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What does it mean for two quantities to be on equal footing? I often see this said for time/space or energy/momentum, but I do not understand what this means beyond a mathematical convenience.

I have taken a look at What does it mean to treat space and time on equal footing? but the answers in that thread are unsatisfactory. Is there some physical interpretation of this statement?

CBBAM
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I use this phrase myself (most recently in this answer on Astronomy SE). The key equation to know is

$$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$$

This equation comes from special relativity and describes flat spacetime (so-called Minkowski space). You don't need to know what it represents; it's good enough to know that $t$ is time, and $x, y, z$ are the three spatial dimensions.

Note you can add time and space. This is a sign they are on equal footing.

To illustrate two physics concepts that are not on equal footing, consider energy $E$ and electric current $I$. The former is measured in $J$ (joules), while the latter is measured in $A$ (amperes). These two units are different, which makes any expression like $E + I$ meaningless. This shows they are not on equal footing.

You might point out that time and space are measured in different units, seconds ($s$) and meters ($m$) respectively, but in relativity, both these units measure the same thing, and you can convert from one to the other using the speed of light. Hence relativity puts time and space on equal footing. Time and space are indeed not on equal footing in Newtonian mechanics.

Allure
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    I think it might be helpful to explicitly write out the factor of $c$ which makes the units of the time-element consistent with the units of the space-elements. – Silly Goose Dec 07 '22 at 08:46
  • So to summarize, for quantities to be on equal footing they have to be in the domain of some inner product and must have the same units? – CBBAM Dec 07 '22 at 20:37
  • @CBBAM why the "inner product"? But same units, yes. – Allure Dec 07 '22 at 23:32
  • @Allure You used the Lorentz inner product as an example of space/time being on an equal footing, that is where I got it from. Have I misunderstood? – CBBAM Dec 08 '22 at 01:10
  • @CBBAM I didn't take any inner products between vectors, no. That equation is the spacetime interval in Minkowski space, also known as the Minkowski metric, with $c = 1$. – Allure Dec 08 '22 at 01:58
  • @Allure It seems your expression for $ds^2$ originates from the Lorentz inner product. In fact that is how Minkowski space is defined, as $\mathbb{R}^4$ equipped with that exact inner product. – CBBAM Dec 08 '22 at 02:12
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To my understanding, for two quantities to be on equal footing, both quantities must be the same type of mathematical object. For example, in textbook Introductory Quantum Mechanics, space coordinates are operators (as well as parameters) and the time coordinate is only a parameter (it is not an operator). This mismatch in object type means that space and time are not on the same footing.

I think the direct consequence of two quantities not being on equal footing is that you cannot manipulate said quantities in the same way. For example, suppose one theory of physics puts space and time on the same footing and thus allows one to manipulate space and time in the same way. Suppose another theory of physics does not put space and time on equal footing so that one cannot manipulate space and time in the same way. Then, you have an inconsistency to solve if you want to combine the two theories of physics.

Silly Goose
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  • Could you elaborate on coordinates being operators? – CBBAM Dec 07 '22 at 20:38
  • By space-coordinates, I mean position. For example, for a single particle in one dimension there exists a position operator which corresponds to the observable value of the position of that particle. There is no such operator for time. @CBBAM – Silly Goose Dec 08 '22 at 00:46
  • In special relativity (not quantum) what is the position operator? – CBBAM Dec 08 '22 at 01:11
  • Special relativity to my understanding does not aim to describe observables (position, etc.). You would be using the machinery of classical mechanics or quantum mechanics to describe such observables. Thus, I feel like it probably doesn't make sense to expect special relativity to have an analogue to a position operator. – Silly Goose Dec 08 '22 at 01:26
  • Yes that is what I had thought as well, which is why I am unclear on how special relativity (not its application in QFT) puts time and space on an equal footing. – CBBAM Dec 08 '22 at 02:13