Arrow of time

[petrushka.googol]

Time is defined in dimensional analysis as L⁰M⁰T¹.

 

Time is measured in seconds and smaller and higher magnitudes.

 

However as we all know time is not absolute. It merely exists in relation to a frame of reference.

 

Also it "moves" along its axis as an object is translated through space. We then say that time has elapsed.

 

How then is time a scalar? It behaves like a vector and as we all know is irreversible.

 

Why then do we not interpret time as a vector in classical physics in the light of this knowledge?

 

Also if time were absolute and scalar throughout the universe then the twins paradox would have been impossible.

 

Different rates of motion elapse the same time at different intervals and this produces an effect like that to a vector in geometry.

 

Please advise.

 

Thanks in advance.

 

 

 

[swansont]

Length is a scalar, too. Why is time being a scalar a problem?

[petrushka.googol]

Length is a scalar, too. Why is time being a scalar a problem?

 

Time can elapse independent of length (the same object may have constant length from t_{0 to} t₁₎ but length cannot elapse independent of time. (for change from L₀ to L₁) time must elapse to instantiate this change.

[pears]

When we talk about time we usually specify the direction (e.g. in an hour's time or an hour ago) not sure if that makes it a scalar or a vector though :/ Would it be a scalar if it was never used in reference to it's direction? What would be an example of such a usage?

[swansont]

 

Time can elapse independent of length (the same object may have constant length from t_{0 to} t₁₎ but length cannot elapse independent of time. (for change from L₀ to L₁) time must elapse to instantiate this change.

 

A length does not "elapse". A length L is simply that: a length L. We change it to a displacement when that's necessary as part of the context of a problem, in which case it's a vector.

 

Time is occasionally used in the context of a vector — you can solve a problem and get a negative value for it, much like you can have a negative value for a displacement. It implies a direction relative to the coordinate system you've adopted.

 

In short, the problem you describe doesn't exist.