 9, 2020 at 14:03
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Though time maybe continuous, you can discretize time into small intervals to solve the problem. This is how we solve differential equations numerically. But i think you mean to question the exact results obtained by calculus. This is valid because if we can discretize time much smaller than least count of our apparatus that measures time, these 'exact results' show correct answers inside the error margin. Also, it's important to note that when we solve a physics problem, we idealise and make it into a mathematical problem. In physics, we assume that time is continuous. If time is indeed quantised, the "exact" results obtained by idealising are wrong.
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Whenever you encounter things such as $dx$ or $dt$, you have to understand that these do not represent actual values that you can talk about on their own. They are merely part of a larger notation. For example, with a derivative, the thing that has a meaning is the entire fraction itself: $$ \frac{dx}{dt} $$ If you evaluated this fraction in the point $t_1$, its value is defined to be $$ \lim_{h \to 0} \frac{x(t_1 + h) - x(t_1)}{h} $$ So, the value that is going to zero is the $h$ in the expression above. And indeed, $h$ is some nonzero real number.
In physics you will very often see $dx$ and $dt$ being on their own, usually on either side of an equality sign, but mathematically speaking, this is incorrect and in fact meaningless. They have to be part of either a derivative or integral, and they are just part of the notation of how you write down a derivative or integral, just like the notation for a fraction includes a straight line, which has no meaning on its own.
The reason it always works when you separate them like that is because limits can be exchanged with most operations, i.e. you can exchange the order of taking the limit with everything from addition and multiplication, to literally any other continuous function. This means that once you put them back together, you get the same results as if you had worked rigorously with limits.
(there are some ways to rigorously give them a value, but those values would not act like ordinary real numbers do, because they are not real numbers)
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1$\begingroup$ When $dx,dy$ etc. stand alone then most of the times they are viewed as differential forms (en.wikipedia.org/wiki/Differential_form). These are perfectly rigorous. $\endgroup$ Commented Aug 9, 2020 at 18:34