# Does this statement make any sense?

I am asking this question completely out of curiosity. The other day, my roommate, by mistake, used 'Light year' as a unit of time instead of distance. When I corrected him (pedantic, much), he said the following:

"Units are relative. And according to Fourier Transforms, units can be changed so Light year is a unit of time."

That got me thinking and I read up Fourier Transforms on wikipedia but couldn't find anything about using a unit in one domain as a unit for another measurement. I do agree that units (particularly, base units are relative. eg: the meter), but does his statement make any sense?

EDIT Thank you everyone for all the answers. It isn't so much to in it in or prove a point as it is to understand the concept better. Anyways this is his response after I showed him this thread. Any comments would be appreciated.

His response: Nevermind, for the first time I accept I was wrong. BUT using lightyears to measure time is possible. My example didn't make sense bacause I was wrong when I meantioned that I'm still measuring dist. If you have a signal in time domain and ...take the FT, I get a signal which DOES NOT HAVE to be in frequency domain. Clarify this to the guy who posted last. Now the new signal is in a domain defined by me and so is its units. This signal although not equal to the original signal, still represents that if ya take an inverse FT. So, the idea of time will still be there. Now coming back to our case: lightyears here is not the lightyears you are used to read when dealing with distance. It represents time.

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I love when people use technobabble they know nothing about to work around their ignorance :) –  Stefano Borini Nov 15 '10 at 22:13
I am answering your friend in an actual answer. –  Stefano Borini Nov 16 '10 at 13:35
@Stefano: Well done but I think it is useless to argue more with that kind of people :p –  Cedric H. Nov 16 '10 at 14:14
@Cedric: maybe, maybe not. I think his friend is in good faith. He is just confusing things and assumes he knows technical facts. in reality, he is putting together a partial and incorrect vision to form something which is, in the end, improper. I think it's our duty to clarify their misunderstanding, if they want to listen. –  Stefano Borini Nov 16 '10 at 14:26

This doesn't make much sense: light year is in any case a unit of distance.

What is common is to use "reduced units", for examples units where $c=1$ (speed of light) or $h=2\pi$. But in these cases the opposite would happen: you would say "year" to mean a distance. Or for example you say "has a mass of xyz MeV" instead of "$MeV/c^2$".

About the Fourier transform: this allow to go from the so-called "time domain" (even if "time" is not always the usual time) to the "frequency domain" involving ... frequencies.

But as you can see this cannot change the definition of light-year.

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I suppose one could use the light-year as a unit of time in reduced units, although the "light" part of it would be kind of redundant. –  David Z Nov 15 '10 at 22:20
@David: I guess it would be valid but really strange. –  Cedric H. Nov 15 '10 at 22:27
@DavidZ and Cedric, actually, I was thinking of "convenient units" like in special relativity. I suppose you could imagine a frame of reference (a photon?) where all distances are in proper time, so distance does become equivalent to time. However, what the friend of "Xbonez" said strongly suggests he is not thinking this way. –  Mark C Nov 16 '10 at 14:26

No, the definition of light year is "the distance that light travels in a year in vacuum"

Short of redefining English, the statement is incorrect!

By changing the Units, you would simply change the numeric value of a l.y., but not its definition.

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His response: Nevermind, for the first time I accept I was wrong. BUT using lightyears to measure time is possible.

It's possible if you rape physics, definitions, and just organize things so that you actually, in the end, obtain time from dimensional analysis. The idea would go along this line of reasoning:

• I know a man walks 5 km/h, hence
• space is a unit of time because I can say 10 km is equivalent to two hours, hence
• I can measure time by specifying a distance. Let's meet in 5 kilometers at the pub 200 meters from here.

But that's not a unit definition. it's just a deep misrepresentation of concepts: using this approach, you can define everything in terms of anything else, assuming there's a relationship among them. you could define time in terms of weight of apples a man can move from the tree to a box.

Also, the whole setup is pretty circular in definition. You define time as space traveled by something at a constant speed for a defined time, so in the end your pulled you own bootstraps.

My example didn't make sense bacause I was wrong when I meantioned that I'm still measuring dist. If you have a signal in time domain and ...take the FT, I get a signal which DOES NOT HAVE to be in frequency domain.

A Fourier transform is nothing but finding the coefficients of a linear combination of plane waves. When you find these coefficients, you can express the original function as the linear combination of these coefficients and a plane wave. If you note, there's a unit relationship between the domain before the FT and after. seconds -> seconds^(-1) = Hz. So even if you want to do fourier decomposition of a space-based periodic or aperiodic system, the resulting domain will be in meters^(-1), which is eventually a wave number.

Clarify this to the guy who posted last. Now the new signal is in a domain defined by me and so is its units.

Nope, it just turns out from the dimensional analysis that this is not the case. Clearly, you can always transform your frobbles units into Hertz through an ad-hoc transformation you invent (see the man-apples above) but that would still not change the final dimensional analysis of your FT, and you would, in any case, introduce an arbitrary constant (in our case above, the walking speed) which, in the end, likely produces a circular definition.

This signal although not equal to the original signal, still represents that if ya take an inverse FT. So, the idea of time will still be there. Now coming back to our case: lightyears here is not the lightyears you are used to read when dealing with distance. It represents time.

Lightyears is a measure of distance. It is a product of two well defined constants: the speed of light and a well defined amount of time. Simple dimensional analysis tells you it's a distance.

Edit: non-tech note. there's nothing wrong not to know things. Tell your friend it's not my intention to mock him. There's, however, something wrong to pretend to be right through misunderstood justification. It looks like he is mature enough to understand he is wrong, which is the spirit we should all live with. I hope my answers clarifies his doubts. Note that I could be wrong myself. I don't know the technicalities of standard unit definitions, and my exposition could be patched by someone who knows more than me on this field. The approximation I present here is good enough for the purpose of explanation, but it is wrong nevertheless when we go down to the gritty details, and I am ready to accept criticism on this regard. This is how we progress.

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Thanks a lot for the explanation. I will point him to our answer. And I am certain he will take it in the right spirit. His reasoning is probably based on misunderstanding the class/class notes in which case I'm sure he'd be glad to be corrected. –  xbonez Nov 16 '10 at 14:41

If you FT from time domain, you end up in frequency domain, not in space domain. Of course you can use the relativistic interpretation by dividing a lightyear by a factor of c to end up with - oh, a year! But instead of that FT explanation one could of course just say "it's a figure of speech as a synonym to half an eternity"...

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He's BS-ing you to the extent that an explanation of the fourier transform is not necessary to point this out ("units are relative" - what like length and time are the same thing? then all unit definitions would be circular as length is defined in terms of time at present. is mass also length? can you give my height, weight, and age in one number and define each as the same number in different units? =) ). I'll be specific though, anyway. The fourier transform of a scalar number is simply not defined (no such thing as a fourier transform of a distance, for instance). Typically fourier transforms give you a function of frequency for every function of time or of space. Under certain conditions, where there is a relationship between spatial frequency and temporal frequency (like waves propagating in a nondispersive medium here wavelength*frequency = speed of waves) you can then get a function of time for any function of space using a fourier transform, indirectly. But this is a special case, involves functions, not numbers, and this fourier transform of the original function is NOT THE SAME as the original function, so it's irrelevant to your original argument. So your friend is screwing with you.

Have a good one

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