One of the assumptions of Newtonian mechanics is that "time is absolute". Absolute, as I understand, implies that it is the same for all observers. But it's not quite true because if Tom's watch is not synchronised with that of Jerry's they will record different times for some event A. However, both will measure the same time interval between two specific events A and B. This implies that Jerry's time ($t^\prime$) can at best be related to Tom's time ($t$) by a linear transformation of the form $t^\prime=at+b$ where $a,b$ are constants. So a better statement is: "time intervals are absolute".

I intuitively feel that this assumption of absoluteness of time intervals for the observers is built into the kinematics of the Newtonian mechanics. But it is perhaps used in an obvious way so that I have overlooked it. I want to have a clearer idea how exactly is this physical input of absoluteness of time intervals gets used all the time in the Newtonian paradigm. Any help?

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    $\begingroup$ Jerry's time will be related to Tom's by a transformation of the form $t'=t+b$. Otherwise, neither the time interval between the events $A$ and $B$ is absolute. $\endgroup$
    – Ozz
    Sep 14, 2018 at 12:54
  • $\begingroup$ @GK Well. I allowed the possibility that one can record the time in seconds while the other in minutes. Otherwise, you are correct. If both use the same unit to measure time, then $a=1$. $\endgroup$
    – SRS
    Sep 14, 2018 at 12:57
  • $\begingroup$ If one measures the time in seconds and the other in minutes there is no point in comparing the time intervals $t'_B-t'_A$ and $t_B-t_A$. Anyway, I get your point but I just commented to make this thing absolutely clear :) $\endgroup$
    – Ozz
    Sep 14, 2018 at 13:00
  • $\begingroup$ The answer here may be helpful. Loosely related: physics.stackexchange.com/a/97749/191954 $\endgroup$
    – user191954
    Sep 14, 2018 at 13:13
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    $\begingroup$ time is a universal parameter that ticks the same at all points in euclidian space, even though Jerry's watch is wrong. $\endgroup$
    – JEB
    Sep 14, 2018 at 15:49

1 Answer 1


You are mixing in the issues of measuring time with some sort of unit and of setting the zero of the measurement. You have exactly the same issue for distance. This should not be a problem. If I'm using feet and you meters, then we won't get the same number for a distance. And if I choose one location as the origin, say my house, and you another, say your house, then we won't agree on the numbers involved in coordinates, even if we use the same units of measure.

These should not be particularly difficult. It is no more a problem than trying to understand that getting to school is a different path for you to what it is for your friend who lives on the other side of town. It does not indicate anything strange about space.

This is not what absolute time is referring to.

The thing to look at would be the path from the school to the public library. This path will be the same for both you and your friend, even if you use meters with your house as the origin, and he uses feet with his house as the origin. You may completely disagree about the specific coordinate numbers of the path. This is not a problem for absolute space and absolute time.

Suppose we make observations of some physical process that involves a bunch of events. I will observe those events to happen in some order in time. With absolute space and absolute time we will agree on the ordering. This event happens before that event, and we will agree from our observations. I will observe that the interval between events A and B is longer than the interval between C and D, and in some ratio. You will agree from your observations, including the ratio.

Determining some such ratio is what is most usually meant by measuring time. Suppose we can agree on a standard for what one unit of time is. Then in an absolute space and absolute time, we will agree on the ratio between the size of our standard and the size of some observed interval between events.

Similarly, I will observe the events to be ordered in a direction. Say we pick a direction, say left and right, and we agree on this axis. Then one is to the left of the other one. You will agree from your observations. Event A and B are farther apart than event C and D, and by some ratio. And you will agree from your observations, including the ratio. And if we can agree on a distance unit we will get the same number of that unit for any distance.

Note that part of the idea of absolute space and time is that we can agree on a direction. In some geometries that is a challenge.

In relativity, for example, these are not always true. Sometimes we will disagree about the ordering of events. Sometimes we will disagree about the relative size of the interval between them. For example, if we are moving relative to each other, and if the events are space-like separated, then we may disagree on their ordering in time.

  • $\begingroup$ "You are mixing in the issues of measuring time with some sort of unit and of setting the zero of the measurement." I don't get this point. How will you measure anything without using some unit? "If I'm using feet and you meters, then we won't get the same number for a distance." Of course. In which sentence did I say that I'll get the same number? That's the whole point of introducing the conversion factor '$a$' in $t^\prime=at+b$. And '$b$' is the factor taking care of the fact that our clocks are not synchronized. @puppetsock $\endgroup$
    – SRS
    Sep 15, 2018 at 12:00
  • $\begingroup$ If you wish you can assume $a=1$ i.e., time is measured in the same unit by both. But $b\neq 0$ since the zero of time may be different. Then the question is this: how is the absoluteness of time intervals for two observers is built into the Newtonian equations. It appears from the opening sentences that you perhaps misunderstood the question or I've failed to state it clearly. Thanks for a detailed answer though I'm afraid it doesn't answer my question. You're telling something which I already know. @puppetsock $\endgroup$
    – SRS
    Sep 15, 2018 at 14:30
  • $\begingroup$ Please read the whole answer. $\endgroup$
    – user93146
    Sep 16, 2018 at 1:39

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